Learning Objectives

After going through this, unit you will be able to understand, appreciate and apply the following concept:

• Understand the methods, or ways, to induce emf in a coil.
• Understand the Faraday’s law of induction and Lenz’s Law.
• Calculate emf, current and magnetic flux using Faradays’ Law.
• Explain the physical meaning of Lenz’s Law.
• Derive the formula for motional emf and power spent by external agency in moving a coil with constant velocity.
• Explain the magnitude and direction of an induced eddy current, and the effect this will have on the object it is induced in.
• Describe the several applications of eddy currents.
• Explain the self induction and back emf.
• Calculate the inductance, energy stored and emf generated in a coil/ inductor.
• Explain the mutual induction between the neighbouring coils.
• Calculate the mutual inductance of coaxial solenoids.
• Sketch (draw phasor diagrams) voltage and current versus time in simple inductive, capacitive and resistive circuits.
• Calculate the reactance, impedance, current and voltage in simple series a.c. circuits having L, C and R.
• Derive the phase angle, resonant frequency, power, power factors in series LCR circuit.
• Explain the significance of series resonant circuit, resonant freqeuncy and quality factor.

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Electromagnetic Induction

Magnetic Flux

The number of magnetic field lines, or magnetic flux ’ ${\varphi }_B{}^{\prime }$, which pass through a given cross-sectional area ’ $\mathrm{A}$ ‘, placed in a uniform magnetic field ’ $\mathrm{B}$ ‘, can be written as

$${\varphi }_{\mathrm{B}}=\mathbf{B}\cdot \mathbf{A}=\mathrm{BA}\cos \theta$$

where $\theta$ is the agnle between the direction the magnetic field and the (outward) unit vector normal aera $(\mathrm{A}=\mathrm{A}\hat{\mathrm{n}})$. The SI unit of magnetic flux is weber $(\mathrm{Wb})$. It also equals tesla metere squared $\left({\mathrm{Tm}}^2\right)$. It is a scalar quantity.

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The figures, given above, show the positions (of the area A, with respect to the direction of a uniform magnetic field) for which the magnetic flux, linked with the area, has its

(i) maximum value $(=\mathrm{BA})$

(ii) minimum value $(=0)$

Electromagnetic Induction

We knwo that moving electric charges, or currents, can produce magnetic fields. Is the conserve effect possible? Can moving magnets produce electric currents? In the year 1830, Faraday conducted many experiments to demonstrade that electric currents can be induced in closed coils when they are present in a changing magnetic field or when changing the magnetic flux, linked with them, is made to change.

From the experiemental observations of Faraday, it was concluded that whenever the number of magnetic field lines or magnetic flux, passing through a coil changes, an emf is induced in that coil. If the circuit is closed, a current flows through it. The emf, and the current so produced, are called. ‘induced emf’ and ‘induced current’ respectivley. They last only as long as the (linked) magnetic flux is changing. This phenomenon is now known as ’electromagnetic induction’.

Ways of Changing the (Linked) Magnetic Flux

The magnetic flux, through a circuit, may be changed in a number of ways. This can be done (i) by moving a magnet relative to the circuit (ii) by changing the current in a neighbouring circuit (iii) by changing the current in the same circuit and (iv) by rotating a coil in a uniform magnetic field.

Let us no consider two experiments demonstrating the phenomenon of electromagnetic.

Experiment-1 :

When a bar magnet is pushed towards or pulled away, from the coil, the pointer in the galvanometer deflects. This indicates the presence of electric current in the coil.

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The deflection, in the two cases are, however, in opposite senses.

• It is also observed that faster the motion of the magnet, larger the deflection.
• When the bar magnet is held fixed and the coil is moved towards or away from the magnet, similar effects are observed.
• This shows that it is the existence of a relative motion, between the magnet and the coil, that is responsible for inducing electric current in the coil.

Experiment-2 :

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• It is observed that the galvanometer shows a deflection (momentary) when the current in ${\mathrm{C}}_2$ is started by closing key ’ $\mathrm{K}$ ‘. The pointer in the galvnometer in $\left({\mathrm{C}}_1\right)$ returns to zero immediately.
• When the key ’ $\mathrm{K}$ ’ is released, there is again a momentary curent in ${\mathrm{C}}_1$ but in opposite direction.
• Similar effects are observed by increasing or decreasing the current in the coil ${\mathrm{C}}_2$.
• It is also observed that the deflection increases, quite significantly, when an iron rod is inserted into the coils along their axis.
• Hence a changing current in one coil, is a source of induced current in another coil (neighbouring).

Faraday’s Law of Induction

Faraday stated these experimental observations in the form of a Law, called ‘Faraday’s Law of Electricmagnetic Induction’. The law is stated as follows:

“The magnitude of the induced emf, in a circuit, is equal to the time rate of change of the magnetic flx linked with the circuit.” Mathematically, the magnitude of induced emf is given by

$$|\epsilon |=\frac{\mathrm{d}{\varphi }_{\mathrm{B}}}{\mathrm{dt}}$$

For a closely wound coil of $\mathrm{N}$ turns, the total induced emf is given by

$$|\epsilon |=\mathrm{N}\frac{\mathrm{d}{\mathrm{\Phi }}_{\mathrm{B}}}{\mathrm{dt}}$$

where ${\mathrm{\Phi }}_{\mathrm{B}}=$ (Average) magnetic flux lined with ‘one turn’.

The product $\left(\mathrm{N}{\varphi }_{\mathrm{B}}\right)$ is called the ’number of magnetic flux linages’ in the coil. Its unit is ‘weber-turns’.

Lenz’s Law

The polarity, of the induced emf, is given by a rule, known as lenz’s law. The statement of the law is as follows:

“The polarity of induced emf is always such that it tends to produce a curernt, which opposes the change in the magnetic flux that induced it.

We can, therefore, combine Faraday’s Law and Lenz’s Law to write

$$\epsilon =-\frac{\mathrm{d}{\varphi }_{\mathrm{B}}}{\mathrm{dt}}$$

The negative sign indicates the direction of the induced emf ’ $\epsilon$ ‘.

It the circuit is closed, a current $(\mathrm{i}=\frac{\epsilon }{\mathrm{R}})$ is set up in it. Here $\mathrm{R}$ is the resistance of the circuit.

$$\therefore \mathrm{i}=-\frac{1}{\mathrm{R}}\frac{\mathrm{d}{\varphi }_{\mathrm{B}}}{\mathrm{dt}}$$

Lenz’s law is in accordance with the principle of conservation of energy. It enables us to realize that it is the mechanical work, needed to be done to ‘move the magnet (towards the coil)’, that gets converted into the induced) eletrical energy.

Induced Current and Induced Charge

If, in a coil of $N$ turns, the rate of change of magnetic flux (linked with each turn) be $\frac{d{\varphi }_B}{dt}$, then

$$\mathrm{i}=-\frac{1}{\mathrm{R}}\left(\frac{\mathrm{Nd}{\varphi }_{\mathrm{B}}}{\mathrm{dt}}\right)$$

Here $\mathrm{R}$ is the total resistance of the circuit. The charge, flowing through the circuit, in a time-interval, will be given by $\mathrm{\Delta }\mathrm{q}=\mathrm{idt}$

$$\begin{array}{rl}& =\frac{1}{\mathrm{R}}\left(\frac{\mathrm{Nd}{\varphi }_{\mathrm{B}}}{\mathrm{dt}}\right)\times \mathrm{dt}\\ & \Rightarrow \mathrm{\Delta }\mathrm{q}=\frac{\mathrm{N}}{\mathrm{R}}\left(\mathrm{d}{\varphi }_{\mathrm{B}}\right)\end{array}$$

The above equation shows that the total induced charge (that flows) does not depend upon the time interval ’ $\mathrm{dt}$ ‘. Whether the change in magnetic flux, is rapid or slow, the total charge that flows in the circuit, remains the same.

Example-1:

The magnetic flux, through a coil, present in a magnetic field, directed perpendicular to its plane, varies with time (in second) according to the relation.

$${\varphi }_{\mathrm{B}}=6{\mathrm{t}}^2+7\mathrm{t}+1\left({\varphi }_{\mathrm{B}}\text{ in miliweber }\right)$$

Find the magnitude of the induced emf in the coil at $t=2\mathrm{s}$. Also find the current, and its direction, in the resistance $\mathrm{R}$, if $\mathrm{R}=10\mathrm{\Omega }$

Show Answer

Solution :

We have $|\epsilon |=\frac{\mathrm{d}{\varphi }_{\mathrm{B}}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}(6{\mathrm{t}}^2+7\mathrm{t}-1)$

$=(12\mathrm{t}+7)\mathrm{mV}$

$\therefore At\mathrm{t}=2\mathrm{s}$, we have

$|\epsilon |=(12\times 2+7)\mathrm{mV}=31\mathrm{mV}$

Also $\mathrm{i}=\frac{|\epsilon |}{\mathrm{R}}=\frac{31}{10}\mathrm{mA}=3.1\mathrm{mA}$

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The direction of current, in the resistance $\mathrm{R}$, will be as shown, according to lenz’s rule, i.e., from left to right. (As the magnetic flux is increasing with time, induced current has to oppose its cause. For this, it has to set up its own magnetic field, in a direction opposite to the applied field. This would require the induced current to flow in an anticlockwise sense in the coil).

Example-2:

A rectangular loop, and a circular loop, are moving out of a uniform magnetic field region, to a field free region, with a constant velocity ’ $\mathrm{V}$ ‘. In which loop would the induced emf be constant during its pass age out of field region. (The magnetic field is directed normal to the plane of both the loops).

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Show Answer

Solution :

In case of circular loop, the rate of change of area of the loop, during its passage out of field region, is not constant. Hence induced emf will varry accordingly. In the case of the rectangular loop $\left(\frac{\mathrm{dA}}{\mathrm{dt}}\text{and, therefore}\right)\frac{\mathrm{d}{\varphi }_{\mathrm{B}}}{\mathrm{dt}}$ is constant. Hence induced emf can be expected to be constant for the rectangular loop.

Motion of a Conductor in a Uniform Magnetic Field

Let us consider a straight conductor, PQ, present in a, uniform and time independent magnetic field. As shown in the figure conductor ’ $\mathrm{PQ}$ ’ is free to move on a rectangular conductor frame work, $\mathrm{PQRS}$. Let the rod, PQ, moved towards right with a constant velocity V. As PQRS forms a closed circuit, there in a change in area as the rod PQ moves.

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If the length $\mathrm{RQ}=x$ and $\mathrm{RS}=\ell$, then the instantaneous magnetic flux, linked with $\mathrm{PQRS}$, is

$${\varphi }_{\mathrm{B}}=\mathrm{BA}=\mathrm{B}\ell x$$

Since $x$ is changing with time, the change of flux, will induce an emf, $\epsilon$, given by

$$\epsilon =-\frac{\mathrm{d}{\varphi }_{\mathrm{B}}}{\mathrm{dt}}=-\mathrm{B}\ell \frac{\mathrm{d}x}{\mathrm{dt}}$$

But $\left(\frac{\mathrm{d}x}{\mathrm{dt}}\right)=\mathrm{v}$ the velocity of the conductor $\mathrm{PQ}$, towards right.

Hence $\epsilon =\mathrm{B}\ell \mathrm{v}$

The induced emf, BV $\ell$, is called motional emf. Thus we can also produce an induced emf by moving a conductor (instead of varying the magnetic field) in a uniform magnetic field.

We can consider the above rod ’ $\mathrm{PQ}$ ’ as a battery of emf, $\epsilon$, having some internal resistance ’ $\mathrm{r}$ ‘. ’ $\mathrm{R}$ ’ be the total resistance of the (other) three arms (RS, RQ, PS).

Then, the induced current in the circuit will then be

$$i=\frac{\epsilon }{R+r}=\frac{BV\ell }{R+r}$$

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From the above, we conclude that when a straight conductor, of length ’ $\ell$ ‘, moves with a velocity ’ $\mathrm{V}$ ‘, perpendicular to a uniform magnetic field $\mathrm{B}$, (this ‘cutting’ its magnetic flux lines) an induced emf, of magnitude $\mathrm{BV}\ell$, is induced between the ends of that conductor.

Special Case :

If the direction of the velocity $\mathbf{v}$, of the conductor, makes an angle $\theta$ with the direction of the magnetic field $\mathbf{B}$, the component of $\mathbf{V}$, perpendicular to $\mathbf{B}$, will be $\mathrm{V}\sin \theta$. Therefore in this case the emf induced between the ends of the conductor, will be $\mathrm{BV}\ell \sin \theta$. It follows that if a conductor moves parallel to the field, $(\theta =0)$ no emf will be induced between its ends. It can be shown that: in general, motional (induced) emf, can be written as $\epsilon =(\mathbf{V}\times \mathbf{B})\cdot \mathit{\ell }$.

Situation-1 :

Consider a conducting rod $\mathrm{PQ}$, of length $\ell$, being rotated about its one end (in the plane of the page) in a uniform magnetic field B (directed normal to the plane of the page), with an angular speed ’ $\omega$ ’ as shown.

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Consider a small element, of length $\mathrm{d}x$, at a distance $x$ from the confined end, i.e. the end $\mathrm{P}$ (about which the rod is rotating).

The induced emf ’ $\mathrm{d}\epsilon$ ‘, across this elemental length, will be $=\mathrm{B}(\mathrm{d}x)\mathrm{V}$

where $\mathrm{V}=\omega x$

Thus, $\epsilon ={\int }_{x=0}^{x=\ell }\mathrm{d}\epsilon =\mathrm{B}\omega {\int }_{x=0}^{x=\ell }x\mathrm{d}x$

$=\frac{1}{2}\mathrm{B}{\ell }^2\omega$

If the rod moves about its centre, we would have

$\begin{array}{c}\epsilon ={\int }_{-\ell /2}^{+\ell /2}\mathrm{d}\epsilon =\frac{1}{2}\mathrm{B}\omega {\left|x^2\right|}_{-\ell /2}^{+\ell /2}\\ =0\end{array}$

Energy Considerations : Let ’ $r$ ’ be the resistance of the moveable arm PQ, (the moving rectangular condutor) in the above figure. We assume that the remaining arms, QR, RS and SP, have negligible resistance. In that case

$$\mathrm{i}=\frac{\epsilon }{\mathrm{r}}=\frac{\mathrm{BV}\ell }{\mathrm{r}}$$

On account of the presence of the magnetic field, there will be a lorentz force, on the arm ’ $\mathrm{PQ}$ ‘, given by $\mathrm{F}=\mathrm{Bi}\ell$. This force would be directed inwards in a direction opposite to the velocity of the rod.

The magnitude of this force is

$$\mathrm{F}=\mathrm{B}\left(\frac{\mathrm{BV}\ell }{\mathrm{r}}\right)\ell$$

$$=\frac{{\mathrm{B}}^2{\ell }^2\mathrm{V}}{\mathrm{r}}$$

This force acts like a retarding force. Hence power is spend by an external agency to move the rod with a constant velocity ’ $\mathrm{V}$ ‘.

The power required to do this is given by

$$\mathrm{P}=\mathbf{F}.\mathbf{V}$$

$$=\frac{{\mathrm{B}}^2{\ell }^2{\mathrm{V}}^2}{\mathrm{r}}\text{ (This is in the form of mechanical work) }$$

This power, spent by an external agent, is dissipated as Joule heat. The Joule heat produced is given by

$$\mathrm{P}={\mathrm{i}}^2\mathrm{r}={\left(\frac{\mathrm{BV}\ell }{\mathrm{r}}\right)}^2\mathrm{r}=\frac{{\mathrm{B}}^2{\ell }^2{\mathrm{V}}^2}{\mathrm{r}}$$

Example-3:

A 10 meter long wire is kept in east west direction. It is falling down with a speed of $5\mathrm{m}/\mathrm{s}$, perpendicular to the horizontal component of earth’s magnetic field of $0.30\times {10}^{-4}\mathrm{Wb}/{\mathrm{m}}^2$. (i) what is the momentary p.d induced between the ends of the wire? (ii) which end of the wire will be at a higher potential?

Show Answer

Solution :

(i) If a wire, of length $\ell$, move perpendicular to a magnetic field $\mathrm{B}$, with a velocity $\mathrm{v}$, the emf induced will be given by

$$\begin{array}{rl}\epsilon & =\mathrm{BV}\ell \\ \therefore \epsilon & =0.30\times {10}^{-4}\times 5\times 10\mathrm{V}=1.5\mathrm{mV}\end{array}$$

(ii) According to fleming left hand rule, the magnetic force on the electrons, in the wire, in the magnetic field will act towards west. Hence electrons will more to the western end of the wire, therefore the eastern end of the wire will be a higher (positive) potential.

Example-4:

Ahorizontal metal frame, $\mathrm{ABCD}$, moves with a uniform (horizontal) velocity of $0.2\mathrm{m}/\mathrm{s}$, into a uniform field of ${10}^{-2}\mathrm{T}$, acting vertically downwards. $\mathrm{AB}=0.1\mathrm{m}$ and $\mathrm{AD}=0.2\mathrm{m}$ and the resistance $\mathrm{R}$ of the frame is $5\mathrm{\Omega }$. As shown, the sides $AB$ and $CD$ enter the field in a direction normal to the field boundary. Find the magnitude of current induced in the metal frame when (i) $\mathrm{CD}$ just enters the field (ii) the whole frame is moving through the field (iii) $\mathrm{CD}$ just moves out of the field on the other side? Draw a sketch graph showing the variation of the induced current, the force required to move the frame with a constant velocity, and the power spend by the external agency.

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Show Answer

Solution :

(i) The sides $\mathrm{AD}$ and $\mathrm{BC}$ are parallel to $\mathbf{V}$. Hence no emf in induced in them. For the side $\mathrm{CD}$, moving in the field,

$$|\epsilon |=\mathrm{BV}\ell$$

$$\begin{array}{rl}=& {10}^{-2}\times 0.1\times 0.2\mathrm{V}\\ & 2\times {10}^{-4}\mathrm{V}\\ \therefore \mathrm{i}& =\frac{|\epsilon |}{\mathrm{R}}=\frac{2\times {10}^{-4}}{5}\mathrm{A}=4\times {10}^{-5}\mathrm{A}\end{array}$$

(ii) As the whole frame in moving through the field from time $t_1$ to $t_2$ say the flux through the frame is constant. Hence there is no induced current during this interval. [Alternatively the induced emf in the arms $\mathrm{AB}$ and $\mathrm{CD}$ are in opposite directions, the resultant emf in the frame is zero.]

(iii) When $\mathrm{CD}$ just moves out of the field region, on the other side, the magnitude of the induced emf, in $\mathrm{AB}$, moving through the field, is again equal to $2\times {10}^{-4}\mathrm{V}$. Hence magnitude of induced current, in the

frame, is same, but its direction is now opposite to that in the first case (The flux is now decreasing with time). The required sketches, therefore, have the forms shown.

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Example-5:

An air plane, with $20\mathrm{m}$ total wing span is flying with a speed of $250\mathrm{m}/\mathrm{s}$, straight south, parallel to the earth’s surface. The earth’s magnetic field has a horizontal component of $2\times {10}^{-5}\mathrm{Wb}{\mathrm{m}}^{-2}$ and the dip angle is ${60}^{\circ }$. Calculate the emf induced between the plane tips.

Show Answer

Solution :

Here $\epsilon ={\mathrm{B}}_{\mathrm{v}}\ell \mathrm{V}$ where ${\mathrm{B}}_{\mathrm{v}}$ is vertical component of earth’s magnetic field.

We have $\frac{{\mathrm{B}}_{\mathrm{V}}}{{\mathrm{B}}_{\mathrm{H}}}=\tan \delta$

$\therefore {\mathrm{B}}_{\mathrm{V}}={\mathrm{B}}_{\mathrm{H}}\tan \delta$

$\therefore \epsilon =({\mathrm{B}}_{\mathrm{H}}\tan \delta )\ell \mathrm{V}$

$=2\times {10}^{-5}\times (\tan {60}^0)\times 20\times 250\mathrm{V}$

$={10}^{-1}\times \sqrt{3}\mathrm{V}=0.173\mathrm{V}$

Example-6:

A copper rod, of length $0.19\mathrm{m}$, kept parallel to a long straight wire carrying a current of $5\mathrm{A}$ is with a uniform velocity of $10\mathrm{m}/\mathrm{s}$. The rod is perpendicular to the wire. The two ends of rod are at distance of $0.01\mathrm{m}$ and $0.2\mathrm{m}$, from the wire, (as shown in the figure). Calculate the emf induced in the rod.

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Show Answer

Solution :

The magnetic field due to the long straight wire, at a distance ’ $x$ ‘, from it is

$${\mathrm{B}}_x=\frac{{\mu }_0\mathrm{i}}{2\pi x}$$

This field is directed into the page (downwards). If ’ $\mathrm{d}\epsilon$ ’ is the emf induced, over the element ’ $\mathrm{d}x$ ’ of the rod, we have

$$\begin{array}{rl}\mathrm{d}\epsilon =\mathrm{BV}& (\mathrm{d}x)\\ =& \frac{{\mu }_0\mathrm{i}}{2\pi x}\mathrm{Vd}x\end{array}$$

$\therefore$ Total emf, $\epsilon =\int \mathrm{d}\epsilon =\frac{{\mu }_0\mathrm{iV}}{2\pi }{|{\log }_{\mathrm{e}}x|}_{0.01\mathrm{m}}^{0.02\mathrm{m}}$

$$\begin{array}{rl}& =\frac{4\pi \times {10}^{-7}\times 5\times 10}{2\pi }\times (2.303{\log }_{10}20)\\ & =3\times {10}^{-5}\mathrm{V}\end{array}$$

Flemings’s Right Hand Rule

The direction of the induced current, when a conductor moves in a uniform magnetic field, is given by Fleming’s right hand rule. This rule states that if we stretch the right hand thumb and the first two fingers, perpendicular to one another, and if the forefinger points in the direction of magnetic field, and the thumb in the direction of motion of conductor, the middle finger will point in the direction of the induced current.

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Eddy Currents

Facoult, in 1895, discoverd that when a solid mass of metal is placed in a changing magnetic field, (or moved in a magnetic field), causing a change in the magnetic flux linked with it, induced currents are set up throughout the volume of the metal. These currents are known as ’eddy currents’ or ‘facault currents’. The direction of circulation of these currents is such as to oppose the motion of the metal or change in magnetic flux (in accordance with Lenz’s law).

The resistance of thick metal being very low, the eddy currents are generally quite large in magnitude and produce considerable heating effect in accordance with ‘Joule’s heating’ law. This heating effect of eddy

currents in undesirable in the interior of iron cores of rotating armatures of motors, dynamos and transformers etc. To minimize these currents, the cores are not taken as a single piece of soft iron but are made of many thin (insultated from each other) laminas of soft iron. This type of core is called a ’laminated core’.

Eddy currents are, however, also used to advantage in certain applications. These include:

(i) Electromagnetic Damping : Certain galvanometers have a fixed core made of a non magnetic metallic material. When the coil oscillates, the eddy currents, generated in the core opposes the motion and bring the coil to rest quickly.

(ii) Induction Furance : The metal to be heated, or melted, is placed in a high frequency changing magnetic field. Strong eddy currents, induced in it, cause heating.

(iii) Magnetic Brakes in Trains : Strong electromagnets are kept above the rails in some electrically powered trains. When the electromagnets are activated the eddy currents, induced in the rails, opposes the motion of the train.

Self Induction

We know that, when a current flows through a coil, it produces a magnetic field and hence a magnetic flux, which can be linked with the coil itself. It is possible that emf is induced in a single isolated coil due to change of flux through the coil itself by means of varying the current through the same coil. This phenomenon is called self-induction. The induced emf is called ‘back emf’.

Thus, when the current in a coil is switched on, the (induced) back emf opposes the growth of current. Similarly, when the current is switched off, the back emfopposes the decay of the current.

Let us consider a coil of $\mathrm{N}$ turns carrying a current i. Let ${\varphi }_{\mathrm{B}}$ the magnetic flux linked with each turn of the coil. The total flux, linked with the coil, is proportional to the current ithrough the coil, and can be expressed as

$$\begin{array}{rl}& \mathrm{N}{\varphi }_{\mathrm{B}}\propto \mathrm{i}\\ & \mathrm{N}{\varphi }_{\mathrm{B}}=\mathrm{Li}\end{array}$$

Here $\mathrm{L}$ is a constant called the coefficient of self induction, or ‘self inductance’, of the coil. Physically, the self inductance plays the role of ‘inertia’ in electrical circuits. It is defined as the magnetic flux, linked with the coil, when a unit current is flowing through the coil itself.

When the current in the coil is varied, the flux linked with the coil also changes and an emf is induced in the coil. This is given by

$$\epsilon =-\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{N}{\varphi }_{\mathrm{B}}\right)$$

$$=-\mathrm{L}\frac{\mathrm{di}}{\mathrm{dt}}$$

The negative sign indicates that the selfinduced emf (or back emf) always opposes any change (increase or decrease) of current in the coil. From the above formula, we have

$$\mathrm{L}=\frac{\epsilon }{\left(\frac{\mathrm{di}}{\mathrm{dt}}\right)}$$

Hence the coefficient of self induction or self inductance of a coil is numerically also equal to the emf induced in coil when the rate of change of current, in the coil in unity.

The S.I unit of ’ $L$ ’ is 1 Henry $=1\mathrm{H}$

We have $\mathrm{H}=\frac{\text{ Volt }}{(\text{ ampere }/\text{ second })}=$ ohm - second

Also $\mathrm{H}=\mathrm{Wb}$ /ampere (from $\varphi =\mathrm{Li}$ )

Calculation of Self Inductance

It is possible to calculate the self in ductance for circuits with simple geometries.

Let us consider a long straight solenoid of cross section area $A$ and length $\ell$, having ’ $n$ ’ turns per unit length. When the solenoid carries a current I, the total flux linking with the solenoid

${\varphi }_{\mathrm{B}}=\mathrm{N}{\varphi }_{\mathrm{B}}=\mathrm{NBA}$

$$=(\mathrm{n}\ell )\left({\mu }_0\mathrm{ni}\right)\mathrm{A}$$

By definition, self inductance $\mathrm{L}=\frac{\left(\mathrm{N}{\varphi }_{\mathrm{B}}\right)}{i}$

Hence $\mathrm{L}={\mu }_0{\mathrm{n}}^2\mathrm{A}\ell$

If we fill the inside of the solenoid with a material of relative permability ${\mu }_r$, then, $L={\mu }_0{\mu }_r{n}^{2}A\ell$ This, the self inductance of the coil depends on its geometry and on the permability of the medium. The corresponding expression for a (small) plane circular coil of radius ’ $r$ ’ will be

$$\mathrm{L}=\frac{{\mu }_0{\mathrm{n}}^2{\ell }^2\pi \mathrm{r}}{2}$$

Energy Associated with Self Inductance

Let $\mathrm{d}\omega$ be the work done in establishing a current ’ $i$ ’ in the solenoid, in a time dt. Work has to be done, against the back emf, in establishing this current. This work is stored as magnetic potential energy.

$$\begin{array}{r}\mathrm{d}\omega =\mathrm{Vdq}=(-\epsilon )(\mathrm{idt})\\ =(-\mathrm{L}\frac{\mathrm{di}}{\mathrm{dt}})\mathrm{dt}\mathrm{i}\end{array}$$

$\mathrm{d}\omega =-\mathrm{Lidi}$

$\therefore$ Total amount of work done in establishing the current $\mathrm{I}$ is, then

$$\omega =\frac{1}{2}{\mathrm{LI}}^2$$

This work is stored in the form of magnetic potential energy $(U_m=\frac{1}{2}{L}^2)$ in the magnetic field of the solenoid. This expression can be considered as equalivalent to $\frac{1}{2}{\mathrm{mV}}^2$ for mechanical kinetic energy of a particle of mass M. It shows that $\mathrm{L}$ is analogus to $\mathrm{m}$ ( $\mathrm{L}$ is the ’electrical inertia’ which opposes the growth and decay of current in the circuit).

The above expression for the magnetic energy, stored in the magnetic field of solenoid, can be rewritten as

${\mathrm{U}}_{\mathrm{m}}=\frac{1}{2}{\mathrm{Li}}^2$

$\begin{array}{ll}=\frac{1}{2}\mathrm{L}{\left(\frac{\mathrm{B}}{{\mu }_0\mathrm{n}}\right)}^2& (\therefore (\mathrm{B}={\mu }_0\mathrm{n}i)\text{ for a solenoid })\\ =\frac{1}{2}\left({\mu }_0{\mathrm{n}}^2\mathrm{A}\ell \right){\left(\frac{\mathrm{B}}{{\mu }_0\mathrm{n}}\right)}^2& (\because \mathrm{L}={\mu }_0{\mathrm{n}}^2\mathrm{A}\ell )\\ =\frac{1}{2}\frac{{\mathrm{B}}^2}{{\mu }_0}(\mathrm{A}\ell )\end{array}$

$\therefore$ The magnetic energy, per unit volume, is

$${\mathrm{U}}_{\mathrm{B}}=\frac{{\mathrm{U}}_{\mathrm{m}}}{\mathrm{V}}=\frac{1}{2}\frac{{\mathrm{B}}^2}{{\mu }_0}$$

This expression is similar to electrostatic potential energy stored per unit volume in the electric field. In both cases energy is propotional to the square of the (corresponding) field strength.

An ideal solenoid, or coil, whose inductance $(\mathrm{L})$ is much greater than its winding resistance $(\mathrm{R})[\mathrm{L}>>\mathrm{R}$ ) is considered to be an inductor. Hence inductor is a circuit element (like the capacitor) which stores magnetic energy in the magnetic field of a current carrying coil.

We have derived the formula for energy density, for the special case of the uniform magnetic field of an ideal solenoid (inductor), the result is a geenral one and is valid for any region of space in which a magnetic field exists.

Mutual Induction

Let there be two coils in close proximity of each other. When the current in one of these coils changes, the magnetic field due to it over the other coil changes. Hence the magnetic flux linked with the other coil also changes and an emf is induced in this neighbouring coil. Such a phenomenon is known as mutual induction. The (first) coil, in which the current changes (leading to change in magnetic flux linked with the second coil), is called the primary coil. The second coil, in which induction takes place, is called the secondary coil.

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Let ${\varphi }_{\mathrm{B}}$ be the magnetic flux linked with the secondary coil, when a current ${\mathrm{i}}_{\mathrm{p}}$ flows through the primary coil. Then

$${\varphi }_{\mathrm{B}}=\left({\mathrm{N}}_{\mathrm{S}}{\varphi }_{\mathrm{S}}\right)\propto {\mathrm{i}}_{\mathrm{P}}$$

or ${\varphi }_{\mathrm{B}}=\left({\mathrm{N}}_{\mathrm{S}}{\varphi }_{\mathrm{S}}\right)={\mathrm{Mi}}_{\mathrm{p}}$

where $\mathrm{M}$ is a constant of proportionlity.

’ $\mathrm{M}$ ’ is called the ‘Mutual inductance’ or ‘Coefficient of mutual inductance’, for the given pair of coils. The emf induced in the secondary coil is given by

$$\begin{array}{rl}& {\epsilon }_{\mathrm{S}}=-\frac{\mathrm{d}{\varphi }_{\mathrm{B}}}{\mathrm{dt}}=-\frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{Mi})\\ & \therefore {\epsilon }_{\mathrm{S}}=-\mathrm{M}\frac{\mathrm{di}}{\mathrm{dt}}\end{array}$$

$$\therefore \mathrm{M}=-\frac{{\epsilon }_{\mathrm{S}}}{\left(\frac{\mathrm{di}}{\mathrm{dt}}\right)}$$

Hence mutual inductance, for a pair of coil, is equal to the induced emf, set up in one coil due to unit rate of change of current in the other coil.

The SI Unit of $\mathrm{M}=1\mathrm{H}=1$ Henry

$$=1(\text{ ohm-second })=1(\text{ Wb/ampere })$$

The value of mutual inductance of a given pair of coils, depends upon the number of turns of the coils their geometrical shape and size, their relative separation as well as their relative orientation. It also depends upon the permiability of the medium in the space between the two coils.

Consider two long co-axial solenoids each of length ’ $\ell$ ‘.

$$

When a current $i_2$ is set up in the outer coil $S_2$, the corresponding flux linkage with $S_1=N_1{\varphi }_1=M_{12}{i}_2$

Also ${\mathrm{N}}_1{\varphi }_1={\mathrm{N}}_1{\mathrm{B}}_2{\mathrm{A}}_1$

$=\left({\mathrm{n}}_1\ell \right)\left({\mu }_0{\mathrm{n}}_2{\mathrm{i}}_2\right)\left(\pi {\mathrm{r}}_1^2\right)$

This implies that

${\mathrm{M}}_{12}={\mu }_0{\mathrm{n}}_1{\mathrm{n}}_2{\mathrm{i}}_2\pi {\mathrm{r}}_1^2\ell$

If we consider the reverse case, the flux linkage, due to current $i_1$ in $S_1$, can be assumed to be confined solely inside ${\mathrm{S}}_1$. Since solenoids are (usually) very large. The corresponding flux linkage with ${\mathrm{S}}_2$,

${\mathrm{N}}_2{\varphi }_2={\mathrm{M}}_{21}{\mathrm{i}}_1$

${\mathrm{N}}_2{\varphi }_2={\mathrm{N}}_2{\mathrm{B}}_1{\mathrm{A}}_1$

This implies that $=\left({\mathrm{n}}_2\ell \right)\left({\mu }_0{\mathrm{n}}_1{\mathrm{i}}_1\right)\pi {\mathrm{r}}_1^2$

We thus see that $M_{21}={\mu }_0{n}_1{n}_2\pi r_1^2\ell$

$${\mathrm{M}}_{12}={\mathrm{M}}_{21}=\mathrm{M}$$

Hence, in general one can write that, for a pair of solenoidal coils

$$\mathrm{M}=\frac{{\mu }_0{\mu }_{\mathrm{r}}{\mathrm{N}}_{\mathrm{S}}{\mathrm{N}}_{\mathrm{P}}\mathrm{A}}{\ell }$$

Here ${\mathrm{N}}_{\mathrm{S}}$ and ${\mathrm{N}}_{\mathrm{p}}$ are number of turns in the primary and secondary coils, having a cross sectional area $\mathrm{A}$, and a length $\ell$.

Special Case-I :

Consider the general case of currents flowing simultaneously in two nearby coils. The flux linked with one coil will be the sum of two fluxes which exist independently. We can write

$${\mathrm{N}}_1{\varphi }_1={\mathrm{M}}_{11}{\mathrm{I}}_1+{\mathrm{M}}_{12}{\mathrm{I}}_2$$

Here ${\mathrm{M}}_{11}$ represent the inductance effects, due to the same coil. It is therefore, its self inductance ; ’ ${\mathrm{L}}_1$ '

$$\therefore {\mathrm{N}}_1{\varphi }_1={\mathrm{L}}_1{\mathrm{I}}_1+{\mathrm{M}}_{12}{\mathrm{I}}_2$$

Hence the total induced emf, in coil ’ $\mathrm{I}$ ‘, will be its self induced emf(back emf) plus the ‘mutual induced emf’, due to change in current in coil ’ 2 ‘. This can be expressed as ${\epsilon }_1=(-L_1\frac{{\mathrm{dI}}_1}{\mathrm{dt}})+(-{\mathrm{M}}_{12}\frac{{\mathrm{dI}}_2}{\mathrm{dt}})$

Special Case-II :

(a) When two ideal solenoids (inductors), of self inductances ${\mathrm{L}}_1$ and ${\mathrm{L}}_2$ are connected in seires, then their equivalent self inductance will be

$${\mathrm{L}}_{\mathrm{eq}}={\mathrm{L}}_1+{\mathrm{L}}_2\pm 2\mathrm{M}\text{ (If the coils are coaxial or very close) }$$

If flux, of one coil, favours the other, then ${\mathrm{L}}_{\mathrm{eq}}={\mathrm{L}}_1+{\mathrm{L}}_2+2\mathrm{M}$

and if flux, of one coil, opposes the other, then ${\mathrm{L}}_{\mathrm{eq}}={\mathrm{L}}_1+{\mathrm{L}}_2-2\mathrm{M}$

(b) If the coils are seperated by a large distances, $\mathrm{M}$ becomes negligible. Hence, in such a case, for the two coils.

$${\mathrm{L}}_{\mathrm{eq}}={\mathrm{L}}_1+{\mathrm{L}}_2$$

(c) If two inductor are connected in parallel, turns out that

$${\mathrm{L}}_{\mathrm{eq}}=\frac{{\mathrm{L}}_1{\mathrm{L}}_2-{\mathrm{M}}^2}{{\mathrm{L}}_1+{\mathrm{L}}_2\pm 2\mathrm{M}}\text{ (For coaxial coils or nearby coils) }$$

For isolated coils (or for large separation of two coils)

$${\mathrm{L}}_{\mathrm{eq}}=\frac{{\mathrm{L}}_1{\mathrm{L}}_2}{{\mathrm{L}}_1+{\mathrm{L}}_2}$$

(d) The coefficient of coupling, $\mathrm{K}$, of two coils is a measure of the magnetic coupling of the coils. It is given by $K=\sqrt{\frac{M}{L_1{L}_2}}$. Here $L_1$ and $L_2$ are the self inductances of the two coils and $M$ is their mutual inductance. $\mathrm{K}$ is always less than one.

Example-7:

A small circular loop, of radius $3\mathrm{cm}$, and a bigger loop, of radius $20\mathrm{cm}$, are placed coaxially, at a distance of $15\mathrm{cm}$ as shown in the figure. A current of 2 A flows through the smaller loop. Find the flux linked with the bigger loop and the mutual inductance of the two loops.

$$

Show Answer

Solution :

By symmetry, the flux linked with the bigger loop, due to current in the smaller loop, will be same as the flux linked with the smaller loop, due to the same current in the bigger loop ( $\therefore {\mathrm{M}}_{12}={\mathrm{M}}_{21}=\mathrm{M})$.

Now the magnetic field, at the centre of smaller loop, due the to a current, i, in the bigger loop, is

$$\begin{array}{rl}& {\mathrm{B}}_1=\frac{{\mu }_0{\mathrm{ir}}_1^2}{2{({\mathrm{r}}_1^2+x^2)}^{3/2}}\\ & =\frac{4\pi \times {10}^{-7}\times 2\times (0.20{)}^2}{2{((0.20{)}^2+(0.15{)}^2)}^{3/2}}\\ & =3.2\times {10}^{-6}{\mathrm{Wbm}}^{-2}\end{array}$$

The field $\mathrm{B}$, is along the axis. The flux linkage, with the smaller loop due to ${\mathrm{B}}_1$ this field will be (nearly) ${\varphi }_2={\mathrm{B}}_1{\mathrm{A}}_2={\mathrm{B}}_1\pi {\mathrm{r}}_2^2$

$$\begin{array}{rl}& =3.2\times {10}^{-6}\times 3.14(0.030{)}^2\mathrm{Wb}\\ & =9\times {10}^{-9}\mathrm{Wb}\end{array}$$

This is the same as the flux ${\varphi }_1$, linked with the bigger coil, due to same current in the smaller loop. We, therefore, have

$$\begin{array}{rl}\mathrm{M}=\frac{{\varphi }_2}{{\mathrm{i}}_1}& =\frac{9\times {10}^{-9}\mathrm{Wb}}{2\mathrm{A}}=4.5\times {10}^{-9}\mathrm{H}\\ \\ & =4.5\mathrm{nH}\end{array}$$

Example-8:

A very small circular loop of area $5\times {10}^{-4}{\mathrm{m}}^2$, resistance 2 ohm and negligible inductance, is initially coplanar and cocentric, with a much larger fixed circular loop of radius $0.1\mathrm{m}$. A constant current of 1 ampere is passed through the bigger loop and the smaller loop is rotated with an angular velocity ’ $\omega$ ’ $\mathrm{rad}/\mathrm{s}$, about a diameter. Calculate the induced emf and induced current, as a function of time, in the smaller loop.

Show Answer

Solution :

The magnetic field, at the centre of larger loop, is

$${\mathrm{B}}_1=\frac{{\mu }_0\mathrm{i}}{2\mathrm{r}}=\frac{4\pi \times {10}^{-7}\times 1}{2\times 0.1}\mathrm{T}=2\pi \times {10}^{-6}\mathrm{T}$$

Magnetic flux, linked with the smaller loop (of area ${\mathrm{A}}_2$ ), placed at centre will be (nearly)

$$\begin{array}{rl}{\varphi }_2={\mathrm{B}}_1{\mathrm{A}}_2& \cos \theta \\ & ={\mathrm{B}}_1{\mathrm{A}}_2\cos \omega \mathrm{t}\\ & =(2\pi \times {10}^{-6}\times 5\times {10}^{-4})\cos \omega \mathrm{t}\\ & =(\pi \times {10}^{-9})\cos \omega \mathrm{t}\end{array}$$

The induced emf, in the smaller

$${\epsilon }_2=\frac{-\mathrm{d}{\varphi }_2}{\mathrm{dt}}=-\pi \times {10}^{-9}(\frac{\mathrm{d}}{\mathrm{dt}}\cos \omega \mathrm{t})$$

$$=[(\pi \times {10}^{-9}\omega )\sin \omega \mathrm{t}]\text{ Volt }$$

$\therefore$ Induced current

$${\mathrm{i}}_2=\frac{{\epsilon }_2}{\mathrm{R}}$$

$$\therefore {\mathrm{i}}_2=[(\frac{\pi }{2}\times {10}^{-9}\omega )\sin \omega \mathrm{t}]\text{ ampere }$$

Induced emf in a planer coil,due to a change in its orientation in a uniform magnetic field

Let a planer coil, having $\mathrm{N}$ turns, of area $\mathrm{A}$ each, rotate in a uniform magnetic field, $\mathrm{B}$, with a constant angular velocity ’ $\omega$ ‘, about an axis coinciding with the plane of the coil and perpendicular to the uniform field. As a result, the effective area of the coil, exposed to the magnetic field lines, changes with time. The flux, at any time $t$, is

$${\varphi }_{\mathrm{B}}=\mathrm{N}(\mathrm{BA}\cos \omega \mathrm{t})(\because \text{ effective area }=\mathrm{A}\cos \theta =\mathrm{A}\cos \omega \mathrm{t})$$

From Faraday’s law, induced emf in the rotating coil, is

$$\begin{array}{c}\epsilon =-\frac{\mathrm{d}{\varphi }_{\mathrm{B}}}{\mathrm{dt}}=-\mathrm{NBA}\frac{\mathrm{d}}{\mathrm{dt}}(\cos \omega \mathrm{t})\\ =\mathrm{NBA}\omega \sin \omega \mathrm{t}\end{array}$$

This is the instantaneous value of induced emf. Clearly ${\epsilon }_0(=\mathrm{NBA}\omega )$ in the maximum value of this induced emf; this occurs when $\sin \omega t=\pm 1$. We can write the above equation as

$\epsilon ={\epsilon }_0\sin \omega \mathrm{t}$ where ${\epsilon }_0=\mathrm{NBA}\omega$.

Hence induced emf, in the coil, varies with time in a sinusoidal way. When the plane of the coil is parallel to B $(\sin \omega t=1)$, the induced emf is maximum. When this plane is perpendicular to $B(\sin \omega t=0)$, the induced emf is zero.

This method, of producing a flux change, due to a change in the loop’s orientation, $(\text{or change in its effective area }(A\cos \theta ))$ is the principle of operation of a simple a.c. generator.

An a.c. generator converts mechanical energy into electrical energy. In commercial generators, the mechanical energy, required for rotation of the coil, is provided either by water falling from heights in dams (hydroelectric generators) or by steam at high pressure, produced in thermal generators. Nuclear energy is now also being increasingly used, for this purpose, in the so called nuclear energy powered electric generators.

Alternating Current

Alternating current is that current which continuously varies in magnitude and periodically reverse its direction. The same is true for alternating emf’s or voltages.

The simples type of alternating current is one which varies with time is a simple harmonic (sinusoidal) way. It is represented by $\mathrm{i}={\mathrm{i}}_0\sin \omega \mathrm{t}$

$$

where is the instantaneous value of curernt, at time $\mathrm{t},{\mathrm{i}}_0$ is the maximum (or peak) value of the curernt. ${\mathrm{i}}_0$ may be called as the ‘current amplitude’. ’ $\omega$ ’ is called the angular frequency, of the alternating current (a.c.), and in given by

$$\omega =\frac{2\pi }{T}=2\pi f$$

Here $\mathrm{T}$ is the time period and $\mathrm{f}$ is the (linear) frequency of the alternating current. In the same manner, the instantaneous value of a sinusoidal alternating emf(or voltage) is represent by

$$\mathrm{v}={\mathrm{v}}_0\sin \omega \mathrm{t}$$

Mean (or Average) Value of A.C.

An alternating current (or voltage) flows, during one half cycle, in one direction, and during the other half cycle in opposite direction. Hence for one complete cycle the mean value of the a.c. would be zero. This is why there is no (net) deflection in a moving coil galvanometer when an alternating current passes through it. Average value over full cycle

$${\mathrm{i}}_{\mathrm{av}}=\frac{{\int }_0^{\mathrm{T}}{\mathrm{i}}_0\sin \omega \mathrm{tdt}}{{\int }_0^{\mathrm{T}}\mathrm{dt}}=0$$

However, the mean value, of a sinusoidal a.c, over a half cycle, has a finite value.

Mean, or average value over half a cycle, is given by

$\begin{array}{r}{\mathrm{i}}_{av}=\frac{{\int }_0^{T/2}{\mathrm{i}}_0\sin \omega tdt}{{\int }_0^{T/2}\mathrm{dt}}=\frac{2{\mathrm{i}}_0}{\pi }\end{array}$

$=0.637{\mathrm{i}}_0$

$($ Similarly ${\mathrm{v}}_{\mathrm{av}}=\frac{2{\mathrm{v}}_0}{\pi }=0.637{\mathrm{v}}_0)$

We thus notice that the mean value, of a simusoidal a.c., over a half cycle, is $63.7\mathrm{\%}$ of its peak value.

Root mean Square Value of a.c

The root mean square (r.m.s) value of a.c is defined as the square root of the average value of ${\mathrm{i}}^2$, over a complete cycle.

Now, average value of ${\mathrm{i}}^2$, over a full cycle is given by,

$$\overline{{\mathrm{i}}^2}=\frac{1}{\mathrm{T}}{\int }_0^{\mathrm{T}}{\mathrm{i}}^2\mathrm{dt}$$

$$=\frac{1}{\mathrm{T}}{\int }_0^{\mathrm{T}}{\mathrm{i}}_0^2{\sin }^2\omega \mathrm{tdt}$$

This comes out to be $\overline{{\mathrm{i}}^2}=\frac{{\mathrm{i}}_0^2}{2}$

Hence, the root mean square value of a sinosoidal a.c, is given by

$$\begin{array}{r}{\mathrm{i}}_{\text{r.m.s }}=\sqrt{\overline{{\mathrm{i}}^2}}=\frac{{\mathrm{i}}_0}{\sqrt{2}}\\ \\ =0.707{\mathrm{i}}_0\end{array}$$

Thus, the root mean square value of an a.c, is $70.7\mathrm{\%}$ of its peak value.

In a similar way,

$${\mathrm{v}}_{\mathrm{r}.\mathrm{m}.\mathrm{s}}=\frac{{\mathrm{v}}_0}{\sqrt{2}}$$

Physical Significance of r.m.s Value

If an a.c, $i=i_0\sin \omega t$, passes through a resistance $R$, the instantaneous rate of heat production is

$$\mathrm{P}={\mathrm{i}}^2\mathrm{R}$$

The ‘average’ rate of production of heat, over one complete cycle of current, is then given by

$${\mathrm{P}}_{\mathrm{av}}=\overline{{\mathrm{i}}^2}\mathrm{R}={\mathrm{i}}_{\mathrm{rms}}^2\mathrm{R}$$

Now if one passes a direct current, of strength ${\mathrm{i}}_{\mathrm{rms}}$, in a resistance $\mathrm{R}$, the rate of heat production will again be ${\mathrm{i}}_{\mathrm{rms}}^2\mathrm{R}$. Thus the root mean square value, of an a.c., is equal to that direct current, which produces the same (total) heat, in a given resistance (in a time equal to the time period of the a.c.) as the given alternating current does over its one complete cycle.

Hence, the root mean square value, of an a.c., is also called as the ’effective value’ or the ‘virtual value’ of the given a.c. The measurable value of an a.c. is thus its r.m.s value, or ’effective (or virtual) value’. The ammeter or voltmeter designed to measure a.c., would directly give the relevant r.m.s value. In our houses, alternating current is supplied at $220\mathrm{V}$. It means that

${\mathrm{V}}_{\mathrm{rms}}=220\mathrm{V}$. The peak value $\left({\mathrm{v}}_0\right)=\sqrt{2}{\mathrm{v}}_{\mathrm{rms}}=311\mathrm{V}$

‘Phasors’ or Phasor Diagrams

In an a.c. circuit, the frequency of the alternating current and alternating emf is same. However, it is not necessary that they both be in the same phase. Usually, when the current in the circuit in maximum, the emf is not maximum and vice-versa. The phase difference between the two, depends upon the type of circuit. Phase of a.c may be defined as the time period that has elapsed. Since the current (or emf) last passed its zero value in the positive direction.

The study of a.c circuits is much simplified if we treat alternative current or alternating emf as (kind of) vectors. The angle between the ‘vectors’ representing current and voltage is taken equal to the phase difference between them. The current and emf ‘vectors’ are better called ‘phasors’. Adiagram, representing alternating current and emf as ‘phasors’, showing the phase angle between them, is called a ‘Phasor Diagram’.

AC Voltage applied to a Resistor

Let an alternating voltage $\mathrm{v}={\mathrm{v}}_0\sin \omega \mathrm{t}$ be applied across a pure resistance ’ $R$ ‘. The applied voltage will be equal to the $p$.d across $R$ at any instant of time t. Thus

$$v(t)=i(t)R$$

$$

$\therefore {\mathrm{v}}_0\sin \omega \mathrm{t}=\mathrm{iR}$

Hence $\mathrm{i}=\frac{{\mathrm{v}}_0}{\mathrm{R}}\sin \omega \mathrm{t}={\mathrm{i}}_0\sin \omega \mathrm{t}$

where ${\mathrm{i}}_0(=\frac{{\mathrm{v}}_0}{\mathrm{R}})$ is the peak value of current in the circuit.

The above expression shows that in a pure resistance, the current is always in phase with the applied voltage.

$$

The above graphical representation shows this. It shows that both $v$ and i reach their zero, minimum and maximum values at the same time; there is thus no phase difference between $v$ and $i$. The instantaneous power disipated in resistor is

$$\mathrm{P}={\mathrm{i}}^2\mathrm{R}=({\mathrm{i}}_0^2{\sin }^2\omega \mathrm{t})\mathrm{R}$$

The average value of $\mathrm{P}$, over a complete cycle, is

$$\begin{array}{rl}& {\mathrm{P}}_{\mathrm{av}}=\frac{({\int }_0^{\mathrm{T}}{\mathrm{i}}_0^2{\sin }^2\omega t\mathrm{dt})}{\mathrm{T}}\mathrm{R}=\frac{{\mathrm{i}}_0^2\mathrm{R}}{\mathrm{T}}{\int }_0^{\mathrm{T}}{\sin }^2\omega t\mathrm{dt}\\ & =\frac{1}{2}{\mathrm{i}}_0^2\mathrm{R}={\left(\frac{{\mathrm{i}}_0}{\sqrt{2}}\right)}^2\mathrm{R}\\ & \therefore {\mathrm{P}}_{\mathrm{av}}={\mathrm{i}}_{\mathrm{rms}}^2\mathrm{R}=\frac{{\mathrm{v}}_{\mathrm{rms}}^2}{\mathrm{R}}\\ & (\because {\mathrm{v}}_{\mathrm{rms}}={\mathrm{i}}_{\mathrm{rms}}\mathrm{R})\end{array}$$

AC Voltage applied to a Pure Inductor

Let an alternating voltage, $\mathrm{v}={\mathrm{v}}_0\sin \omega \mathrm{t}$, be applied across a pure inductor of inductance L (it is supposed to have zero resistance). As the current i, in the inductor varies with time, an opposing back $\mathrm{emf}(=\mathrm{L}\frac{\mathrm{di}}{\mathrm{dt}})$ (self induced), gets induced in the inductor. We, therefore, have, as per Kirchoff’s law.

$$

$$\begin{array}{c}\mathrm{v}-\mathrm{L}\frac{\mathrm{di}}{\mathrm{dt}}=0\\ \therefore {\mathrm{v}}_0\sin \omega \mathrm{t}=\mathrm{L}\frac{\mathrm{di}}{\mathrm{dt}}\\ \therefore \mathrm{di}=(\frac{{\mathrm{v}}_0}{\mathrm{L}}\sin \omega \mathrm{t})\mathrm{dt}\end{array}$$

On integrating both sides, we get an expression for the instantaneous current in the circuit. We have

$\mathrm{i}=(-\frac{{\mathrm{v}}_0}{\omega \mathrm{L}})\cos \omega \mathrm{t}$

or $\mathrm{i}=\left(\frac{{\mathrm{v}}_0}{\omega \mathrm{L}}\right)\sin (\omega \mathrm{t}-\frac{\pi }{2})$

or $\mathrm{i}={\mathrm{i}}_0\sin (\omega \mathrm{t}-\frac{\pi }{2})$

Here ${\mathrm{i}}_0(=\frac{{\mathrm{v}}_0}{\omega \mathrm{L}})$ is the maximum, or peak, value of curernt in the circuit.

A comparison of this equation with the voltage equation, shows that, in a pure inductor, the current lags behind the applied voltage by a phase angle of $\frac{\pi }{2}$. Voltage thus reaches its maximum value $\left(\frac{T}{4}\right)$ seconds earlier than the current in the circuit.

As $i_0=\frac{v_0}{\omega \mathrm{L}}$, we find that the product $\omega \mathrm{L}$ must have the same dimension as those of resistance. This term, therefore, represents the effective opposition of the inductor to the flow of alternating current. It is known as the ‘reactance’ of the inductance, or inductive reactance. It is denoted by the symbol ${\mathrm{X}}_{\mathrm{L}}$. Thus ${\mathrm{X}}_{\mathrm{L}}=\omega \mathrm{L}=2\pi \mathrm{fL}$

$$

The inductive reactance increases with frequency, in a linear, or proportinoal way.

We can also write, ${\mathrm{v}}_{\mathrm{rms}}={\mathrm{i}}_{\mathrm{rms}}{\mathrm{X}}_{\mathrm{L}}$.

The instantaneous power supplied to the inductor is

$$\begin{array}{rl}\mathrm{P}=\mathrm{vi}=& ({\mathrm{v}}_0\sin \omega \mathrm{t})({\mathrm{i}}_0\sin (\omega \mathrm{t}-\frac{\pi }{2}))\\ & =-{\mathrm{v}}_0{\mathrm{i}}_0\sin \omega \mathrm{t}\cos \omega \mathrm{t}\end{array}$$

$$

The average power, over a complete cycle, is given by

$${\mathrm{P}}_{\mathrm{av}}=\frac{1}{\mathrm{T}}{\int }_0^{\mathrm{T}}\mathrm{vidt}$$

This comes out to be zero i.e. ${\mathrm{P}}_{\mathrm{av}}=0$. This means that in alternate quater cycles, energy is absorbed from the source by the inductor and then returned back to the source. This is because the ‘back emf’ opposes both the growth and decay of the current in the circuit. Over a complete cycle, the inductor thus acts as a ‘wattless’ circuit element.

AC Voltage applied to a Pure Capacitor

Let alternating voltage, $\mathrm{v}={\mathrm{v}}_0\sin \omega \mathrm{t}$, be applied to a pure capacitor of capacitance ’ $\mathrm{C}$ ‘. The capacitor gets alternatively charged, and discharged, as the current reverses each half cycle.

$$

Let $q$ be the charge on the capacitor plates, at any instant of time ’ $t$ ‘. We have

$\mathrm{q}(\mathrm{t})=\mathrm{cv}(\mathrm{t})$

$\therefore \mathrm{v}(\mathrm{t})=\frac{\mathrm{q}}{\mathrm{c}}(\mathrm{t})$

$\therefore \frac{\mathrm{dv}}{\mathrm{dt}}=\frac{1}{\mathrm{c}}\frac{\mathrm{dq}}{\mathrm{dt}}$

The instaneous current in the circuits is, therefore, given by

$$\mathrm{i}=\frac{\mathrm{dq}}{\mathrm{dt}}=\mathrm{c}\frac{\mathrm{d}}{\mathrm{dt}}({\mathrm{v}}_0\sin \omega \mathrm{t})$$

$\therefore \mathrm{i}={\mathrm{cv}}_0\omega \cos \omega \mathrm{t}$

or $\mathrm{i}=\frac{{\mathrm{v}}_0}{\left(\frac{1}{\omega \mathrm{c}}\right)}\cos \omega \mathrm{t}=\frac{{\mathrm{v}}_0}{\left(\frac{1}{\omega \mathrm{c}}\right)}\sin (\omega \mathrm{t}+\frac{\pi }{2})$

where $(=\frac{{\mathrm{v}}_0}{\left(\frac{1}{\omega \mathrm{c}}\right)})$ is the maximum, or peak value, of current in the circuit.

A comparison of this equation with the voltage equation shows that, in a pure capacitor, current is ahead of voltage by a phase angle of $\frac{\pi }{2}$. The current, therefore, reaches its maximum value $\frac{T}{4}$ seconds earlier than the voltage applied across the capacitor.

$$

As ${\mathrm{i}}_0=\frac{{\mathrm{v}}_0}{\left(\frac{1}{\omega \mathrm{c}}\right)}$, we find that the product $\frac{1}{\omega \mathrm{c}}$ must have the same dimension as those of resistance. This term thus represents the ’effective opposition’ of the capacitor to the flow of alternating current. It is known as ‘reactance of the circuit’ or ‘capacitive reactance’. It is denoted by the symbol ${\mathrm{X}}_{\mathrm{C}}$.

$${\mathrm{X}}_{\mathrm{C}}=\frac{1}{\omega \mathrm{c}}=\frac{1}{2\pi \mathrm{fc}}$$

Thus capacitive reactance decreases with increase in freqeuncy. As $X_C\propto \frac{1}{\omega }$, the graph of $X_C$, against $\omega$ (or f), has the form shown here.

$$

We can also write, ${\mathrm{v}}_{\mathrm{rms}}={\mathrm{i}}_{\mathrm{rms}}{\mathrm{X}}_{\mathrm{C}}$.

The instantaneous power supplied to the capacitor is

$$\begin{array}{rl}P=vi& =(v_0\sin \omega t)(i_0\sin (\omega t+\frac{\pi }{2}))\\ & ={\mathrm{v}}_0{\mathrm{i}}_0\sin \omega \mathrm{t}\cos \omega \mathrm{t}\end{array}$$

The average power, over a complete cycle,

$${\mathrm{P}}_{\mathrm{av}}=\frac{1}{\mathrm{T}}{\int }_0^{\mathrm{T}}\mathrm{vidt}$$

This comes out to be zero.

This means that energy is absorbed from the source during one quater cycle (when capacitor is getting charged). In the next quater cycle, energy gets returned to the source, (as capacitor gets discharged over a complete cycle), the capacitor thus acts like a ‘wattless’ element.

To summarise:

Pure R	Pure L	Pure C
$\mathrm{v}={\mathrm{v}}_0\sin \omega \mathrm{t}$	$\mathrm{v}={\mathrm{v}}_0\sin \omega \mathrm{t}$	$\mathrm{v}={\mathrm{v}}_0\sin \omega \mathrm{t}$
$\mathrm{i}={\mathrm{i}}_0\sin \omega \mathrm{t}$	$\mathrm{i}={\mathrm{i}}_0\sin (\omega \mathrm{t}-\frac{\pi }{2})$	$\mathrm{i}={\mathrm{i}}_0\sin (\omega \mathrm{t}+\frac{\pi }{2})$
${\mathrm{v}}_{\mathrm{rms}}={\mathrm{i}}_{\mathrm{rms}}\mathrm{R}$	${\mathrm{v}}_{\mathrm{rms}}={\mathrm{i}}_{\mathrm{rms}}{\mathrm{X}}_{\mathrm{L}}$	${\mathrm{v}}_{\mathrm{rms}}={\mathrm{i}}_{\mathrm{rms}}{\mathrm{X}}_{\mathrm{C}}$
${\mathrm{P}}_{\mathrm{av}}={\mathrm{i}}_{\mathrm{rms}}^2\mathrm{R}$	${\mathrm{P}}_{\mathrm{av}}=0$	${\mathrm{P}}_{\mathrm{av}}=0$
$\mathrm{v},\mathrm{i}$ are in same phase	$\mathrm{v}$ is ahead ofi (in phase) by $\frac{\pi }{2}$	v lags behind i (in phase) by $\frac{\pi }{2}$

AC Voltage applied to Inductor and Resistor in Series (LR Series Circuit)

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Let an alternating voltage, $\mathrm{v}={\mathrm{v}}_0\sin \omega \mathrm{t}$ be applied to a circuit containing a pure $\mathrm{L}$ and a pure $\mathrm{R}$, joined in series. The same current ’ $i$ ’ will flow both in L and R. However, in R, it is in phase with the applied a.c voltage but in $\mathrm{L}$, it lags behind the applied a.c voltage by a phase angle of $\frac{\pi }{2}$. Thus rms potential drop across $\mathrm{R},{\left({\mathrm{V}}_{\mathrm{R}}\right)}_{\mathrm{rms}}$, and the rms potential drop across $\mathrm{L},\left({\mathrm{V}}_{\mathrm{L}}\right)\mathrm{rms}$, must be shown at right angles to each other in a phasor diagram. The phasor diagram is, therefore, drawn as shown.

$$

$\therefore {\mathrm{V}}_{\mathrm{rms}}=\sqrt{{\left({\mathrm{V}}_{\mathrm{R}}\right)}_{\mathrm{rms}}^2+{\left({\mathrm{V}}_{\mathrm{L}}\right)}_{\mathrm{rms}}^2}$

or ${\mathrm{V}}_{\mathrm{rms}}={\mathrm{i}}_{\mathrm{rms}}\sqrt{{\mathrm{R}}^2+{\mathrm{X}}_{\mathrm{L}}^2}$

The term $\sqrt{R^2+X_L^2}$ is therefore, the ’effective resistance’ (series combination of the resistance of $R$ and the reactance of $\mathrm{L}$ ) of the circuit. It has been given the name ‘impedance of the circuit’. We use the symbol $\mathrm{z}$ for impedance.

Thus, here $\mathrm{z}=\sqrt{{\mathrm{R}}^2+{\mathrm{X}}_{\mathrm{L}}^2}$

$=\sqrt{{\mathrm{R}}^2+(\omega \mathrm{L}{)}^2}$

Also, ${\mathrm{v}}_{\mathrm{rms}}={\mathrm{i}}_{\mathrm{rms}}\mathrm{z}$

The phasor diagram shows that, in a series LR circuit, the voltage (still) leads the current, now by a phase angle $\varphi (0<\varphi <\frac{\pi }{2})$. This phase angle is given by

$$\tan \varphi =\frac{{\left({\mathrm{V}}_{\mathrm{L}}\right)}_{\mathrm{rms}}}{{\left({\mathrm{V}}_{\mathrm{R}}\right)}_{\mathrm{rms}}}=\frac{{\mathrm{i}}_{\mathrm{rms}}\left({\mathrm{X}}_{\mathrm{L}}\right)}{{\mathrm{i}}_{\mathrm{rms}}(\mathrm{R})}=\frac{\omega \mathrm{L}}{\mathrm{R}}$$

The inductive effect dominates in a series LR circuit. This means that the voltage reaches its maximum value $\left(\frac{T}{2\pi }\varphi \right)$, or $\left(\frac{\varphi }{\omega }\right)$, seconds earlier than the current in the circuit. (There is a time $-\mathrm{lag}$ of $\mathrm{\Delta }\mathrm{t}=\frac{\mathrm{T}\varphi }{2\pi }=\frac{\varphi }{\omega }$ seconds, between the current and the voltage, vis-a-vis reaching their peak values).

Example-9:

A students connects a long air cored coil, of manganin wire, to a $100\mathrm{V}$ d.c source and records a current of $1.5\mathrm{A}$. When the same coil is connected across. $(100\mathrm{V};50\mathrm{Hz})$ a.c source, the current reduces to $1\mathrm{A}$. Why? Also calculate the reactance of the coil.

Show Answer

Solution :

In d.c circuit the impedance of the coil, is only due to its resistance ’ $R$ ‘. Hence

$$\begin{array}{c}{\mathrm{i}}_{\mathrm{d}.\mathrm{c}}=\frac{\mathrm{V}}{\mathrm{R}}\\ \therefore \mathrm{R}=\frac{100}{1.5}\mathrm{\Omega }=66.67\mathrm{\Omega }\end{array}$$

When the same coil is connected to an a.c source, the impedance of the coil is due to both its resistance $\mathrm{R}$ and its inductance L. The coil now acts like a LR series circuit, having a impedance

$$\mathrm{z}=\sqrt{{\mathrm{R}}^2+(\omega \mathrm{L}{)}^2}$$

The value of current, in the a.c circuit decreases, because of this increase in impedance of the coil.

Now $i_{\text{rms }}=\frac{v_{\text{rms }}}{\sqrt{R^2+(\omega L{)}^2}}$

$\therefore 1=\frac{(100{)}^2}{\sqrt{(66.67{)}^2+(2\pi \mathrm{fL}{)}^2}}$

$\therefore (2\pi \mathrm{fL}{)}^2=(5555)$

$\Rightarrow {\mathrm{X}}_{\mathrm{L}}(=2\pi \mathrm{fL})=\sqrt{5555}\mathrm{\Omega }$

$=74.53\mathrm{\Omega }$

Example-10:

An inductor ’ $\mathrm{L}$ ‘, and a resistor ’ $\mathrm{R}$ ‘, are connected in series to an a.c source. The potential drop across inductor is 160 watts and that across the resistance ’ $R$ ’ is 120 volt. Find the virtual value of the applied voltage. Is algebric sum of the potential drops across $\mathrm{L}$ and $\mathrm{R}$ more than this value? If yes, resolve the paradox.

Show Answer

Solution :

We have ${\mathrm{v}}_{\mathrm{rms}}=\sqrt{{\left({\mathrm{V}}_{\mathrm{R}}\right)}_{\mathrm{rms}}^2+{\left({\mathrm{V}}_{\mathrm{L}}\right)}_{\mathrm{rm}}^2}$

$$\begin{array}{rl}& =\sqrt{(160{)}^2+(120{)}^2}\\ & =200\mathrm{V}\end{array}$$

Hence algebric sum of p.d. across $\mathrm{L}$ and $\mathrm{R}((160+120)\mathrm{V}=280\mathrm{V})$ is more than the virtual or $\mathrm{rms}$ value of applied voltage. This is due to the phase relation between ${\left({\mathrm{V}}_{\mathrm{L}}\right)}_{\mathrm{ms}}$ and ${\left({\mathrm{V}}_{\mathrm{R}}\right)}_{\mathrm{rms}}$. The potential drop, across the inductor, is ahead of potential drop across resistor by a phase angle of $\frac{\pi }{2}$. Therefore, they cannot be added like pure numbers. We need to add them by using the rule for ‘vector’. This type of ‘addition’ gives

$$\begin{array}{c}\mathrm{v}=\left[\sqrt{(120{)}^2+(160{)}^2}\right]\mathrm{V}\\ =200\mathrm{V}\end{array}$$

AC Voltage applied to a Capacitor and a Resistor in Series (RC Series Circuit)

$$

Let an alternating voltage, $\mathrm{v}={\mathrm{v}}_0\sin \omega \mathrm{t}$, be applied to a circuit containing a pure $\mathrm{C}$ an da Pure $\mathrm{R}$, joined in series. The same current $\mathrm{i}$ will flow in both $\mathrm{C}$ and $\mathrm{R}$. However in $\mathrm{R}$ it is in phase with the applied a.c voltage, whereas in $\mathrm{C}$, it leads the a.c.voltage in phase by $\frac{\pi }{2}$. Thusrms potential drop, across the resistance, is ahead ot the rms potential drop across the capacitor by a phase angle of $\frac{\pi }{2}$. The phasor diagram is, therefore drawn as shown.

Thus, ${\mathrm{v}}_{\mathrm{rms}}=\sqrt{{\left({\mathrm{V}}_{\mathrm{R}}\right)}_{\mathrm{rms}}^2+{\left({\mathrm{V}}_{\mathrm{C}}\right)}_{\mathrm{ms}}^2}$

or ${\mathrm{V}}_{\mathrm{rms}}={\mathrm{i}}_{\mathrm{rms}}\sqrt{{\mathrm{R}}^2+{\mathrm{X}}_{\mathrm{C}}^2}$

Here $\sqrt{{\mathrm{R}}^2+{\mathrm{X}}_{\mathrm{C}}^2}$ is the ’effective resistance’ of the (series CR) circuit.

$$

It is called the impedance of this circuit and is denoted by $z$.

$$\therefore \mathrm{z}=\sqrt{{\mathrm{R}}^2+{\mathrm{X}}_{\mathrm{C}}^2}$$

$$=\sqrt{{\mathrm{R}}^2+{\left(\frac{1}{\omega \mathrm{c}}\right)}^2}$$

and ${\mathrm{v}}_{\mathrm{rms}}={\mathrm{i}}_{\mathrm{rms}}\mathrm{z}$

The phasor diagram shows that in the $\mathrm{RC}$ circuit, the voltage lags behind the current by a phase angle $\varphi (0<\varphi <\frac{\pi }{2})$. As seen from the phasor diagram, the phase angle, $\varphi$, is given by

$$\tan \varphi =\frac{{\left({\mathrm{V}}_{\mathrm{C}}\right)}_{\mathrm{rms}}}{{\left({\mathrm{V}}_{\mathrm{R}}\right)}_{\mathrm{rms}}}=\frac{{\mathrm{X}}_{\mathrm{C}}}{\mathrm{R}}=\frac{1}{\omega \mathrm{CR}}$$

The capacitive effect dominates in this circuit. This means that the current reaches its maximum value $\left(\frac{T}{2\pi }\varphi \right)$, or $\left(\frac{\varphi }{\omega }\right)$, seconds earlier value the voltage in the circuit. There is a time lag of $\mathrm{\Delta }t=\frac{T}{2\pi }=\frac{\varphi }{\omega }$ seconds between the voltage and the current, vis-a-vis reaching their peak values.

Example-11:

A $100\mu \mathrm{F}$ capacitor is connected in seires with a $40\mathrm{\Omega }$ resistor to a ( $110\mathrm{V},60\mathrm{Hz}$ ) supply. Find the time lag between the instants at which the current and voltage reach their, respective, maximum values.

Show Answer

Solution :

Time lag $\mathrm{\Delta }\mathrm{t}=\frac{\mathrm{T}}{2\pi }\varphi$

Here $\varphi ={\tan }^{-1}\left(\frac{{\mathrm{X}}_{\mathrm{C}}}{\mathrm{R}}\right)={\tan }^{-1}\left(\frac{1}{\omega \mathrm{CR}}\right)$

$$\begin{array}{rl}& =[{\tan }^{-1}\left(\frac{1}{2\times 3.14\times 60\times 40}\right)]\\ & ={33.56}^{\circ }\end{array}$$

Now $\mathrm{\Delta }\mathrm{t}=\frac{\mathrm{T}}{{360}^{\circ }}\times \left({33.56}^{\circ }\right)(\mathrm{T}=\frac{1\mathrm{s}}{60})$

$$\begin{array}{rl}\therefore \mathrm{\Delta }\mathrm{t}& =[\frac{{33.56}^{\circ }}{{360}^{\circ }}\times \frac{1}{60}]\mathrm{s}=1.55\times {10}^{-3}\mathrm{s}\\ & =1.55\mathrm{ms}\end{array}$$

AC Voltage applied to a Pure Inductor and a Pure Capacitor in Series (LC Series Circuit)

$$

In this case, voltage across $L$ will lead the current in the circuit by $\frac{\pi }{2}$; while the voltage across Cwill lag behind the current by $\frac{\pi }{2}$. Thus phase difference between the two voltages is $\pi$.

$$

Hence ${\mathrm{v}}_{\mathrm{rms}}=|{\left({\mathrm{V}}_{\mathrm{L}}\right)}_{\mathrm{rms}}-{\left({\mathrm{V}}_{\mathrm{C}}\right)}_{\mathrm{rms}}|$

$={\mathrm{i}}_{\mathrm{rms}}|{\mathrm{X}}_{\mathrm{L}}-{\mathrm{X}}_{\mathrm{C}}|$

Here $|X_L-X_C|$ is the effective reactance of the circuit.

Special Case:

(i) If, in this circuit, $X_L=X_C$, the amplitude of current in the circuit, would tend to be infinite. It is therefore, the condition of ’electrical resonance in this circuit.

When ${\mathrm{X}}_{\mathrm{L}}={\mathrm{X}}_{\mathrm{C}}$, we have

$$\omega \mathrm{L}=\frac{1}{\omega \mathrm{C}}$$

or

$$\omega =\frac{1}{\sqrt{\mathrm{LC}}},\text{ or }f=\frac{1}{2\pi \sqrt{\mathrm{LC}}}$$

This frequency value ( $=$ f), is called the ‘resonant frequency’ of the given (series) LC circuit.

(ii) If $X_L>X_C$, the inductive effect dominates. The current lags behind the voltage by a phase angle of $\frac{\pi }{2}$ in this a.c circuit. However the peak value, or amplitude of current, increases as effective value of reactance is lower than that of either $\mathrm{L}$ or $\mathrm{C}$.

(iii) If $X_C>X_L$, the capacitive effect dominates. The current leads the voltage by a phase angle of $\frac{\pi }{2}$ in the a.c circuit. However the peak value of amplitude, of current increases as the effective value of reactance is lower than that of either $\mathrm{L}$ or $\mathrm{C}$.

AC Voltage applied to Series LCR Circuit

$$

Let an alternating voltage, $\mathrm{v}={\mathrm{v}}_0\sin \omega \mathrm{t}$, be applied to an a.c circuit, having $\mathrm{L},\mathrm{C}$ and $\mathrm{R}$, joined in series. The same current i, will flow in all the three elements. As discussed earlier, the phasor diagram for this circuit will be drawn as shown here.

$$

Case (i) :

If ${\left({\mathrm{V}}_{\mathrm{L}}\right)}_{\mathrm{rms}}>{\left({\mathrm{V}}_{\mathrm{C}}\right)}_{\mathrm{rms}}$, the resultant, of these two, will be $[{\left({\mathrm{V}}_{\mathrm{L}}\right)}_{\text{rms }}-{\left({\mathrm{V}}_{\mathrm{C}}\right)}_{\mathrm{rms}}]$.

We, therefore, have

$$\begin{array}{rl}{\mathrm{v}}_{\mathrm{rms}}=& \sqrt{{\left({\mathrm{V}}_{\mathrm{R}}\right)}_{\mathrm{rms}}^2+({\left({\mathrm{V}}_{\mathrm{L}}\right)}_{\mathrm{rms}}-{\left({\mathrm{V}}_{\mathrm{C}}\right)}_{\mathrm{rms}})2}\\ & ={\mathrm{i}}_{\mathrm{rms}}\sqrt{{\mathrm{R}}^2+{({\mathrm{X}}_{\mathrm{L}}-{\mathrm{X}}_{\mathrm{C}})}^2}\\ & ={\mathrm{i}}_{\mathrm{rms}}\mathrm{z}\end{array}$$

$$

Here, the impedance of the circuit, ’ $z$ ’ is given by

$$\mathrm{z}=\sqrt{{\mathrm{R}}^2+{({\mathrm{X}}_{\mathrm{L}}-{\mathrm{X}}_{\mathrm{C}})}^2}$$

Here ${\left({\mathrm{V}}_{\mathrm{L}}\right)}_{\text{rms }}>{\left({\mathrm{V}}_{\mathrm{C}}\right)}_{\text{rms }}$

$$\Rightarrow {\mathrm{i}}_{\mathrm{rms}}{\mathrm{X}}_{\mathrm{L}}>{\mathrm{i}}_{\mathrm{rm}}{\mathrm{X}}_{\mathrm{C}}$$

$\therefore {\mathrm{X}}_{\mathrm{L}}>{\mathrm{X}}_{\mathrm{C}}$

Hence the inductive effect dominates in such a circuit. The applied voltage, $\mathrm{V}$, leads the current, $i$, by a phase angle $\varphi$ where

$$\begin{array}{rl}\tan \varphi =& \frac{[{\left({\mathrm{V}}_{\mathrm{L}}\right)}_{\mathrm{rms}}-{\left({\mathrm{V}}_{\mathrm{C}}\right)}_{\mathrm{rms}}]}{{\left({\mathrm{V}}_{\mathrm{R}}\right)}_{\mathrm{rms}}}\\ & =\left(\frac{{\mathrm{X}}_{\mathrm{L}}-{\mathrm{X}}_{\mathrm{C}}}{\mathrm{R}}\right)\end{array}$$

This means that the voltage reaches its maximum value $\left(\frac{\mathrm{T}}{2\pi }\varphi \right)$ or $\left(\frac{\varphi }{\omega }\right)$ seconds eariler than the current reaching its maximum value, in the circuit. The value of instantaneous current in the circuit is, $i={\mathrm{i}}_0\sin (\omega t-\varphi )$

Case (ii) :

If ${\left({\mathrm{V}}_{\mathrm{C}}\right)}_{\mathrm{ms}}>{\left({\mathrm{V}}_{\mathrm{L}}\right)}_{\mathrm{rms}}$, the resultant of the two will be $[{\left({\mathrm{V}}_{\mathrm{C}}\right)}_{\mathrm{mms}}-{\left({\mathrm{V}}_{\mathrm{L}}\right)}_{\mathrm{ms}}]$

We, therefore, have

$$\begin{array}{rl}& {\mathrm{V}}_{\mathrm{rms}}=\sqrt{{\left({\mathrm{V}}_{\mathrm{R}}\right)}_{\mathrm{rms}}^2+{({\left({\mathrm{V}}_{\mathrm{C}}\right)}_{\mathrm{rms}}-{\left({\mathrm{V}}_{\mathrm{L}}\right)}_{\mathrm{rms}})}^2}\\ & ={\mathrm{i}}_{\mathrm{rms}}\sqrt{{\mathrm{R}}^2+{({\mathrm{X}}_{\mathrm{C}}-{\mathrm{X}}_{\mathrm{L}})}^2}\\ & ={\mathrm{i}}_{\mathrm{rms}}\mathrm{Z}\end{array}$$

$$

Here the impedance of the circuit, ’ $Z$ ‘, is given by

$\mathrm{Z}=\sqrt{{\mathrm{R}}^2+{({\mathrm{X}}_{\mathrm{C}}-{\mathrm{X}}_{\mathrm{L}})}^2}$

Here ${\left({\mathrm{V}}_{\mathrm{C}}\right)}_{\mathrm{mms}}>{\left({\mathrm{V}}_{\mathrm{L}}\right)}_{\mathrm{ms}}$

$\Rightarrow {\mathrm{i}}_{\text{rms }}{\mathrm{X}}_{\mathrm{C}}>{\mathrm{i}}_{\text{rms }}{\mathrm{X}}_{\mathrm{L}}$

or ${\mathrm{X}}_{\mathrm{C}}>{\mathrm{X}}_{\mathrm{L}}$ Hence capacitive effect dominates in such a circuit. The applied voltage, $\mathrm{V}$, lags behind the current i, by a phase angle $\varphi$; where

$$\begin{array}{c}\tan \varphi =\frac{[{\left({\mathrm{V}}_{\mathrm{C}}\right)}_{\mathrm{rms}}-{\left({\mathrm{V}}_{\mathrm{L}}\right)}_{\mathrm{rms}}]}{{\left({\mathrm{V}}_{\mathrm{R}}\right)}_{\mathrm{rms}}}\\ =\frac{[{\mathrm{X}}_{\mathrm{C}}-{\mathrm{X}}_{\mathrm{L}}]}{\mathrm{R}}\end{array}$$

This means that voltage reaches its maximum value $\left(\frac{T}{2\pi }\varphi \right)$ or $\left(\frac{\varphi }{\omega }\right)$ seconds later than the current reaching its maximum value in the circuit.

The instantaneous current $\mathrm{i}$ in such circuit is, $\mathrm{i}={\mathrm{i}}_0\mathrm{Sin}(\omega \mathrm{t}+\varphi )$

Case (iii) :

An interesting characteristics of the series LCR circuit is the phenomenon of resonance.

For an LCR circuit, driven with applied voltage of amplitude $V_0$ and (angular) frequency $\omega$, we know that the current amplitude equals

$$i_0=\frac{V_0}{Z}=\frac{V_0}{\sqrt{R^2+{(X_L-X_C)}^2}}$$

Here $X_L=\omega L$ and $X_C=\frac{1}{\omega C}$. If $\omega$ is varied, at a particular frequency ${\omega }_0$, we can have $X_C=X_L$. [or ${\left({\mathrm{V}}_{\mathrm{C}}\right)}_{\mathrm{rms}}={\left({\mathrm{V}}_{\mathrm{L}}\right)}_{\mathrm{rms}}]$

The impedance of the circuit is then at its minimum value $(z_{\text{min }}=\sqrt{R^2+0}=R)$. This (angular) frequency ${\omega }_0$, is called the resonant (angular) frequency. It is given by as ${\mathrm{X}}_{\mathrm{C}}={\mathrm{X}}_{\mathrm{L}},\frac{1}{{\omega }_0\mathrm{C}}={\omega }_0\mathrm{L}$

$$\therefore {\omega }_0=\frac{1}{\sqrt{\mathrm{LC}}}$$

At resonant frequency the current amplitude is maximum $\left({\mathrm{i}}_0\right)=\frac{{\mathrm{V}}_0}{{\mathrm{Z}}_{\text{min }}}=\frac{{\mathrm{V}}_0}{\mathrm{R}}$

For values of $\omega$, other than ${\omega }_0$, the amplitude of current in less than its maximum value. The maximum value of current, however, depends on the resistance $\mathrm{R}$ in the circuit.

Resonant circuits have a variety of applications. These include, among others, the tunning circuits of a radio or T.V. set, and the working of metal detectors.

Sharpness of Resonance

Suppose we choose a value of $\omega$ for which the current amplitude is $\frac{1}{\sqrt{2}}$ times its maximum value. At this frequency, the power, dissipated by the circuit, becomes half of its maximum value.

$$

From the graph between $\mathrm{i}$ and $\omega$, we see that there are two frequncies, ${\omega }_1$ and ${\omega }_2$, one greaer and the other smaller than ${\omega }_0$. These two frequencies are located symeterically about ${\omega }_0$; for the angular frequency, the current amplitude is $\frac{{\mathrm{i}}_0}{\sqrt{2}}$. These two frequencies are referred to as the half power frequencies.

$$\text{We can write}{\omega }_1={\omega }_0+\mathrm{\Delta }\omega$$

$${\omega }_2={\omega }_0-\mathrm{\Delta }\omega$$

The difference, $({\omega }_1-{\omega }_2)(=2\mathrm{\Delta }\omega )$, is often called the ‘bandwidth’ of the circuit. The quantity $\left(\frac{{\omega }_0}{2\mathrm{\Delta }\omega }\right)$ is regarded as a measure of the sharpness of resonance. The smaller $\mathrm{\Delta }\omega$ is the sharper, or narrower, is the resonance. The value of $\mathrm{\Delta }\omega$ comes out to be equal to $\frac{R}{2L}$. (by solving the expression for impedances for $\omega ={\omega }_0$ and $\omega ={\omega }_2$ or ${\omega }_1$ )

The sharpness of resonance is, therefore, given by

$$\frac{{\omega }_0}{2\mathrm{\Delta }\omega }=\frac{{\omega }_{0}L}{R}$$

The ratio $\left(\frac{{\omega }_{0}L}{R}\right)$ is called the ‘quality factor’, $Q$ of the circuit. Thus

$\mathrm{Q}=\frac{{\omega }_0}{({\omega }_2-{\omega }_1)}=\frac{{\omega }_0}{\mathrm{\Delta }\omega }=\frac{{\omega }_0\mathrm{L}}{\mathrm{R}}=\frac{1}{{\omega }_0\mathrm{CR}}=\frac{1}{\mathrm{R}}\sqrt{\frac{\mathrm{L}}{\mathrm{C}}}(\because {\omega }_0=\frac{1}{\sqrt{\mathrm{LC}}})$

Example-12:

An a.c. source of frequency $50\mathrm{Hz}$ is applied to a series circuit of resistance $20\mathrm{\Omega }$, an inductance of 100 $\mathrm{mH}$ and a capacitance of $30\mu \mathrm{F}$. Does the current lag or lead the applied voltage? If so, by what (phase) angle?

Show Answer

Solution :

Here inductive reactance $\mathrm{XL}=\omega \mathrm{L}$

$$\begin{array}{rl}& =2\pi \mathrm{fL}\\ & =2\times 3.14\times 50\times 0.1\mathrm{\Omega }=31.4\mathrm{\Omega }\end{array}$$

Capacitive reactance ${\mathrm{X}}_{\mathrm{C}}=\frac{1}{\omega \mathrm{C}}$

$$\begin{array}{rl}& =\frac{1}{2\pi \mathrm{fC}}=\frac{1}{2\times 3.14\times 60\times 30\times {10}^{-6}}\mathrm{\Omega }\\ & =106.2\mathrm{\Omega }\end{array}$$

Thus ${\mathrm{X}}_{\mathrm{C}}>{\mathrm{X}}_{\mathrm{L}}$. Hence the current leads the applied voltage by an angle $\varphi$ where

$$\begin{array}{rl}\tan \varphi =& \frac{({\mathrm{X}}_{\mathrm{C}}-{\mathrm{X}}_{\mathrm{L}})}{\mathrm{R}}=\frac{106.2-13.4}{20}\\ & =3.74\\ \therefore \varphi & ={\tan }^{-1}(3.74)\simeq 750\end{array}$$

Example-13:

AnLCR series circuit, with resistance $100\mathrm{\Omega }$ is connected to an a.c. source of $200\mathrm{V}$ and angular frequency $300\mathrm{rad}/\mathrm{s}$. When the capacitance is removed from the circuit, current lags behind the voltage by ${60}^{\circ }$. When the inductance is removed from the circuit, the current leads the voltage by ${60}^{\circ }$. Calculate the current in the circuit.

Show Answer

Solution :

We have $\tan {60}^{\circ }=\frac{\omega \mathrm{L}}{\mathrm{R}}$ (when ’ $\mathrm{C}$ ’ is removed) and

$\tan {60}^{\circ }=\frac{1}{\omega \mathrm{CR}}($ when ’ $\mathrm{L}$ ’ is removed)

$\therefore \omega \mathrm{L}=\frac{1}{\omega \mathrm{C}}$

or $\omega =\frac{1}{\sqrt{\mathrm{LC}}}$

This implies that circuit is at resonance.

$\therefore {\mathrm{i}}_{\mathrm{rms}}=\frac{{\mathrm{V}}_{\mathrm{rms}}}{{\mathrm{Z}}_{\text{min }}}=\frac{{\mathrm{V}}_{\mathrm{rms}}}{\mathrm{R}}$

$$=\frac{200}{100}\mathrm{A}=2\mathrm{A}$$

Example-14:

Sketch the variation of impedance (z) with angular frequency $(\omega )$, for a series LCR (a.c.) circuit.

$$

Show Answer

Solution :

We have ${\mathrm{Z}}^2={\mathrm{R}}^2+{(\omega \mathrm{L}-\frac{1}{\omega \mathrm{C}})}^2$

At $\omega ={\omega }_{\mathrm{r}},\mathrm{Z}=\mathrm{R}$

$\mathrm{Z}$ is more than $\mathrm{R}$ both for $\omega <{\omega }_{\mathrm{R}}$ and for $\omega >{\omega }_{\mathrm{R}}$. The curve is symmetrical.

Further,

For $\omega <{\omega }_{\mathrm{r}}$, capacitive effect dominates in the circuit.

$\omega >{\omega }_{\mathrm{r}}$, inductive effect dominates in the circuit.

Power in a Series LCR AC Circuit

When a voltage, $\mathrm{v}={\mathrm{v}}_0\sin \omega \mathrm{t}$, is applied to a series LCR circuit, the current in the circuit is given by

$\mathrm{i}={\mathrm{i}}_0\sin (\omega \mathrm{t}\mp \varphi )$. Where ${\mathrm{i}}_0=\frac{{\mathrm{v}}_0}{\mathrm{z}}$ and $\varphi ={\tan }^{-1}\left(\frac{|{\mathrm{X}}_{\mathrm{L}}-{\mathrm{X}}_{\mathrm{C}}|}{\mathrm{R}}\right)$

[negative sign for inductive effect and positive sign for capacitive effect]

Therefore, the instantneous power ’ $\mathrm{P}$ ’ supplied, by the source, is

$$\begin{array}{rl}& \mathrm{P}=\mathrm{vi}\\ & =({\mathrm{v}}_0\sin \omega \mathrm{t})({\mathrm{i}}_0\sin (\omega \mathrm{t}\mp \varphi ))\end{array}$$

The average power over a full cycle is given by

\

$$\begin{array}{rl}& {\mathrm{P}}_{\mathrm{av}}=\frac{1}{\mathrm{T}}{\int }_0^{\mathrm{T}}\mathrm{vidt}\\ & =\frac{{\mathrm{v}}_0{\mathrm{i}}_0}{\mathrm{T}}{\int }_0^{\mathrm{T}}(\sin \omega \mathrm{t})({\mathrm{i}}_0\sin (\omega \mathrm{t}\mp \varphi ))\mathrm{dt}\\ & =\frac{1}{2}{\mathrm{v}}_0{\mathrm{i}}_0\cos \varphi \\ & ={\mathrm{v}}_{\mathrm{rms}}{\mathrm{i}}_{\mathrm{rms}}\cos \varphi \end{array}$$

This can be written as

$${\mathrm{P}}_{\mathrm{av}}\equiv {\mathrm{i}}_{\mathrm{rms}}^2\mathrm{z}\cos \varphi$$

So, the average power dissipated in the circuit depends not only on the rms values of voltage and current but also on the phase angle, $\varphi$, between the two.

The quantity ’ $\cos \varphi$ ’ is called the power factor (for the series a.c.) circuit.

Case (i)

Resistive Circuit : If the circuit contains only a pure resistor, there voltage and current are in same phase i.e. $\varphi =0$. The power factor ’ $\cos \varphi$ ‘, is therefore unity. There is maximum power dissipation in such a pure resistive circuit.

Case (ii)

Purely inductive or capacitive or a series LC circuit: If the circuit contains only a pure reactance ’ $\mathrm{X}$ ‘, the phase diffrence, between voltage and current, is $(\pm \frac{\pi }{2})$. Therefore, power factor ’ $\cos \varphi$ ‘, is zero. Hence no net power is dissipated, over a complete cycle, though current is flowing through the circuit all the time. This current is therefore also referred to as a ‘watless current’.

Case (iii)

LCR Circuit : In a series LR, RC or a series LCR circuit, $\varphi$ is non zero $(0<\varphi <\frac{\pi }{2})$. Even in such cases, power is dissipated only in the resistor. In ideal case ${\mathrm{P}}_{\mathrm{av}}={\mathrm{v}}_{\mathrm{rms}}{\mathrm{i}}_{\mathrm{rms}}\cos \varphi ={\mathrm{i}}_{\mathrm{rms}}^2\mathrm{R}\cos \varphi$

Case (iv)

LCR circuit at resonance : At resonance $X_L=X_C$ and $\varphi =0$. Therefore ’ $\cos \varphi$ ’ is unity. As $Z_{\text{min }}=R$, ${\mathrm{P}}_{\mathrm{av}}={\mathrm{i}}_{\mathrm{rms}}^2\mathrm{R}$. Thus maximum power is dissipated in such a circuit (through $\mathrm{R}$ ) at resonance.

Example-15:

A series LR circuit draws a power of $550\mathrm{W}$ from a $220\mathrm{V}-50\mathrm{Hz}$ source. The power factor of the circuit in 0.8 . The current in the circuit lags behind the voltage. Show that a capacitor, of value about $(\frac{1}{42\pi }\times {10}^{-2})\mathrm{F}$, will have to be connected in the circuit to bring its power factor to unity.

Show Answer

Solution :

We are given that $\cos \varphi =0.8$

$\therefore \tan \varphi =\left(\frac{\omega \mathrm{L}}{\mathrm{R}}\right)=\frac{\sqrt{1-{\cos }^2\varphi }}{\cos \varphi }=\frac{0.6}{0.8}=\frac{3}{4}$

$\therefore \omega \mathrm{L}=\frac{3}{4}\mathrm{R}$

$\therefore$ Inpedance of the LR circuit $=\mathrm{z}=\sqrt{{\mathrm{R}}^2+(\omega \mathrm{L}{)}^2}=\frac{5}{4}\mathrm{R}$

Power dissipated $=\frac{{\mathrm{v}}_{\text{rms }}^2}{\mathrm{z}}\cdot \cos \varphi$

$\therefore 550=\frac{(220{)}^2}{(5/4\mathrm{R})}\cdot (0.8)$

This gives $\mathrm{R}=56.32\mathrm{\Omega }$

and $\omega \mathrm{L}=\frac{3}{4}\times 56.3\mathrm{\Omega }$

The power factor would become unity if a capacitor, $\mathrm{C}$, is connected in the circuit, for which

$$\omega \mathrm{L}=\frac{1}{\omega \mathrm{C}}$$

$$\text{This given }\mathrm{C}=\frac{1}{{\omega }^2\mathrm{L}}=\frac{1}{\omega \cdot \omega \mathrm{L}}$$

$$=\frac{1}{(2\pi \times 50)\times 3\times 56.32}\mathrm{F}\simeq \frac{1}{42\pi }\times {10}^{-2}\mathrm{F}$$

Transformer

It consists of two coils wound on the same core. The alternating current passing through the primary creates a continuously changing flux through the core. This changing flux induces an alternating emf in the secondary.

$$

(1) Transofrmer works on ac only and never on d.c.

(2) It can increase or decrease either voltage or current but not both simultaneously.

(3) Transformer does not change the frequency of input a.c.

(4) There is no electrical connection between the winding but they are linked magnetically.

(5) Effective resistance between primary and secondary winking is infinite.

(6) The flux per turn of each coil must be same i.e. ${\varphi }_{\mathrm{s}}={\varphi }_{\mathrm{p}};-\frac{\mathrm{d}{\varphi }_{\mathrm{s}}}{\mathrm{dt}}=-\frac{\mathrm{d}{\varphi }_{\mathrm{p}}}{\mathrm{dt}}$

(7) If ${\mathrm{N}}_{\mathrm{P}}=$ number of turns in primary, ${\mathrm{N}}_{\mathrm{S}}=$ number of turns in secondary ${\mathrm{V}}_{\mathrm{P}}=$ applied (input) voltage to primary, ${\mathrm{V}}_{\mathrm{S}}=$ Voltage across secondary (load voltage or output), ${\mathrm{e}}_{\mathrm{p}}=$ induced emf in primary; ${\mathrm{e}}_{\mathrm{S}}$ $=$ induced emf in secondary, $\varphi =$ flux linked with primary as well as secondary, ${\mathrm{i}}_{\mathrm{p}}=$ current in primary, ${\mathrm{i}}_{\mathrm{s}}=$ current in secondary (or load current).

As in an ideal transformer there is no loss of power i.e. ${\mathrm{P}}_{\text{out }}={\mathrm{P}}_{\text{in }}$ so ${\mathrm{V}}_{\mathrm{S}}{\mathrm{i}}_{\mathrm{s}}={\mathrm{V}}_{\mathrm{P}}{\mathrm{i}}_{\mathrm{p}}$ and ${\mathrm{V}}_{\mathrm{P}}\simeq {\mathrm{e}}_{\mathrm{P}},{\mathrm{V}}_{\mathrm{S}}\approx {\mathrm{e}}_{\mathrm{S}}$. Hence $\frac{{\mathrm{e}}_{\mathrm{S}}}{{\mathrm{e}}_{\mathrm{P}}}=\frac{{\mathrm{N}}_{\mathrm{S}}}{{\mathrm{N}}_{\mathrm{P}}}=\frac{{\mathrm{V}}_{\mathrm{S}}}{{\mathrm{V}}_{\mathrm{P}}}=\frac{{\mathrm{i}}_{\mathrm{p}}}{{\mathrm{i}}_{\mathrm{s}}}=\mathrm{k};\mathrm{k}=$ transformation ratio (or turn ratio)

Types of Transformer

Step up transformer	Step down transformer
It increases voltage and decreases current	It decreases voltage and increases current
${\mathrm{V}}_{\mathrm{S}}>{\mathrm{V}}_{\mathrm{P}}$	${\mathrm{V}}_{\mathrm{S}}<{\mathrm{V}}_{\mathrm{P}}$
${\mathrm{N}}_{\mathrm{S}}>{\mathrm{N}}_{\mathrm{P}}$	${\mathrm{N}}_{\mathrm{S}}<{\mathrm{N}}_{\mathrm{P}}$
${\mathrm{E}}_{\mathrm{S}}>{\mathrm{E}}_{\mathrm{p}}$	${\mathrm{E}}_{\mathrm{S}}<{\mathrm{E}}_{\mathrm{p}}$
${\mathrm{i}}_{\mathrm{S}}<{\mathrm{i}}_{\mathrm{p}}$	${\mathrm{i}}_{\mathrm{S}}<{\mathrm{i}}_{\mathrm{p}}$
${\mathrm{R}}_{\mathrm{S}}>{\mathrm{R}}_{\mathrm{P}}$	${\mathrm{R}}_{\mathrm{S}}<{\mathrm{R}}_{\mathrm{P}}$
${\mathrm{t}}_{\mathrm{s}}>{\mathrm{t}}_{\mathrm{p}}$	${\mathrm{t}}_{\mathrm{s}}>{\mathrm{t}}_{\mathrm{p}}$
${\mathrm{k}}_{\mathrm{s}}>1$	${\mathrm{k}}_{\mathrm{s}}<1$

Efficiency of transformer (h): Efficiency is defined as the ratio of output power to input power i.e.

$\eta \mathrm{\%}=\frac{{\mathrm{P}}_{\text{out }}}{{\mathrm{P}}_{\text{in }}}\times 100=\frac{{\mathrm{V}}_{\mathrm{S}}{\mathrm{i}}_{\mathrm{s}}}{{\mathrm{V}}_{\mathrm{p}}{\mathrm{i}}_{\mathrm{P}}}\times 100$

For an ideal transformer ${\mathrm{P}}_{\text{out }}={\mathrm{P}}_{\text{in }}$ so $\eta =100\mathrm{\%}$ (But efficiency of practical transformer $70\mathrm{\%}-90)$

For practical transformer ${\mathrm{P}}_{\text{in }}={\mathrm{P}}_{\text{out }}+{\mathrm{P}}_{\text{losses }}$

so $\eta \mathrm{\%}=\frac{{\mathrm{P}}_{\text{out }}}{{\mathrm{P}}_{\text{in }}}\times 100=\frac{{\mathrm{P}}_{\text{out }}}{({\mathrm{P}}_{\text{out }}+{\mathrm{P}}_{\mathrm{L}})}\times 100=\frac{{\mathrm{P}}_{\text{in }}-{\mathrm{P}}_{\mathrm{L}}}{{\mathrm{P}}_{\text{in }}}\times 100$

(9) Losses in transformer : In transformers some power is always lost due to, heating effect, magnetic flux leakage, eddy currents, hystersis and humming.

(i) $\mathbf{C}\mathbf{u}\mathrm{loos}\left({\mathbf{i}}^2\mathbf{R}\right)$ : When current flows through the transformer windings some power is lost in the form of heat $(\mathrm{H}={\mathrm{I}}^2\mathrm{Rt}$ ). To minizie this loss windings are made of thick wires (to reduce resistance)

(ii) Eddy Current loss : Some electrical power is wasted in the form of heat due to eddy currents induced in core, to minimize this loss transformers core are laminated and silicon is used as the core material as it increases the resistivity. The material of the core is then called alloy of iron (steel).

(iii) Hystersis loss: The alternating current flowing through the coils magnetises and demagneties the iron core again and again. Therefore, during each cycle of magnetisation, some energy is lost due to hysteresis. However, the loss of energy can be minimised by selecting the suitable of core, which has a narrow hysterisis loop. Therefore core of transformer is of soft iron. Now a days it is made of “Permalloy” $(Fe-22\mathrm{\%},Ni-78\mathrm{\%})$.

(iv) Magnetic flux Leakage : Magnetic flux product in the primary winding is not completely linked with secondary becuase few magnetic field lines complete their path in air. To minimize this loss secondary winding is kept inside the primary winding.

(v) Humming losses : Due to the passage of alternating current, the core of the transformer starts vibrating and produces humming sound. Thus, some part (may be very small fraction) electircal energy is wasted in the form of humming sound produced by the vibrating core of the transformer.

(10) Uses of transformer : A transformer is used in almost all a.c. operation e.g.

(i) In voltage regulators for TV, refregerator, computer, air conditioner etc.

(ii) In the induction furnances.

(iii) Step down transformer is used for welding purposes.

(iv) In the transmission of a.c. over long distance.

PROBLEMS FOR PRACTICE

1. A long solenoid of radius $0.01\mathrm{m}$ has 600 turns per meter. A smaller coil of 100 turns is wrapped closely around the centre of the bigger solenoid. If the current in the solenoid rises fromzero to $2\mathrm{A}$ in $1\mathrm{ms}$. Find the emf induced in smaller coil.

Show Answer

Answer: $47.3\mathrm{mV}$

2. A coil of area $500{\mathrm{cm}}^2$ and 2000 turns is kept normal to a magnetic field of $4\times {10}^{-5}\mathrm{Wb}/{\mathrm{m}}^2$. It is rotated through ${180}^{\circ }$ in 0.1 second. (i) Calculate the average induced emf(ii) if the total resistance of circuit be $20\mathrm{ohm}$, calculate the induced charge.

Show Answer

Answer: (i) $0.08\mathrm{V}$, (ii) $4\times {10}^{-4}\mathrm{C}$

3. In an aeroplane, the distnace between the edges of its wings being $30\mathrm{m}$. During landing down its velocity is $300\mathrm{Km}/\mathrm{hr}$ and wings of the aeroplane be east-west. Find out the potential difference between the edges of the wings. What happens if wings be north-south then? [Hint : (In first case wings will cut ${\mathrm{B}}_{\mathrm{H}}$ and in second case neither ${\mathrm{B}}_{\mathrm{H}}$ or $\mathrm{BV}$ ) (Give ${\mathrm{B}}_{\mathrm{H}}=0.4$ gauss and angle of dip at the place ${60}^{\circ })$ ]

Show Answer

Answer: (i) $01.V$, (ii) zero

4. A conducting loop of area $5.0{\mathrm{cm}}^2$ is placed in a magnetic field which varies sinosoudally with time as $\mathrm{B}={\mathrm{B}}_0\sin \omega \mathrm{t}$ where ${\mathrm{B}}_0=0.20\mathrm{T}$ and $\omega 300/\mathrm{s}$, such that the normal to the coil makes an angle of ${60}^{\circ }$ with the field.

Find (a) the maximum emf induced in the coil (b) the emf induced at $\mathrm{t}=\left(\frac{\pi }{600}\right)\mathrm{s}$.

Show Answer

Answer: (a) $15\mathrm{mV}$, (b) zero

5. An air cored coil has a self inductance of $0.1\mathrm{H}$. A soft iron core of relative permability 1000 is introduced and the number of turns is reduced to $\frac{1}{10}$ th of the previous value. Find the new value of self inductance.

Show Answer

Answer: $1\mathrm{H}$

6. In the given circuit, the resistance ’ $R$ ’ is equal to the resistance of the coil that makes the inductance $\mathrm{L}$, and the bulbs ${\mathrm{B}}_1$ and ${\mathrm{B}}_2$ are identical. (i) which of the bulbs lights up earlier when $\mathrm{k}$ is closed? (ii) Will the bulbs be equally bright after same time?

$$

7. A conducting circular loop of area $1{\mathrm{mm}}^2$ is placed coplanarly with a long, straight wire at a distance of $20\mathrm{cm}$ from it. The straight wire carries an electric current which changes from $10\mathrm{A}$ to zero in $0.1\mathrm{s}$. Find the average emf induced in the loop in $0.1\mathrm{s}$.

Show Answer

Answer: ${10}^{-10}\mathrm{V}$

8. A uniform magnetic field $\mathrm{B}$ exists in a cylindrical region of radius $0.1\mathrm{m}$ as shown in the figure. A uniform wire of length $0.80\mathrm{m}$ and resistance $4.0\mathrm{\Omega }$ is bent into a square frame and is placed with one side along a diameter of the cylindrical region. If the magnetic field increases at a constant rate of $0.010{\mathrm{TS}}^{-1}$. Calculate the current induced in the frame.

$$

$[\text{Hint}:|\epsilon |=\frac{\mathrm{d}{\varphi }_{\mathrm{B}}}{\mathrm{dt}}=\frac{1}{2}{\pi }^2\frac{\mathrm{dB}}{\mathrm{dt}}]$

Show Answer

Answer: $3.925\times {10}^{-5}\mathrm{A}$

9. Figure shows a bar magnet $\mathrm{M}$ falling under gravity through an aircored coil $\mathrm{C}$ plot a graph showing the variation of induced emf ’ $\epsilon$ ’ with time ’ $t$ ‘. What does the area enclosed by the $\epsilon -t$ curve depict.

$$

10. Two coils $m$ and $n$ having 10 turns and 15 turns respectively are placed close to each other. When $2\mathrm{A}$ current is passing through coil $\mathrm{m}$, then flux linked with coil $\mathrm{n}$ is $1.8\times {10}^{-4}\mathrm{wb}$ per turn. If $3\mathrm{A}$ current passed through coil $\mathrm{n}$, then calculate the flux linked per turn of coil $\mathrm{m}$.

Show Answer

Answer: $4.05\times {10}^{-4}\mathrm{wb}$

11. An alternating voltage $\mathrm{v}=200\sin 300\mathrm{t}$ is applied across a series combination of $\mathrm{R}=10\mathrm{\Omega }$ and $\mathrm{L}=$ $800\mathrm{mH}$. Calculate (i) impedance of the circuit (ii) peak current in the circuit and (iii) power factor of the circuit.

Show Answer

Answer: (i) $240.2\mathrm{\Omega }$ (ii) $0.83\mathrm{A}$ (iii) 0.041

12. When and inductor $\mathrm{L}$ and a resistor $\mathrm{R}$ in series are connected across a $12\mathrm{V}-50\mathrm{Hz}$ supply, a current of 0.5 A flows through the circuit. The current differs in phase from the applied voltage by $\frac{\pi }{3}$ radian. Calculate the value of $\mathrm{L}$ and $\mathrm{R}$.

Show Answer

Answer: $0.066\mathrm{H}$ (ii) $23\mathrm{\Omega }$

13. An alternating emf $110\mathrm{V}$ is applied to a circuit containing a resistance of $40\mathrm{\Omega }$ and an inductance $\mathrm{L}$ in series. The current lags behind the emf by an angle ${\tan }^{-1}\left(\frac{3}{4}\right)$. Find the inductive reatance, impedance of the circuit and current in the circuit. If $=0.1\mathrm{H}$. Find the frequency of applied emf.

14. The voltage $\epsilon$ and current $i$ of an alternating source are represented by

$$\epsilon =100\sin (\omega \mathrm{t}+\frac{\pi }{6})\text{ and }\mathrm{i}=5\sin (\omega \mathrm{t}+\frac{\pi }{6})$$

Dtermine the phase difference between $\epsilon$ and $i$ and impedance of the circuit.

Show Answer

Answer: $\mathrm{\Delta }\varphi =\frac{\pi }{3},\mathrm{z}=22\mathrm{\Omega }$

15. In the given LCR series circuit fed by $200\mathrm{V}-50\mathrm{Hz}$ a.c. mains. Find (i) current in the circuit (ii) value of inductor $\mathrm{L}$ (iii) value of capacitor $\mathrm{C}$ and (iv) value of $\mathrm{C}$ (for same $\mathrm{L}$ ) to have power factor of the circuit unity.

$$

Show Answer

Answer: (i) $0.65\mathrm{A}$, (ii) $1\mathrm{H}$, (iii) $5\mu \mathrm{F}$ (iv) $10\mu \mathrm{F}$

16. A $80\mathrm{V},600\mathrm{W}$ heater is to be operated on a $100\mathrm{V}-50\mathrm{Hz}$ supply. Calculate the inductance of the choke required.

Show Answer

Answer: $0.02\mathrm{H}$

17. A current of $4\mathrm{A}$ flows in a coil when connected to $12\mathrm{V}\mathrm{dV}$ source. If the same coil is connected to a $12\mathrm{V},50\mathrm{rad}/\mathrm{s}$ a.c. source, a current of $2.4\mathrm{A}$ flows in the circuit. Determine the inductance of the coil. Also find the power developed in the circuit if a $2500\mu \mathrm{F}$ capacitor is connected in series with the coil.

Show Answer

Answer: $0.08\mathrm{H},17.28\mathrm{W}$

18. An $\mathrm{LCR}$ circuit has $\mathrm{L}=10\mathrm{mH},\mathrm{R}=3\mathrm{\Omega }$ and $\mathrm{C}=1\mu \mathrm{F}$ connected in series to a source $15\cos \omega \mathrm{t}$ volt. Calculate the current amplitude and average power dissipated per cycle at a frequency $10\mathrm{\%}$ lower than the resonant frequency.

Show Answer

Answer: $0.744\mathrm{A},0.744\mathrm{W}$

19. A step down transformer having a power output of $10\mathrm{kW}$ and efficiency of $90\mathrm{\%}$ downs the voltage from $11\mathrm{kV}$ to $220\mathrm{V}$. Find the number of turns in the primary if the secondary has 100 turns and the current in the primary.

Show Answer

Answer: 5000, 1.01 A

20. An a.c. source of internal resistance $9000\mathrm{\Omega }$ is is to supply current to a load resistor of $10\mathrm{\Omega }$. Find the ratio of the currents passing through the load and the source?

[Hint: ${\mathrm{V}}_{\mathrm{P}}{\mathrm{i}}_{\mathrm{P}}={\mathrm{V}}_{\mathrm{S}}{\mathrm{i}}_{\mathrm{s}}$. Also if transformer is ideal then power dilivered is constant i.e. $\frac{{\mathrm{V}}_{\mathrm{S}}^2}{{\mathrm{R}}_{\mathrm{S}}}=\frac{{\mathrm{V}}_{\mathrm{P}}^2}{{\mathrm{R}}_{\mathrm{P}}}$ ]

Show Answer

Answer: 30

Question Bank

Key Learning Points

1. The magnetic flux through a surface of area $A$, placed in a uniform magnetic field $B$ is defined as

$${\varphi }_{\mathrm{B}}=\bar{\mathrm{B}}\cdot \bar{\mathrm{A}}=\mathrm{BA}\cos \theta$$

where $\theta$ is the angle between $\mathbf{B}$ and the normal to the plane of the area.

2. The emf, induced in a coil of $\mathrm{N}$ turns, is directly proportional to the rate of change of magnetic flux through it

$$\epsilon =-\mathrm{N}\frac{\mathrm{d}{\varphi }_{\mathrm{B}}}{\mathrm{dt}}$$

where ${\varphi }_B$ is the flux linked with one turn. The negative sign indicates that polarity of induced emf is such that it opposes its own cause.

3. When a metal rod of length $\ell$ is placed normal, to a uniform magnetic field $\mathbf{B}$, and moved with velocity $\mathbf{V}$, perpendicular to the field, the emfinduced (Motional emf) across its ends is $\epsilon =\mathrm{BV}\ell$.

4. For a rod rotating about its one end, in a uniform magnetic filed, about an axis perpendicular to the rod, with the magnetic field parallel to the axis, the emf induced between its two ends, $\epsilon =\frac{1}{2}\mathrm{B}{\ell }^2\omega$.

5. Changing magnetic fields can set up current loops in nearby metal bodies. They dissipate electrical energy as heat. Such currens are called eddy currents.

6. When a current in a coil changes, it induces back emf in the same coil. The self induced emf is given by

$$\epsilon =-\mathrm{L}\frac{\mathrm{di}}{\mathrm{dt}}$$

where $L$ is the self inductance of the coil.

7. The self inductance (or inductancee) of a long (ideal) solenoid (the core of which has a permability) is given by $L={\mu }_0{\mu }_{\mathrm{r}}{\mathrm{n}}^2\mathrm{A}\ell$. Here $\mathrm{A}$ is the cross-section area of the solenoid $\ell$ its length and ’ $\mathrm{n}$ ’ the number of turns per unit length.

8. A changing current, in one coil, can induce an emf in another (nearly) coil. The emf induced, in the second coil, in given by ${\epsilon }_2=M_{12}\frac{{\mathrm{di}}_1}{\mathrm{dt}}$. Here ${\mathrm{M}}_{12}={\mathrm{M}}_{21}=\mathrm{M}$ is called the mutual inductance between the two coils.

For two coaxial (long) solenoids placed such that one is completely placed well inside the other, the mutual inductance is given by

$$\mathrm{M}={\mu }_0{\mu }_{\mathrm{r}}{\mathrm{n}}_1{\mathrm{n}}_2\pi {\mathrm{r}}^2\ell$$

where $n_1$ and $n_2$ are the number of turns, per unit length, in the two coils. The coils have a radius $r$ and a length ’ $\ell$ ‘.

9. When a coil, of $\mathrm{N}$ turns, and area $\mathrm{A}$, is rotated, at $\mathrm{v}$ revolutions per second, in a uniform magnetic field $\mathbf{B}$, then the emf induced is given by

$$\epsilon =\mathrm{NBA}2\pi v\sin (2\pi v\mathrm{t})$$

Here, $at=0$, the coil is considered tobe perpendicular to the field. This is the working principle of an a.c generator.

10. The mean, or average, value ofan alternating voltage (or current) over a complete cycle is zero. Over a half cycle, the mean value is $63.7\mathrm{\%}$ of its peak value.

$$\left({\mathrm{i}}_{\mathrm{AV}}\right)\text{ Half cycle }=\frac{2{\mathrm{i}}_0}{\pi }=0.637{\mathrm{i}}_0$$

11. The r.m.s value of current is given by

$${\mathrm{i}}_{\mathrm{rms}}=\frac{{\mathrm{i}}_0}{\sqrt{2}}=0.707{\mathrm{i}}_0$$

Similarly ${\mathrm{v}}_{\mathrm{rms}}=\frac{{\mathrm{v}}_0}{\sqrt{2}}=0.707{\mathrm{v}}_0$

12. An alternating voltage $(\mathrm{v}={\mathrm{v}}_0\sin \omega \mathrm{t})$, applied to a pure resistor $\mathrm{R}$, drives a current, $\mathrm{i}={\mathrm{i}}_0\sin \omega \mathrm{t}$, where $i_0=\frac{v_0}{R}$. The current is in phase with the applied voltage.

13. An a.c voltage, $v=v_0\sin \omega t$, applied to a pure inductor, drives a current $\mathrm{i}={\mathrm{i}}_0\sin (\omega \mathrm{t}-\frac{\pi }{2})$ where ${\mathrm{i}}_0=\frac{{\mathrm{v}}_0}{{\mathrm{x}}_{\mathrm{L}}}$. Here ${\mathrm{x}}_{\mathrm{L}}(=\omega \mathrm{L})$, is called inductive reactance. The current in the inductor lags behind the voltage by $\frac{\pi }{2}$. The average power supplied to a pure inductor over a complete cycle is zero.

14. An a.c voltage, $\mathrm{v}={\mathrm{v}}_0\sin \omega \mathrm{t}$, applied to pure capacitor, drives a current in the capacitor, $\mathrm{i}={\mathrm{i}}_0\sin (\omega \mathrm{t}-\frac{\pi }{2})$, where ${\mathrm{i}}_0=\frac{{\mathrm{v}}_0}{{\mathrm{x}}_{\mathrm{C}}}$. Here ${\mathrm{x}}_{\mathrm{C}}(=\frac{1}{\omega \mathrm{C}})$, is called the capacitive reactance. The current through the capacitor is $\frac{\pi }{2}$ ahead of the applied voltage. Average power, supplied to a pure capacitor, over a complete cycle, is zero.

15. In a series LCR circuit, if a voltage $(\mathrm{v}={\mathrm{v}}_0\sin \omega \mathrm{t})$, is applied, then current in the circuit is given by $\mathrm{i}={\mathrm{i}}_0^{\prime }\sin (\omega \mathrm{t}\mp \varphi )$ where ${\mathrm{i}}_0=\frac{{\mathrm{v}}_0}{\mathrm{z}}$. Here $(=\sqrt{{\mathrm{R}}^2+{({\mathrm{x}}_{\mathrm{L}}-{\mathrm{x}}_{\mathrm{C}})}^2})$ is called the impedance of the circuit. The phase angle, between the voltage and current, is $\varphi$. Here $\varphi ={\tan }^{-1}\frac{(x_L-x_C)}{R}$.

Case (i): ${\mathrm{x}}_{\mathrm{L}}<{\mathrm{x}}_{\mathrm{C}}$

or $\omega >\frac{1}{\sqrt{\mathrm{LC}}},\omega >\omega \mathrm{r}(\omega \mathrm{r}=\frac{1}{\sqrt{\mathrm{LC}}})$

Inductive effect dominates in the circuit. Voltage leads the current by phase angle $\varphi$.

Case (ii): ${\mathrm{x}}_{\mathrm{C}}>{\mathrm{x}}_{\mathrm{L}}$

or $\omega >\frac{1}{\sqrt{\mathrm{LC}}},(\omega <\omega \mathrm{r})$

Capacitive effect dominates in the circuit. Current leads the voltage by phase angle $\varphi$.

16. An interesting characterstics of a series LCR circuit is the phenomenon of resonance. $({\mathrm{x}}_{\mathrm{L}}={\mathrm{x}}_{\mathrm{C}})$. The circuit exhibits resonance at the (angular) frequency ${\omega }_0>\frac{1}{\sqrt{\mathrm{LC}}}$. The impedance of the circuit is minimum. The amplitude of the current is maximum at resonance. At resonance, maximum power is supplied to the series LCR circuit.

17. The quality factor, $\mathrm{Q}$ is defined as

$$\mathrm{Q}=\frac{{\omega }_0\mathrm{L}}{\mathrm{R}}=\frac{1}{{\omega }_0\mathrm{CR}}=\frac{1}{\mathrm{R}}\sqrt{\frac{\mathrm{L}}{\mathrm{C}}}$$

’ $Q$ ’ is an indicator of the sharpness of the (i) vs ( $\omega$ ) curve around its resonance value.

18. The average power dissipation, or loss of power over a complete cycle, is given by

$${\mathrm{P}}_{\mathrm{av}}={\mathrm{v}}_{\mathrm{rms}}{\mathrm{i}}_{\mathrm{rms}}\cos \varphi$$

The term, $\cos \varphi$ is called the power factor. For pure resistive circuit, $\cos \varphi$ is unity. For pure reactance (Pure L or pure L), the cos $\varphi$ term is zero.

19. In pure inductive, or capacitive or a series (LC) circuit, the power factor, $\cos \varphi$, is zero. Hence no power is dissipated even though current is flowing in the circuit. In such cases current is referred as a ‘wattlless current’.

20. A transformer consists of an iron core on which are bound a primary coil of ${\mathrm{N}}_{\mathrm{P}}$ turns and a secondary coil of ${\mathrm{N}}_{\mathrm{s}}$ tunrs. If the primary coil is connected to an a.c cource the primary and secondary voltages are related by

$${\mathrm{V}}_{\mathrm{S}}=\left(\frac{{\mathrm{N}}_{\mathrm{S}}}{{\mathrm{N}}_{\mathrm{P}}}\right){\mathrm{V}}_{\mathrm{P}}$$

The currents are related by

$$I_S=\left(\frac{N_P}{N_S}\right)I_P$$

In step up transformer ${\mathrm{N}}_{\mathrm{S}}>{\mathrm{N}}_{\mathrm{P}}$

and in step down transformer ${\mathrm{N}}_{\mathrm{P}}<{\mathrm{N}}_{\mathrm{S}}$

Easy

Self Induction

1. When the current in a coil changes from $+2\mathrm{A}$ to $-2\mathrm{A}$ in 0.05 second, an emf of $8\mathrm{V}$ is induced in a coil. The coefficient of self induction of the coil is:

(1) $0.1\mathrm{H}$

(2) $0.2\mathrm{H}$

(3) $0.4\mathrm{H}$

(4) $0.8\mathrm{H}$

Show Answer

Correct answer: (1)

Solution:

$$\begin{array}{rl}& \epsilon =-\mathrm{L}\frac{\mathrm{di}}{\mathrm{dt}}\\ & \delta =-\mathrm{L}\frac{(-2-2)}{0.5}\\ & \Rightarrow \mathrm{L}=0.1\mathrm{H}\end{array}$$

Average

Induced emf

2. A conducting circular loop is placed in a uniform magnetic field $B=0.25\mathrm{T}$, with its plane perpendicular to the field. The radius of the loop is made to shrink at a constant rate of 1 $\mathrm{mm}/\mathrm{s}$. The induced emf, when the radius is $2\mathrm{cm}$, would be

(1) $\left(\frac{\pi }{2}\right)\mu \mathrm{V}$

(2) $(\pi )\mu \mathrm{V}$

(3) $(2\pi )\mu \mathrm{V}$

(4) $(2.5)\pi \mu \mathrm{V}$

Show Answer

Correct answer: (2)

Solution:

$$\begin{array}{rl}|\epsilon |=& \frac{\mathrm{d}{\varphi }_{\mathrm{B}}}{\mathrm{dt}}\\ & =\frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{B}.\pi {\mathrm{r}}^2)\\ & =\mathrm{B}\pi (2\mathrm{r})\frac{\mathrm{dr}}{\mathrm{dt}}\\ & =0.025\times \pi \times 2\times 2\times {10}^{-2}\times (1\times {10}^{-3})\\ & =\pi \times {10}^{-6}\mu \mathrm{V}\\ & =(\pi )\mu \mathrm{V}\end{array}$$

Difficult

Change in Magnetic Flux

3. A magnetic field, directed along $z$ axis, varies as $B=B_0\frac{x}{a}$. where a is a positive constant. A square loop (of side $\ell$ ), made of copper, is placed with its edges parallel to the $x$ and $y$ axes. If the loop is made to move with a constant velocity $V_0$, directed along $x$ axis; the emf induced in it, would be

(1) ${\mathrm{B}}_0{\mathrm{V}}_0\frac{{\ell }^2}{\mathrm{a}}$

(2) ${\mathrm{B}}_0{\mathrm{V}}_0\ell$

(3) ${\mathrm{B}}_0{\mathrm{V}}_0\frac{{\ell }^2}{2\mathrm{a}}$

(4) ${\mathrm{B}}_0{\mathrm{V}}_0\frac{{\ell }^3}{{\mathrm{a}}^2}$

Show Answer

Correct answer: (2)

Solution:

We have,

$$\begin{array}{rl}& \mathrm{d}\varphi =\mathbf{B}\cdot \mathrm{d}\mathbf{A}\\ & =\mathrm{BdA}(\text{ where }\mathrm{dA}=\ell \mathrm{d}x)\end{array}$$

Initial total flux $\varphi ={\int }_0^{\ell }d\varphi$

$$=\frac{{\mathrm{B}}_0}{\mathrm{a}}\ell {\int }_0^{\ell }\mathrm{d}x=\frac{{\mathrm{B}}_0\ell }{\mathrm{a}}{\left|\frac{x^2}{2}\right|}_0^{\ell }$$

$${\varphi }_{\text{initial }}=\frac{{\mathrm{B}}_0{\ell }^3}{2\mathrm{a}}$$

In one second the loop covers a distance of ${\mathrm{V}}_0$, so that the final flux can be found by integrating $\mathrm{d}\varphi$ between ${\mathrm{V}}_0$ and $({\mathrm{V}}_0+\ell )$.

This flux comes out to be $[\frac{{\mathrm{B}}_0{\mathrm{V}}_0{\ell }^2}{\mathrm{a}}+\frac{{\mathrm{B}}_0{\ell }^3}{2\mathrm{a}}]$

${\varphi }_{\text{initial }}=\frac{{\mathrm{B}}_0{\mathrm{V}}_0{\ell }^2}{\mathrm{a}}+\frac{{\mathrm{B}}_0{\ell }^3}{2\mathrm{a}}$

$\Rightarrow$ Change in flux $\varphi =\frac{{\mathrm{B}}_0{\mathrm{V}}_0{\ell }^2}{\mathrm{a}}$

Difficult

Induced emf

4. A conducting ring, of radius $r$, is placed in a varying magnetic field perpendicular to the plane of the ring. If the rate at which the magnetic field, varies, is $x$, the (average) electric field intensity, at any point of the ring, would be

(1) $\mathrm{r}x$

(2) $\frac{\mathrm{r}x}{2}$

(4) $\frac{4r}{x}$

Show Answer

Correct answer: (2)

Solution:

Let $\mathrm{E}$ be the field strength at any point on the circumference of the ring. Then, the emf induced

$$\epsilon =\oint \mathbf{E}\cdot \mathrm{d}\ell =\mathrm{E}2\pi \mathrm{r}\dots .\dots (1)$$

$$$$

$$\begin{gathered} \text{Also induced emf}|\epsilon |=\frac{\mathrm{d}{\varphi }_{\mathrm{B}}}{\mathrm{dt}}=\pi {\mathrm{r}}^2\frac{\mathrm{dB}}{\mathrm{dt}} \\ =\pi r^{2}x\dots .\dots (2) \end{gathered}$$

From (1) and (2)

$$\mathrm{E}=\frac{\mathrm{r}x}{2}$$

Difficult

Power Loss

5. Magnetic flux, through a stationary loop, with a resistance $R$, varies during the time internal ’ $\tau$ ‘, as $\varphi =at(\tau -t)$ where a is a constant. The amount of of heat generated in the loop, during time internal, $\tau$, would be

(1) $\frac{a^2{\tau }^3}{6R}$

(2) $\frac{a^2{\tau }^3}{4R}$

(3) $\frac{a^2{\tau }^3}{3R}$

(4) $\frac{a^2{\tau }^3}{2R}$

Show Answer

Correct answer: (3)

Solution:

We have, ${\varphi }_{\mathrm{B}}=$ at $\tau +{\mathrm{at}}^2$ (given)

$\therefore$ Instantaneous $|\epsilon |=\frac{\mathrm{d}{\varphi }_{\mathrm{B}}}{\mathrm{dt}}=\mathrm{a}\tau -2\mathrm{at}$

Instantaneous power loss / dissipation $=\frac{|\epsilon {|}^2}{\mathrm{R}}$

$$\begin{array}{rl}& =\frac{(\mathrm{a}\tau -2\mathrm{at}{)}^2}{\mathrm{R}}\\ \Rightarrow \mathrm{H}& ={\int }_0^{\tau }\mathrm{Pdt}\\ & =\frac{1}{\mathrm{R}}{\left|\frac{(\mathrm{a}\tau -2\mathrm{at}{)}^3}{(3-2\mathrm{a})}\right|}_0^{\tau }\\ & ={\left|\frac{(\mathrm{a}\tau -2\mathrm{at}{)}^3}{-6\mathrm{aR}}\right|}_0^{\tau }\\ & =\frac{{\mathrm{a}}^2{\tau }^3}{3\mathrm{R}}\end{array}$$

Average

Self Induction

6. The current, through a coil of self inductance. $L=2\mathrm{mH}$, is given by $I=t^2{e}^{-t}$ at time $t$. The time taken, to make the induced emf zero, will be

(1) $1\mathrm{S}$

(2) $2\mathrm{S}$

(3) $3\mathrm{S}$

(4) $4\mathrm{S}$

Show Answer

Correct answer: (2)

Solution:

We want to have $|\epsilon |=\mathrm{L}\frac{\mathrm{di}}{\mathrm{dt}}=0$

$$\begin{array}{rl}& \Rightarrow \mathrm{L}\frac{\mathrm{d}}{\mathrm{dt}}\left({\mathrm{t}}^2{\mathrm{e}}^{-\mathrm{t}}\right)=0\\ & \Rightarrow -{\mathrm{t}}^2{\mathrm{e}}^{-\mathrm{t}}+2{\mathrm{te}}^{-\mathrm{t}}=0\\ & \Rightarrow {\mathrm{te}}^{-\mathrm{t}}(-\mathrm{t}+2)=0\\ & \Rightarrow \mathrm{t}=2\mathrm{s}\end{array}$$

Difficult

Induced emf

7. The magnetic flux $\varphi$, through a stationary loop of wire, having a resistance $R$, varies with time as $\varphi =({\mathbf{a}\mathbf{t}}^2+\mathbf{b}\mathbf{t})$ where $a$ and $b$ are positive constants. The average emf, and the total charge flowing in the loop in the time interval $t=0$ to $t=\tau$, would be equal, respectively, to

(1) $(\mathrm{a}\tau +\mathrm{b}),\left(\frac{\mathrm{a}{\tau }^2+\mathrm{b}\tau }{\mathrm{R}}\right)$

(2) $(a\tau +b),\left(\frac{a{\tau }^2+b\tau }{2R}\right)$

(3) $\left(\frac{a\tau +b}{2}\right),\left(\frac{a{\tau }^2+b\tau }{R}\right)$

(4) $2(\mathrm{a}\tau +\mathrm{b}),\left(\frac{\mathrm{a}{\tau }^2+\mathrm{b}\tau }{2\mathrm{R}}\right)$

Show Answer

Correct answer: (1)

Solution:

We have $|\epsilon |=\frac{\mathrm{d}\varphi }{\mathrm{dt}}=2\mathrm{at}+\mathrm{b}$

$\begin{array}{c}\therefore {\epsilon }_{\text{average }}=\frac{{\int }_0^{\tau }|\epsilon |dt}{\tau {\int }_0^{dt}}={\left|\frac{a{t}^2+bt}{t}\right|}_0^{\tau }\\ =\frac{a{\tau }^2+b\tau }{\tau }=a\tau +b\\ q={\int }_0^{\tau }idt=\frac{\mathrm{\Delta }\varphi }{R}=\frac{a{\tau }^2+b\tau }{R}\end{array}$

Average

LC Oscillations

8. An LC circuit contains a $20\mathrm{mH}$ inductor, and a $50\mu \mathrm{F}$ capacitor, of the circuit is negligible. Let the instant, the circuit is closed, be $t=0$. The (first instant of) time, at which energy stored would be completely magnetic, is

(1) $t=0$

(2) $\mathrm{t}=1.57\mathrm{ms}$

(3) $\mathrm{t}=3.14\mathrm{ms}$

(4) $\mathrm{t}=6.28\mathrm{ms}$

Show Answer

Correct answer: (2)

Solution:

Total energy, in a LC circuit, is completely magnetic (at first) at $t=\frac{T}{4}$

Here $\mathrm{T}=2\pi \sqrt{\mathrm{LC}}(\omega =\frac{1}{\sqrt{\mathrm{LC}}})$

$\begin{array}{rl}& =2\times 3.14\sqrt{50\times {10}^{-6}\times 20\times {10}^{-3}}\mathrm{s}\\ & =2\times 3.14\sqrt{{10}^{-6}}\mathrm{s}\end{array}$

$=6.28\mathrm{ms}$

$\therefore \frac{\mathrm{T}}{4}=1.57\mathrm{ms}$

Easy

Induced emf due to Change in Orientation

9. A rectangular coil, of 300 turns, has an average area of $25\mathrm{cm}\times 10\mathrm{cm}$. The coil rotates, with a speed of 50 cycles per second, in a uniform magnetic field of $4\times {10}^{-2}\mathrm{T}$, about an axis perpendicular to the field. The peak value of induced emf (in volt), is

(1) $3\pi$

(2) $30\pi$

(3) $300\pi$

(4) $3000\pi$

Show Answer

Correct answer: (2)

Solution:

$$

$$\begin{array}{rl}{\epsilon }_0& =\mathrm{NBA}\omega \\ & =(300\times 4\times {10}^{-2}\times 250\times {10}^{-4}\times 2\pi \times 50)\mathrm{V}\\ & =(30\pi )\mathrm{Volt}\end{array}$$

Easy

Change in Magnetic Flux / Lenz’s Law

10. Two identical coaxial coils, $P$ and $Q$, carrying equal amount of current, in the same direction, are brought nearer. The current in

(1) $P$ increases while in $Q$ decreases

(2) $\mathrm{Q}$ increases while in $\mathrm{P}$ decreases

(3) both P and Q increases

(4) both P and Q decreases

Show Answer

Correct answer: (4)

Solution:

When both the coils are brought closer. The linked magnetic flux, increases for both of then. Hence induced current (set up in then) will try to decreases the flux (as per Lenz’s law). Hence current in both the coils will decrease.

Easy

Induced emf

11. A 50 turns circular coil has as average a radius of $3\mathrm{cm}$. It is kept in a magnetic field acting normal to the area of the coil. The magnetic field $B$ is increased from $(0.10\mathrm{T})$ to $(0.35\mathrm{T})$ in $2\mathrm{ms}$. The average induced emf in the coil, would be

(1) $0.177\mathrm{V}$

(2) $1.77\mathrm{V}$

(3) $17.7\mathrm{V}$

(4) $177\mathrm{V}$

Show Answer

Correct answer: (3)

Solution:

Change in magnetic flux, linking the coil, is

$$\begin{array}{rl}\mathrm{d}{\varphi }_{\mathrm{B}}& =(\mathrm{NA}\cos 0^0)(\mathrm{dB})\\ & =50\times 3.14\times {(3\times {10}^{-2})}^2(0.35-0.10)\text{ Weber }\end{array}$$

$$\begin{array}{rl}\therefore \text{Induced emf}|\epsilon |& =\frac{\mathrm{d}{\varphi }_{\mathrm{B}}}{\mathrm{dt}}\\ & =\frac{50\times 3.14\times {(3\times {10}^{-2})}^2(0.35-0.10)}{2\times {10}^{-3}}\mathrm{V}\\ & =\frac{50\times 3.14\times 9\times 0.25\times {10}^{-4}\times {10}^3}{2}\mathrm{V}\\ & =17.7\mathrm{V}\end{array}$$

Average

Induced emf

12. A conducting circular loop is placed in a uniform magnetic filed (of induction ’ $B$ ’ tesla), with its plane normal to the field. If the radius of the loop were to start shrinking at a constant rate $\frac{\mathrm{dr}}{\mathrm{dt}}$. The induced emf, at the instant when its radius is ’ $r$ ‘, would be

(1) $-\pi \mathrm{rB}\left(\frac{\mathrm{dr}}{\mathrm{dt}}\right)$

(2) $-2\pi \mathrm{rB}\left(\frac{\mathrm{dr}}{\mathrm{dt}}\right)$

(3) $-\pi r^{2}B\left(\frac{dr}{dt}\right)B$

(4) $-\frac{1}{2}\pi {\mathrm{r}}^2\mathrm{B}\left(\frac{\mathrm{dr}}{\mathrm{dt}}\right)\mathrm{B}$

Show Answer

Correct answer: (2)

Solution:

The magnetic flux, linking the loop, (when its radius is $\mathrm{r}$ ) is

$\varphi -\mathbf{B}\cdot \mathbf{A}=\mathrm{BA}\cos 0^{\circ }$

$=\mathrm{B}\left(\pi {\mathrm{r}}^2\right)$

$\therefore$ The emf, induced at this instant is

$$\begin{array}{rl}\therefore \epsilon & =-\frac{\mathrm{d}\varphi }{\mathrm{dt}}\\ & =-\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{B}\pi {\mathrm{r}}^2\right)\equiv \mathrm{B}\pi \frac{\mathrm{d}}{\mathrm{dt}}\left({\mathrm{r}}^2\right)\\ & =-2\pi \mathrm{rB}\frac{\mathrm{dr}}{\mathrm{dt}}\end{array}$$

Average

Motion of a Conductor in Magnetic Field

13. A conducting rod of length $L$ lies along the $y$-axis with one end at origin and the other end at $(0,L,0)$. There is a magnetic field $B=B_{0}y\hat{k}$ in that region. If the rod moves with a velocity $V=V_0\hat{i}$, the e.m.f induced between the ends of the rod will be

(1) $\frac{{\mathrm{B}}_0{\mathrm{V}}_0\mathrm{L}}{2}$

(2) $\frac{{\mathrm{B}}_0{\mathrm{V}}_0{\mathrm{L}}^2}{2}$

(3) ${\mathrm{B}}_0{\mathrm{V}}_0\mathrm{Ld}$

(4) $\frac{{\mathrm{B}}_0{\mathrm{V}}_0{\mathrm{L}}^2}{4}$

Show Answer

Correct answer: (2)

Solution:

The induced emf, across a small element dy of the rod is

$$\mathrm{d}\epsilon =\left({\mathrm{B}}_0\mathrm{y}\right)(\mathrm{dy}){\mathrm{V}}_0[\therefore \epsilon =\mathrm{BV}\ell ]$$

$\therefore$ The induced emf across the rod of length $L$ will be

$$\begin{array}{rl}& \epsilon ={\int }_0^{\mathrm{L}}\left({\mathrm{B}}_0\mathrm{y}\right)(\mathrm{dy}){\mathrm{V}}_0\\ & ={\mathrm{B}}_0{\mathrm{V}}_0{\int }_0^{\mathrm{L}}\mathrm{ydy}\\ & ={\mathrm{B}}_0{\mathrm{V}}_0{\left|\frac{{\mathrm{y}}^2}{2}\right|}_0^{\mathrm{L}}=\frac{1}{2}{\mathrm{B}}_0{\mathrm{V}}_0{\mathrm{L}}^2\end{array}$$

Average

Induced emf and Current

14. The magnetic flux, through a circular coil of resistance $\mathbf{R}$, changes by an amount $\mathrm{\Delta }\varphi$ in a time $\mathrm{\Delta }t$. Then the total quantity of electric charge ’ $Q$ ‘, that passes any point in the coil, during the time $\mathrm{\Delta }\mathrm{t}$, is represented by

(1) $\mathrm{Q}=\frac{1}{\mathrm{R}}\frac{\mathrm{\Delta }\varphi }{\mathrm{dt}}$

(2) $\mathrm{Q}=\frac{\mathrm{\Delta }\varphi }{\mathrm{R}}$

(3) $\mathrm{Q}=\frac{\mathrm{R}\mathrm{\Delta }\varphi }{\mathrm{dt}}$

(4) $\mathrm{Q}=-\frac{\mathrm{\Delta }\varphi }{\mathrm{R}}$

Show Answer

Correct answer: (4)

Solution:

The induced emf in the coil will be given by

$\epsilon =-\frac{\mathrm{d}\varphi }{\mathrm{dt}}$

$\Rightarrow \mathrm{i}=-\frac{\epsilon }{\mathrm{R}}=-\frac{1}{\mathrm{R}}\frac{\mathrm{\Delta }\varphi }{\mathrm{\Delta }\mathrm{t}}$

$\therefore \mathrm{Q}=\mathrm{i}\mathrm{\Delta }\mathrm{t}=-\frac{\mathrm{\Delta }\varphi }{\mathrm{R}}$

Average

Induced Current

15. A cylindrical bar magnet is rotated about its axis as shown in the diagram. One end of a wire is connected to apoint on the axis and its other end is made to touch the cylindrical surface through an ammeter. We would then observe

$$

(1) That a direct current flowing in the ammeter $\mathrm{A}$

(2) That no current flows through the ammeter A

(3) That an alternating simusoidal current flows through the ammeter A

(4) Atime varying, non-simusoidal, current flows through the ammeter A

Show Answer

Correct answer: (2)

Solution:

As there is no change in the magnetic flux linked with the circuit, no emf gets induced in the circuit. The ammeter ’ $A$ ’ would, therefore, not show any deflection.

Average

Induced emf and Current

16. A magnet NS is suspended from a spring. While it oscillates simple harmonically, it moves in and out a coil ’ $C$ ‘. The coil is connected to a galvanometer $G$.

$$

As the magnet oscillates (Simple harmonically)

(1) G shows no deflection at all

(2) G shows deflection, alternately, to the left and right, but its amplitude keeps on changing

(3) G shows deflection, alternately, to the left and right, with a constant amplitude

(4) G shows a deflection on one side only

Show Answer

Correct answer: (2)

Solution:

As the magnet moves in and out of the coil, there is a continuous change in the magnetic flux linked with the coil. The flux first increases (as magnet moves in) and then decreases (as magnet moves out). Hence the induced emf keeps on changing.

$$\epsilon =-\frac{\mathrm{d}{\varphi }_{\mathrm{B}}}{\mathrm{dt}}$$

The motion of the magnet is S.H.M. (an accelerated motion). Hence $\frac{\mathrm{d}{\varphi }_{\mathrm{B}}}{\mathrm{dt}}$ would not be a constant, hence $\frac{\mathrm{d}{\varphi }_{\mathrm{B}}}{\mathrm{dt}}$ varies.

Difficult

Magnetic Flux

17. A loop, made of straight edges has six corners: at $A(0,0,0),B(L,0,0),C(L,L,0),D(0,L$, 0) E $(\mathbf{0},\mathbf{L},\mathbf{L})$ and $F(0,0,L)$. A uniform magnetic field, $B=B_0(\hat{i}+\hat{k})$ is present in the region. The magnetic flux, linking the whole loop (‘ABCDEFA’) is

(1) ${\mathrm{B}}_0{\mathrm{L}}^2\mathrm{wb}$

(2) $2{\mathrm{B}}_0{\mathrm{L}}^2\mathrm{wb}$

(3) $\sqrt{2}{\mathrm{B}}_0{\mathrm{L}}^{2}wb$

(4) $4{\mathrm{B}}_0{\mathrm{L}}^2\mathrm{wb}$

Show Answer

Correct answer: (2)

Solution:

Given $\mathbf{B}={\mathrm{B}}_0(\hat{\mathrm{i}}+\hat{\mathrm{k}})$

$$

The area vector of part $\mathrm{ABCD}$ is

$${\mathbf{A}}_1={\mathrm{L}}^2\hat{\mathrm{k}}$$

and that of part DEFA is

$${\mathbf{A}}_2={\mathrm{L}}^2\hat{\mathrm{k}}$$

Total flux linking to the loop will be

$$\begin{array}{rl}\varphi & =\mathbf{B}({\mathbf{A}}_1+{\mathbf{A}}_2)\\ & =[{\mathrm{B}}_0(\hat{\mathrm{i}}+\hat{\mathrm{k}})]\cdot [{\mathrm{L}}^2(\hat{\mathrm{i}}+\hat{\mathrm{k}})]\\ & ={\mathrm{B}}_0{\mathrm{L}}^2(1+0+0+1)\\ & =2{\mathrm{B}}_0{\mathrm{L}}^2\end{array}$$

Average

Induced emf

18. A conducting wire $xy$, of mass $m$, (and negligible resistance) slides smoothly on two parallel conducting wires asshown in the diagram. The closed circuit has resistance ’ $R$ ’ due to side AC. Sides $AB$ and $CD$ are perfect conductors. If a time varying magnetic field $B=B(t)\hat{k}$, is present over a large region, the acceleration of the wire $xy$ will be

$$

(1) $-\frac{{\ell }^2{\mathrm{B}}^2}{\mathrm{mR}}\left(\frac{\mathrm{d}x}{\mathrm{dt}}\right)$

(2) $-\frac{{\ell }^2{\mathrm{B}}^2}{\mathrm{mR}}\frac{\mathrm{d}x}{\mathrm{dt}}+\frac{{\ell }^2\mathrm{B}}{\mathrm{mR}}\frac{\mathrm{dB}}{\mathrm{dt}}(x(\mathrm{t}))$

(3) $-\frac{{\ell }^2{\mathrm{B}}^2}{\mathrm{mR}}\frac{\mathrm{d}x}{\mathrm{dt}}-\frac{{\ell }^2\mathrm{B}}{\mathrm{mR}}\frac{\mathrm{dB}}{\mathrm{dt}}x(\mathrm{t})$

(4) $-\frac{{\ell }^2{\mathrm{B}}^2}{\mathrm{mR}}\frac{\mathrm{d}x}{\mathrm{dt}}+\frac{{\ell }^2\mathrm{B}}{\mathrm{mR}}\frac{\mathrm{dB}}{\mathrm{dt}}x(\mathrm{t})$

Show Answer

Correct answer: (3)

Solution:

Let, at time $\mathrm{t}$, the wire $\mathrm{xy}$ wire be at $x=x(\mathrm{t})$.

The magnetic flux associated with the area ACYXA, (at time t) will be

$\varphi =\bar{\mathrm{B}}\cdot \bar{\mathrm{A}}$

$=\mathrm{B}(\mathrm{t})[\ell x(\mathrm{t})]$

This magnetic flux changes with time due to two reasons. There are:

(i) The change of the magnetic field itself with time.

(ii) The motion of the wire on the rods.

We can, therefore, write

$$\begin{array}{c}\epsilon =-\frac{\mathrm{d}\varphi }{\mathrm{dt}}\\ \\ =-\{[\frac{\mathrm{dB}}{\mathrm{dt}}\cdot \ell x(\mathrm{t})]+[\mathrm{B}(\mathrm{t})\cdot \ell \cdot \frac{\mathrm{dd}}{\mathrm{dt}}\cdot x(\mathrm{t})]\}\dots .\dots (1)\end{array}$$

The force, on wire $xy$, at this instant, will be

$$\mathrm{F}=\mathrm{iB}\ell$$

$$=\frac{\epsilon }{\mathrm{R}}\mathrm{B}\ell \dots .\dots (2)$$

From (1) & (2), we get

$$\begin{array}{rl}& \mathrm{F}=\mathrm{ma}=\frac{\mathrm{B}\ell }{\mathrm{R}}[-\frac{\mathrm{dB}}{\mathrm{dt}}\cdot \ell x(\mathrm{t})]-[\mathrm{B}(\mathrm{t})\cdot \ell \cdot \frac{\mathrm{d}x}{\mathrm{dt}}]\\ & \Rightarrow \mathrm{a}=\frac{{\ell }^2\mathrm{B}}{\mathrm{mR}}\frac{\mathrm{dB}}{\mathrm{dt}}(\mathrm{t})-\frac{{\ell }^2{\mathrm{B}}^2}{\mathrm{mR}}\frac{\mathrm{d}x}{\mathrm{dt}}\end{array}$$

Average

Motion of a Conductor in Uniform Magnetic Field

19. In a uniform magnetic field of induction $B$, a wire, in the form of semicircular of radius ’ $R$ ‘, rotates about the diameter of the circle with an angular speed ’ $\omega$ ‘. The axis of rotation is perpendicular to the field. If the total resistance of the circuit is $R^{\prime }$, the near power generated per period of rotation is

(1) $\frac{B\pi R^2\omega }{2R^{\prime }}$

(2) $\frac{{\left(B\pi R^2\omega \right)}^2}{8R^{\prime }}$

(3) $\frac{(\mathrm{B}\pi \mathrm{R}\omega {)}^2}{8{\mathrm{R}}^{\prime }}$

(4) $\frac{\mathrm{B}\pi {\mathrm{R}}^2{\omega }^2}{8{\mathrm{R}}^{\prime }}$

Show Answer

Correct answer: (2)

Solution:

We have $\varphi =\mathrm{BA}\cos \theta =\left(\frac{\mathrm{B}\pi {\mathrm{R}}^2}{2}\right)\cos \omega \mathrm{t}$

$\therefore$ Induced emf, $\epsilon =-\frac{\mathrm{d}\varphi }{\mathrm{dt}}=-\frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{\mathrm{B}\pi {\mathrm{R}}^2\cos \omega \mathrm{t}}{2}\right)$

$=-\frac{\mathrm{B}\pi {\mathrm{R}}^2\omega \sin \omega \mathrm{t}}{2}$

$\therefore$ Instantaneous power generated $=\frac{{\epsilon }^2}{R^{\prime }}=\frac{{\left(B\pi R^2\omega \right)}^2}{4R^{\prime }}{\sin }^2\omega t$

The average value of ${\sin }^2\omega t$, over one complete cycle, is $\frac{1}{2}$.

$\therefore$ Average power over one cycle $=\frac{{\left(\mathrm{B}\pi {\mathrm{R}}^2\omega \right)}^2}{8{\mathrm{R}}^{\prime }}$

Average

Motion of a Conductor in Uniform Magnetic Field

20. A straight rod $PQ$, of length $L$, is rotating about an axis passing through ’ $O$ ’ and perpendicular to its plane. The rod rotates in a uniform (and normal) magnetic field $\mathrm{B}$, with an angular speed $\omega$ (The piont ’ $O$ ’ is at a perpendicular distance $\frac{L}{3}$ from $P$ ). The potential difference between $P$ and $Q$, would be the equal to

(1) $\frac{\mathrm{B}\omega {\mathrm{L}}^2}{6}$

(2) $\frac{\mathrm{B}\omega {\mathrm{L}}^2}{4}$

(3) $\frac{\mathrm{B}\omega {\mathrm{L}}^2}{3}$

(4) $\frac{1}{2}\mathrm{B}\omega {\mathrm{L}}^2$

Show Answer

Correct answer: (1)

Solution:

The potential difference, between two ends of a rod, confined at one end, is given by $\frac{1}{2}\mathrm{B}\omega {\mathrm{L}}^2$

( $L=$ distance of the other end from the confined end)

$\therefore {\mathrm{V}}_{\mathrm{Q}}-{\mathrm{V}}_{\mathrm{O}}=\frac{1}{2}\mathrm{B}\omega {\left(\frac{2\mathrm{L}}{3}\right)}^2$

and ${\mathrm{V}}_{\mathrm{P}}-{\mathrm{V}}_{\mathrm{O}}=\frac{1}{2}\mathrm{B}\omega {\left(\frac{\mathrm{L}}{3}\right)}^2$

$\therefore ({\mathrm{V}}_{\mathrm{Q}}-{\mathrm{V}}_{\mathrm{P}})=\frac{1}{2}\mathrm{B}\omega {\mathrm{L}}^2(\frac{4}{9}-\frac{1}{9})$

$=\frac{B\omega L^2}{6}$

Difficult

Induced emf

21. A thin semicircular conducting ring, of radius $R$, in falling with its plane vertical in a uniform horizontal magnetic field of strength $B$. At the position MNQ the speed of the ring is $V$. The potential difference, developed across the ring, would than be

$$

(1) $\frac{1}{2}\mathrm{BV}\pi {\mathrm{R}}^2$ with $\mathrm{M}$ at highter potential

(2) $\pi \mathrm{RBV}$, with $\mathrm{Q}$ at higher potential

(3) $2\mathrm{RBV}$, with $\mathrm{Q}$ at higher potential

(4) $2R^{2}BV$, with $M$ at higher potential

Show Answer

Correct answer: (3)

Solution:

Rate of decrease of area, of the semicircular ring, in the magnetic field, for the position MNQ, will be

$\frac{\mathrm{dA}}{\mathrm{dt}}=(2\mathrm{R})\mathrm{V}$

Now $|\epsilon |=\frac{\mathrm{d}\varphi }{\mathrm{dt}}=\frac{\mathrm{BdA}}{\mathrm{dt}}=\mathrm{B}(2\mathrm{RV})$

As per Lenz’s law, the induced current in the ring must produce magnetic field directed ‘outwards’. Thus Q must be at a higher potential (so as to send the required anticlockwise (induced) current in the ring).

Easy

Self Induction

22. A conducting coil of 600 turns has a self inductance of $108\mathrm{mH}$. The self inductance of a second identical coil, having of 500 turns, will be

(1) $75\mathrm{mH}$

(2) $90\mathrm{mH}$

(3) $130\mathrm{mH}$

(4) $155\mathrm{mH}$

Show Answer

Correct answer: (1)

Solution:

We know that

$\mathrm{L}=\frac{{\mu }_0{\mathrm{N}}^2\mathrm{A}}{\ell }$

$\Rightarrow \mathrm{L}\propto {\mathrm{N}}^2$ (for identical coils)

$\Rightarrow \frac{{\mathrm{L}}_2}{{\mathrm{L}}_1}={\left(\frac{{\mathrm{n}}_2}{{\mathrm{n}}_1}\right)}^2$

${\mathrm{L}}_2={\left(\frac{500}{600}\right)}^2\times 108\mathrm{mH}$

$=\frac{25}{36}\times 108\mathrm{mH}$

$=75\mathrm{mH}$

Average

Self Induction

23. The current ’ $i$ ‘, in a coil of self inductance ’ $\ell$ ’ varies with time as shown in figure (1). The variation of the induced emf ’ $\epsilon$ ’ with time would than be represented by the graph labelled as graph

$$

(1) $\mathrm{A}$

(2) $\mathrm{B}$

(3) $\mathrm{C}$

(4) $\mathrm{D}$

Show Answer

Correct answer: (1)

Solution:

We have ${\epsilon }_{\text{induced }}=-\mathrm{L}\frac{\mathrm{di}}{\mathrm{dt}}$

During 0 to $\frac{\mathrm{T}}{4},\frac{\mathrm{di}}{\mathrm{dt}}=$ a constant, as $(\mathrm{i}-\mathrm{t})$ graph is a straight line with positive slope $(={\mathrm{i}}_0)$. Hence ${\epsilon }_{\text{induced }}$ is negative and constant $(=(-{\mathrm{i}}_0\mathrm{L}))$.

For time $\frac{\mathrm{T}}{4}\le \mathrm{t}\le \frac{\mathrm{T}}{2}\frac{\mathrm{di}}{\mathrm{dt}}=0$

$\Rightarrow \left|{\epsilon }_{\text{induced }}\right|=0$

For time $\frac{\mathrm{T}}{2}\le \mathrm{t}\le \frac{3\mathrm{T}}{4}\frac{\mathrm{di}}{\mathrm{dt}}$ is again a constant $(=-{\mathrm{i}}_0)$.

Hence ${\epsilon }_{\text{induced }}$ is positive and constant $(={\mathrm{i}}_0\mathrm{L})$.

Average

Self Induction

24. A (solenoid) coil is wound on a core of square cross-section. If all the linear dimensions of core are increased 2 times, but the number of turns per unit length of the coil, remains same, the self inductance would increases by a factor of

(1) 2

(2) 4

(3) 8

(4) 16

Show Answer

Solution:

The self inductance of the coil is given by

$\mathrm{L}={\mu }_0{\mu }_{\mathrm{r}}{\mathrm{n}}^2\mathrm{A}\ell$

For a square core $\mathrm{A}={\ell }^2$

$\Rightarrow \mathrm{L}\propto {\ell }^3$

$\therefore \frac{{\mathrm{L}}_2}{{\mathrm{L}}_1}=(2{)}^3=8$

The self inductance, therefore, increases by a factor of 8 .

Average

Mutual Induction

25. A pair of (insulated) coils having of $n_1$ and $n_2$ tunrs, are kept close together. When the current, passing through the first, is reduced at a rate $r$, an emf of $3\mathrm{mV}$ is developed across the other coil. If the second coil were to carry a current which gets reduced at a rate $2\mathrm{r}$, then induced emf developed across the first coil will be

(1) $1.5\mathrm{mV}$

(2) $2.0\mathrm{mV}$

(3) $3\mathrm{mV}$

(4) $6\mathrm{mV}$

Show Answer

Correct answer: (4)

Solution:

Induced emf, due to mutual induction between in two coils, is given by

$|\epsilon |=+\mathrm{M}\frac{\mathrm{di}}{\mathrm{dt}}$

We are given that ${\epsilon }_2={\mathrm{M}}_{21}\frac{{\mathrm{di}}_1}{\mathrm{dt}}=3\mathrm{mV}$

$\therefore {\mathrm{M}}_{21}=\frac{3\mathrm{mV}}{(\mathrm{r})}$

We know that ${\mathrm{M}}_{12}={\mathrm{M}}_{21}$

$\begin{array}{rl}\therefore {\epsilon }_1& ={\mathrm{M}}_{12}\frac{{\mathrm{di}}_2}{\mathrm{dt}}\\ & =\frac{3\mathrm{mV}}{4}\times 24=6\mathrm{mV}\end{array}$

Average

Mutual Induction

26. The coefficient of mutual inductance of two circuits $A$ and $B$ is $3\mathbf{m}\mathbf{H}$ and their respective resistances are $100\mathrm{hm}$ and $4\mathrm{ohm}$. How much should the currents change, in 0.02 second, in circuit $A$, so that an induced current 0.006 ampere flows through the circuit B?

(1) $0.16\mathrm{A}$

(2) $0.18\mathrm{A}$

(3) $0.24\mathrm{A}$

(4) $1.6\mathrm{A}$

Show Answer

Correct answer: (1)

Solution:

We have

$$\begin{array}{rl}& \left|{\epsilon }_{\mathrm{B}}\right|=\mathrm{M}\frac{{\mathrm{di}}_{\mathrm{A}}}{\mathrm{dt}}\\ & \therefore {\mathrm{i}}_{\mathrm{B}}{\mathrm{R}}_{\mathrm{B}}=\mathrm{M}\frac{{\mathrm{di}}_{\mathrm{A}}}{\mathrm{dt}}\\ & \Rightarrow 0.006\times 4=\frac{(3\times {10}^{-3})\times {\mathrm{di}}_{\mathrm{A}}}{0.02}\\ & \Rightarrow {\mathrm{di}}_{\mathrm{A}}=\frac{48\times {10}^{-5}}{3\times {10}^{-3}}=10\times {10}^{-2}\\ & =0.16\mathrm{A}\end{array}$$

Average

Mutual Induction

27. Consider two coplanar, co-centric circular coils having radii $r_1$ and $r_2(r_2>r_1)$ as shown in the figure. The mutual inductance between these two coils will be

$$

(1) $\frac{{\mu }_0\pi r_1^2}{2r_2}$

(2) $\frac{{\mu }_0\pi r_1}{2r_2}$

(3) $\frac{{\mu }_0\pi r_2^2}{2r_2}$

(4) $\frac{{\mu }_0\pi r_1^2}{2r_1}$

Show Answer

Correct answer: (1)

Solution:

We will consider the bigger coil as the primary coil. In the expression

$${\varphi }_{\mathrm{S}}={\mathbf{B}}_{\mathrm{p}}\cdot {\mathbf{A}}_{\mathrm{S}}$$

we put the value of $\mathrm{B}$ at the centre of the bigger coil.

$$\begin{array}{rl}& \Rightarrow \varphi =\left(\frac{{\mu }_0{i}_1}{2r_2}\right)\pi r_1^2\\ & \Rightarrow \varphi =\left(\frac{{\mu }_0\pi r_1^2}{2r_2}\right)i_1\end{array}$$

$$\mathrm{But}\varphi =\mathrm{MI}$$

$$\therefore M=\frac{{\mu }_0\pi r_1^2}{2r_2}$$

Difficult

Mutual Induction

28. A small square loop of wire, of side $\ell$, is placed inside a large square loop of wire of side $L$ $(\mathrm{L}>{>}_{\ell })$. The loops are coplanar and their centres coincide. The mutual inductance of the system is proportional to

(1) $\frac{\ell }{\mathrm{L}}$

$$

(1) $\frac{\ell }{\mathrm{L}}$

(2) $\frac{{\ell }^2}{\mathrm{L}}$

(3) $\frac{\ell }{\mathrm{L}}$

(4) $\frac{L^2}{\ell }$

Show Answer

Correct answer: (2)

Solution:

The bigger loop may be taken as primary coil and the smaller one as the secondary coil. (We need to use the expressions for the magnetic field due to a finite wire). The bigger coil can be considered as equivalent to four current carrying long wires. All these wires, it can be seen produce magnetic fields in the same sense. Hence net magnetic field at the centre, will be

$$\begin{array}{rl}{\mathrm{B}}_{\text{net }}& =4\frac{{\mu }_0\mathrm{i}}{4\pi \mathrm{R}}(\sin {45}^{\circ }+\sin {45}^{\circ })\\ & =\frac{2\sqrt{{\mu }_0\mathrm{i}}}{\pi \mathrm{L}}(\therefore \text{ Here }\mathrm{R}=\frac{\mathrm{L}}{2})\end{array}$$

The flux, linked with the smaller loop, will be

$${\varphi }_{\mathrm{S}}={\mathrm{Mi}}_{\mathrm{S}}\dots .\dots (1)$$

Also ${\varphi }_{\mathrm{S}}=\sum \left({\mathrm{B}}_{\mathrm{P}}\left(\mathrm{\Delta }{\mathrm{A}}_{\mathrm{S}}\right)\right)$

We can assume the magnetic field to $b$ constant over the whole area of the smaller loop.

or $\therefore {\varphi }_{\mathrm{S}}=\frac{2\sqrt{2}{\mu }_0\mathrm{i}\times {\ell }^2}{\pi \mathrm{L}}\dots .\dots (2)$

From (1) & (2), we get

$\mathrm{M}=\frac{2\sqrt{2}{\mu }_0}{\pi }\frac{{\ell }^2}{\mathrm{L}}$

$\mathrm{M}\propto \frac{{\ell }^2}{\mathrm{L}}$

Easy

Self + Mutual Induction

29. Two (insulated) coils are wound, (one over the other), on a solenoid core. If the number of turns, per unit length, of the first ciol and of the second coil, were to be increased by a factors of 3 and 4 , respectively, the self inductance of the two coils $(L_1$ and $L_2)$ and their mutual inductance (M), would increase (respectively) by factors of

(1) 3,4 and 12

(2) 3, 4 and $\frac{4}{3}$

(3) 9,16 and $\frac{16}{9}$

(4) 9,16 and 12

Show Answer

Correct answer: (4)

Solution:

We know that the self inductance, of a solenoidal coil, is proportional to $n^2$ ( $n=$ number of turns per unit length).

Hence the self inductances, of the two coils will increase by factors of $3^2$ and $4^2$, i.e. 9 and 16.

]

The mutual inductance of the two solenoid coils (one over the other) is proportional to the product $n_1{n}_2(n_1$, ${\mathrm{n}}_2$ are the number of turns per unit length in the two coils).

Hence M, in the given case, would increase by a factor of $3\times 4$, i.e. (12).

Average

Energy Associated with an Inductor

30. A spring, of spring constant $100\frac{\mathrm{N}}{\mathrm{M}}$, is compressed through a distance of $4\mathrm{cm}$. The potential energy, stored in this compressed spring, equals the work done, by the ‘source’, in establishing a current of $2\mathrm{A}$ in a given coil.

The self inductance of the coil equals

(1) $20\mathrm{mH}$

(2) $40\mathrm{mH}$

(3) $60\mathrm{mH}$

(4) $80\mathrm{mH}$

Show Answer

Correct answer: (2)

Solution:

The P.E., of a compressed spring $=\frac{1}{2}\mathrm{k}{x}^2$

$(\mathrm{k}=$ spring constant; $x=$ distance through which the spring has been compressed $)$.

The work, needed to be done, to establish a current I, in a coil of self inductance L, equals

$$\mathrm{w}=\frac{1}{2}{\mathrm{LI}}^2$$

We, therefore, have

$$\begin{array}{rl}& \frac{1}{2}{\mathrm{LI}}^2=\frac{1}{2}\mathrm{k}{x}^2\\ \therefore \mathrm{L}& =\frac{\mathrm{kx}}{{\mathrm{I}}^2}=\frac{100\times {(4\times {10}^{-2})}^2}{2^2}\mathrm{H}\\ & =\frac{16\times {10}^{-2}}{4}\mathrm{H}=40\times {10}^{-3}\mathrm{H}=40\mathrm{mH}\end{array}$$

Average

Induced emf

31. A circular ciol, of average radius $R$, having $N$ turns and carrying a current $I$, is set-up, horizontally with its plane in the $x-y$ plane and its axis along the $z-axis$, so shown.

A small rectangular lamina of sides $\mathrm{\Delta }\mathrm{a}$ and $\mathrm{\Delta }\mathrm{b}$, is set up with its centre on the $\mathrm{z}$-axis and its plane parallel to the $xy$ plane. If this lamina is flipped over through an angle $\pi$, in time $\mathrm{\Delta }t$, the magnitude of the emf, momentairly induced, in it, would equal (nearly)

$$

(1) zero

(2) $\frac{(\mathrm{\Delta }\mathrm{a})(\mathrm{\Delta }\mathrm{b})}{\mathrm{\Delta }\mathrm{t}}N\frac{{\mu }_0\mathrm{I}}{(\mathrm{R})}$

(3) $\frac{(\mathrm{\Delta }\mathrm{a})(\mathrm{\Delta }\mathrm{b})}{\mathrm{\Delta }\mathrm{t}}N\frac{{\mu }_0{\mathrm{R}}^2\mathrm{I}}{{({\mathrm{R}}^2+{\mathrm{z}}^2)}^{3/2}}$

(4) $\frac{(\mathrm{I})(\mathrm{\Delta }\mathrm{a})(\mathrm{\Delta }\mathrm{b})}{2\mathrm{\Delta }\mathrm{t}}\mathrm{N}\frac{{\mu }_0{\mathrm{R}}^2}{({\mathrm{R}}^2+{\mathrm{z}}^2{)}^{3/2}}$

Show Answer

Correct answer: (3)

Solution:

The magnetic field, due to the current carrying circular coil, at the centre of the rectangular lamina

$$\mathrm{B}=\frac{\mathrm{N}\mu \mathrm{o}}{2}\frac{{\mathrm{R}}^2\mathrm{I}}{{({\mathrm{R}}^2+{\mathrm{Z}}^2)}^{3/2}}$$

The field can be assumed to have this (nearly) constant magnitude all over the plane of the small lamina. Hence the change in flux through it, when it is ‘flipped over’ is

$$|\mathrm{\Delta }\varphi |=2\cdot (\mathrm{\Delta }\mathrm{a}\mathrm{\Delta }\mathrm{b})\cdot \mathrm{N}\frac{{\mu }_0}{2}\frac{{\mathrm{R}}^2\mathrm{I}}{{({\mathrm{R}}^2+{\mathrm{z}}^2)}^{3/2}}$$

$$\therefore \text{Magnitude of emf induced}=\frac{(\mathrm{\Delta }\mathrm{a})(\mathrm{\Delta }\mathrm{b})}{\mathrm{\Delta }\mathrm{t}}=\frac{\mathrm{N}{\mu }_0{\mathrm{R}}^2\mathrm{I}}{{({\mathrm{R}}^2+{\mathrm{z}}^2)}^{3/2}}$$

Average

Induced Current

32. A rod, $\mathrm{PQ}$, of length $\ell$, slides with a constant velocity on two conducting rails. The system of the rod and conducting rails, is present in a region where a uniform (normal) magnetic field, of strength $B$, is present. The two ends of the rod $PQ$, (rod’s resistance $=R$ ) are joined to the wheatstone bridge, abcd, shown here. If the current in the arms and and dc, of the wheatstone bridge, equals $i_0$, the velocity, $v$, of the sliding rod, equals

$$

(1) $\frac{16{\mathrm{Ri}}_0}{3\mathrm{B}\ell }$

(2) $\frac{16{\mathrm{Ri}}_0}{\mathrm{B}\ell }$

(3) $\frac{\mathrm{B}\ell }{\left(16{\mathrm{Ri}}_0\right)}$

(4) $\frac{3\mathrm{B}\ell }{\left(16{\mathrm{Ri}}_0\right)}$

Show Answer

Correct answer: (1)

Solution:

The emf, induced in the rod, equal $\mathrm{B}\ell \mathrm{v}$ where, $\mathrm{v}$, is the (constant) velocity of the sliding rod.

The wheatstone bridge, abcd, as shown is a balanced bridge. Its equivalent resistance, $\mathrm{r}$, is given by

$$\frac{1}{\mathrm{r}}=\frac{1}{(6\mathrm{R}+6\mathrm{R})}+\frac{1}{(2\mathrm{R}+2\mathrm{R})}$$

This gives $\mathrm{r}=3\mathrm{R}$

The total resistance of the circuit is, therefore $(R+3R)=4R$

$\therefore$ The induced current, in rod $\mathrm{PQ}$, equals $\frac{\mathrm{B}\ell \mathrm{v}}{4\mathrm{R}}$

In the wheatstone bridge, the part, $i_0$, going into the arms $(\mathrm{ad}$ and $(\mathrm{dc})$, is $(\frac{1}{(4\mathrm{R}+12\mathrm{R})}\times 12\mathrm{R})$, (i.e.) $\frac{3}{4}$ of this current.

Hence ${\mathrm{i}}_0=\frac{3}{4}\times \frac{\mathrm{B}\ell \mathrm{v}}{4\mathrm{R}}$

This given $\mathrm{v}=\frac{16{\mathrm{Ri}}_0}{3\mathrm{B}\ell }$

Average

Force due to Induced Current

33. The set-up, shown in the figure, is present in a uniform magnetic field, $B$, directed perpendicular to the plane of this ‘set-up’. The rod, $PQ$, of length $L$, is allowed to slide down vertically, under its own weight, on the (shaded) conducting rods. The terminal velocity, acquired by the sliding rod (of mass $m$ ), would equal

$$

(1) $\left(\frac{{\mathrm{m}}^2{\mathrm{g}}^2\mathrm{R}}{\mathrm{B}\ell }\right)$

(2) $\left(\frac{\mathrm{B}\ell }{{\mathrm{m}}^2{\mathrm{g}}^2\mathrm{R}}\right)$

(3) $\left(\frac{{\mathrm{B}}^2{\ell }^2}{\mathrm{mgR}}\right)$

(4) $\left(\frac{\mathrm{mgR}}{{\mathrm{B}}^2{\ell }^2}\right)$

Show Answer

Correct answer: (4)

Solution:

As the rod PQ slides down, the magnitude of the induced emf, in it, at an instant when its velocity is v, equals $\mathrm{B}\ell \mathrm{v}$. Due to this an induced current $\mathrm{i}(=\frac{\mathrm{B}\ell \mathrm{v}}{\mathrm{R}})$ flows in it. Because of this induced current, the sliding rod experiences an (mg). This (upward) force equals $\mathrm{Bi}\ell =\frac{{\mathrm{B}}^2{\ell }^{2}v}{R}$. When this upward force becomes equal to the weight of the sliding rod, the net force on the rod becomes zero and the rod, subsequently, slides down with a constant velocity, ${\mathrm{v}}_{\mathrm{T}}$, its terminal velocity. We thus have

$$\frac{{\mathrm{B}}^2{\ell }^2}{\mathrm{R}}{\mathrm{V}}_{\mathrm{T}}=\mathrm{mg}$$

$$\text{or}{\mathrm{v}}_{\mathrm{T}}=\left(\frac{\mathrm{mgR}}{{\mathrm{B}}^2{\ell }^2}\right)$$

Average

Induced Charge Flow

34. A small coil, having $n$ turns each of (everage) radius $r$, is held normal to the magnetic field liens in the region between the (flat) pole pieces of a horse shoe magnet. The terminals ofthe coil are connected to a special galvanometer that can measure the total charge fleming in the circuit. The coil is taken out of the field region in a (small) time $\mathrm{\Delta }t$ and it is observed that a charge $Q$ flows through the coil-galvanometer circuit.

The graph that best shows the dependence of the charge flowing $(Q)$ on the time of withdrawl $(\mathrm{\Delta }t)$ of the coil (from the region of the magnetic field), is the graph labelled as graph.

$$

(1) $\mathrm{K}$

(2) $\mathrm{L}$

(3) $\mathrm{M}$

(4) $\mathrm{N}$

Show Answer

Correct answer: (1)

Solution:

Let $\mathrm{B}$ be the strength of the magnetic field between the poles of the horse shoes magnet. The magnetic flux, linked with the small coil, when it is present in the field region, is

$$\varphi =(\mathrm{B}.\pi {\mathrm{r}}^2)\mathrm{n}$$

The linked magnetic flux becomes zero when it is withdrawn from the field. The instantaneous induced emf, would be

$$\epsilon =-\frac{\mathrm{d}\varphi }{\mathrm{dt}}$$

The instantaneous induced current $(\mathrm{k})$ is $\frac{\epsilon }{\mathrm{R}}$ where $\mathrm{R}$ is the resistance of the coil-galvanometer circuit. The total charge flowing is given by

$$\begin{array}{rl}Q=& {\int }_0^{\mathrm{\Delta }t}idt=\frac{-\pi r^{2}n}{R}{\int }_0^{\mathrm{\Delta }t}\frac{dB}{dt}dt\\ & =\frac{-\pi r^{2}n}{R}[(B{)}_{\mathrm{\Delta }t}-B(0)]=\frac{-\pi r^{2}n}{R}B\end{array}$$

$\mathrm{Q}$, therefore, does not depend on $\mathrm{\Delta }\mathrm{t}$. This is shown in graph $\mathrm{K}$.

Average

Self Inductance of a Solenoid

35. A solenoid, having a total of $\mathrm{N}$ turns, of insulated wire, would uniformly on an iron core has a (long) length a and a radius $r(r<<L)$. The total flux, linked with its cross section, when a current I flows through it, equals $\varphi$. The relative permeability $\left({\mu }_r\right)$, of iron and the self inductance, $\mathbf{L}$, of the solenoid, expressed in terms, of its dimensions and $\varphi$, I and ${\mu }_0$, would then equal (nearly) respectively

(1) $({\mu }_{\mathrm{r}}=\frac{{\mu }_0{\mathrm{N}}^2\cdot \pi {\mathrm{r}}^2\mathrm{I}}{\varphi {\mathrm{a}}^2})$ and $(\mathrm{L}=\frac{\varphi }{\mathrm{I}})$

(2) $({\mu }_{\mathrm{r}}=\frac{\varphi {\mathrm{a}}^2}{{\mu }_0{\mathrm{N}}^2\cdot \pi {\mathrm{r}}^2\mathrm{I}})$ and $(\mathrm{L}=\frac{\varphi }{\mathrm{I}})$

(3) ${\mu }_{\mathrm{r}}=\left(\frac{{\mu }_0\cdot {\mathrm{a}}^2}{\pi {\mathrm{r}}^2{\mathrm{N}}^2.\mathrm{I}}\right)$ and $(\mathrm{L}=\frac{\varphi }{\mathrm{NI}})$

(4) ${\mu }_{\mathrm{r}}=\left(\frac{{\mu }_0\cdot \varphi {\mathrm{a}}^2{\mathrm{IN}}^2}{\pi {\mathrm{r}}^2}\right)$ and $\mathrm{L}=\left(\frac{\mathrm{N}.\varphi }{\mathrm{I}}\right)$

Show Answer

Correct answer: (2)

Solution:

The magnetic field, B, along the axis of this (long) solenoid, equals

$$\mathrm{B}={\mu }_0{\mu }_{\mathrm{r}}\frac{\mathrm{N}.\mathrm{I}}{\mathrm{a}}$$

The flux, $\varphi$, linked with its cross section, is given by

$$\varphi \simeq \mathrm{B}\cdot \pi \cdot {\mathrm{r}}^2\cdot \frac{\mathrm{N}}{\mathrm{a}}$$

We also have

$$\varphi =\mathrm{LI}$$

We thus get

$${\mu }_{\mathrm{r}}=\frac{\varphi {\mathrm{a}}^2}{({\mu }_0\cdot {\mathrm{N}}^2\cdot \pi {\mathrm{r}}^2\mathrm{I})}$$

$$\text{and }\mathrm{L}=[{\mu }_0\cdot {\mu }_{\mathrm{r}}\frac{\pi {\mathrm{r}}^2}{{\mathrm{a}}^2}{\mathrm{aN}}^2]=\left(\frac{\varphi }{\mathrm{I}}\right)$$

Average

Slef + Mutual Inductance

36. A (long thin) solenoid core has an insulated primary coil, (of $n_1$ turns per unit length) and an insulated secondary coil, (of $n_2$ turns per unit length) wound on top of the primary coil. The cross sectional area of the solenoid, equals $A$.

If $L_1$ and $L_2$ are the self inductances of the two coils and $M$ is their coefficient of mutual inductance, we would have (in the ideal case).

(1) $\mathrm{M}=\frac{\left({\mathrm{L}}_1^3\right)}{\left({\mathrm{L}}_2^2\right)}$

(2) $\mathrm{M}=\sqrt{{\mathrm{L}}_1{\mathrm{L}}_2}$

(3) $\mathrm{M}=\left(\frac{{\mathrm{L}}_1^2}{{\mathrm{L}}_2}\right)$

(4) $\mathrm{M}=\left(\frac{{\mathrm{L}}_2^2}{{\mathrm{L}}_1}\right)$

Show Answer

Correct answer: (2)

Solution:

The self inductance, ${\mathrm{L}}_1$ and ${\mathrm{L}}_2$, of the two solenoid coils, are given by

${\mathrm{L}}_1=\mu {\mathrm{n}}_1^2\mathrm{A}\ell$

and ${\mathrm{L}}_2=\mu {\mathrm{n}}_2^2\mathrm{A}\ell$

when $\mu$ is the magnetic permeability of the core material and $\ell$ is the length of the solenoid. Also M, the coefficient of mutual industance, is going by

$\mathrm{M}=\mu {\mathrm{n}}_1{\mathrm{n}}_2\mathrm{A}\ell$

We, therefore, have ${\mathrm{L}}_1{\mathrm{L}}_2={\mathrm{M}}^2$ or $\mathrm{M}=\sqrt{{\mathrm{L}}_1{\mathrm{L}}_2}$

[Alternatively: We know that, in the dieal case, the coefficient of coupling between the two coils, equals unity.

Now $\mathrm{k}$ is defined as $\mathrm{K}=\frac{\mathrm{M}}{\sqrt{{\mathrm{L}}_1{\mathrm{L}}_2}}\therefore$ when $\mathrm{k}=1$, we have $\mathrm{M}=\sqrt{{\mathrm{L}}_1{\mathrm{L}}_2}$ ]

Average

Mutual Inductance of a Pair of Coils

37. Two circular loops, of radii $R$ and $r(R>>r)$, respectively, are positioned, parallel to each other, in the yz plane. The centres and axis, of both the loops, lie on the $x$-axis, a distance $x$ apart. The larger loop has $\mathrm{N}$ turns, while the smaller one has $\mathrm{n}$ turns, all of the same average radius, in each case. The mutual inductance $M$, of this pair of loops, for $x>>R$, equals (nearby)

(1) $\frac{{\mu }_0\pi }{2x^3}\left(\frac{{\mathrm{r}}^2{\mathrm{R}}^2}{\mathrm{nN}}\right)$

(2) $\frac{{\mu }_0\pi }{2x^2}\left({\mathrm{nNrR}}^2\right)$

(3) $\frac{{\mu }_0\pi }{2x^2}(\mathrm{nN})\left({\mathrm{Rr}}^2\right)$

(4) $\frac{{\mu }_0\pi }{2x^3}\left({\mathrm{nr}}^2\right)\left({\mathrm{NR}}^2\right)$

Show Answer

Correct answer: (4)

Solution:

We imagine a current I to flow through the larger loop. The magnetic field, due to it, at the centre of the small loop, would then equal

$$\mathrm{B}=\mathrm{N}\cdot \frac{{\mu }_0}{2}\frac{{\mathrm{R}}^2}{{({\mathrm{R}}^2+x^2)}^{3/2}}\mathrm{I}$$

$$

The small coil has a very small radius. We may, therefore, assume that the magnetic flux, linked with it, is (nearly)

$$\varphi \simeq \mathrm{B}\cdot \left(\pi {\mathrm{r}}^2\right)\cdot \mathrm{n}=\frac{\mathrm{N}{\mu }_0}{2}\frac{{\mathrm{R}}^2\left(\pi {\mathrm{r}}^2\right)\cdot \mathrm{n}}{{({\mathrm{R}}^2+x^2)}^{3/2}}\mathrm{I}$$

But $\varphi =\mathrm{MI}$

Hence $\mathrm{M}\simeq \frac{{\mu }_0(\mathrm{nN})}{2}\frac{\pi {\mathrm{r}}^2{\mathrm{R}}^2}{{({\mathrm{R}}^2+x^2)}^{3/2}}$

For $x>>\mathrm{R}$, we can write

$${\mathrm{M}}^{\prime }\simeq {\mu }_0\left[\frac{\mathrm{nN}\cdot \pi {\mathrm{r}}^2{\mathrm{R}}^2}{2x^3}\right]$$

Difficult

Mutual Inductance

38. A long straight wire and a planer rectangular loop are present in the vicinity of each other in the $xy$ plane. The coefficient of mutual inductance, for this pair, would equal

$$

(1) $\frac{\frac{{\mu }_0\ell }{2\pi }}{[\ln (1+\frac{\omega }{a})]}$

(2) $\frac{\frac{{\mu }_0\omega }{2\pi }}{[\ell \ln (1+\frac{\ell }{a})]}$

(3) $\frac{{\mu }_0\ell }{2\pi }\ln (1+\frac{\omega }{a})$

(4) $\frac{{\mu }_0\omega }{2\pi }\ln (1+\frac{\ell }{a})$

Show Answer

Correct answer: (3)

Solution:

We imagine a current I to flow through the long straight wire. The magnetic field, due to it at any point (in the xy plane), distant $x$ from it, has a magnitude $\frac{{\mu }_{0}I}{2\pi x}$ and it is directed perpendicular to the plane. Hence the magnetic flux, linked with a small segmental strip, (of with $\mathrm{d}x$ ), of the rectangular loop, would be

$$\mathrm{d}\varphi =(\ell \mathrm{d}x)\cdot \frac{{\mu }_0\mathrm{I}}{2\pi x}$$

$$

The total magnetic flux, linked with the rectangular loop, is

$\varphi =\frac{{\mu }_0\mathrm{I}\ell }{2\pi }{\int }_{x=\mathrm{a}}^{x=a+\omega }\frac{\mathrm{d}x}{x}$

$=\frac{{\mu }_0\mathrm{I}\ell }{2\pi }\ln (1+\frac{\omega }{\mathrm{a}})$

But $\varphi =$ MI

Hence $M=\frac{{\mu }_0\ell }{2\pi }\ln (1+\frac{\omega }{a})$

Difficult

Motional emf

39. A (infinite) long straight wire, carrying a current $I$, is placed parallel to the $x$-axis. If a straight conductor, of length $\ell$, positioned along the $y$-axis, as shown, starts moving with a uniform velocity $v=\hat{v}$, the magnitude of the emf, induced between its two ends, $P$ and $Q$, would be

$$

(1) zero

(2) $\frac{2{\mu }_0\mathrm{Iv}\ell }{(2\mathrm{a}+\ell )}$

(3) $\frac{{\mu }_0\mathrm{Iv}}{2\pi }\ln (1+\frac{\ell }{\mathrm{a}})$

(4) $\frac{{\mu }_0\mathrm{Iv}}{2\pi }\frac{\ell }{(\mathrm{a}+\frac{\ell }{2})}$

Show Answer

Correct answer: (3)

Solution:

The magnetic field, due to the long straight current carrying wire, would have different magnitudes on the different elements of the straight conductor PQ. However its direction, at all these elements, is directed along $(-\hat{k})$, as seen from the Biot-Savart rule. Hence the magnitude of the (motional) emf, induced in an element of length dy, situated at a distance below the straight wire, is

$$\mathrm{d}\epsilon =|\mathrm{By}|(\mathrm{dy})\mathrm{v}=\frac{{\mu }_0\mathrm{I}}{2\pi \mathrm{y}}(\mathrm{dy})\mathrm{v}$$

$y$ varies from $y=a$ to $y=(a+\ell )$. Hence the emf induced, between the ends $P$ and $Q$, is

$$\epsilon ={\int }_{\mathrm{a}}^{\mathrm{a}+\ell }\frac{{\mu }_0\mathrm{Iv}}{2\pi \mathrm{y}}\mathrm{dy}=\frac{{\mu }_0\mathrm{Iv}}{2\pi }\ln (1+\frac{\ell }{\mathrm{a}})$$

Difficult

Induced emf

40. A circular coil, of radius $R$, is set up with its plane in the yz plane; its centre at the origin and its axis along the $x$-axis. It carries a constant current $I$.

A small rectangular lamina, of area $\mathrm{\Delta }\mathrm{A}$, is set-up with its plane parallel to the yz plane and its ‘centre’ on the $x$-axis. It is made to move, with a uniform velocity $\mathbf{v}(=v\hat{i})$. The dependence of the instantaneous induced emf $(\epsilon (x))$, in it (when its centre is to a distance $x$ from the origin $(x>>R)$, on $x$, is best represented by the graph, labelled as graph

$$

$$

(1) $\mathrm{P}$

(2) $\mathrm{Q}$

(3) $\mathrm{R}$

(4) $\mathrm{S}$

Show Answer

Correct answer: (1)

Solution:

The set-up, of the ‘system’, is as shown here.

The magnetic field, due to the current carrying circular coil, at the centre of the rectangular lamina (when this centre is at a distance $x$ from the origin), is given by

$\mathrm{B}(x)=\frac{{\mu }_0}{2}\frac{{\mathrm{R}}^2}{{({\mathrm{R}}^2+x^2)}^{3/2}}\mathrm{I}$

The rectangular lamina, being a small sized one, the magnetic flux, linked with it, can be written as

$$

$\begin{array}{rl}\varphi (x)& =\mathrm{\Delta }\mathrm{AB}(x)\\ & =\frac{{\mu }_0}{2}\frac{{\mathrm{R}}^2(\mathrm{\Delta }\mathrm{A})\mathrm{I}}{{({\mathrm{R}}^2+x^2)}^{3/2}}\end{array}$

We are given that $x>>R$. Hence

$\varphi (x)\simeq \frac{{\mu }_0}{2}\frac{{\mathrm{R}}^2(\mathrm{\Delta }\mathrm{A})\mathrm{I}}{x^3}$

The instantaneous induced emf, $\epsilon (x)$, would equal

$\epsilon (x)=-\frac{\mathrm{d}\varphi (x)}{\mathrm{dt}}$

$=-\frac{\mathrm{d}\varphi (x)}{\mathrm{d}x}\cdot \frac{\mathrm{d}x}{\mathrm{dt}}$

$=-v\frac{{\mu }_0}{2}\frac{R^2(\mathrm{\Delta }A)I}{x^4}(-3)$

$=+\frac{3}{2}\frac{{\mu }_0{\mathrm{R}}^2(\mathrm{\Delta }\mathrm{A})\mathrm{Iv}}{x^4}$

$=($ const $)\frac{1}{x^4}$

$\therefore \epsilon (x)$ varies as $\left(\frac{1}{x^4}\right)$. This corresponds to graph $(\mathrm{P})$.

Difficult

Power Factor and Quality Factor in a Series LCR Circuit

41. A voltage source, $v=v_0\sin \omega t$, is connected across a series $LCR$ circuit. The value of $L$ and $C$, in the circuit, are equal to $\frac{nR}{\sqrt{3}\omega }$ and $\frac{\sqrt{3}}{(n-1)R\omega }$, respectively. The power factor and the quality factor, of the circuit, are equal, respectively, to

(1) $\frac{(n-1)}{(2n^2-2n+1)}$ and $\sqrt{\frac{3}{n(n-1)}}$

(2) $\left(\frac{\pi }{6}\right)$ and $\sqrt{\frac{\mathrm{n}(\mathrm{n}-1)}{3}}$

(3) $\left(\frac{\pi }{6}\right)$ and $\sqrt{\frac{3}{\mathrm{n}(\mathrm{n}-1)}}$

(4) $\frac{(n-1)}{{(2n^2-2n+1)}^{1/2}}$ and $\sqrt{\frac{n(n-1)}{3}}$

Show Answer

Correct answer: (2)

Solution:

The phase angle, Q, (current lagging), between the current flowing and the applied voltage, in a series LCR circuit, is given by

$$\tan \theta =\left(\frac{\omega \mathrm{L}-\frac{1}{\omega \mathrm{C}}}{\mathrm{R}}\right)$$

In the given case, $\omega \mathrm{L}=\frac{\mathrm{nR}}{\sqrt{3}}$ and $\omega \mathrm{C}=\frac{\sqrt{3}}{(\mathrm{n}-1)\mathrm{R}}$

$\therefore \tan \theta =\frac{\mathrm{R}}{\sqrt{3}}\cdot \frac{1}{\mathrm{R}}=\frac{1}{\sqrt{3}}$

$\therefore \theta =\left(\frac{\pi }{6}\right)$

The quality factor, $\mathrm{Q}$, of a series $\mathrm{LCR}$ circuit, equals $\frac{{\omega }_0\mathrm{L}}{\mathrm{R}}=\frac{\mathrm{L}}{\mathrm{R}}\frac{1}{\sqrt{\mathrm{LC}}}=\frac{1}{\mathrm{R}}\sqrt{\frac{\mathrm{L}}{\mathrm{C}}}$. In the given case

$$\begin{array}{rl}\frac{L}{C}& =\frac{nR}{\sqrt{3}\omega }\cdot \frac{(n-1)R\omega }{\sqrt{3}}=\frac{n(n-1)}{3}{R}^2\\ \therefore & Q=\frac{\sqrt{n(n-1)}}{3}\end{array}$$

Average

Voltage Magnification in a Series LCR Circuit

42. A series LCR circuit can bring about a ‘voltage magnification’, of the voltage of the a.c. source connected across it, provided the frequency, of the a.c. voltage source, equals

(1) $\frac{1}{\sqrt{\mathrm{LC}}}$; the voltage magnification factor, then equals $\frac{1}{\mathrm{R}}\sqrt{\frac{\mathrm{L}}{\mathrm{C}}}$

(2) $\frac{1}{\sqrt{\mathrm{LC}}}$; the voltage magnification factor, then equals $\left(\frac{\mathrm{L}}{\mathrm{C}}\right)\frac{1}{{\mathrm{R}}^2}$

(3) $\frac{1}{2\pi \sqrt{\mathrm{LC}}}$; the voltage magnification factor, then equals $\frac{1}{\mathrm{R}}\sqrt{\frac{\mathrm{L}}{\mathrm{C}}}$

(4) $\frac{1}{2\pi \sqrt{\mathrm{LC}}}$; the voltage magnification factor, then equals $\left(\frac{\mathrm{L}}{\mathrm{C}}\right)\frac{1}{{\mathrm{R}}^2}$

Show Answer

Correct answer: (3)

Solution:

The resonant frequency, for the series LCR circuit, equals $\frac{1}{2\pi \sqrt{\mathrm{LC}}}$.

For a voltage source, having this frequency, the net impedance of the circuit equals R. Hence the source voltage equals the voltage drop ${\mathrm{V}}_{\mathrm{R}}(=\mathrm{i})$, across the resistor.

The total voltage drop, across the combination of $\mathrm{L}$ and $\mathrm{C}$, is zero but the individual voltage drops, across $\mathrm{L}$ and $\mathrm{C}$, are NOT zero.

At the resonant frequency, voltage drop across $\mathrm{L}$ (equal and opposite to the voltage drop across $\mathrm{C}$ ) $=$ $i{\omega }_0\mathrm{L}$

$\therefore \frac{{\mathrm{V}}_{\mathrm{L}}}{{\mathrm{V}}_{\mathrm{R}}}=(\frac{\mathrm{i}{\omega }_0\mathrm{L}}{\mathrm{iR}}=\frac{{\omega }_0\mathrm{L}}{\mathrm{R}})=\left(\frac{{\mathrm{V}}_{\mathrm{L}}}{\text{ Voltage of source }}\right)=$ Voltage magnification

$\therefore$ Voltage magnification factor $=\frac{{\omega }_0\mathrm{L}}{\mathrm{R}}$

$$=\frac{\mathrm{L}}{\mathrm{R}}\cdot \frac{1}{\sqrt{\mathrm{LC}}}=\frac{1}{\mathrm{R}}\sqrt{\frac{\mathrm{L}}{\mathrm{C}}}$$

Easy

Sharpness of Resonance

43. In A.C.voltage source, $v=v_0\sin \omega t$; having a fixed value for $v_0$, but of variable frequency, is connected across a given series LCR circuit. The variation of the current I, flowing in the circuit, with $\omega$, (the angular frequency ofthe source) is as shown. The power consumption, in the circuit, at (angular) frequencies ${\omega }_1$ and ${\omega }_2$, is

$$

(1) $50\mathrm{\%}$ of that at (angular) frequnecy ${\omega }_0$ and the difference, $({\omega }_2-{\omega }_1)$, increases with an increase in the Q-factor of the circuit

(2) $50\mathrm{\%}$ of that at (angular) frequnecy ${\omega }_0$ and the difference, $({\omega }_2-{\omega }_1)$, decreases with an increase in the Q-factor of the circuit

(3) (nearly) $71\mathrm{\%}$ of that at (angular) frequency ${\omega }_0$ and the difference, $({\omega }_2-{\omega }_1)$, decreases with an increase in the ’ $\mathrm{Q}$-factor’ of the circuit

(4) (nearly) $71\mathrm{\%}$ of that at (angular) frequency ${\omega }_0$ and the difference, $({\omega }_2-{\omega }_1)$, increases with an increase in the ’ $\mathrm{Q}$-factor’ of the circuit

Show Answer

Correct answer: (2)

Solution:

The current value, for ${\omega }_1$ and ${\omega }_2$, is $\frac{1}{\sqrt{2}}$ times the current value at ${\omega }_0$. Hence the power consumption $\left(\alpha {\mathrm{I}}^2\right)$ would become $\left(\frac{1}{2}\right)$ or $50\mathrm{\%}$ of that at ${\omega }_0$.

Further, the more the value of $\mathrm{Q}$, the sharper is the resonance of the circuit. This implies that there is a decrease, in the difference, $({\omega }_2-{\omega }_1)$ when the Q-further increases.

Difficult

Power Factor and Quality Factor of a Series LCR Circuit

44. A voltage source, $v=v_0\sin \omega t$; is connected across a series $LCR$ circuit. The value of $L$ and $C$, in the circuit, equal $\frac{5R}{\sqrt{3}\omega }$ and $\frac{\sqrt{3}}{4R\omega }$, respectively. The power factor and the quality factor, of the circuit, are equal, respectively, to

(1) $\frac{\sqrt{3}}{2}$ and $\sqrt{\frac{20}{2}}$

(2) $\frac{1}{\sqrt{3}}$ and $\sqrt{\frac{14}{3}}$

(3) $\frac{1}{\sqrt{3}}$ and $\sqrt{\frac{14}{3}}$

(4) $\frac{\sqrt{3}}{2}$ and $\sqrt{\frac{20}{3}}$

Show Answer

Correct answer: (1)

Solution:

The power factor, of a series LCR circuit, equals $\cos \theta$ where

$$\tan \theta =\frac{(\omega \mathrm{L}-\frac{1}{\omega \mathrm{C}})}{\mathrm{R}}$$

For the given circuit : $\omega \mathrm{L}=\frac{5}{\sqrt{3}}\mathrm{R}$ and $\omega \mathrm{C}=\frac{3}{4\mathrm{R}}$

$\therefore \tan \theta =\frac{\frac{1}{\sqrt{3}}R}{R}=\frac{1}{\sqrt{3}}$

$\therefore \theta =\frac{\pi }{6}$

Hence $\cos \theta =\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2}$

Also $\mathrm{Q}=\frac{{\omega }_0\mathrm{L}}{\mathrm{R}}=\frac{\mathrm{L}}{\mathrm{R}}\cdot \frac{1}{\sqrt{\mathrm{LC}}}=\frac{1}{\mathrm{R}}\sqrt{\frac{\mathrm{L}}{\mathrm{C}}}$

Here $\frac{\mathrm{L}}{\mathrm{C}}=\frac{5\mathrm{R}}{\sqrt{3}\omega }\cdot \frac{4\mathrm{R}\omega }{\sqrt{3}}=\left(\frac{20}{3}{\mathrm{R}}^2\right)$

$\therefore \mathrm{Q}=\frac{1}{\mathrm{R}}\sqrt{\frac{\mathrm{L}}{\mathrm{C}}}=\frac{1}{\mathrm{R}}\cdot \mathrm{R}\cdot \sqrt{\frac{20}{3}}=\sqrt{\frac{20}{3}}$

Average

Q-Factor

45. For a given series LCR circuit, the term

$$\frac{1}{\mathrm{R}}\sqrt{\frac{\mathrm{L}}{\mathrm{C}}}$$

(1) is a dimensionless term that provides a measure of (i) the quality factor (Q) and (ii) the power factor of the circuit at resonance

(2) is a dimensionless term that provides a measure of (i) the sharpness of resonance (S) and (ii) the voltage magnification of the circuit at resonance

(3) is a dimensional term that provides a measure of the (i) voltage magnification of the circuit at resonance and (ii) the sharpness of resonance (S) of the ciruit

(4) is a dimensionless term that provides a measure of (i) the sharpness (S) of resonance the circuit and (ii) the quality factor $(Q)$ of the circuit

Show Answer

Correct answer: (4)

Solution:

For a series LCR circuit, the sharpness of resonance $(\mathrm{S})$, the quality factor $(\mathrm{Q})$ and the voltage magnification, at resonance, are all measureing through the term $\frac{{\omega }_0\mathrm{L}}{\mathrm{R}}$. The resonant frequency ${\omega }_0$, being equal to $\frac{1}{\mathrm{LC}}$, this term is equivalent to the term $\frac{1}{\mathrm{R}}\sqrt{\frac{\mathrm{L}}{\mathrm{C}}}$.

This term is a dimensionless term (because ${\omega }_0\mathrm{L}$ as well as $\mathrm{R}$ both have the dimensions of impedance). (The power factor, of the circuit, at resonance, equals 1 ; it is not equal to this term).

Average

Q-Factor and Voltage Magnification Factor

46. A series LCR circuit is connected across a given a.c. source, $v=v_0\sin \omega t$, (where ${\omega }_0=\frac{1}{\sqrt{\mathrm{LC}}}$. Keeping $L$ and $C$ fixed, the value of $R$ is varied. The graph, showing the relation between the $Q$-factor of the circuit, and its voltage magnification $\left({\mathrm{V}}_{\mathrm{MF}}\right)$, is the graph labelled as graph

$$

(1) $\mathrm{A}$

(2) $\mathrm{B}$

(3) $\mathrm{C}$

(4) $\mathrm{D}$

Show Answer

Correct answer: (1)

Solution:

The Q-factor, for a series LCR circuit, equals $\frac{{\omega }_0\mathrm{L}}{\mathrm{R}}(=\frac{1}{{\omega }_0\mathrm{CR}})$

The voltage magnification factor, is,

$${\mathrm{V}}_{\mathrm{MF}}=\frac{{\mathrm{V}}_{\mathrm{L}}(={\mathrm{V}}_{\mathrm{C}})}{{\mathrm{V}}_{\mathrm{R}}}=\frac{{\mathrm{V}}_{\mathrm{L}}(={\mathrm{V}}_{\mathrm{C}})}{{\mathrm{V}}_{\text{source }}}=\frac{{\omega }_0\mathrm{L}}{\mathrm{R}}(=\frac{1}{{\omega }_0\mathrm{CR}})$$

Hence ${\mathrm{V}}_{\mathrm{MF}}=\mathrm{Q}$

The graph, between the two, is, therefore, a straight line of slope $=1$. The straight line graph should, therefore, be inclined equally $($ at ${45}^{\circ }$ ) to either of the two axes. Of the four straight line graphs shown here, it is only graph A that has the straight line graph of slope equal to 1.

Average

Q-factor and Resonant Frequency

47. A sinusoidal A.C. voltage source, of adjustable frequency, is connected across a series ‘LCR’ combination. The inductance $(L)$ and the capacitance $(C)$, present in the circuit, can be expresssed in terms of the quality factor $(Q)$ the resonant frequency $\left({\omega }_0\right)$ (of the circuit), and the resistance $R$, through the relations.

(1) $\mathrm{L}=\left(\frac{1}{{\omega }_0\mathrm{QR}}\right)$ and $\mathrm{C}=\left(\frac{\mathrm{QR}}{{\omega }_0}\right)$

(2) $\mathrm{L}=\frac{\mathrm{QR}}{{\omega }_0}$ and $\mathrm{C}=\frac{1}{{\omega }_0\mathrm{QR}}$

(3) $\mathrm{L}=\frac{{\mathrm{Q}}^2{\mathrm{R}}^2}{{\omega }_0}$ and $\mathrm{C}=\frac{1}{{\omega }_0{\mathrm{Q}}^2{\mathrm{R}}^2}$

(4) $\mathrm{L}=\frac{1}{{\omega }_0{\mathrm{Q}}^2{\mathrm{R}}^2}$ and $\mathrm{C}=\left(\frac{{\mathrm{Q}}^2{\mathrm{R}}^2}{{\omega }_0}\right)$

Show Answer

Correct answer: (2)

Solution:

The resonant frequency, ${\omega }_0$, and the quality factor, $\mathrm{Q}$, are related, to $\mathrm{L}$ and $\mathrm{C}$, as

$\mathrm{Q}=\frac{1}{\mathrm{R}}\sqrt{\frac{\mathrm{L}}{\mathrm{C}}}\text{ and }{\omega }_0=\frac{1}{\sqrt{\mathrm{LC}}}$

These give $\mathrm{L}={\mathrm{Q}}^2{\mathrm{R}}^2\mathrm{C}$ and $\mathrm{L}=\frac{1}{{\omega }_0^2\mathrm{C}}$

$\therefore \mathrm{C}=\left(\frac{1}{{\omega }_0^2\mathrm{QR}}\right)$

$\therefore \mathrm{L}=\frac{1}{{\omega }_0^2\mathrm{C}}=\frac{1}{{\omega }_0^2}\cdot \frac{{\omega }_0\mathrm{QR}}{1}=\left(\frac{\mathrm{QR}}{{\omega }_0}\right)$

Average

Phase Angle and Quality Factor in a Series LCR Circuit

48. A voltage source, $v=v_0\sin \omega t$, is connected across a series $LCR$ circuit. The values of $L$ and $C$, in the circuit, equal $\frac{9R}{\omega }$ and $\frac{1}{10R\omega }$, respectively. The phase angle (between the current and voltage in the circuit) and the quality factor, $Q$, of the circuit, are then equal, respectively, to

(1) $\frac{\pi }{4}$ (Current lagging) and $2\sqrt{10}$

(2) $\frac{\pi }{4}$ (Current lagging) and $3\sqrt{10}$

(3) $\frac{\pi }{4}$ (Current lagging) and $3\sqrt{10}$

(4) $\frac{\pi }{4}$ (Current lagging) and $2\sqrt{10}$

Show Answer

Correct answer: (3)

Solution:

The phase angle, $\theta$, (current lagging), between the current flowing and the voltage applied, in a series LCR circuit (connected to a A.C. voltage source) is given by

$\tan \theta =\frac{(\omega \mathrm{L}-\frac{1}{\omega \mathrm{C}})}{\mathrm{R}}=\frac{9\mathrm{R}-10\mathrm{R}}{\mathrm{R}}=(-1)$

$\therefore \theta =(-\frac{\pi }{4})$ or $\left(\frac{3\pi }{4}\right)$

The quality factor, $\mathrm{Q}$, for a series LCR circuit, equals

$$\frac{{\omega }_0\mathrm{L}}{\mathrm{R}}=\frac{\mathrm{L}}{\mathrm{R}}\cdot \frac{1}{\sqrt{\mathrm{LC}}}=\frac{1}{\mathrm{R}}\sqrt{\frac{\mathrm{L}}{\mathrm{C}}}$$

Hence, for the given case,

$$\mathrm{Q}=\frac{1}{\mathrm{R}}\sqrt{90{\mathrm{R}}^2}=3\sqrt{10}$$

Average

Quality Factor Series LCR Circuits

49. The following table gives the value of $L,C$ and $R$, used in three different series $LCR$ circuits.

Circuits No.		Value
	$\mathbf{L}$	C		$\mathbf{R}$
1.	$x$ henry	$\left(\frac{1}{x^3}\right)$ farad	$\left(x^3\right)$ ohm
2.	$3x$ henry	$\left(\frac{27}{x^3}\right)$ farad	$\left(\frac{1}{3}{x}^3\right)$ ohm
3.	$5x$ henry	$\left(\frac{125}{x^5}\right)$ farad	$\left(\frac{x^4}{5}\right)$ ohm

$$

The quality factors, $Q_1,Q_2,Q_3$ of these circuits would be in the ratio

(1) $1:x:x^2$

(2) $x:1:1$

(3) $1:x:x$

(4) $x^2:x:1$

Show Answer

Correct answer: (3)

Solution:

We know that $\mathrm{Q}=\frac{1}{\mathrm{R}}\sqrt{\frac{\mathrm{L}}{\mathrm{C}}}$

$\therefore {\mathrm{Q}}_1=\frac{1}{x^3}\sqrt{\frac{x}{\left(\frac{1}{x^3}\right)}}=\frac{1}{x^2}$

${\mathrm{Q}}_2=\frac{3}{x^3}\sqrt{\frac{3x}{\left(\frac{27}{x^3}\right)}}=\frac{3}{x^2}\times \frac{1}{3}{x}^2=\frac{1}{x}$

${\mathrm{Q}}_3=\frac{5}{x^4}\sqrt{\frac{5x}{\left(\frac{125}{x^5}\right)}}=\frac{5}{x^4}\times \frac{1}{5}{x}^3=\frac{1}{x}$

$\therefore {\mathrm{Q}}_1:{\mathrm{Q}}_2:{\mathrm{Q}}_3=\frac{1}{x^2}:\frac{1}{x}:\frac{1}{x}$

$=1:x:x$

Easy

Quality Factor and Resonant Frequency

50. The inductance, capacitance and resistance in a given series LCR circuit, have values of (n) henry, $\left(\frac{1}{{\mathrm{n}}^3}\right)$ farad and $\left({\mathrm{n}}^2\right)\mathrm{ohm}$. The quality factor, $Q$, and the resonant frequency $\left(v_0\right)$, of this circuit, equal, respectively,

(1) $\mathrm{Q}=\frac{1}{2};v_0=\frac{\mathrm{n}}{4\pi }$

(2) $\mathrm{Q}=1;v_0=\left(\frac{\mathrm{n}}{2\pi }\right)$

(3) $\mathrm{Q}=\frac{4{\pi }^2}{{\mathrm{n}}^2};v_0=\left(\frac{2\pi }{\mathrm{n}}\right)$

(4) $\mathrm{Q}=\frac{8{\pi }^2}{{\mathrm{n}}^2};v_0=\left(\frac{4\pi }{\mathrm{n}}\right)$

Show Answer

Correct answer: (2)

Solution:

The quality factor, $\mathrm{Q}$, of a series LCR circuit, equals

$$\frac{1}{\mathrm{R}}\sqrt{\frac{\mathrm{L}}{\mathrm{C}}}$$

$\therefore$ In the given case,

$$\mathrm{Q}=\frac{1}{{\mathrm{n}}^2}\sqrt{\frac{\mathrm{n}}{\left(\frac{1}{{\mathrm{n}}^3}\right)}}=1$$

Also the resonant frequency, $v_0$, is given by

$$\begin{array}{c}{v}_0=\frac{1}{2\pi }\cdot \frac{1}{\sqrt{\mathrm{LC}}}\\ \therefore v_0=\frac{1}{2\pi }\cdot \frac{1}{{\left(\mathrm{n}\frac{1}{{\mathrm{n}}^3}\right)}^{1/2}}=\left(\frac{\mathrm{n}}{2\pi }\right)\end{array}$$

[Note: We can check the consistency of these results by noting that $Q=\frac{{\omega }_{0}L}{R}$

$\therefore$ In the case, $Q=\frac{\left(2\pi \frac{n}{2\pi }n\right)}{n^2}=1$, as obtained above].

Easy

Sharpness of Resonance of a Series LCR Circuit

51. It is required to improve the sharpness of resonance of a series LCR circuit by a factor of 4. This requirement can be met by

(1) increasing the resistance to 4 times its present value; leaving $L$ and $C$ values unchanged

(2) increasing the inductance to eight times to present value; decreasing the capacitance by a factor of 2; and leaving $R$ unchanged

(3) increasing the inductance by a factor of 4 , leaving $\mathrm{C}$ and $\mathrm{R}$ unchanged

(4) decreasing the capacitance by a factor or 4; leaving $\mathrm{L}$ and $\mathrm{R}$ unchanged

Show Answer

Correct answer: (2)

Solution:

The sharpness of resonance, of a series LCR circuit, improves by a factor of 4 when its quality factor goes up by a factor of 4 . Now

$$\mathrm{Q}=\frac{1}{\mathrm{R}}\sqrt{\frac{\mathrm{L}}{\mathrm{C}}}$$

It is easy to see that it is only the changes, suggested in option (2), that increase the value of the Q-factor by a factor of 4 .

Hence we can improve the sharpness of resonance by a factor of (4) by following the changes suggested in option(2).

[Note: The sharpness of resonance improves by a factor of 4 when for the same value of the resonant frequency, the band width, $({\omega }_2-{\omega }_1)$, goes down by a factor of 4 . The ratio $\left(\frac{{\omega }_2-{\omega }_1}{{\omega }_0}\right)$ equals $\frac{1}{Q}$ ].

Average

LCR Circuit

52. A series LC circuit is connected across an a.c. voltage source : $(v=v_0\sin \omega t)$, for which $\omega \ne \frac{1}{\sqrt{\mathrm{LC}}}$ and $\omega \mathrm{L}>\frac{1}{\omega \mathrm{C}}$. The instantneous current and the instantaneous power dissipated, in the circuit, are then equal respectively, to

(1) $\mathrm{\infty }$ and 0

(2) $\begin{array}{r}\left[\frac{{\mathrm{v}}_0}{(\omega \mathrm{L}-\frac{1}{\omega \mathrm{C}})}\right]\sin (\omega \mathrm{t}-\frac{\pi }{2})\text{and}\frac{-{\mathrm{v}}_0^2}{(\omega \mathrm{L}-\frac{1}{\omega \mathrm{C}})}(\sin \omega \mathrm{t}\cos \omega \mathrm{t})\end{array}$

(3) $\begin{array}{r}\left[\frac{{\mathrm{v}}_0}{(\omega \mathrm{L}-\frac{1}{\omega \mathrm{C}})}\right]\sin (\omega \mathrm{t}+\frac{\pi }{2})\text{ and }\frac{+{\mathrm{v}}_0^2}{(\omega \mathrm{L}-\frac{1}{\omega \mathrm{C}})}(\sin \omega \mathrm{t}\cos \omega \mathrm{t})\end{array}$

(4) $\begin{array}{r}\left[\frac{{\mathrm{v}}_0}{(\omega \mathrm{L}-\frac{1}{\omega \mathrm{C}})}\right]\sin (\omega \mathrm{t}+\frac{\pi }{2})\text{ and zero }\end{array}$

Show Answer

Correct answer: (2)

Solution:

Since $\omega \ne \frac{1}{\sqrt{\mathrm{LC}}}$, the net impedance of the circuit is not zero. Also since $\omega \mathrm{L}\ne \frac{1}{\omega \mathrm{C}}$, the net impedance of the circuit, is $(\omega \mathrm{L}-\frac{1}{\omega \mathrm{C}})$. The inductive impedance being greater the current, in the circuit LAGS in phase with respect to the voltage. Hence the instantaneous current, is the circuit, is given by

$$I=\frac{v_0}{(\omega \mathrm{L}-\frac{1}{\omega \mathrm{C}})}\sin (\omega \mathrm{t}-\frac{\pi }{2})=\frac{-{\mathrm{v}}_0\cos \omega \mathrm{t}}{(\omega \mathrm{L}-\frac{1}{\omega \mathrm{C}})}$$

$\therefore$ The instantaneous power, dissipated in the circuit, is

$${\mathrm{P}}_{\mathrm{inst}}=\mathrm{VI}=\frac{-{\mathrm{v}}_0^2}{(\omega \mathrm{L}-\frac{1}{\omega \mathrm{C}})}\sin \omega \mathrm{t}\cos \omega \mathrm{t}$$

Easy

Behaviour of $L$ and $C$ in A.C. Circuits

53. A student writes the following statements about the ‘inductors’ and the ‘capacitors’.

(a) An inductor behaves almost like an ‘open’ for very low frequency a.c. sources as well as d.c. sources.

(b) A capacitor behaves like an ‘open’ for very low frequency a.c. sources as well as d.c. sources.

(c) For very high frequency a.c. sources, a capacitor behaves as an ‘open’ while an inductor behaves like a ‘short’.

(d) For very law frequency a.c. sources (as well as d.c. sources), an inductor behaves like an ‘open’ while a capacitor behaves like a short.

(1) a

(2) b

(3) c

(4) d

Show Answer

Correct answer: (2)

Solution:

An ‘open’ implies an (almost) infinite impedance while a ‘short’ implies an impedance tending to zero. Now

(i) Impedance of inductor $=\omega \mathrm{L}$

$\therefore$ The inductor is (almost) a ‘short’ for very low frequency a.c. sources as well as d.c. sources. It behaves like an ‘open’ for very high frequency a.c. sources.

(ii) Impedance of capacitor $=\frac{1}{\omega \mathrm{C}}$

Hence a capacitor behaves in a way opposite to the behaviour of an inductor.

Hence statement (B) is the only ’logically valid’ statement, out of the four given statement.

Average

LC Circuits

54. A circuit is made up of an (almost) pure inductor and an (almost) pure capacitor, connected together. The circuit (i) has a resonance frequency $v_0$ when it is put in an evacuated container and (ii) has a resonance frequency $v_0^{\prime }$ when the container contains a non-reacting gas. The dielectric constant of the gas equals.

(1) $\frac{{\left(v_0^{\prime }\right)}^2}{\left(v_0^2\right)}$

(2) $\frac{{\left(v_0\right)}^2}{{\left(v_0^{\prime }\right)}^2}$

(3) $[\frac{{\left(v_0^{\prime }\right)}^2}{\left(v_0^2\right)}-1]$

(4) $(\frac{v_0^2}{v_0^{\prime 2}}-1)$

Show Answer

Correct answer: (2)

Solution:

Let $L$ be the value of the inductance and $C$ that of the capacitor. We then have

$$v_0=\frac{1}{2\pi \sqrt{\mathrm{LC}}}\text{ or }\mathrm{LC}=\frac{1}{4{\pi }^2{v}_0^2}$$

In the presence of the gas, of dielectric constant, say, K, the capacitance changes to KC. Hence

$$\begin{array}{rl}& v_0^{\prime }=\frac{1}{2\pi \sqrt{\mathrm{L}\cdot \mathrm{KC}}}\text{ or }\mathrm{KLC}=\frac{1}{4{\pi }^2{v}_0^2}\\ & \therefore \mathrm{K}=\frac{v_0^2}{{\left(v_0^{\prime }\right)}^2}\end{array}$$

Average

Series LCR Circuit

55. When a given (fixed frequency, fixed voltage) a.c. source is connected across a series LCR circuit, the current in the circuit has its maximum value when

(1) $\mathrm{C}=\frac{1}{{\omega }^2\mathrm{L}}$; it decreases in a (nearly) symmetrical way when $\mathrm{C}>\frac{1}{{\omega }^2\mathrm{L}}$ or $\mathrm{C}<\frac{1}{{\omega }^2\mathrm{L}}$

(2) $\mathrm{C}=\frac{1}{{\omega }^2\mathrm{L}}$; it decreases in a non-symmetrical way when $\mathrm{C}>\frac{1}{{\omega }^2\mathrm{L}}$ and when $\mathrm{C}<\frac{1}{{\omega }^2\mathrm{L}}$

(3) $\mathrm{C}=\frac{{\omega }^2}{\mathrm{L}}$;it decreases in a (nearly) symmetrical way when $\mathrm{C}>\frac{{\omega }^2}{\mathrm{L}}$ or $\mathrm{C}<\frac{{\omega }^2}{\mathrm{L}}$

(4) $\mathrm{C}=\frac{{\omega }^2}{\mathrm{L}}$; it decreases in a non-symmetrical way when $\mathrm{C}>\frac{{\omega }^2}{\mathrm{L}}$ and when $\mathrm{C}<\frac{{\omega }^2}{\mathrm{L}}$

Show Answer

Correct answer: (1)

Solution:

The impedance of a series $LCR$ circuit $(z={[R^2+{(\omega L-\frac{1}{\omega C})}^2]}^{1/2})$ has its minimum value $(=R)$ when $\omega \mathrm{L}=\frac{1}{\omega \mathrm{C}}$ or $\mathrm{C}=\frac{1}{{\omega }^2\mathrm{L}}$. The current in the circuit is, therefore, maximum for this value of $\mathrm{C}$.

Because of the presence of the squared term $\left[{(\omega \mathrm{L}-\frac{1}{\omega \mathrm{C}})}^2\right]$ in $\mathrm{Z}$, we can expect a (nearly) identical increase in $Z$ when $C$ increases / decreases, around the value $\left(\frac{{\omega }^2}{L}\right)$, by equal amounts. Hence the decrease in current, in either case, is likely to take palce in a (nearly) symmetrical way.

Average

LCR Circuit

56. For the LCR series circuit shown here, the peak value of the voltage of the source, the phase difference between the applied voltage and the current flowing, and the power factor of the circuit, equal, respectively.

$$

(1) $\sqrt{175}$ volt; $+\frac{\pi }{3};\frac{1}{2}$

(2) $\sqrt{175}$ volt $;-\frac{\pi }{3};\frac{1}{2}$

(3) 10 volt; $0^0;1$

(4) $\sqrt{200}$ volt $;\frac{\pi }{4};\frac{1}{\sqrt{2}}$

Show Answer

Correct answer: (3)

Solution:

The impedance, $z$, of the circuit is given by

$\mathrm{z}={[{\mathrm{R}}^2+{(\omega \mathrm{L}-\frac{1}{\omega \mathrm{C}})}^2]}^{1/2}$

Here $\mathrm{R}=10\mathrm{\Omega },\omega ={\left({10}^4\right)}^{-1}\mathrm{s},\mathrm{L}=10\mathrm{mH},\mathrm{C}=1\mathrm{pF}$

$\therefore \omega \mathrm{L}={10}^4\times 10\times {10}^{-3}\mathrm{\Omega }={10}^2\mathrm{\Omega }$

and $\frac{1}{\omega \mathrm{L}}=\frac{1}{{10}^4\times {10}^{-6}}\mathrm{\Omega }={10}^2\mathrm{\Omega }$

$\therefore \omega \mathrm{L}-\frac{1}{\omega \mathrm{C}}=0$

$\therefore \mathrm{z}=\mathrm{R}=10\mathrm{\Omega }$

$\therefore$ Peak value of voltage $=({\mathrm{i}}_{\text{rms }}\times \sqrt{2})\times (\mathrm{z})$ volt

$$=(\frac{1}{\sqrt{2}}\times \sqrt{2}\times 10)\text{ volt }=10\text{ volt }$$

Phase difference between current and voltage

$\theta ={\tan }^{-1}\left(\frac{\omega \mathrm{L}-\frac{1}{\omega \mathrm{C}}}{\mathrm{R}}\right)={\tan }^{-1}0=0^0$

$\therefore$ Power factor $=\cos \theta =\cos 0^{\circ }=1$

Difficult

Adjusting the Value of C in a given LCR Circuit

57. The graph drawn here, shows, simultaneously the variation of the (alternating) voltage (applied) and the current (flowing) for a given series LCR circuit.

Without making any other change, the power factor of this circuit, can be made unity if another capacitor, $C^{\prime }$, of value

$$

(1) $\frac{\mathrm{C}}{({\omega }^2\mathrm{LC}-1)}$ is put in parallel with a given $\mathrm{C}$

(2) $\frac{\mathrm{C}}{({\omega }^2\mathrm{LC}-1)}$ is put in series with a given $\mathrm{C}$

(3) $\frac{({\omega }^{2}LC-1)}{C}$ is put in series with a given $C$

(4) $\frac{({\omega }^{2}LC-1)}{C}$ is put in parallel with a given C

Show Answer

Correct answer: (2)

Solution:

The given graph shows the current lagging in phase with respect to the (applied) voltage. The power factor of the circuit becomes unity if the circuit is in its resonance state. This requires the capacitor to have a value, C’, such that

$$\frac{1}{\omega C^{\prime }}=\omega L\text{, i.e. }{C}^{\prime }=\frac{1}{{\omega }^{2}L}$$

We need to increase the capacitative impedance to make it equal to $\omega \mathrm{L}$. This means that the capacitance needs to be decreases. We, therefore, need to put another capacitor, C", in SERIES with the given capacitor (C) such that

$\frac{1}{\mathrm{C}}+\frac{1}{{\mathrm{C}}^{\prime \prime }}=\frac{1}{{\mathrm{C}}^{\prime }}={\omega }^2\mathrm{L}$

This gives $\frac{1}{{\mathrm{C}}^{\prime \prime }}=\frac{({\omega }^2\mathrm{LC}-1)}{\mathrm{C}}$

$\therefore C^{\prime \prime }=\frac{C}{({\omega }^{2}LC-1)}$

Difficult

Adjusting the Value of L in a given LCR Circuit

58. A given $AC$ source, $v=v_0\sin \omega t$, is connected across a given series LCR circuit. The graph, showing simultaneously, the variation of the voltage and current in the circuit, with time, has the form shown.

$$

The current, flowing in this circuit, has to be made to have its maximum value $(=\frac{\mathrm{V}}{\mathrm{R}})$. This aim can be achieved if, without making any other change in the circuit, we join another inductor ( $L^{\prime \prime })$, of value

(1) $\left(\frac{1-{\omega }^{2}LC}{{\omega }^{2}C}\right)$ in series with the given inductor (L)

(2) $\left(\frac{1-{\omega }^{2}LC}{{\omega }^{2}C}\right)$ in parallel with the given inductor (L)

(3) $\left(\frac{{\omega }^2\mathrm{C}}{1-{\omega }^2\mathrm{LC}}\right)$ in series with the given inductor (L)

(4) $\left(\frac{{\omega }^2\mathrm{C}}{1-{\omega }^2\mathrm{LC}}\right)$ in parallel with the given inductor (L)

Show Answer

Correct answer: (1)

Solution:

The given graph shows the current leading in phase with respect to the applied voltage. The given condition (current has its maximum value $(=\frac{\mathrm{V}}{\mathrm{K}})$ ) is met when the circuit is in its resonance state. For this, the net inductance, has to have a value, $L^{\prime }$, such that $\omega L^{\prime }=\frac{1}{\omega C}$. This requires $L^{\prime }=\frac{1}{{\omega }^{2}C}$.

Since the current is leading in phase, we need to increase, the value of the inductance. We, therefore, need to add another inductor (of value L") in series with the given inductor, such that

$$\begin{array}{c}\mathrm{L}+{\mathrm{L}}^{\prime \prime }={\mathrm{L}}^{\prime }=\frac{1}{{\omega }^2\mathrm{C}}\\ \therefore {\mathrm{L}}^{\prime \prime }=[\frac{1}{{\omega }^2\mathrm{C}}-\mathrm{L}]=\left(\frac{1-{\omega }^2\mathrm{LC}}{{\omega }^2\mathrm{C}}\right)\end{array}$$

Average

Changes in LCR Circuit

59. A series $LCR$ circuit is connected to an a.c. voltage source $(v=v_0\sin \omega t)$ and the value of $\mathrm{C}$ is such that $\frac{1}{\omega \mathrm{C}}\simeq 0$. The phase angle $(\theta )$, between the current flowing and the voltage applied, then equals.

(1) ${\tan }^{-1}\left(\frac{\omega \mathrm{L}}{\mathrm{R}}\right)$; current leading; $\theta$ can be made zero by taking a value of $\mathrm{C}$ equal to $\left(\frac{1}{{\omega }^2\mathrm{L}}\right)$

(2) ${\tan }^{-1}\left(\frac{1}{\omega \mathrm{LR}}\right)$; current lagging; $\theta$ can be made zero by taking a value of $C$ equal to $\left(\frac{1}{{\omega }^2\mathrm{L}}\right)$

(3) ${\tan }^{-1}\left(\frac{1}{\omega \mathrm{LR}}\right)$; current leading; $\theta$ can be made zero by taking a value of $C$ equal to $\left(\frac{1}{{\omega }^2\mathrm{L}}\right)$

(4) ${\tan }^{-1}\left(\frac{\omega \mathrm{L}}{\mathrm{R}}\right)$; current lagging; $\theta$ can be made zero by taking a value of $\mathrm{C}$ equal to $\left(\frac{1}{{\omega }^2\mathrm{L}}\right)$

Show Answer

Correct answer: (4)

Solution:

The phase angle (correct lagging) between the current flowing, and the voltage applied, in a series LCR circuit is given by

$$\theta ={\tan }^{-1}\frac{(\omega \mathrm{L}-\frac{1}{\omega \mathrm{C}})}{\mathrm{R}}$$

Here $\frac{1}{\omega \mathrm{C}}=0$.

Hence $\theta ={\tan }^{-1}\left(\frac{\omega \mathrm{L}}{\mathrm{R}}\right)$

We can make $\theta =0$ by taking $\frac{1}{\omega \mathrm{C}}=\omega$ L, i.e., $\mathrm{C}=\frac{1}{{\omega }^2\mathrm{L}}$.

The required value of $\mathrm{C}$ is, therefore,

$$\mathrm{C}=\left(\frac{1}{{\omega }^2\mathrm{L}}\right)$$

Average

Sereis LCR Circuits

60. A series LCR circuit is connected across a given a.c. voltage source for which $v=v_0\sin \omega t$. The threee phasor diagrams, for this circuit (having a total impedance $z$ ), correspond, respectively, to the cases

$$

(1) $(\mathrm{L}>\frac{1}{{\omega }^2\mathrm{C}});(\mathrm{L}=\frac{1}{{\omega }^2\mathrm{C}})$ and $(\mathrm{L}<\frac{1}{{\omega }^2\mathrm{C}})$

(2) $(\mathrm{L}<\frac{1}{{\omega }^2\mathrm{C}});(\mathrm{L}=\frac{1}{{\omega }^2\mathrm{C}})$ and $(\mathrm{L}>\frac{1}{{\omega }^2\mathrm{C}})$

(3) $(\mathrm{L}<\frac{1}{{\omega }^2\mathrm{C}});(\mathrm{z}=\mathrm{R})$ and $(\mathrm{L}>\frac{1}{{\omega }^2\mathrm{C}})$

(4) $(\mathrm{L}<\frac{1}{{\omega }^2\mathrm{C}});(\mathrm{L}=\frac{1}{{\omega }^2\mathrm{C}})$ and $(\mathrm{z}=\mathrm{R})$

Show Answer

Correct answer: (1)

Solution:

For the series LCR circuit, (since $\mathrm{i}$ has the same value for all three elements), implies that ${\mathrm{V}}_{\mathrm{L}}>{\mathrm{V}}_{\mathrm{C}}$ is equivalent to saying $X_L>X_C$. In the given diagram, the upward phasor corresponds to $V_L$ while the downward phasor corresponds to ${\mathrm{V}}_{\mathrm{C}}$. Thus diagram (i), (ii), (iii), correspond to $({\mathrm{X}}_{\mathrm{L}}>{\mathrm{X}}_{\mathrm{C}})$; $({\mathrm{X}}_{\mathrm{L}}={\mathrm{X}}_{\mathrm{C}})$ and $(X_C<X_L)$, respectively. Since $X_L=\omega L$ and $X_C=\frac{1}{\omega C}$, we have $X_L>X_C$ when $L>\frac{1}{{\omega }^{2}C}$ and vicecrsa. Further $\mathrm{z}=\mathrm{R}$ is true only when ${\mathrm{X}}_{\mathrm{L}}={\mathrm{X}}_{\mathrm{C}}$. It is only in option (1), that the given information is in agreement with these requirements (for all the three bits of given information).

Average

Series LCR Circuit

61. The graphs, (A) and (B), drawn here, show, respectively, the phase relationship between the voltage and current, in a series LCR circuit (connectd to an a.c. source) when

(1) (A) $\mathrm{R}$ and $\mathrm{L}$ are both shorted and (B) $\mathrm{R}$ and $\mathrm{C}$ are both shorted

(2) (A) $\mathrm{R}$ and $\mathrm{C}$ are both shorted and (B) $\mathrm{R}$ and $\mathrm{L}$ are both shorted

(3) (A) $\mathrm{C}$ alone is shorted (B) Lalone is shorted

(4) (A) alone is shorted (B) $\mathrm{C}$ alone is shorted

Show Answer

Correct answer: (4)

Solution:

For a series LCR circuit, the (phase) angle $\theta$, by which the voltage $(\mathrm{V})$ leads the current (I), is given by

$$\tan \theta =\frac{(\omega \mathrm{L}-\frac{1}{\omega \mathrm{C}})}{\mathrm{R}}$$

(i) When $\mathrm{L}$ alone is shorted, $\theta$ becomes a negative angle, i.e. the voltage lags behind the voltage in phase. This corresponds to graph A.

$$

(A)

(ii) When $\mathrm{C}$ alone is shorted, $\theta$ becomes a positive angle; i.e. the voltage leads the current in phase. This corresponds to graph(B).

$$

(B)

[Option (1) and (2) are incorrect becuase when $\mathrm{R}$ and $\frac{\mathrm{L}}{\mathrm{C}}$ are both shorted, the circuit becomes a pure $\frac{\mathrm{C}}{\mathrm{L}}$ circuit. For both these cases, $\theta$ would be $\mp \frac{\pi }{2}$; this is not so in the given graph].

Average

Resonant Frequency of a Series Combination of Two LCR Circuits

62. The resonance frequency for each of the two individual (LCR) series circuits, [(A) and (B)], shown here, has the same value, say ${\omega }_0$. For the series LCR circuit, obtained by having the ‘series combination’ shown, of the circuits (A) and (B), the resonant frequency will be

$$

(1) ${\omega }_0$

(2) ${\omega }_0^2$

(3) $2{\omega }_0$

(4) $\frac{{\omega }_0}{2}$

Show Answer

Correct answer: (1)

Solution:

We are given that

$\frac{1}{\sqrt{{\mathrm{L}}_1{\mathrm{C}}_1}}={\omega }_0\text{ and }\frac{1}{\sqrt{{\mathrm{L}}_2{\mathrm{C}}_2}}={\omega }_0$

For the ‘combined circuit’, the equivalent inductance (L), and the equivalent capacitance (C), have the values given by

$\mathrm{L}={\mathrm{L}}_1+{\mathrm{L}}_2\text{ and }\frac{1}{\mathrm{C}}=\frac{1}{{\mathrm{C}}_1}+\frac{1}{{\mathrm{C}}_2}(\therefore \mathrm{C}=\frac{{\mathrm{C}}_1{\mathrm{C}}_2}{{\mathrm{C}}_1+{\mathrm{C}}_2})$

The resonant frequency, say $\omega$ ‘, of the ‘combined circiuit’ is given by

${\omega }^{\prime }=\frac{1}{\sqrt{\mathrm{LC}}}$

Now $\mathrm{LC}=\frac{({\mathrm{L}}_1+{\mathrm{L}}_2)\left({\mathrm{C}}_1{\mathrm{C}}_2\right)}{{\mathrm{C}}_1+{\mathrm{C}}_2}=\frac{\left({\mathrm{L}}_1{\mathrm{C}}_1\right){\mathrm{C}}_2+\left({\mathrm{L}}_2{\mathrm{C}}_2\right){\mathrm{C}}_1}{({\mathrm{C}}_1+{\mathrm{C}}_2)}$

$=\frac{\frac{1}{{\omega }_0^2}[{\mathrm{C}}_2+{\mathrm{C}}_1]}{{\mathrm{C}}_1+{\mathrm{C}}_2}=\frac{1}{{\omega }_0^2}$

$\therefore {\omega }^{\prime }=\sqrt{{\omega }_0^2}={\omega }_0$

Average

Phase Angle and Power Factor in a Series LCR Circuit

63. An A.C. voltage source $(v=v_0\sin \omega t)$, is connected across a series LCR Circuit. If the values of $L$ and $C$, in the circuit, equal $\left(\frac{2\sqrt{3}R}{\omega }\right)$ and $\left(\frac{1}{\sqrt{3}(R\omega )}\right)$, respectively; the phase angle (between the current and voltage) and the power factor of the circuit, are equal, respectively, to

(1) $\left(\frac{\pi }{6}\right)$ and $\frac{\sqrt{3}}{2}$

(2) $\left(\frac{\pi }{3}\right)$ and $\left(\frac{1}{2}\right)$

(3) $\left(\frac{\pi }{4}\right)$ and $\left(\frac{1}{\sqrt{2}}\right)$

(4) $\left(\frac{\pi }{3}\right)$ and $\left(\frac{\sqrt{3}}{2}\right)$

Show Answer

Correct answer: (2)

Solution:

The phase angle, $\theta$, between the current and voltage, in a series LCR circuit, is given by

$$\tan \theta =\frac{\omega \mathrm{L}-\frac{1}{\omega \mathrm{C}}}{\mathrm{R}}$$

$\therefore$ In the given case:

$$\begin{array}{rl}\tan \theta & =\frac{1}{2}[\frac{\omega \cdot 2\sqrt{3}R}{\omega }-\frac{1}{\omega \cdot \left(\frac{1}{\sqrt{3}R\omega }\right)}]\\ & =\frac{1}{2}[2\sqrt{3}R-\sqrt{3}R]=\sqrt{3}\end{array}$$

$$\therefore \theta =\frac{\pi }{3}$$

$$\therefore \text{ Power factor}=\cos \theta =\cos \left(\frac{\pi }{3}\right)=\frac{1}{2}$$

Average

Induced emf

64. A series $LCR$, connected to an a.c.voltage source $(v=v_0\sin \omega t)$, has $L=0$. The phase angle, $\theta$, between the current flowing, and the voltage applied, then equals

(1) ${\tan }^{-1}\left(\frac{1}{\omega \mathrm{CR}}\right)$; current leading; $\theta$ can be made zero by taking a value of $L$ equal to $\left(\frac{1}{{\omega }^2\mathrm{C}}\right)$

(2) ${\tan }^{-1}\left(\frac{1}{\omega \mathrm{CR}}\right)$; current lagging; $\theta$ can be made zero by taking a value of $L$ equal to $\left({\omega }^2\mathrm{C}\right)$

(3) ${\tan }^{-1}\left(\frac{\omega \mathrm{C}}{\mathrm{R}}\right)$; current leading; $\theta$ can be made zero by taking a value of $L$ equal to $\left({\omega }^2\mathrm{C}\right)$

(4) ${\tan }^{-1}\left(\frac{\omega \mathrm{C}}{\mathrm{R}}\right)$; current lagging; $\theta$ can be made zero by taking a value of $L$ equal to $\left(\frac{1}{{\omega }^2\mathrm{C}}\right)$

Show Answer

Correct answer: (1)

Solution:

The phase angle (current lagging), $\theta$, between the current flowing andthe voltage applied, in a series LCR circuit (connected to an a.c. voltage source) is given by

$$\theta ={\tan }^{-1}\left[\frac{(\omega \mathrm{L}=\frac{1}{\omega \mathrm{C}})}{\mathrm{R}}\right]$$

Here $\mathrm{L}=0$. Hence $\theta ={\tan }^{-1}\left(\frac{1}{\omega \mathrm{CR}}\right)$; The $(-\mathrm{ve})$ sign implies that the current now leads the voltage.

We can make $\theta =0$ if we select a value of $L$ for which $\omega \mathrm{L}=\frac{1}{\omega \mathrm{C}}$. The required value of $\mathrm{L}$ is, therefore, $(\mathrm{L}=\frac{1}{{\omega }^2\mathrm{C}})$.

Average

LCR Circuit

65. A given series LCR circuits is connected across a simusoidal AC voltage source.

$$\mathrm{v}={\mathrm{v}}_0\sin (\omega \mathrm{t})$$

The inductance, $L$, in the circuit, has a value equal to $\frac{1}{{\omega }^2\mathrm{C}}$. The rms voltage drop, across $\mathbf{L},\mathbf{C}$, and $\mathrm{R}$, of this circuit, are equal, respectively, to

(1) $\frac{{\mathrm{v}}_0}{\sqrt{2}\mathrm{R}}\cdot \frac{1}{\omega \mathrm{C}};\frac{{\mathrm{v}}_0}{\mathrm{R}}\cdot \frac{1}{\omega \mathrm{C}}$ and ${\mathrm{v}}_0$

(2) $\frac{{\mathrm{v}}_0}{\mathrm{R}}\cdot \frac{1}{\omega \mathrm{C}};\frac{{\mathrm{v}}_0}{\mathrm{R}}\cdot \omega \mathrm{L}$ and ${\mathrm{v}}_0$

(3) 0,0 and $\frac{{\mathrm{v}}_0}{\sqrt{2}}$

(4) $\begin{array}{r}\frac{{\mathrm{v}}_0}{(\sqrt{2})\mathrm{R}}\cdot \frac{1}{\omega \mathrm{C}};\frac{{\mathrm{v}}_0}{(\sqrt{2})\mathrm{R}}\cdot \frac{1}{\omega \mathrm{C}};\frac{{\mathrm{v}}_0}{\sqrt{2}}\end{array}$

Show Answer

Correct answer: (4)

Solution:

We are given that $\mathrm{L}=\frac{1}{{\omega }^2\mathrm{C}}$. This implies that $\omega =\frac{1}{\sqrt{\mathrm{LC}}}$, i.e. the given circuit is in its resonance state. At resonance, net impedance of the circuit equals R. Hence, the rms value of the current in the circuit, is $\frac{{\mathrm{v}}_0}{\sqrt{2}\cdot \mathrm{R}}$. The impedance of the inductance and capacitance are both equal to $\frac{1}{\omega \mathrm{C}}($ or $\omega \mathrm{L})$. Hence the rms voltage drop, across the inductor and the capacitor, are each equal to $\frac{v_0}{\sqrt{2}R}\cdot \frac{1}{\omega C}$. The total of these two voltage drop, at resonance, is zero. The rms voltage drop, across the resistor, equals the rms value of the supply voltage, i.e. $\frac{{\mathrm{v}}_0}{\sqrt{2}}$.

Difficult

Adjusting the Value of L in a given LCR Circuit

66. A given series LCR circuit has an alternating voltage ( $\mathrm{v}={\mathrm{v}}_0\sin \omega \mathrm{t})$, applied across it. The current, flowing in the circuit, is observed to lag in phase with respect to the applied voltage.

Without making any other change, the value of the (net) inductance, present in the circuit, is to be adjusted such that the voltage of the source, and the voltage drop across the resistor in the circuit, have identical values. For this, we need to put another inductor, $L^{\prime \prime }$. of value

(1) $\frac{({\omega }^2\mathrm{LC}-1)}{(\mathrm{L})}$ in parallel with the given inductor (L), already present in the circuit

(2) $\frac{({\omega }^2\mathrm{LC}-1)}{(\mathrm{L})}$ in series with the given inductor (L), already present in the circuit

(3) $\frac{(\mathrm{L})}{({\omega }^2\mathrm{LC}-1)}$ in parallel with the given inductor $(\mathrm{L})$, already present in the circuit $$

(4) $\frac{(\mathrm{L})}{({\omega }^2\mathrm{LC}-1)}$ in series with the given inductor (L), already present in the circuit

Show Answer

Correct answer: (3)

Solution:

The voltage of the source, and the voltage drop across the resistor, have identical values only when the circuit is in its resonance state. Hence the net inductance (L’), in the circuit must be such that

$$\omega L^{\prime }=\frac{1}{\omega C}\text{ or }{L}^{\prime }=\frac{1}{{\omega }^{2}C}$$

Since the current is lagging in phase, the inductance value needs to be decreased. For this, we need to put an inductance $\mathrm{L}$ " in Parallel with the given inductance L, such that

$$\frac{1}{\mathrm{L}}+\frac{1}{{\mathrm{L}}^{\prime }}=\frac{1}{{\mathrm{L}}^{\prime \prime }}={\omega }^2\mathrm{C}$$

This gives $L^{\prime \prime }=\frac{L}{({\omega }^{2}LC-1)}$

Difficult

Adjusting the Value of C in a LCR Circuit

67. Graph A, drawn here, shows simulteneously the variation of the (alternating) voltage (applied) and the current (flowing), with time, for a given senses LCR circuit.

Without making any other change in the circuit, the (voltage/current) vs (time) graph, for the circuit, would take the form shown, in graph B, if another capacitor C”, of value

$$

(1) $\left(\frac{1-{\omega }^2\mathrm{LC}}{{\omega }^2\mathrm{L}}\right)$ is put in parallel with the given $\mathrm{C}$

(2) $\left(\frac{1-{\omega }^{2}LC}{{\omega }^{2}L}\right)$ is put in series with the given $C$

(3) $\left(\frac{{\omega }^2\mathrm{L}}{1-{\omega }^2\mathrm{LC}}\right)$ is put in parallel with the given $\mathrm{C}$

(4) $\left(\frac{{\omega }^2\mathrm{L}}{1-{\omega }^2\mathrm{LC}}\right)$ is put in series with the given $\mathrm{C}$

Show Answer

Correct answer: (1)

Solution:

Graph A shows the current leading in phase w.r.t the voltage. In graph B, the voltage and current are in phase. This means that we need a net value, $C^{\prime }$, of capacitance, for which

$$\frac{1}{\omega C^{\prime }}=\omega L\text{ or }{C}^{\prime }=\frac{1}{{\omega }^{2}L}$$

Since the current is leading in phase, it can be brought ‘in phase’, by increasing the given value of C. For this, we put a capacitor, C", in parallel with the given capacitance ( $=\mathrm{C}$ ) such that

$$\begin{array}{c}\mathrm{C}+{\mathrm{C}}^{\prime \prime }={\mathrm{C}}^{\prime }\\ \therefore {\mathrm{C}}^{\prime \prime }={\mathrm{C}}^{\prime }-\mathrm{C}=\frac{1}{\left({\omega }^2\mathrm{L}\right)}-\mathrm{C}=\left(\frac{1-{\omega }^2\mathrm{LC}}{{\omega }^2\mathrm{L}}\right)\end{array}$$

Average

LCR Circuits

68. The impedance of a given series LCR circuit, at its resonance (angular) frequency $(={\omega }_0)$ is $\mathrm{X}$ ohms. its impedance becomes $\mathrm{Y}\mathrm{ohm}$ when the source voltage has a (angular) frequency $\omega$. The values of $L,C$ and $R$, in the given circuit, are equal, respectively, to

(1)$\begin{array}{r}\mathrm{L}=\frac{{\omega }^2}{{\omega }_0}\frac{\sqrt{{\mathrm{Y}}^2-{\mathrm{X}}^2}}{({\omega }^2-{\omega }_0^2)};\mathrm{C}=\frac{({\omega }^2-{\omega }_0^2)}{\omega {\omega }_0^2\sqrt{{\mathrm{Y}}^2-{\mathrm{X}}^2}};(\mathrm{R}=\mathrm{X})\end{array}$

(3)$\begin{array}{r}\mathrm{L}=\frac{{\omega }_0^2}{{\omega }_0}\frac{\sqrt{{\mathrm{Y}}^2-{\mathrm{X}}^2}}{({\omega }^2-{\omega }_0^2)};\mathrm{C}=\frac{{\omega }^2-{\omega }_0^2}{\omega {\omega }_0^2\sqrt{{\mathrm{Y}}^2-{\mathrm{X}}^2}};(\mathrm{R}=\mathrm{X})\end{array}$

(3)$\begin{array}{r}\mathrm{L}=\left(\omega {\omega }_0\right)\frac{\sqrt{{\mathrm{Y}}^2-{\mathrm{X}}^2}}{({\omega }^2-{\omega }_0^2)};\mathrm{C}=\frac{{\omega }^2-{\omega }_0^2}{\omega {\omega }_0^2\sqrt{{\mathrm{Y}}^2-{\mathrm{X}}^2}};(\mathrm{R}=\mathrm{X})\end{array}$

(4)$\begin{array}{r}\mathrm{L}=\frac{1}{\left(\omega {\omega }_0\right)}\frac{\sqrt{{\mathrm{Y}}^2-{\mathrm{X}}^2}}{({\omega }^2-{\omega }_0^2)};\mathrm{C}=\frac{({\omega }^2-{\omega }_0^2){\omega }^2{\omega }_0}{\sqrt{{\mathrm{Y}}^2-{\mathrm{X}}^2}};(\mathrm{R}=\mathrm{X})\end{array}$

Show Answer

Correct answer: (1)

Solution:

The impedance, of a series LCR circuit, at its resonance (angular) frequency $\left({\omega }_0\right)$, equals R. Hence

$$\mathrm{R}=\mathrm{X}$$

We also have ${\omega }_0\mathrm{L}-\frac{1}{{\omega }_0\mathrm{C}}=0$

and ${\mathrm{R}}^2+{(\omega \mathrm{L}-\frac{1}{\omega \mathrm{C}})}^2={\mathrm{Y}}^2$

$\therefore \omega \mathrm{L}-\frac{1}{\omega \mathrm{C}}=\sqrt{{\mathrm{Y}}^2-{\mathrm{R}}^2}=\sqrt{{\mathrm{Y}}^2-{\mathrm{X}}^2}$

Solving the two equations, we get

$\mathrm{L}=[\frac{{\omega }^2}{{\omega }_0}\cdot \frac{\sqrt{{\mathrm{Y}}^2-{\mathrm{X}}^2}}{({\omega }^2-{\omega }_0^2)}]$

and $\mathrm{C}=\frac{({\omega }^2-{\omega }_0^2)}{\left[\omega {\omega }_0^2\sqrt{{\mathrm{Y}}^2-{\mathrm{X}}^2}\right]}$

Difficult

Analysing LCR Circuit

69. An electrical circuit is made up of an inductance ( $L)$, a capacitor $(C)$ and a resistor $(R)$, connected in series. When it is connected across the mains (Voltage $=111.8\mathrm{V}\simeq \sqrt{12500}\mathrm{V}$ ), it draws $125\mathrm{W}$ power from these mains. It is also observed that the current lags in phase, with respect to the applied voltage by $\frac{\pi }{4}$. If the inductive impedance is twice the capacitance impedance, the three impedances in the circuit: (i) its resistive impedance; (ii) its inductive impedance and (iii) its capacitative impedance, are (nearly) equal, respectively, to

(1) $142\mathrm{\Omega },284\mathrm{\Omega },142\mathrm{\Omega }$

(2) $106.5\mathrm{\Omega },213\mathrm{\Omega },106.5\mathrm{\Omega }$

(3) $89\mathrm{\Omega },178\mathrm{\Omega },89\mathrm{\Omega }$

(4) $71\mathrm{\Omega },142\mathrm{\Omega },71\mathrm{\Omega }$

Show Answer

Correct answer: (4)

Solution:

Let $\mathrm{z}$ be the total impedance of the circuit. We then have

Power $=\mathrm{P}=\frac{{\mathrm{V}}^2}{\mathrm{z}}$

$\therefore \mathrm{z}=\frac{{\mathrm{V}}^2}{\mathrm{P}}=\frac{12500}{125}=100\mathrm{\Omega }$

Now phase angle, $\theta$, is given by

$\tan \theta =\frac{X_L-X_C}{R}$

Here $\theta =\frac{\pi }{4}$. This means that $({\mathrm{X}}_{\mathrm{L}}-{\mathrm{X}}_{\mathrm{C}})=\mathrm{R}$

Also ${\mathrm{z}}^2={\mathrm{R}}^2+{({\mathrm{X}}_{\mathrm{L}}-{\mathrm{X}}_{\mathrm{C}})}^2$. Hence ${100}^2={\mathrm{R}}^2+{\mathrm{R}}^2$

$\therefore {\mathrm{R}}^2=5000(\mathrm{\Omega }{)}^2$

$\therefore \mathrm{R}=\sqrt{5000}\mathrm{\Omega }\simeq 71\mathrm{\Omega }$

Also ${\mathrm{X}}_{\mathrm{L}}-{\mathrm{X}}_{\mathrm{C}}=\mathrm{R}=70\mathrm{\Omega }$

$\therefore 2{\mathrm{X}}_{\mathrm{C}}-{\mathrm{X}}_{\mathrm{C}}=\mathrm{R}$.

Hence ${\mathrm{X}}_{\mathrm{C}}=71\mathrm{\Omega }$

$\therefore {\mathrm{X}}_{\mathrm{L}}=142\mathrm{\Omega }$

Average

Impedance Calculation

70. The circuit, shown in the figure here, has been connected to an a.c. voltage source $(v=v_0\sin \omega t)$ of a very high frequency. The effective impedance of this circuit, and the current flowing in it, would then be given by

$$

(1) $4\mathrm{R},\mathrm{I}=\frac{{\mathrm{V}}_0}{4\mathrm{R}}\sin \omega \mathrm{t}$

(2) $7\mathrm{R},\mathrm{I}=\frac{{\mathrm{v}}_0}{7\mathrm{R}}\sin \omega \mathrm{t}$

(3) $7\mathrm{R},\mathrm{I}=\frac{{\mathrm{v}}_0}{7\mathrm{R}}\cos \omega \mathrm{t}$

(4) $4\mathrm{R},\mathrm{I}=\frac{{\mathrm{V}}_0}{4\mathrm{R}}\cos \omega \mathrm{t}$

Show Answer

Correct answer: (1)

Solution:

At very high frequencies, the effective impedance of the capacitor tends to zero while that of the inductor tends to infinity. The branches, containing inductors, are, therefore, (effectively) not parts of the circuit while those conatining the capacitor, are (effectively) like a simple connecting wire. The circuit, therefore, looks effectively like the one shown here.

$$

The effective impedance is, therefore, $[R+\frac{4\times 12}{16}R]$ i.e. $4R$. The current flowing, in the circuit, is therefore, in phase with the applied voltage and is given by

$$I=\left(\frac{v_0}{4R}\right)\sin \omega t$$

Average

Induced emf in a Rotating Coil

71. A planar circular coil, of radius $a$, has $\mathbf{N}$ turns and it can be made to rotate about its diameter as axis. The coil has a resistance $R$ and is present in a region where a uniform magnetic field, B, directed perpendicular to the axis of rotation of the coil, exists. If the coil has to be kept rotating with a constant frequency $v$, (assuming friction and other dissipative effects to be negligible)

(1) an external torque, of magnitude $2\pi vNB{\sin }^2(2\pi vt)$ must be made to act on it

(2) an external torque, of magnitude $2\pi \mathrm{vN}[\pi {\mathrm{a}}^2\mathrm{B}\sin (2\pi \mathrm{vt})]$ must be made to act on it

(3) an external torque, of magnitude $2\pi \mathrm{vN}{\left[\pi {\mathrm{a}}^2\mathrm{B}\right]}^2\sin (2\pi \mathrm{vt})$ must be made to act on it

(4) it just needs to be given the necessary ‘spin’ at start, it would then keep on rotating at its initial rate, due to its ‘rotational inertia’.

Show Answer

Correct answer: (2)

Solution:

Let the coil rotate about a diameter, taken as the $\mathrm{z}$-axis. The magnetic field isdirected along the $x$-axis. The flux, linked with the coil, at any instant, is

$$

$\begin{array}{rl}& \varphi =\mathrm{N}\left(\pi {\mathrm{a}}^2\right)\mathrm{B}\cos \theta \\ & =\mathrm{N}\pi {\mathrm{a}}^2\mathrm{B}\cos (2\pi \mathrm{t})\end{array}$

$\therefore$ Induced emf $=\epsilon -\frac{\mathrm{d}\varphi }{\mathrm{dt}}=2\pi \mathrm{vN}\cdot \pi {\mathrm{a}}^2\mathrm{B}\sin (2\pi \mathrm{t})$

Instantaneous induced current $=\mathrm{i}=\frac{\epsilon }{\mathrm{R}}$

The magnetic moment, associated with this induced current,

$\mathrm{M}=\mathrm{i}\cdot \left(\pi {\mathrm{a}}^2\right)$

The magnetic field exerts a torque, which has an instantneous value, $\tau =\mathrm{MB}\sin \theta$

The external torque is needed to oppose this torque on the coil due to the induced current. Hence external torque $=\tau =\mathrm{MB}\sin \theta =2\pi \nu \mathrm{N}\pi {\mathrm{a}}^2\mathrm{B}\sin (2\pi \nu t)$

Average

AC Generator

72. A circular coil, having $\mathbf{N}$ turns of wire, each of mean diameter $D$, is set up so that it can rotate about a vertical axis, at the rate of $p$ rotations per second. The horizotal component, of the earth’s magnetic field, at that place, equals $B_H$.

The magnitudes of the instantaneous emf, induced in the coil, when its plane

(i) is at right angles to the magnetic meridian.

(ii) makes an angle of ${30}^{\circ }$ with the magnetic meridian.

(iii) coincides with the magnetic merician, would equal, respectively.

(1) $$ (0), $\left(\frac{\sqrt{3}}{4}{\pi }^2{\mathrm{D}}^2{\mathrm{NB}}_{\mathrm{H}}\mathrm{p}\right);\left(\frac{1}{2}{\pi }^2{\mathrm{D}}^2{\mathrm{NB}}_{\mathrm{H}}\mathrm{p}\right)$

(2) $$ $\left(\frac{1}{2}{\pi }^2{\mathrm{D}}^2{\mathrm{NB}}_{\mathrm{H}}\mathrm{p}\right);\left(\frac{\sqrt{3}}{4}{\pi }^2{\mathrm{D}}^2{\mathrm{NB}}_{\mathrm{H}}\mathrm{p}\right),(0)$

(3) $$ (0), $\left(\frac{1}{4}{\pi }^2{\mathrm{D}}^2{\mathrm{NB}}_{\mathrm{H}}\mathrm{p}\right);\left(\frac{1}{2}{\pi }^2{\mathrm{D}}^2{\mathrm{NB}}_{\mathrm{H}}\mathrm{p}\right)$

(4) $$ $\begin{array}{r}(0),\left(\frac{1}{4}{\pi }^2{D}^2{\mathrm{NB}}_{\mathrm{H}}\mathrm{p}\right);\left(\frac{1}{2}{\pi }^2{\mathrm{D}}^2{\mathrm{NB}}_{\mathrm{H}}\mathrm{p}\right)\end{array}$

Show Answer

Correct answer: (1)

Solution:

When the normal, to the plane of the coil, makes an angle $\theta$ with the direction of the magnetic field, the instantaneous flux linked with it is

$$\varphi =\mathrm{N}\left(\frac{\pi {\mathrm{D}}^2}{4}\right){\mathrm{B}}_{\mathrm{H}}\cos (2\pi \mathrm{pt})(\theta =\omega \mathrm{t}=2\pi \mathrm{pt})$$

$\therefore$ Instantaneous induced emf is

$$\begin{array}{rl}\epsilon & =-\frac{\mathrm{d}\varphi }{\mathrm{dt}}=\frac{\mathrm{N}\pi {\mathrm{D}}^2}{4}\cdot {\mathrm{B}}_{\mathrm{H}}(2\pi \mathrm{p})\sin (2\pi \mathrm{pt})\\ & =\frac{{\pi }^2{\mathrm{D}}^2{\mathrm{NB}}_{\mathrm{H}}\mathrm{p}}{2}\sin \theta \end{array}$$

In the cases (i), (ii), (iii), $\theta$ (angle betwwen the normal to the plane of the coil and the magnetic field) equals $0^{\circ },{60}^{\circ }$ and ${90}^{\circ }$, respectively.

Hence the three values of $\epsilon$ are

$0,\frac{\sqrt{3}}{4}\pi {\mathrm{DNB}}_{\mathrm{H}}\mathrm{p}$ and $\frac{{\pi }^2{\mathrm{d}}^2{\mathrm{NB}}_{\mathrm{H}}\mathrm{p}}{2}$, respectively

Average

AC Generator

73. A mini generator has a coil of 1000 turns, each of area ${10}^{-2}{\mathrm{m}}^2$. The coil is placed with its plane perpendicular to a uniform magnetic field of intensity $25\mathrm{mT}$ and is rotated at a uniform rate of 100 rotations per second. The magnitudes of the average of the emf generated, during the four quarters of its rotation, equal, respectively

(1) 0 volt, 0 volt, 0 volt, 0 volt

(2) 0 volt, 100 volt, 0 volt, 100 volt

(3) 100 volt, 0 volt, 100 volt, 0 volt

(4) 100 volt, 100 volt, 100 volt, 100 volt

Show Answer

Correct answer: (4)

Solution:

The flux, linked with the coil, at any instant is

$\varphi =\mathrm{NAB}\cos \omega \mathrm{t}$

$\therefore$ Instantaneous induced emf, $\epsilon$, is given by

$\epsilon =\frac{\mathrm{d}\varphi }{\mathrm{dt}}=\mathrm{NAB}\omega \cos \omega \mathrm{t}$

Here $\mathrm{N}=100,\mathrm{A}={10}^{-2}{\mathrm{m}}^2,\mathrm{B}=25\times {10}^{-3}\mathrm{T};\omega =(100\times 2\pi ){\mathrm{s}}^{-1}$

$\begin{array}{rl}& \therefore \text{ NAB }\omega ={10}^3\times {10}^{-2}\times 25\times {10}^{-3}\times 2\pi \times {10}^2\text{ volt }\\ & =50\pi \text{ volt }\\ & \therefore \epsilon =(50\pi )\sin \omega \mathrm{t}\end{array}$

The average, of this, over success four quarters of the cycle, are given by $\frac{4}{\mathrm{T}}$ times the integral of the above, from (i) (0) to $\left(\frac{T}{4}\right)$ (ii) $\left(\frac{\mathrm{T}}{4}\right)$ to $\left(\frac{\mathrm{T}}{2}\right)$ (iii) $\left(\frac{T}{2}\right)$ to $\left(\frac{3T}{4}\right)$ and (iv) $\left(\frac{3T}{4}\right)$ to (T). Each of there is seen to have a magnitude.

$(\frac{4}{\left(\frac{1}{100}\right)}\times 50\pi \times \frac{\cos 0^0}{200\pi })$ volt $=100\mathrm{V}$

Easy

Transformer

74. A step down transformer (having windings of negligible resistance) is used to step down the mains voltage of $220\mathrm{V}$, to22 $\mathrm{V}$ to operate a device of impedance $220\mathrm{V}$. Assuming an efficienty of $80\mathrm{\%}$, the current flowing in the primary coil of two transformer is

(1) $8\mathrm{nA}$

(2) $10.25\mathrm{mA}$

(3) $12.5\mathrm{mA}$

(4) $14.75\mathrm{mA}$

Show Answer

Correct answer: (3)

Solution:

The secondary, supplying a voltage of $22\mathrm{V}$, operates a device of impedance $200\mathrm{Ohms}$.

Hence current in the secondary $=\frac{22}{220}\mathrm{A}=0.1\mathrm{A}$

$\therefore$ Power consumed in the secondary circuit $=22\times 0.1,\mathrm{W}=2.2\mathrm{W}$

Efficiency $=80\mathrm{\%}$

Power available in the primary circuit $=2.2\times \frac{100}{80}\mathrm{W}=2.75\mathrm{W}$

$\therefore$ If I is the current in the primary coil, we have

$220\times \mathrm{I}=2.75$

$\therefore \mathrm{I}=\frac{2.75}{220}\mathrm{A}=\frac{2750}{220}\mathrm{mA}$

$=12.5\mathrm{mA}$

Average

Transformer

75. The diagram, given here, shows a ‘step-up’ that may be used to supply power from a ‘power plant’ to a ’town’ via transmission lines of a total (to and fro) length of $50\mathrm{kM}$. The power requirement, of the town is $2.2\mathrm{MW}$. Assuming that the two transformers, used at the ‘power plant’ and the ’town substation’ are ideal ones and the resistance of the transmission line is $1\mathrm{ohm}/\mathrm{km}$, the stepped up voltage, available at the secondary coil of the transformer, used at the power plant, should be

$$

(1) $4400\mathrm{V}$

(2) $16900\mathrm{V}$

(3) $20600\mathrm{V}$

(4) $29400\mathrm{V}$

Show Answer

Correct answer: (4)

Solution:

The power available at the primary coil of the town substation transformer = power requirement of town

$$=2.2\times {10}^6\mathrm{W}$$

Voltage available at its primary $=4400\mathrm{V}$

$\therefore$ rms current in the line $=\frac{2.2\times {10}^6\mathrm{W}}{4.4\times {10}^3\mathrm{V}}=500\mathrm{A}$

Total resistance of the line $=50\mathrm{\Omega }$

Voltage drop across the line $=500\times 50\mathrm{V}=25000\mathrm{V}$

$\therefore$ Voltage that needs to be available at the secondary coil of the transformer of the power plant

$$=29400\mathrm{V}$$

Average

Transformer

76. A power station, located at a distance of $(3\pi )\mathrm{km}$ from a town, has to deliver 8.8 of MW of power to the town. The transmission wires are made from a metal of resistivity $2\times {10}^{-8}\mathrm{SI}$ units and have a diameter of $1\mathrm{cm}$.

The power station uses a step-up transformer to boost the voltage to $22\mathrm{kV}$; transmitts power at this voltage and uses a step down transformer, near the town, to bring down voltage to $220\mathrm{V}$. The percentage of ohmic losses, during transmission, in terms of the power transmitted, equal (nearly)

(1) $13\mathrm{\%}$

(2) $8.7\mathrm{\%}$

(3) $6.5\mathrm{\%}$

(4) $4.4\mathrm{\%}$

Show Answer

Correct answer: (2)

Solution:

The total resistance, of the transmission lines, is

$$\begin{array}{c}\mathrm{R}=\rho \frac{\ell }{\mathrm{A}}=\frac{2\times {10}^{-8}\times 2\times 3\pi \times {10}^3}{\pi {(0.5\times {10}^{-2})}^2}\\ \\ =48\times {10}^{-1}\mathrm{\Omega }=4.8\mathrm{\Omega }\end{array}$$

The current in the transmission wires

$$\begin{array}{c}\mathrm{I}=\frac{\text{ Power }}{\text{ Voltage }}=\frac{8.8\times {10}^6}{22000}\mathrm{A}\\ =400\mathrm{A}\end{array}$$

$\therefore$ Ohmic losses $=I^2\mathrm{R}={(4\times {10}^2)}^2\times 4.8\mathrm{W}$

$=76.8\times {10}^4\mathrm{W}=0.768\times {10}^6\mathrm{W}$

$\therefore$ Percentage of ohmic losses, in terms of power transmitted $=\frac{0.768\times {10}^6}{8.8\times {10}^6}\times 100\simeq 8.7\mathrm{\%}$

Average

RMS Value

77. The alternating current, in a given set up, varies with time in the manner shown. The rms value of this current, over one cycle, is

$$

(1) $3\mathrm{A}$

(2) $3.17\mathrm{A}$

(3) $3.28\mathrm{A}$

(4) $3.62\mathrm{A}$

Show Answer

Correct answer: (2)

Solution:

The rms value, by definition, is obtained by taking the average of the squares of all values (over one complete cycle) and then taking the square root of this average.

For the given case, we have

For $(t=0)$ to $(t=1)$; sum of the squares of all values $=(2{)}^3=4$

For $(t=1)$ to $(t=3)$; sum of the squares of all values $=(-4{)}^2\times 2=32$

For $(t=3)$ to $(t=4)$; sum of the squares of all values $=(2{)}^2\times 1=4$

$\therefore$ Sum of the squares of all values over one complete cycle $(=4\mathrm{s})=(4+32+4){\mathrm{A}}^2=40{\mathrm{A}}^2$

$\therefore$ Average of the squared values $=\left(\frac{40}{4}\right){\mathrm{A}}^2=10{\mathrm{A}}^2$

$\therefore$ rms value $=\sqrt{10}\mathrm{A}$

$\simeq 3.17\mathrm{A}$

Difficult

Voltage (in Volt)

78. The rms value, of the saw tooth alternating voltage, shown here, equals (nearly)

$$

(1) 1.732 Volt

(2) 1.00 Volt

(3) 1.15 Volt

(4) 1.42 Volt

Show Answer

Correct answer: (3)

Solution:

The rms value, by definition, is obtained by taking the average of the squares of all values (over one complete cycle) and then taking the square root of this average.

We see that for the time period of the alternating voltage is $4\mathrm{V}$.

For $(t=0)$ to $(t=2)$, we have $v(t)=1\times t($ slope $=1)$

$\therefore$ Sum of the squares of all values $={\int }_0^2[\mathrm{v}(\mathrm{t}){]}^2\mathrm{dt}$

$={\int }_0^2{t}^{2}dt=(8/3){\mathrm{volt}}^2$

For $(t=2)$ to $t=4$, we have

$v(t)=1\times t-4$

$\therefore$ Sum of the squares of all values $={\int }_2^4(\mathrm{t}-4{)}^2\mathrm{dt}$

$={\left|\frac{(\mathrm{t}-4{)}^3}{3}\right|}_2^4=[0-\left(\frac{-8}{3}\right)]=\frac{8}{3}{\mathrm{v}}^2$

$\therefore$ Sum of the squares of all values over one complete cycle

$=(\frac{8}{3}+\frac{8}{3}){\text{ volt }}^2=\frac{16}{3}{\text{ volt }}^2$

$\therefore$ Average of the squared values $=\frac{16}{3}{\mathrm{v}}^2\times \frac{1}{4}$

$=\frac{4}{3}{\mathrm{v}}^2$

$\therefore$ rms value $-\frac{\sqrt{4}}{2}\mathrm{v}=\frac{2}{\sqrt{3}}\mathrm{v}\simeq 1.15\mathrm{v}$

Average

Induced Current and Resulting Power Loss

79. A circular coil, of radius $a$, (having $N$ turns) is made to rotate about its vertical diameter with an angular speed $\omega$. The coil is present in a region where a uniform horizontal magnetic field $B$ is present. If the coil has a resistance $R$, the rms value of the induced current and the resulting power loss, in it, are given, respectively, by

(1) $\frac{\mathrm{N}\pi {\mathrm{a}}^2\omega \mathrm{B}}{\mathrm{R}}$ and $\frac{{\left(\mathrm{N}\pi {\mathrm{a}}^2\right)}^2\omega \mathrm{B}}{\mathrm{R}}$

(2) $\frac{N\pi a^2\omega B}{\sqrt{2}R}$ and $\frac{2\left(N^2\pi a^2\omega B\right)}{R}$

(3) $\frac{N\pi a^2\omega B}{\sqrt{2}R}$ and $\frac{{\left(N\pi a^2\omega B\right)}^2}{2R}$

(4) $\frac{\mathrm{N}\pi {\mathrm{a}}^2\omega \mathrm{B}}{\mathrm{R}}$ and $\frac{\left(\mathrm{N}\pi {\mathrm{a}}^2\omega \mathrm{B}\right)}{\mathrm{R}}$

Show Answer

Correct answer: (2)

Solution:

The instantaneous flux, linked with the coil; is

$\varphi =\mathrm{N}\pi {\mathrm{a}}^2\mathrm{B}\cos (\omega \mathrm{t})$

$\therefore$ Instantaneous induced emf is

$\epsilon =-\frac{\mathrm{d}\varphi }{\mathrm{dt}}=\mathrm{N}\pi {\mathrm{a}}^2\omega \mathrm{B}\sin \omega \mathrm{t}$

$={\epsilon }_0\sin \omega \mathrm{t}$

where ${\epsilon }_0=\mathrm{N}\pi {\mathrm{a}}^2\omega \mathrm{B}$

$\therefore$ The rms of the induced current $=\frac{{\epsilon }_{\mathrm{rms}}}{\mathrm{R}}=\frac{{\epsilon }_0}{\sqrt{2}\mathrm{R}}$

$=\frac{\mathrm{N}\pi {\mathrm{a}}^2\omega \mathrm{B}}{\sqrt{2}\mathrm{R}}$

$\therefore$ Instantaneous power loss $={\epsilon }_{\mathrm{rms}}{\mathrm{I}}_{\mathrm{rms}}$

$\begin{array}{rl}& =\frac{\mathrm{N}\pi {\mathrm{a}}^2\omega \mathrm{B}}{\sqrt{2}}\cdot \frac{\mathrm{N}\pi {\mathrm{a}}^2\omega \mathrm{B}}{\sqrt{2}\mathrm{R}}\\ \\ & =\frac{1}{2\mathrm{R}}{[\mathrm{N}\cdot \pi {\mathrm{a}}^2\omega \mathrm{B}]}^2\end{array}$

Easy

Wattless Curents

80. For a pure inductor, or a pure capacitor, connected to an a.c. voltage source, one can say that the instantnaeous power.

(1) as well as the average, over a cycle, are both zero

(2) may not be zero but the average power, over a cycle, is zero

(3) can zero but the average power, over a cycle, is hot zero

(4) as well as the average power, over a cycle, are both not zero

Show Answer

Correct answer: (2)

Solution:

The current, in a pure inductor, or a pure capacitor, has a phase lag of $\pm \frac{\pi }{2}$ with respect to the applied voltage. Hence if the applied voltage is $(v_0\sin \omega t)$, the current is $I_0\sin (\omega t\pm \frac{\pi }{2})$, is $\pm I_0\cos \omega t$. The instantaneous power is, therefore,

$$\begin{array}{rl}{P}_{\text{inst }}& =\pm {\mathrm{v}}_0{\mathrm{I}}_0\sin \omega \mathrm{t}\cos \omega \mathrm{t}\\ & =\pm \frac{{\mathrm{v}}_0{\mathrm{I}}_0}{2}\sin 2\omega \mathrm{t}\end{array}$$

This is not zero at all instant. However, its average, over a complete cycle, is zero.