SATHEE: UNIT 12 CURRENT ELECTRICITY

UNIT 12 CURRENT ELECTRICITY

Learning Objectives

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

Electrical energy from different type of the cells.
Meaning of emf and the essential difference between emf and terminal potential difference.
Current and sign convention for the current.
Drift velocity and the physics of the Ohm’s law.
Concept of resistance and its variation will temeperature.
Combination of resistance in series and parallel.
Joule’s law and the concept of electrical power.
Kirchoff’s laws for circuit analysis.
Wheatstone bridge: its practical form the meter bridge.
The principle and applications of potentiometer.

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Electric Current

It is a well known fact that the electron is an atomic particle carrying a negative charge. Electrons can be detached from the atoms of certain elements quite easily. If we connect a conductor to something which supply or absorb charge as required to maintain a applied electric field (cells, batteris etc.), this maintained field than can act on the free electrons in the conductor and will give them a resultant motion in a direction opposite to that of the field. Charges in motion constitute an electric current. If a net charge q passes through any cross section of the conductor in time $t$ the average current is given by.

$$ I=\frac{q}{t} $$

When equals quantities of charge $q$ passes a given cross section of the conductor in successive equal interval time of t. The current $I$ is said to be constant in magnitude or STEADY.

In general the rate of flow of charge with time is not constant and the current varies with time. It is then given by differential limit viz.

$$ \mathrm{I}=\frac{\mathrm{dq}}{\mathrm{dt}} $$

The current $\mathrm{I}$ is the same for all cross section of a conductor even though the cross sectional area may be different at different points as there can be no continued accumulation of charge at any point along the conductor.

Figure shows a portion of a wire in which there is a field towards the left and consequently the motion of the free electron towards the right. Let the electrons move with a constant drift velocity $\mathrm{v} _{\mathrm{d}}$ (discussed later in the chapter). Then in time there dt each advances a distance $\mathrm{v} _{\mathrm{d}} \mathrm{t}$. In this time the number of electrons crossing any place such as shaded is the number contained in a section of length $\mathrm{v} _{\mathrm{d}} t$ or volume $\mathrm{Av} _{\mathrm{d}} \mathrm{t}$ where $A$ is the area of cross section of the wire. Let $n$ be the number of free force electrons per unit volume. Then the number of electrons crossing the plane in time $\mathrm{dt}^{2} \mathrm{nAv} _{\mathrm{d}} \mathrm{dt}$ and if e represents the clarge of each electron, the total clarge crossing in time dt is

$$ \mathrm{dq}=\mathrm{neAv} \mathrm{d} _{\mathrm{d}} \mathrm{dt} $$

Hence the current $I=\frac{d q}{d t}=n e A v _{d}$

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The Direction of a Current

A logic way to define the direction of a current is the direction of motion of free electrons. However in electrolytic and gaseous conduction, the current is due to the motion of positive and negative ions. Historically the sign convention for the current was defined taking positive charge carriers as current carriers. Since these would be a incosistancy whether we define a sign convention taking positive or negative particles as charge carriers we continue to work with historical sign convention. In conductors the free electrons move in the opposite direction to the conventional current. If required we can label the current flowing in a conductor due to the motion of free electrons as Electronic Current.

Resistance: Proper understanding of resistance is possible only through Ohm’s law. Qualitatively speaking resistance of a material is a measure of its ability to resist the flow of electricity through it.

Ohm’s Law

Verified experimentally Ohm’s law states that the current flowing through a conductor is proportional to the voltage $\mathrm{V}$ across its ends when its physical conditions (temperature etc.) remains unchanged Expressed mathmatically we can write.

$$ \frac{\mathrm{V}}{\mathrm{I}}=\mathrm{R} \quad \text { (Constant of proportionally) } $$

The constant $\mathrm{R}$ is known as the resistance of the conductor.

Ohm’s law, a experimental law is valid only for metallic conduction. We now know that there are quite a number of substances (e.g. semi conductors) for which Ohm’s law is not always valid. For such cases the current voltage graphs are not linear. We illustrate this piont by showing the plot of current voltage graph for some interesting cases:

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Electrical Resistivity

The resistance of a conductor depends on the nature of its material, its shape and size, Simple consideration show that the resistance of wires of a given length increases in direct proportion to their length and for wires of a given area of cross section the resistance increases in inverse proportion of their area of cross section.

We thus have,

$\mathrm{R} \propto \mathrm{L}$ and $\mathrm{R} \propto \frac{1}{\mathrm{~A}}$

or $\mathrm{R}=\rho \frac{\mathrm{L}}{\mathrm{A}}$

$\rho$ is a constant characterstic of the material and is known as resistivity.

Colour Code for Resistances

Resistances are an integral path of practically all electrical and electronic circuits. The values of resistances needed in different circuits very over a wide range and for convenience a colour code has been developed.

The colour code works as follows:

1. We associate a colour with each digit $0,1,2, \ldots . .9$. These colours and associated digits are given in table below:

Colour	Black	Brown	Red	Orange	Yellow	Green	Blue	Violet	Gray	White
Associated Digit	0	1	2	3	4	5	6	7	8	9

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2. There are three colour bands on each resistance to indicate its value. The first two bands from one end indicate their corresponds digits while the third band colour gives the power of ten with which the number obtained by first two colours must be multiplied to get the resistance value in ohm. For example if a resistance has three colours as yellow, orange and blue, the resistance value is $43 \times 10^{6} \Omega$.

3. In addition to the above colours, two additional colours - SILVER and GOLD are also used. When used as fourth colours band they indicate the TOLERANCE or percentage realibility in the value of the resistanc obtained.

A silver band implies a tolerance of $10 \%$ while a gold band implies a tolerance of $5 \%$.

4. Silver band used as the third band indicate a multiplier value of $10^{-2}$ while a gold band used as third band indicate a multiplies value of $10^{-1}$.

Physics of the Oh’m Law

Ohm’s law relate the current flowing in a conductor with the potential difference across its ends. We know that current is a flow of free elctrons under a maintained field. This could imply that free clarges in a conductor should continuously get accelerated which imply that the current should go on increasing with time. However only a stready current governed by Ohm’s law flows. This implies that in a conductor, a constant electric field produces a constant velocity, a contradication of Newton’s second law. Let us try to understand the reason for this observed behaviour.

We note that an electric field exerts a force $\mathrm{n}$ eE on a free electron of charge $\mathrm{e}$. This force gives; the electron an acceleration ’ $a$ ’ in the direction of the field where

$$ \mathrm{a}=\frac{\mathrm{e} \mathbf{E}}{\mathrm{m}} $$

Now the electrons in a metal at any temperature are, by themselves, moving will fairly high velocities that are randomly distributed. Hence the average velocity of any free electron in the direction of the field, at the instant the field is switched on is zero. The field imports the electron an acceleration in its own direction

where

$$ \mathrm{a}=\left(\frac{\mathrm{e} \mathbf{E}}{\mathrm{m}}\right) $$

However this acceleration lasts but for a short time as the electron on its way encounters the metal ions that are vibrating about their mean position. Since the magnitude and direction of these ions vibration at the instant of collision can have arbitrary value, they will produce a random deflection in the path of the electron when it undergoes a collision with one of these. On the average these random collisions of the free electron may occur after a time $\tau$ so that the velocity of the electron in the direction of the field just before a collision is

$$ \mathrm{v}=0+\mathrm{a} \tau=\frac{\mathrm{eE} \tau}{\mathrm{m}} $$

This additional velocity imparted by the field to a freely and otherwise randamly moving electron is known its DRIFT velocity. After the collision the electron may again be deflected in any random direction so that as we can assume that the (average) distribution of free electtron velocities after a time $\tau$ is the same as it was before the field was applied. Each collision causes the electron to loose the energy it gained from the field back to the atoms or ions with which it collided (cause for heating effects of current). Now the drift velocity just before and just afer a collision are $\frac{\mathrm{eE} \tau}{\mathrm{m}}$ and zero respectively. Hece the average drift velocity of the electron is $\mathrm{v} _{\mathrm{d}}=\frac{1}{2} \frac{\mathrm{eE} \tau}{\mathrm{m}}=\frac{1}{2} \frac{\mathrm{e}}{\mathrm{m}} \tau \frac{\mathrm{V}}{\mathrm{L}}$

$$ \mathrm{v} _{\mathrm{d}}=\frac{1}{2} \frac{\mathrm{eE} \tau}{\mathrm{m}}=\frac{1}{2} \frac{\mathrm{e}}{\mathrm{m}} \tau \frac{\mathrm{v}}{\mathrm{L}} $$

If there are $\mathrm{n}$ free electrions per unit volume the current flowing in a conductor is given by

$$ \begin{aligned} \mathrm{I} & =\mathrm{ne} \mathrm{Av} _{\mathrm{d}} \\ \text { or } \quad \mathrm{I} & =\mathrm{neA} \frac{1}{2} \frac{\mathrm{e} \tau}{\mathrm{m}} \frac{\mathrm{v}}{\mathrm{L}}=\frac{1}{2} \frac{\mathrm{ne}^{2} \tau}{\mathrm{m}} \frac{\mathrm{V}}{(\mathrm{L} / \mathrm{A})} \end{aligned} $$

Putting, $\rho=\frac{2 \mathrm{~m}}{\mathrm{ne}^{2} \tau}$ we have $\mathrm{I}=\frac{\mathrm{V}}{\rho(\mathrm{L} / \mathrm{A})}=\frac{\mathrm{V}}{\mathrm{R}}$

which is Ohm’s Law

$\tau$ is known as the relexation time.

Effect of Temperature on Resistance

We have seen that the resistivity $\rho$ is inversaly proportional to relaxtion time. The frequency of collisions us depends upon:

1. The extent of randomness in the arragement of ionic and atomic sites on the lattice of the given material.

2. The temperature of the given substance.

An increase in temperature will cause the ions and atomc to vibrate with a greater amplitude about their mean position and will increase the frequency of collisions. Hence a rise in temperature cause $\tau$ to decrease and therefore resistively $\rho$ to increase. We thus expact a rise in resistivily with increase in temperature.

For quite a few substances, the increase in resistively with temperature is a linear one and we may write a relation of the form.

$$ \rho _{t}=\rho _{0}(1+\alpha t) $$

where we define $\alpha$ as $\alpha=\frac{\rho _{\tau}-\rho _{0}}{\rho _{0} \mathrm{t}}$

$\propto $ is the known as the co-efficient of resistivity and its units are ${ }^{0} \mathrm{C}^{-1}$.

However the variation of resistivity with temperature is not linear for all type of substances. Even for metals, the resistivity increases as a higher power of $\mathrm{T}$ at low temeperatures.

For insulationis and semi conductors, the resistivity increases exponentially with decreasing temperature, the variation being expressed by a relation of the form.

$$ \rho(T)=\rho _{0}(T) \exp \left(\frac{E _{g}}{k T}\right) $$

Here $\rho _{0}(T)$ is a factor that varies slightly will temperature, $\mathrm{k}$ is the Boltzman constant and $\mathrm{E} _{g}$ is the positive energy known as GAP ENERGY.

It follows that the resistivity of insulators and semi-conductors becomes very large at low temperatures tending to infinite large values at $\mathrm{T} \rightarrow 0$. On the other hand for metals resistivity tends to zero as $\mathrm{T} \rightarrow 0$.

Super Conductors

The property of metals of getting currents flow through them without practically any opposition as $\mathrm{T} \rightarrow 0$ is known as super conductivity. For metals the transtion from their normal resistivity behaviour to a zero resistance state occurs at a particular low temperature charaterstic of the metal considered. A substance in a zero resistance state is reffered as a super conductor. Recentally it has been observed that some substances acquire this super conducting behaviour at relatively higher temperature $\left(\sim 10^{\circ} \mathrm{K}\right)$ for $\mathrm{Nb}$ and $90^{\circ} \mathrm{K}$ for $\mathrm{Y} \mathrm{Ba} _{2} \mathrm{Cu} _{3} \mathrm{O} _{7}$

Super conductors are finding many interesting applications in technology. Some of the prominent applications include.

1. Super conductor electric power transmission lines with no power loss due to heating may soon become a possiblity.

2. Super conductors may be used to product very high speed computers.

3. Powerful magnets made with superconducting coils are being used.

Example-1

$V-I$ graph for a metallic wire at temperatures $T _{1}, T _{2}$, and $T _{3}$ are shown in the figures. Which of temperature is highest among the three?

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Solution:

The slope of a $\mathrm{I}-\mathrm{V}$ graph is equal to $\frac{1}{\mathrm{R}}$.

Hence $\mathrm{R}=\frac{1}{\text { Slope }}$

Since the slope of the graph corresponding to $\mathrm{T} _{3}$ is having the least value the therefore the resistance of wire at temeprature $\mathrm{T} _{3}$ is greather than $\mathrm{T} _{2}$ or $\mathrm{T} _{1}$. As resistance increases with temperature so temperature $\mathrm{T} _{3}$ is greater than $\mathrm{T} _{1}$ or $\mathrm{T} _{2}$.

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Resistance in Series and Parallel

Sereis Arrangment: When two or more resistance are joined in such a way that the second end of the first resistance is connected to the first and of the second resistance and so on as shown the resistance are said to be connected in series.

In series arrangment the current flowing through all the resistance is the same. Thus

$\mathrm{I}=\mathrm{I} _{1}=\mathrm{I} _{2}=\mathrm{I} _{3} \ldots . \ldots=\mathrm{I} _{\mathrm{n}}$

Also voltage drops are added and their summation would equal to total voltage output.

$$ \begin{aligned} \mathrm{V} _{\mathrm{T}} & =\mathrm{V} _{1}+\mathrm{V} _{2}+\ldots . \ldots . \mathrm{V} _{\mathrm{n}} \\ \text { or } \quad \mathrm{V} _{\mathrm{T}} & =\mathrm{I}\left(\mathrm{R} _{1}+\mathrm{R} _{2}+\mathrm{R} _{3}+\ldots . \ldots . \ldots \mathrm{R} _{\mathrm{n}}\right) \\ \frac{\mathrm{V} _{\mathrm{T}}}{\mathrm{I}} & =\mathrm{R} _{\mathrm{T}}=\mathrm{R} _{1}+\mathrm{R} _{2}+\mathrm{R} _{3}+\ldots . \ldots . \ldots . \ldots . \ldots \mathrm{R} _{\mathrm{n}} \end{aligned} $$

Parallel Arragnment: When two or more resistances and joined in such way that the first ends of all are connected to one common point and second ends of all connected to a second common point, the resistance are said to be joined in parallel.

In parallel arrangment the same current should leave the end 2 as enter at the end 1.

Hence $\mathrm{I} _{\mathrm{T}}=\mathrm{I} _{1}+\mathrm{I} _{2}+\mathrm{I} _{3}+\ldots$.

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The voltage that impresses itself on the side 1 of each resistor has be the same. Since the voltage must equalize itself all branches on the left side of parallel arrangement.

$$ \mathrm{V} _{\mathrm{T}}=\mathrm{V} _{1}=\mathrm{V} _{2}=\mathrm{V} _{3} \ldots . \ldots $$

Now, $\quad \mathrm{I} _{\mathrm{T}}=\frac{\mathrm{V} _{\mathrm{T}}}{\mathrm{R} _{\mathrm{T}}} \quad \mathrm{I} _{1}=\frac{\mathrm{V} _{\mathrm{T}}}{\mathrm{R} _{1}} \quad \mathrm{I} _{2}=\frac{\mathrm{V} _{\mathrm{T}}}{\mathrm{R} _{2}}+$ and so on

$$ \mathrm{I} _{\mathrm{T}}=\frac{\mathrm{V} _{\mathrm{T}}}{\mathrm{R} _{\mathrm{T}}}=\frac{\mathrm{V} _{\mathrm{T}}}{\mathrm{R} _{1}}+\frac{\mathrm{V} _{\mathrm{T}}}{\mathrm{R} _{2}}+\ldots . \ldots \frac{\mathrm{V} _{\mathrm{T}}}{\mathrm{R} _{\mathrm{n}}} $$

or $\quad \frac{1}{\mathrm{R} _{\mathrm{T}}}=\frac{1}{\mathrm{R} _{1}}+\frac{1}{\mathrm{R} _{2}}+\frac{1}{\mathrm{R} _{3}}+\ldots . \ldots$

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Example-2

Determine the equivalent resistance of the following networks.

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Show Answer

Solution :

We can regard the network of Figure (a) as a series combination of four resistances in the form of network as shown

This can further be simplified as shown.

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It is clear that the effective resistance of this networks is $\frac{1}{\mathrm{r} _{\mathrm{e}}}=\frac{1}{2 x}+\frac{1}{2 \mathrm{y}}=\frac{2 \mathrm{y}+2 x}{4 x \mathrm{y}}$

$$ r _{e}=\frac{4 x y}{2(x+y)}=\frac{2 x y}{x+y} $$

Hence the total effective resistance of the given network is given by:

$$ \mathrm{R} _{\mathrm{eF}}=4 \times \frac{(x \mathrm{y})}{x+\mathrm{y}}=\frac{8 x \mathrm{y}}{x+\mathrm{y}} $$

The network of figure (b) is simply a series combination of 5 resistances each of value $5 \Omega$. Hence the effective resistance of this network is $\mathrm{R} _{\text {eff }}=5 \times 5=25 \Omega$

Example-3

Find the equivalent resistance between points $a$ and $b$ for the combination shown in adjoining figure.

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Show Answer

Solution :

The $3 \Omega$ and $2 \Omega$ resistors are in series and are equivalent to a $5 \Omega$ resistor. This equivalent is in parallel with the $6 \Omega$ and their equivlent is given by

$\frac{1}{\mathrm{R}^{\prime}}=\frac{1}{5}+\frac{1}{6}=0.20+0.167=0.367 \Omega$

$\therefore \mathrm{R}^{\prime}=\frac{1}{0.367}=2.730$

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The $7 \Omega$ and $2.73 \Omega$ resistors are in series and the their equivalent is $9.73 \Omega$.

Now $5 \Omega, 12 \Omega$ and $9.73 \Omega$ resistance are in parallel.

Let their equivalent be R". The

$\frac{1}{\mathrm{R}^{\prime \prime}}=\frac{1}{5}+\frac{1}{12}+\frac{1}{9.73}=0.386$

$\mathrm{R}^{\prime \prime}=\frac{1}{0.386}=2.6 \Omega$

Finally $2.6 \Omega$ resistor is in series with the $9 \Omega$ resistor. Hence the equivalent resistance of the combination is given by $9.0+2.6=11.6 \Omega$

Example-4

Find the effective resistance between points $A$ and $B$ in the circuit diagram shwon in Figure

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Show Answer

Solution :

Figure shows successive reduction of the circuit.

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It is clear from figure that the effective resistance between points $A$ and $B$ is $2 \Omega$.

Electric Cell

The simplest device to maintain a current in an electric circuit is the electrolytic cell. In this type of cell the chemical energy of a reaction is converted into electrical energy. The electrods in a simple volticcell are copper and zinc with dilute sulphuric acid as electrolyte. The action of a simple voltate cell can be understood as follows. At the zinc electrode $\mathrm{Zn}$ atoms are ionized and pass into solution as $\mathrm{Zn}^{++}$ions leaving behind two electrons on each electrode. For each $\mathrm{Zn}^{++}$ion formed two $\mathrm{H} _{2}^{+}$ions are discharged at the copper electrods taking two electrons from it. The action continue as long as electrons can pass from zinc to copper through some piece of metal.

The simple voltic cell has two main defects viz (s) local action and polarization.

Daniel cell, Leclance cell and Dry cells are example of Primary cells.

Lead accumulator and alkali accumlator are exmaples of Secondary cells.

We can now directly convert the energy of light into electrical energy by using solar cells. The action of solar cells depend on proparties of semiconducting materials.

In nuclear cells the energy of $\beta$ particles given out by radioactive materials in converted directly into electrical energy.

e.m.f. of a Cell

We know that electrostatic field is conservative in nature i.e.

$$ \oint \mathbf{E} \cdot \mathbf{d} \ell=0 $$

Hence electrostatic field by itself is incapable of esatablishing a current in a closed circuit. It follows that for the maintanance of current in a closed circuit a non conservative field force that can spend energy is a basic necessity.

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Any source which can produce such a force is called a source of electromotive force. The electric force due to chemical ractions in a battery is an example of a non conservative field that is used for maintaining current.

Considering a battery as an example, we know that chemical energy is converted into electrical energy by chemical reactions taking place in it. The generation of energy provides the non conservative field which we assume to be directed from $\mathrm{p}$ to $\mathrm{q}$ in the battery. This is so because as a result of chemical reaction positive charges develop on plate $q$ and equal negative charges on plate $p$ of the battery. As a result of development of these charges an electrostatic field $\mathbf{E} _{2}$ which opposes the non conservative field $\mathbf{E} _{1}$ is established. The field $\mathbf{E} _{2}$ grows up a result of development of further charges. Soon the field $\mathbf{E} _{2}$ attains a value so that it becomes equal and opposite to $E$ which was in first instance responsible for the development of these changes.

The battery e.m.f. is defined as

$$ \varepsilon _{\mathrm{B}}=\int _{\mathrm{p}}^{\mathrm{q}} \mathbf{E} _{1} \cdot \mathbf{d} \ell $$

Here the line integral between $p$ and $q$ along the straight path passing through the battary. Thus the emf of a battery is the work done when a unit charge moves from one plate to the other through the battery.

Internal resistance of a cell is defined as the resistance offered by the electrolyte and electrode of a cell when a electric current flows through it.

Terminal Potential Differnece

When a current is drawn from a cell there is a drop of potential across the internal resistance of the cell (equal to ir) where as e.m.f. of a cell is the maximum potential difference between two electrodes of a cell. when current is drawn from the cell.

Hence terminal potential difference is between two electrodes of a cell in a closed circuit.

Thus, it is clear $\mathrm{V}$, the terminal potential difference is given by

$$ \mathrm{V}=\varepsilon-\mathrm{ir} $$

Grouping of Cells

Cells can be connected in series as shown. When $n$ cells of emf’s $\mathrm{E} _{1} \mathrm{E} _{2} \ldots . . \mathrm{E} _{\mathrm{n}}$ with their internal resistance as $\mathrm{r} _{1} \mathrm{r} _{2}, \ldots . . \mathrm{r} _{\mathrm{n}}$ are connected in series.

Then

The equivalent e.m.fof a series combinations $n$ cells is just the sum of their individual e.m.f and

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The equivalent internal resistance of a series combination of $n$ cells $s$ just the sum of their internal resistance when $n$ cells of e.m.f $\varepsilon _{1} \varepsilon _{2}-\varepsilon _{\mathrm{n}}$ and internal resistances are connected in parallel use these equivalent e.m.f and internal resistance all given by

$ \frac{\varepsilon _{\mathrm{eq}}}{\mathrm{r} _{\mathrm{eq}}}=\frac{\varepsilon _{1}}{\mathrm{r} _{1}}+\frac{\varepsilon _{2}}{\mathrm{r} _{2}}+\ldots . \ldots \frac{\varepsilon _{\mathrm{n}}}{\mathrm{r} _{\mathrm{n}}} $

and $\quad \frac{1}{\mathrm{r} _{\mathrm{q}}}=\frac{1}{\mathrm{r} _{1}}+\frac{1}{\mathrm{r} _{2}}+\ldots . \ldots \frac{1}{\mathrm{r} _{\mathrm{n}}}$

We can also have a mixed grouping of cells (say $\mathrm{N}$ ) where we connect $\mathrm{n}$ cells in series in row and then connect $m$ such rows in parallel. Assuming each cell has an e.m.f. $\varepsilon$ and internal resistance. Assuming each cell has an emf $\varepsilon$ and internal resistance $r$, the equivalent e.m.f. and resistance of $n$ cells in series in a row would be given by $\mathrm{n} \varepsilon$ and $\mathrm{nr}$.

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Now there are $m$ rows of cells in parallel therefore total internal resistance of all cells would be given by

$\frac{1}{\mathrm{r} _{\mathrm{p}}}=\frac{1}{\mathrm{n} _{\mathrm{r}}}+\frac{1}{\mathrm{n} _{\mathrm{r}}}+\ldots . \ldots \frac{1}{\mathrm{n} _{\mathrm{r}}}$ up to $\mathrm{n}$ times.

Hence, $\mathrm{r} _{\mathrm{p}}=\frac{\mathrm{nr}}{\mathrm{m}}$

Since the parallel combination of rows of cells does not effect the e.m.f. of each row of cells therfore the effective e.m.f. of all cells is $n \varepsilon$. If we connect an external resistance $R$ to such as arrangement, the current in the external resistance $\mathrm{R}$ is given by:

$$ \mathrm{I}=\frac{\mathrm{n} \varepsilon}{\mathrm{R}+\frac{\mathrm{nr}}{\mathrm{m}}}=\frac{\mathrm{mn} \varepsilon}{\mathrm{mr}+\mathrm{nr}} $$

The condition for $\mathrm{I}$ to be maximum is given by $\mathrm{R}=\frac{\mathrm{nr}}{\mathrm{m}}$

Heating Effect of Curretn (Joule Heat)

We know that in a circuit certaining resistors the flow of current electric result in the production of heat energy. We have already discussed that the metallic conduction is due to free electrons. These free electrons frequently collide with the atoms of metal in lattice. At each collision they lose some of their kinetic energy and give it to the atoms they strike. Thus as the current flows through a wire it increases the kinetic energy of we vibrations of metal atoms and hence it generates heat in the wire, the energy lost per second by the electrons is the electrical power supplied by the battery which maintains the current. Considering a circuit as shown in figure, experimentally we observe the following facts:

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1. The heat produced is proportional to the square of current.

2. The heat produced is proportional to resistance.

3. The heat produced is proportional to the time for which current flows.

Thus, we can write

$ \mathrm{H}=\mathrm{i}^{2} \mathrm{Rt} $

Electrical Power

The energy liberated per second in an electrical device is called electric power.

When an electric current flows through a wire or ‘passive resistance’ all the power which it conveys to the wire appears as heat. If I is he current $\mathrm{R}$ the resistance, when since

$\mathrm{V}=\mathrm{I} R$ we have $\mathrm{P}=\mathrm{I}^{2} \mathrm{R}$

as $I=\frac{V}{R}$

$\therefore \mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}^{2}} \times \mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$

Example-5

An immersion heater which is to operate on $230 \mathrm{~V}$ supply has to raise the temprature of $45.36 \mathrm{~kg}$ of water from $20^{\circ} \mathrm{C}$ to $9^{\circ} \mathrm{C}$ in a hour. Taking the efficiency of opection as $86 \%$, determine the resistance of heating element in the heater.

Show Answer

Solution :

Mass of water $=45.36 \mathrm{Kg}$

Rise of temperature $=95-20=75^{\circ} \mathrm{C}$

Heat used $\mathrm{m}$ raising the temperature $=45.36 \times 4200 \times 75 \mathrm{~J}$

Efficiency of operation $=86 \%=0.86$

Total heat produced $=\frac{45.36 \times 4200 \times 75}{0.86} \mathrm{~J}$

Heat produced in one hour $=x \times 1000 \times 3600 \mathrm{~J}$

$\therefore x \times 1000 \times 3600=\frac{45.36 \times 4200 \times 75}{0.86}$

$ x=\frac{45.36 \times 4200 \times 75}{1000 \times 3600 \times 0.86}=4.6 \mathrm{~kg} $

Now wattage $=\frac{\mathrm{V}^{2}}{\mathrm{R}}$

$\therefore 4600=\frac{230 \times 230}{\mathrm{R}}$

or $\quad \mathrm{R}=\frac{230 \times 230}{4600}=13.2 \mathrm{ohm}$

Example-6

A series bettery of 10 accumulation each of emf $1.5 \mathrm{~V}$ and internal resistance of $0.5 \mathrm{ohms}$ is being charged by a D.C. source of p.d. 120V. What series resistance be connected in series in charging circuit if charging curernt is $10 \mathrm{~A}$. Using this resistance what is the power supplied by d.c. source and energy stored in accumulator in 2 hrs.

Show Answer

Solution :

Let $\mathrm{R}$ be the series resistance used. Then from Ohm’s law

$\mathrm{I}=\frac{120-15}{\mathrm{R}+5}=10$

$\therefore \mathrm{R}=10.5-5=5.5 \Omega$

$\mathrm{P}=$ power supplied by d.c. source $=\mathrm{VI}=120 \times 10=1200 \mathrm{~W}$

Let,

$\mathrm{P}^{\prime}=$ Power dissipicted as heat in external resistance $=\mathrm{I}^{2} \mathrm{R}=(10)^{2} \times 5.5=550 \mathrm{~W}$

$\mathrm{P}^{\prime \prime}$ = Power dissipicted as heat inside accumulator $=\mathrm{I}^{2} \mathrm{r}=(10)^{2} \times 5=500 \mathrm{~W}$

$\mathrm{P} _{1}=$ Power stored inside accumulator $\mathrm{P}-\left(\mathrm{P}^{\prime}+\mathrm{P}^{\prime \prime}\right)=1200-(550+500)=150 \mathrm{~W}$

Energy stored in 2 Hour $=150 \times 2 \times 60 \times 60 \mathrm{~J}=108 \times 10^{4} \mathrm{~J}$

Kirchoff’s Laws

Electrical circuits contains simple arrangment of resistances in series or parallel and with a known source of e.m.f. can be analysed in a straightward manner using Ohm’s law. In many important cases we encounter networks where there may be junctions between two or more conductors at which current can flow along various paths and where one or more sources e.m.f. may be active. In such cases we solve the circuits by making use of Kirchoff’s law. To apply these law we define two terms.

1. A branch point is a point of the circuit where three or more conductors are joined together.

2. A loop is any closed conducting path in the network.

First Law

The algebraic sum of the currents flowing toward a branch point is zero. The convention for sign is.

The current flowing toward a branch point is considered positive whereas a current flowing away from a branch point is considred negative. For example in the figure shown we have:

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$$ \mathrm{I} _{1}-\mathrm{I} _{2}-\mathrm{I} _{3}-\mathrm{I} _{4}+\mathrm{I} _{5}=0 $$

Second Law

The algebraic sum of the emf’s in any loop of the network is euqal to algebraic sum of IR drops in it i.e.

$$ \sum \varepsilon _{\mathrm{i}}=\sum \mathrm{I} _{\mathrm{j}} \mathrm{R} _{\mathrm{j}} $$

To apply the loop equation we must take some direction either clockwise or anticlockwise as the traversal direction. An emf is taken with positive sign if the emf by itself produce a positive current in the sense of traversal direction. Similarly we take an IR turn with the positive sign if the current through the resistor in question is in the direction of the traversal.

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For example in the figure shown taking $\mathrm{ABCDE}$ as the sense of traversal we have.

$$ I _{1} R _{1}-I _{2} R _{2}+I _{3} R _{3}-I _{4} R _{4}+I _{5} R _{5}=0 $$

To solve the networks will the help of Kirchoff’s law we proceed stepwise as follows:

(a) We assign currents with assumed directions to every branch in the network. There direction should be indicated in the circuit diagram. Then making use of Kirchoff’s first law we reduce the number of unknowns as far as possible.

(b) We next write down the voltage equations using Kirchoff’s second law to conveniently choosen loops. There must be the same number of independent equation as are the unknown.

We illustrate the use of Kirchoff’s law by taking the case of a Wheatstone bridge. The circuit diagram is shown in figure. Suppose we want to calculate Ig. Let the currents in all the branches be as indicated. Applying Kirchoff’s fisst law we have

$$ \begin{array}{ll} \mathrm{A}, & \mathrm{I}=\mathrm{I} _{1}+\mathrm{I} _{3} \text { or } \mathrm{I} _{3}=\mathrm{I}-\mathrm{I} _{1} \\ \mathrm{~B}, & \mathrm{I} _{1}=\mathrm{I} _{2}+\mathrm{I} _{\mathrm{g}} \text { or } \mathrm{I} _{2}=\mathrm{I} _{1}-\mathrm{I} _{\mathrm{g}} \\ \mathrm{C}, & \mathrm{I}=\mathrm{I} _{2}+\mathrm{I} _{4} \text { or } \mathrm{I} _{4}=\mathrm{I}-\mathrm{I} _{2}+\mathrm{I} _{\mathrm{g}} \end{array} $$

Applying Kirchoff’s second law to mesh ABDA weget

$~$

$$ \begin{array}{ll} & \mathrm{I} _{1} \mathrm{R} _{1}+\mathrm{I} _{\mathrm{g}} \mathrm{G} _{\mathrm{g}}-\mathrm{I} _{3} \mathrm{R} _{3}=0 \\ \text { or } & \mathrm{I} _{1} \mathrm{R} _{1}+\mathrm{I} _{\mathrm{g}} \mathrm{G} _{\mathrm{g}}-\left(\mathrm{I}-\mathrm{I} _{1}\right) \mathrm{R} _{3}=0 \\ \text { or } & \mathrm{I} _{1}\left(\mathrm{R} _{1}+\mathrm{R} _{3}\right)+\mathrm{I} _{4} \mathrm{G} _{1}-\mathrm{IR} _{3}=0 \qquad \quad \ldots . \ldots (1) \end{array} $$

For mesh BCDB we have

$$ \mathrm{I} _{2} \mathrm{R} _{2}-\mathrm{I} _{4} \mathrm{R} _{4}-\mathrm{Ig} _{\mathrm{g}}^{\mathrm{g}}=0 $$

$$ \left(\mathrm{I} _{1}-\mathrm{I} _{2}\right) \mathrm{R} _{2}-\left(\mathrm{I}-\mathrm{I} _{1}+\mathrm{I} _{\mathrm{g}}\right) \mathrm{R} _{4}-\mathrm{I} _{\mathrm{g}} \mathrm{GR} _{4}=0 $$

$$ \begin{equation*} I _{1}\left(R _{2}+R _{4}\right)-I _{g} G _{g}-I _{g}\left(R _{2}+R _{4}\right)-I R _{4} \qquad \quad \ldots . \ldots (2) \end{equation*} $$

We eliminate $I _{1}$ from these equations by multiplying (1) by $\left(R _{2}+R _{4}\right)$ and (2) by $\left(R _{1}+R _{3}\right)$ and then subracting.

Solving we get,

$$ I _{g}=I \frac{\left(R _{2} R _{3}-R _{1} R _{4}\right)}{G _{g}\left(R _{1}+R _{2}+R _{3}+R _{4}\right)+\left(R _{1}+R _{3}\right)\left(R _{2}+R _{4}\right)} $$

Condition for balance would be obtained by putting $\mathrm{I} _{\mathrm{g}}=0$

This gives

$$ \frac{\mathrm{R} _{1}}{\mathrm{R} _{2}}=\frac{\mathrm{R} _{3}}{\mathrm{R} _{4}} $$

We can also obatin this condition by putting $\mathrm{I} _{\mathrm{g}}=0$ in the equations $\left(1 _{2}\right)$ and obtained from Kirchoff’s laws. Wheatstone bridge is a basic circuit for determination of an unknown resistance. However the balance condition

$$ \frac{\mathrm{R} _{1}}{\mathrm{R} _{2}}=\frac{\mathrm{R} _{3}}{\mathrm{R} _{4}} $$

The above equation an IDEAL condition. This is because $\mathrm{I} _{\mathrm{g}}$ can never be made equal to zero because a galvanometer will always draw some current, howsoever small, for its working. Hence in practice we obatin a value $R _{3}^{\prime}$ instead of $R _{3}$ in the balancing condition. $\left(R _{3}^{\prime}-R _{3}\right)$ is known as WANT OF BALANCE. For a Wheatstone bridge to be sensitive the galvanometer should show as large a deflection as possible for a given want of balance. Broadly speaking we can say that a sensitive cap wheatstonte brdige is reasonable senstive if the resisance in all the arms are of the same order of magnitude.

Example-7

Three cells of emf’s $E _{1}, E _{2}$ and $E _{3}$ and internal resist are $r _{1}, r _{2}, r _{3}$ and connected in parallel through an external resistance $R$. Calculate the current flowing through each cell.

$~$

Show Answer

Solution :

Suppose $\mathrm{i} _{1}, \mathrm{i} _{2}$ and $\mathrm{i} _{3}$ be the currents in given three cells and $\mathrm{r}$ be this current through external resistance $\mathrm{R}$. Applying Kirchoff’s first law to the point, we have A

$$ \begin{equation*} \mathrm{i}=\mathrm{i} _{1}+\mathrm{i} _{2}+\mathrm{i} _{3} \qquad \quad \ldots . \ldots (1) \end{equation*} $$

Applying second law to the mesh $\mathrm{E} _{1} \mathrm{AR} \mathrm{B} _{1}$, we have

$$ \begin{equation*} \mathrm{i} _{1} \mathrm{r} _{1}+\mathrm{iR}=\mathrm{E} _{1} \qquad \quad \ldots . \ldots (2) \end{equation*} $$

Similarly, $ \quad \mathrm{i} _{2} \mathrm{r} _{2} + \mathrm{iR}=\mathrm{E} _{2} \qquad\qquad (\text{Mesh} \mathrm{E} _{2} \mathrm{ARB} \mathrm{E} _{2} )\quad \ldots . \ldots (3)$

and $\quad \qquad \mathrm{i} _{3} \mathrm{r} _{3}+\mathrm{iR}=\mathrm{E} _{3} \qquad\qquad (\text{Mesh} \mathrm{E} _{3} \mathrm{ARB} \mathrm{E} _{3} ) \quad \ldots . \ldots (4)$

Divides equation (2) by $\mathrm{r} _{1}$, equation (3) by $\mathrm{r} _{2}$ and equation (4) by $\mathrm{r} _{3}$ and adding we get

$ \left(i _{1}+i _{2}+i _{3}\right)+i R\left(\frac{1}{r _{1}}+\frac{1}{r _{2}}+\frac{1}{r _{3}}\right)=\frac{E _{1}}{r _{1}}+\frac{E _{2}}{r _{2}}+\frac{E _{3}}{r _{3}} $

or $\quad i+i R\left(\frac{1}{r _{1}}+\frac{1}{r _{2}}+\frac{1}{r _{3}}\right)=\frac{E _{1}}{r _{1}}+\frac{E _{2}}{r _{2}}+\frac{E _{3}}{r _{3}}$

$ \begin{aligned} & \mathrm{i}\left[1+\mathrm{R}\left(\frac{1}{\mathrm{r} _{1}}+\frac{1}{\mathrm{r} _{2}}+\frac{1}{\mathrm{r} _{3}}\right)\right]=\frac{\mathrm{E} _{1}}{\mathrm{r} _{1}}+\frac{\mathrm{E} _{2}}{\mathrm{r} _{2}}+\frac{\mathrm{E} _{3}}{\mathrm{r} _{3}} \\ & \mathrm{i}\left\{\frac{1}{\mathrm{r} _{1} \mathrm{r} _{3} \mathrm{r} _{3}}\left[\mathrm{r} _{1} \mathrm{r} _{2} \mathrm{r} _{3}+\mathrm{R}\left(\mathrm{r} _{2} \mathrm{r} _{3}+\mathrm{r} _{3} \mathrm{r} _{1}+\mathrm{r} _{2} \mathrm{r} _{3}\right)\right]\right\}=\frac{1}{\mathrm{r} _{1} \mathrm{r} _{3} \mathrm{r} _{3}}\left(\mathrm{E} _{1} \mathrm{r} _{2} \mathrm{r} _{3}+\mathrm{E} _{2} \mathrm{r} _{3} \mathrm{r} _{1}+\mathrm{E} _{3} \mathrm{r} _{1} \mathrm{r} _{2}\right) \end{aligned} $

or $\quad i=\frac{E _{1} r _{2} r _{3}+E _{2} r _{3} r _{1}+E _{3} r _{1} r _{2}}{r _{1} r _{3} r _{3}+R\left(r _{2} r _{3}+r _{3} r _{1}+r _{2} r _{3}\right)}$

$ \begin{aligned} & i _{1}=\frac{E _{1}-i R}{1} \\ & i _{1}+\frac{1}{r _{1}}\left[E-R\left\{\frac{\left(E _{1} r _{2} r _{3}+E _{2} r _{3} r _{1}+E _{3} r _{1} r _{2}\right)}{r _{1} r _{2} r _{3}\left[R\left(r _{2} r _{3}+r _{3} r _{1}+r _{1} r _{2}\right)\right.}\right\}\right] \end{aligned} $

Similar expression for $\mathrm{i} _{2}$ and $\mathrm{i} _{3}$ can be written from the symmetry.

Example-8

The resistance of three wires $B C, C A$ and $A B$ of the same material and uniform cross section are $a, b, c$ resectively. Another wire from $A$ of constant resistance can make a contact with BC. If a current enters at $A$ and leaves at the point of contant with BC. Show that the maximum resistance of the network is:

$ \frac{(a+b+c) d}{a+b+c+4 d} $

$~$

Show Answer

Solution :

Let the sliding were make contact with wire $\mathrm{BC}$ at point $\mathrm{D}$. Let the resistance of part BD of wire be $\beta$ and of DC $(\alpha-\beta)$. Let I be the current staring at A and leaving at D. From Kirchoff’s second law.

We have for network ADVA

$$ \begin{equation*} \mathrm{I} _{2} \mathrm{~d}=\mathrm{V} \quad \text { or } \quad \mathrm{I} _{2}=\frac{\mathrm{V}}{\mathrm{d}} \qquad \quad \ldots . \ldots (1) \end{equation*} $$

For network ABDA

$$ \mathrm{I} _{3} \mathrm{c}+\beta \mathrm{I} _{3}-\mathrm{I} _{2} \mathrm{~d}=0 $$

$$ \mathrm{I} _{3}(\mathrm{c}+\beta)=\mathrm{I} _{2} \mathrm{~d}=\mathrm{V} $$

$$\therefore \mathrm{I} _{3}=\frac{\mathrm{V}}{\mathrm{c}+\beta} \qquad \quad \ldots . \ldots (2)$$

Similarly for network ACDA

$$ \begin{align*} & I _{1} b+I _{1}(a-\beta)-I _{2} d=0 \\ & I _{1}(a+b-\beta)=I _{2} d=V \\ & I _{1}=\frac{V}{a+b-\beta} \qquad \quad \ldots . \ldots (3)\\ & I=I _{1}+I _{2}+I _{3}=\frac{V}{a+b-\beta}+\frac{V}{c+\beta}+\frac{V}{d} \end{align*} $$

$$ \text { or } \quad I=V\left[\frac{1}{a+b-\beta}+\frac{1}{d}+\frac{1}{c+\beta}\right] $$

If $R$ is the resistance of equivalnet network then $I=\frac{V}{R}$

Hence combining we get,

$$ \frac{1}{R}=\frac{1}{a+b-\beta}+\frac{1}{d}+\frac{1}{c+\beta} $$

For maximum value of $\mathrm{R}, \frac{\mathrm{dR}}{\mathrm{d} \beta}=0$

or $\quad-\frac{1}{R^{2}} \frac{d R}{d \beta}=\frac{1}{(a+b-\beta)^{2}}-\frac{1}{(c+\beta)^{2}}$

or $\quad-\frac{1}{R^{2}} \frac{d R}{d \beta}=\frac{(c+\beta)^{2}-(a+b-\beta)^{2}}{(a+b-\beta)^{2}(c+\beta)^{2}}$

For $\frac{\mathrm{dR}}{\mathrm{d} \beta}=0$ we must have

$$ a+b-\beta=c+\beta $$

or $\quad \beta=\frac{a+b-c}{2}$

Substituting the value of $\beta$ we get

$$ \begin{aligned} \frac{1}{R _{\max }} & =\frac{1}{a+b-\frac{a+b-c}{2}}+\frac{1}{d}+\frac{1}{c+\frac{a+b-c}{2}} \\ & =\frac{2}{a+b+c}+\frac{1}{d}+\frac{2}{a+b+c} \\ & =\frac{2 d+a+b+c+2 d}{(a+b+c) d}=\frac{a+b+c+4 d}{(a+b+c) d} \end{aligned} $$

Hence, $R _{\max }=\frac{(a+b+c) d}{a+b+c+4 d}$

Exmaple-9 :

Calculate the steady stage current in the $2 \Omega$ resistor shown in the circuit. The internal resistance of battery is neglible and the capacitance of the capacitor is 0.2 microfarad.

$~$

Show Answer

Solution :

The current distribution in the various arms of the network is shown in figure It is important to note the capacitor $\mathrm{C} _{1}$ has infinite resistance for d.c. current. At branch point B, therefore no current passes through the arm $\mathrm{BC} _{1} \mathrm{~F}$

$~$

From loop ABCD H EFGA

$$ \begin{equation*} 2 \mathrm{i} _{1}+2.8 \mathrm{i}=0 \qquad \quad \ldots . \ldots (1) \end{equation*} $$

From loop DHEJD

$$ \begin{array}{ll} & 2 i _{1}-3\left(i-i _{1}\right)=0 \\ \text { or } & 5 i _{1}-3 i=0 \qquad \quad \ldots . \ldots (2) \end{array} $$

From (1) and (2) we have,

$$ \begin{aligned} & 2 \mathrm{i} _{1}+2.8\left[(5 / 3) \mathrm{i} _{1}\right]=0 \\ & \mathrm{i} _{1}=0.9 \mathrm{~A} \end{aligned} $$

Metre Bridge

A meter bridge is a pratical application of a Wheatstone bridge. It consists of a straight and uniform wire one meter long. The internal wire has a high resistivity and low temperature co-efficient of resistance (Manganin is the material used).

The wire is stretched along a meter scale with its ends soldered to two copper strips of neglible resistance between the copper strips and third one fixed along the panel have two gaps $\mathrm{AB}$ and $\mathrm{CD}$.

The unknown resistance is connected in the gap $\mathrm{AB}$ and a suitable resistance is connected in the gap. There is a jockey which series to make contact will the stretched wire and a pointer attached it to indicates we position of the point of contact with scale.

The position of jockey is adjusted until the galvanometer shows no deflection. Since the wire is of uniform thickness, the resistance is proportional to length.

Let $x z=\ell _{1}$ and $\mathrm{zy}=100-\ell _{1}=\ell _{2}$

Then we have,

$$ \frac{\mathrm{R}}{\mathrm{S}}=\frac{x \mathrm{z}}{\mathrm{zy}}=\frac{\ell _{1}}{\ell _{2}} $$

$~$

The determination here may be inaccurate due to the following errors:

(a) Non uniformaty of the wire. To eliminate this error the metar bridge wire must be calibrated in an auxillary experiment.

A program to guve wingst on th stdents

(b) The position of the point where the jockey makes contact with the wire may not be shown accurately by the pointer attached to the scale.

This error may be eliminated by interchanging $\mathrm{R}$ and $\mathrm{S}$ and repeating the experiment. The mean of two gives the accurate balancing length.

(c) The ends of the wire may not coincide will the ends of the scale and there may be a small resistance where the wire is connected to the copper strip. This resistance is known as END correction. To determine we take $\mathrm{R}$ and $\mathrm{S}$ known and obtain a balance point a length $\mathrm{L} _{1}$ of the melor bridge wire. If $\alpha$ and $\beta$ are end corrections at the two ends we have,

$$ \frac{\mathrm{R}}{\mathrm{S}}=\frac{\mathrm{L} _{1}+\alpha}{100-\mathrm{L} _{1}+\beta} $$

Now interchanging the positions of $\mathrm{R}$ and $\mathrm{S}$ we obtain a balance at a length $\mathrm{L} _{2}$ of the meter bridge wire.

$$ \frac{\mathrm{S}}{\mathrm{R}}=\frac{\mathrm{L} _{2}+\alpha}{100-\mathrm{L} _{2}+\beta} $$

Solving these two equations for $\alpha$ and $\beta$ we get

$$ \begin{aligned} & \alpha=\frac{\mathrm{SL} _{1}+\mathrm{RL} _{2}}{\mathrm{R}-\mathrm{S}} \\ & \beta=\frac{\mathrm{RL} _{1}+\mathrm{SL} _{2}}{\mathrm{R}-\mathrm{S}} \end{aligned} $$

(d) If a current is passed for a longer direction of time the point may flicker. To avoid this the current in the wire may be kept as small as possible and the key should be pressed for as short a time as possible while taking a reading.

Example-10

Two parallel resistance $4 \Omega$ and $8 \Omega$ are connected in the right gap of a meler bridge and a resistance of $6 \Omega$ is connected in the left gap. Where would be the balance point on the meter bridge wire?

Show Answer

Solution :

S is the combination of parallel combination of $4 \Omega$ and $8 \Omega$ resistance.

$ \frac{1}{S}=\frac{1}{4}+\frac{1}{8}=\frac{3}{8} $

or $\quad \mathrm{S}=\frac{8}{3} \Omega$

Let the balance point be at a distance of $x$ from the left end of the wire.

$ \begin{aligned} & \frac{\mathrm{R}}{\mathrm{S}}=\frac{x}{100-x} \\ & \frac{6 \times 3}{8}=\frac{x}{100-x} \end{aligned} $

Solving for $x$ we get,

$ x=69.2 \mathrm{~cm} $

Potentiometer

The potentiometer is a very useful instrument which can be used to measure a unknown e.m.f.

The principle of a potentiometer can be understodd by referring to figure.

Here $A B$ is a long uniform wire connected to a battery $E _{1}$. The source of unkonwn e.m.f. $E _{2}$ is connected across $\mathrm{AB}$ through a sliding contact $\mathrm{J}$. The sliding contact is moved along $\mathrm{AB}$ till there is no deflection in the galvanometer. When this happens the emf $\mathrm{E} _{2}$ is equal in magnitude of the potential difference between $A$ and $\mathrm{J}$ (For obatining a balance $\mathrm{E} _{1}$ must be greater than $\mathrm{E} _{2}$ ).

$~$

Let us consider the wire $\mathrm{AB}$ is uniform and the fall of potential across unit length (potential gradient) be $\rho$. If the balancing length for an unknown e.m.f. is $\ell$ than,

$$ \mathrm{e}=\rho \ell $$

Given a choice, we would prefer to use a ten wire potentlometer over a four wire potentiometer for a better accuracy of measurement ( $\rho$ would be smaller for a ten wire potentiometer).

Usually one find various faults in potentiometer circuit. We discuss below various possibilties:

(a) $\quad$ Both defiections are in the same direction the one with the contact maker at the zero end of the scale being the smaller of two. This error arises because the p.d. under test is connected in the wrong way round and the connections are to be reveresed.

(b) $\quad$ Both deflections are in the same direction the one with the contact maker at the zero of the scale being the larger of two.

The reason for this fault is that the accumlator that drives the current through the potentiometer produces too small a p.d. across the wire. To eliminate this error we must either recharge the accumulator or if there is a resistance in series with the wire this must be reduced.

(c) $\quad$ The same deflection is obatined for all position of the contact maker. This happen when there is a broken connection somewhere in the part of the circuit in series with the potentionmeter wire and accumulator.

(d) $\quad$ No deflection is obatined for all positions of the contact maker. This happens when there is a broken connection in the galvanometer circuit.

(e) $\quad$ The deflecion of the galvonometer are in opposite directions for the two positions of the contact maker but the pointer flickers unsteady.

There is a loose connection some where in the circuit.

A potentiometer is a very verstile instrument and can be used for:

(i) Comparison of emf’s of two cells.

(ii) Determination of internal resistance of a cell.

(iii) Determination of a low resistance.

(iv) Measurement of current.

(v) Measurement of high emf.

(vi) Measurement a very low e.m.f. (thermocouple)

Comparision of e.m.f. of Two Cells

Connect point 1 of the two way key and obtain a balance point at a length $L _{1}$ of the potentiometer wire. Than $\mathrm{E} _{1}=\rho \mathrm{L} _{1}$

$~$

Where $\rho$ is the potential gradient of the potentiometer wire.

Now connect contact 2 of the two way key and obtain a balance point at a length $L _{2}$ of the potentiometer wire. Than

$ \mathrm{E} _{2}=\rho \mathrm{L} _{2} $

Hence, $\frac{E _{1}}{E _{2}}=\frac{L _{1}}{L _{2}}$

Determination of Internal Resistance of a Cell

The circuit arrangment is shown in the figure. E is the cell whose internal resistance ir is to be measured A resistance box $\mathrm{R}$ is connected across the cell through a key $\mathrm{K} _{2}$.

By adjusting the rheostat $\mathrm{R} _{1}$ and closing key $\mathrm{K} _{1} \mathrm{a}$ balancing point is obtained is on the potentiometer wire at a point $\mathrm{L} _{1}$ with $\mathrm{K} _{2}$ open.

Then $\mathrm{E} _{1}=\rho \mathrm{L} _{1}$

Where $\rho$ is the potential gradient of the potentiometer wire.

Now key K is closed with a known resistance R out of resistance box. Once again without disturbing rheostat $\mathrm{R} _{1}$ the balancing length $\mathrm{L} _{2}$ is measured. The we have,

$~$

$ \begin{gathered} \mathrm{V}=\mathrm{L} _{2} \rho \\ \therefore \frac{\mathrm{E}}{\mathrm{V}}=\frac{\mathrm{L} _{1}}{\mathrm{~L} _{2}} \end{gathered} $

Let I be the current drawn throught the cell by resistance $\mathrm{R}$. Then the potential difference of the cell is

$ \mathrm{V}=\mathrm{IR} $

Also, $I=\frac{E R}{R+r}$

$\therefore \frac{\mathrm{E}}{\mathrm{V}}=\frac{\mathrm{R}+\mathrm{r}}{\mathrm{R}}$

Sovling we get,

$ \mathrm{r}=\mathrm{R}\left(\frac{\mathrm{L} _{1}-\mathrm{L} _{2}}{\mathrm{~L} _{2}}\right) $

Example-11

The circuit shown in figure is designed to measure an unknown. Resistance using a potentiometer. When there is no deflection $m$ the galvanometer, the balance point is obtained on $P$ of the potentiometer in the galvonometer $G _{1}$. Another balance point is located $\mathbf{Q}$ of the potentiometer wire in the galvanometer $G _{2}$. It is found that

$A P=\frac{A Q}{3}$. What is the value of $x$ ?

$~$

Show Answer

Solution :

We have,

$\frac{5}{5+x} \times 5=\rho(\mathrm{AP})=\rho\left(\frac{\mathrm{AQ}}{3}\right)$

and $\quad \frac{5}{5+x}(5+x)=\rho \mathrm{AQ}$

Hence, $\frac{5}{5+x}=\frac{1}{3}$

$x=10 \Omega$

Example-12

A potentiometer (with potential gradient $\rho$ ) is set up as shown. The ammeter reads 1.0 when the two way key is switched off. When the points 1,2 of the key are connected the balance is obatined at $x \mathrm{~cm}$ and when the points 1 and 3 are connected the balance is obatined at $\mathbf{y ~ c m ~ o f ~ t h e ~ p o t e n t i o m e t e r ~ w i r e . ~ T h e n ~ t h e ~ m a g n i t u d e ~}$ of the resistors $R$ and $Q$ are equals when respectively to $\rho(y-x)$ and $\rho x$.

$~$

Show Answer

Solution :

When the key between 1 and 2 is plugged in $\mathrm{P}$ is in circuit.

$\therefore \mathrm{V} _{1}=\mathrm{IP}=1 \mathrm{P}=\rho x$

When the key between 1 and 3 is plugged in coil $\mathrm{P}$ and $\mathrm{Q}$ are in the circuit.

$\mathrm{V} _{2}=\mathrm{I}(\mathrm{P}+\mathrm{Q})=1(\mathrm{P}+\mathrm{Q}) \mathrm{y}$

$\therefore \mathrm{V} _{2}-\mathrm{V} _{1}=\rho+\mathrm{Q}-\rho=\mathrm{k}(\mathrm{y}-x)$

PROBLEMS FOR PRACTICE

1. The resistivity of the material of a wire is $4 \times 10^{-5} \mathrm{ohm}-\mathrm{cm}$. If the resistance of the wire is $13 \mathrm{ohm}$ and its diameter is $1 \mathrm{~mm}$. Calculate the length of the wire.

Show Answer

Answer: $27.5 \mathrm{~m}$

2. Acarbon resistor of $74 \Omega$ to be marked with rings of different colour for its identification. Write the sequence of colours.

Show Answer

Answer: Violet, yellow and orange

3. Two wires of equal length one of aluminum and the other of copper have the same resistance. Which of the two is lighter?

$\rho _{\mathrm{Al}}=2.63 \times 10^{-8} \Omega \mathrm{m}, \rho _{\mathrm{Cu}}=1.72 \times 10^{-8} \Omega \mathrm{m}$

Relative density of $\mathrm{Al}$ and $\mathrm{Cu}$ are 2.7 and 8.9

Show Answer

Answer: Aluminium

4. A wire with a resistance of $6.0 \Omega$ is drawn out so that its new length is three times its original length. Find the resistance of longer wire assuming that its resistivity and density of meterial are not changing during the process.

Show Answer

Answer: $5 \Omega$

5. At what temeperature would the resistance of a copper conductor be double its resistance at $0^{\circ} \mathrm{C}$.

Show Answer

Answer: $260^{\circ} \mathrm{C}$

6. An iron wire $2 \mathrm{~km}$ long and a copper wire $3 \mathrm{~km}$ long are connected in parallel across a source of voltage $200 \mathrm{~V}$. Both wires carry the same current. The diameter of copper wire is $1 \mathrm{~mm}$ and its temeperature is $100^{\circ} \mathrm{C}$. Obtain the diameter of iron wire. The temperature coefficient of copper and iron at $20^{\circ} \mathrm{C}$ is $3.9 \times 10^{-3}$ and $5 \times 10^{-8}\left({ }^{\circ} \mathrm{C}\right)^{-1}$ respectively. Resistivity of copper and iron is $1.7 \times 10^{-8}$ and $10^{-7} \Omega \mathrm{m}$. Neglect the thermal conductivity of the wires.

Show Answer

Answer: $2.046 \mathrm{~mm}$

7. A heater connected to a $230 \mathrm{~V}$ supply initially draws a current of $3 \mathrm{~A}$ which becomes $2.3 \mathrm{~A}$ after 5 second. What is the steady temeperature of the heating element if the initial temperature is $27^{\circ} \mathrm{C}$. Temperature cofficient is $0.00017 \mathrm{C}^{-1}$

Show Answer

Answer: $1817^{\circ} \mathrm{C}$

8. A transmission line $1 \mathrm{~km}$ long is prepared from $17 \mathrm{~kg}$ of magnin $20^{\circ} \mathrm{C}$. What is the maximum and minimum temperature within which it can operate if its resistance is to be within $0.76 \Omega \mathrm{m}$ of its calibrated value. Given

Coefficient of linear expansion $1.8 \times 10^{-3}\left({ }^{0} \mathrm{C}\right)^{-1}$

Temperature coefficient $=10^{-5}\left({ }^{0} \mathrm{C}\right)^{-1}$

Density at $20^{\circ} \mathrm{C}=8.5 \times 10^{3} \mathrm{~m}^{-3}$

Resistively at $20^{\circ} \mathrm{C}=4.4 \times 10^{-7} \Omega \mathrm{m}$

Show Answer

Answer: -80 or 120

9. Calculate the equivalent resistance between $\mathrm{A}$ and $\mathrm{B}$ for the combination shown in figure.

$~$

Show Answer

Answer: $3.18 \Omega$

10. Find the equivalent resistance between $\mathrm{P}$ and $\mathrm{Q}$ of the figure.

$~$

Show Answer

Answer:

11. Find the effective resistance between points $x$ and $y$ of the hexagonal circuit shown in figure Each resistor has a value of $10 \Omega$.

$~$

Show Answer

Answer: $5 \Omega$

12. Calculate the equivalent resistance between point $\mathrm{A}$ and $\mathrm{B}$ of the network shown.

$~$

Show Answer

Answer: $50 \Omega$

13. Find the equivalent resistance between $A$ and $B$ of the network shown in figure. Neglect the internal resistance of the battery.

$~$

Show Answer

Answer: $11.33 \Omega$

14. Twelve resistance each of value $4 \Omega$ are connected as shown in the figure. Find the effective resistance between $\mathrm{A}$ and $\mathrm{C}$.

$~$

Show Answer

Answer: $3 \Omega$

15. Four identical cells each of e.m.f. E are connected in parallel providing supply of current to external circuit consisting of two resistances of resistance $\mathrm{R}$ in parallel. The terminal voltage of the cells as read by an ideal voltmeter is V. Calculate the internal resistance of each cell.

Show Answer

Answer: $\left(\frac{E-V}{V}\right) \times 2 R$

16. In the circuit shown, calculate the current flowing between $\mathrm{A}$ and $\mathrm{B}$ through $\mathrm{E} _{2}$.

$~$

Show Answer

Answer: $2 \mathrm{~A}$ from B to $\mathrm{A}$

17. In an electric furnace $100 \mathrm{~kg}$ of tin are to be melted in an hour from an initial temperature of $15^{\circ} \mathrm{C}$. Find the power required for this operation.

Given,

Specific heat of tin $=0.056$

Melting point of tin $=235^{\circ} \mathrm{C}$

Latent heat of fusion of $\mathrm{tin}=13.3$ kilocalorie per $\mathrm{kg}$.

Thermal efficienty $=70 \%$

$\mathrm{J}=0.24$ Calorie per joule

Show Answer

Answer: $4.2377 \mathrm{Kwh}$

18. What is the wattage and voltage ratings of a bulb which will produce maximum brightness when connected across a battery of e.m.f. $4.5 \mathrm{~V}$ and internal resistance 20.

Show Answer

Answer: $2.53 \mathrm{~W}$ and $2.25 \mathrm{~V}$

19. Two bulbs rated $100 \mathrm{~W}, 200 \mathrm{~V}, 200 \mathrm{~W}, 200 \mathrm{~V}$ are connected in (1) series (2) in parallel and joined to a source of p.d $200 \mathrm{~V}$. In each case which bulb glow more brightly. What is total power consumed in each case.

Show Answer

Answer: In series: $100 \mathrm{~W}$ bulb glows more brightly then $200 \mathrm{~W}$ bulb and power consumed $=66.6 \mathrm{~W}$. In parallel: $200 \mathrm{~W}$ bulb glows more brightly than $100 \mathrm{~W}$ bulb and power consumed is $300 \mathrm{~W}$.

20. A lead fuse in a circuit consisting of a copper wire with cross sectional area $5 \mathrm{~mm}^{2}$ metls if the wire heat up by $25 \mathrm{~K}$. The initial temperature of the lead fuse is $293 \mathrm{~K}$. What is the cross sectional area of the fuse lead wire.

Show Answer

Answer: $2.5 \mathrm{~mm}^{2}$

21. Ahigh speed lift weighing $15.7 \mathrm{KN}$ goes up at $1.0 \mathrm{~m} / \mathrm{s}$. The supply line is $220 \mathrm{~V}$, the motor efficienty is $92 \%$. What is the power of the electric motor driving the lift.

Show Answer

Answer: $17 \mathrm{Kw}$

22. $\mathrm{ABCD}$ is a quadrileteral of which arms have resistance $\mathrm{AB}=1 \Omega, \mathrm{BC}=2 \Omega, \mathrm{CD}=3 \Omega$ and $\mathrm{DA}=4 \Omega$. A galvanometer of resistance 352 is placed across $\mathrm{BD}$. If a current of $1 \mathrm{~A}$ is passed at $\mathrm{A}$ and leaves at $\mathrm{C}$ calculate the current in the galvonometer.

Show Answer

Answer: $0.00667 \mathrm{~A}$

23. An electric current of 5A is divided into three branches, the lenghts of the wires in three branches are proportional to 1,2 and 3.Find the current in each branch. The wires are of the same material and cross section.

Show Answer

Answer: $0.3075 \mathrm{~A}, 2 \mathrm{~A}, 0.9225 \mathrm{~A}$

24. In the circuit shown in figure, what is the current through $4 \Omega$ resistance.

$~$

Show Answer

Answer:

25. Two squares $\mathrm{ABCD}$ and $\mathrm{AEFC}$ have the side $\mathrm{BC}$ in common. The sides made of conducting wires have resistance as follows.

$\mathrm{AB}, \mathrm{BE}, \mathrm{FC}, \mathrm{CD}$ each $2 \Omega$

$\mathrm{AD}, \mathrm{BC}$ EF each $1 \Omega$

A cell of e.m.f. 2 volt and internal resistance $2 \Omega$ is joined across AD. Find the current in various parts of the circuit.

$~$

Show Answer

Answer: $\mathrm{I} _{2}=\frac{2}{99} \mathrm{~A}$

$ \qquad \mathrm{I} =\frac{70}{99} \mathrm{~A}$

$ \qquad \mathrm{I} _{1}=\frac{12}{99} \mathrm{~A}$

26. In the circuit shown in figure $\mathrm{E}, \mathrm{F}, \mathrm{G}$ and $\mathrm{H}$ are cells of emf 2, 1,3 and 1 volt respectively and their internal resistance are $2,1,3$ and $1 \Omega$ respectively. Calcualte:

(i) The potential difference between $\mathrm{B}$ and $\mathrm{D}$.

(ii) The potential difference across the terminals of the cells $\mathrm{G}$ and $\mathrm{H}$.

$~$

Show Answer

Answer: $\mathrm{V} _{\mathrm{DB}}=\frac{2}{13} \mathrm{~V}, \mathrm{~V} _{\mathrm{G}}=\frac{21}{13} \mathrm{~V}, \mathrm{~V} _{\mathrm{H}}=\frac{19}{13} \mathrm{~V}$

27. In the circuit shown in figure, the current through $3 \Omega$

resistance. $2 \mathrm{~A} _{1} \mathrm{E} _{1}=14 \mathrm{~V}$,

$~$

What is the value of $\mathrm{E}$ ? Internal resistance of each battery is $1 \Omega$.

Show Answer

Answer: $7 \Omega$

28. Twelve equal wires each of resistance $r$ are joined up to form a skelton cube. A current enters at one corner and leaves at the opposite corner. What resistance does the cube offer in this situation.

Show Answer

Answer: $\frac{5}{6} \mathrm{r}$

29. A letter is constructed of a uniform wire of resistance $1 \Omega \mathrm{cm}^{-1}$. The side of the letter are $20 \mathrm{~cm}$ long and the across piece in the middle is $10 \mathrm{~cm}$ long while the apex angle is $60^{\circ}$. Find the current flowing through the different pairs when a battery of emf $15 \mathrm{~V}$ and internal resistance $3.33 \Omega$ is connected across the two ends of the legs.

Show Answer

Answer: $0.167 \mathrm{~A} ; 0.33 \mathrm{~A}$

30. In a meter bridge, the balance point is found to be at $40 \mathrm{~cm}$ from the left end of the meter bridge wire when the resistance in the left gap is $15 \Omega$. Find the balance point when the resistance in left and right gaps are interchanged.

Show Answer

Answer: $60 \mathrm{~cm}$

31. The resistance of a meter bridge wire is $2 \Omega$. The galvonometer used in the circuit has a resistance of $100 \Omega$ and the battery used has an emf of $2 \mathrm{~V}$. If the resistance in the left and right gaps are $1 \Omega$ and $3 \Omega$, where is the balance point obatined.

Show Answer

Answer: $25 \mathrm{~cm}$

32. A resistance of $2 \Omega$ is connected in the left gap of a meter bridge wire which has a length of $1 \mathrm{~m}$. An unknown resistance $\mathrm{X}(\mathrm{X}<x)$ is connected in the right gap. The balance point is found at $\mathrm{L} \mathrm{cm}$ from the positive end of the battery. On interchanging the resistances in two gaps it is found that the balance point further shifts by $20 \mathrm{~cm}$ away. Neglecting the end corrections calculate the value of unknown resistance $x$ used.

Show Answer

Answer: $3 \Omega$

33. $x$ is an unknown resistance to be determined by a meter bridge by using a $10 \Omega$ resistance in the right hand side gap of the meter bridge wire. The balance point is obtained at $52 \mathrm{~cm}$ mark. The end corrections are $1 \mathrm{~cm}$ and $2 \mathrm{~cm}$ respectively for the left and right ends. What is the value of $x$.

Show Answer

Answer: $10.6 \mathrm{~cm}$

34. The resistance of a potentiometer wire is $8 \Omega$ and its length is 8 meler. Ahigh reisistance box and a 92 volt accumlator are placed in series will it. What should be the value of the resistance box if it is desired to have a potential drop of one microvolt per millimeter?

Show Answer

Answer: $912 \Omega$

35. A potentiometer wire of length $100 \mathrm{~cm}$ has a resistance of $10 \Omega$. It is connnected in series with a resistance and a cell of emf 2 volt and neglible internal resistance source of emf 10 milli volt is balanced by a length of $40 \mathrm{~cm}$ of the potentiometer wire. What is the value of the external resistance.

Show Answer

Answer: $790 \Omega$

36. In an experiemnt with a potentiometer to measure the internal resistance of a cell when the secondary circuit is shunted by $5 \Omega$, the null point is at $2.20 \mathrm{~m}$. When the cell is shunted by $20 \Omega$ the null point is at $3.00 \mathrm{~m}$. Calculate the internal resistance of cell.

Show Answer

Answer: $2.76 \Omega$

37. A $1.5 \mathrm{~V}$ poteniometer is used for finding the internal resistance of $1.2 \mathrm{~V}$ cell. The balance point of the cell comes at $65 \mathrm{~cm}$. If a resistor of $10 \Omega$ is used in the external circuit of the cell, the balance point changes to $55 \mathrm{~cm}$. Calculate the internal resistance of the cell.

Show Answer

Answer: $1.82 \Omega$

38. A potentiometer circuit using a standard cell of emf 1.02 volt is setup as shown in Figure If the balancing length with 1,3 connection is $90 \mathrm{~cm}$ and with the 2,3 connection is $45 \mathrm{~cm}$. What is the e.m.f. of thermocouple? Also find the value of R.

$~$

Show Answer

Answer: $0.51 \mathrm{~V}$ and $1.004 \Omega$

Question Bank

Key Learning Points

1. The free electrons in a conducter in a maintained electric field constitute a current.

2. The electric field that acts on the electrons in a conductor does not produce a net acceleration because the electrons keep on colliding with the ctoons leat make up the conductor and thereby acquire a constant drift velocity $v _{\mathrm{d}}$ in a direction opposite to the field.

$$ \mathrm{v} _{\mathrm{d}}=\frac{1}{2} \frac{\mathrm{e}}{\mathrm{m}} \tau \frac{\mathrm{v}}{\mathrm{L}} $$

Where $\tau$ is known as relaxtion time and other symbols have their usual meaning.

3. The general expression for the current is $i=\operatorname{nev} _{d} A$

where $\mathrm{n}$ is the number of free electrons per unit volume and $\mathrm{A}$ is the area of cross section of the conductor.

4. The same current flows through the every cross section of the conductor even of its is not uniform.

5. According to ohm’s law the current is directly proportional to the potational difference provided the physical conditions of the conductor remains unchanged.

i.e. $V=I R$

Where $\mathrm{R}$ is the resistance offered by the conductor.

6. The resistance of any conductor can be expressed in terms of the length $\ell$ and area of cross section A as $\mathrm{R}=\rho \frac{\ell}{\mathrm{A}}$

Where $\rho$ is a constant of the material and is called its resistivity. The receprocal of $\rho$ is known as conductivity any is denoted by $\sigma$.

7. The equivalnce resistance of resistance $\mathrm{R} _{1}, \mathrm{R} _{2}, \mathrm{R} _{3}, \ldots . \ldots . \mathrm{R} _{\mathrm{n}}$, connected.

(i) In series, is $\mathrm{R} _{\text {eq }}=\mathrm{R} _{1}+\mathrm{R} _{2}+\ldots . \ldots . \mathrm{R} _{\mathrm{n}}$

(ii) In parallel, is $\frac{1}{\mathrm{R} _{\mathrm{eq}}}=\frac{1}{\mathrm{R} _{1}}+\frac{1}{\mathrm{R} _{2}}+\ldots . \frac{1}{\mathrm{R} _{\mathrm{n}}}$

8. The variation of resistance with temperature is given by

$$ \mathrm{R} _{\tau}=\mathrm{R} _{0}(1+\alpha \mathrm{t}) $$

where $\alpha$ is the temperature co-efficient of material.

9. The resistance of metals increases with temperature. However for semi conductor and insulators resistivity increases exponnentally with decreasing temperature.

10. Super conductors (metals like $\mathrm{Nb}$ and other meterial likes $\mathrm{Ba} _{2} \mathrm{Cu} _{2} \mathrm{O} _{7}$ ) have zero resistance.

11. For solving complicated network use is made of Kirchoff’s laws.

12. Wheatstone bridge is a special type of resistance network for measuring the resistance (It cannot be used for measurement of very low and very high resistances). The condition of balance is:

$$ \frac{\mathrm{R} _{1}}{\mathrm{R} _{2}}=\frac{\mathrm{R} _{3}}{\mathrm{R} _{4}} $$

13. A Wheatstone bridge has optimum sensitivity when the resistance in its all branches are of the same order of magnitude.

14. A cell is a device in which by means of chemical reactions, chemical energy is converted into electrical energy. A cell provides the necessary potential difference for maintaining a current in an electric circuit.

15. Emf a cell is the potential difference between its terminal when the cell does not draw any current (open circuit) whereas the terminal potential difference is defined as the potential difference between its terminal in a closed circuit.

The two are related as:

$\mathrm{E}=\mathrm{V}-\mathrm{ir}$

where $\mathrm{r}$ is the internal resistance of the cell.

16. When a number of cells of emf’s $\mathrm{E} _{1}, \mathrm{E} _{2} \ldots . . \mathrm{E} _{\mathrm{n}}$ are connected

(i) In series, then, $\mathrm{E} _{\mathrm{eq}}=\mathrm{E} _{1}+\mathrm{E} _{2}+\ldots . . \mathrm{E} _{\mathrm{n}}$

(ii) In parallel, then, $E _{e q}=\frac{E _{1}}{r _{1}}+\frac{E _{2}}{r _{2}}+\ldots . \frac{E _{n}}{r _{n}}$

and $\frac{1}{\mathrm{r} _{\mathrm{eq}}}=\frac{1}{\mathrm{r} _{1}}+\frac{1}{\mathrm{r} _{2}}+\ldots . \ldots \frac{1}{\mathrm{r} _{\mathrm{n}}}$

17. Meter bridge is based on the Wheatstone bridge and is used to measure resistance.

Referring to figure we have,

$ P=Q \frac{L}{100-L} $

For better accuracy we have to take into account end corrections.

$~$

18. Potentiometer is a verstile instrument. Here we balance an unknown e.m.f. $\mathrm{E} _{2}$ against a length $\mathrm{L}$ of the potentiometer wire. We can use potentiometer as:

$~$

(i) To compare two emf’s $\left(\frac{\mathrm{E} _{1}}{\mathrm{E} _{2}}=\frac{\ell _{1}}{\ell _{2}}\right) \quad$ (Reference Text)

(ii) To determine the internal resistance of a cell $\left(\mathrm{r}=\left(\frac{\mathrm{L} _{1}-\mathrm{L} _{2}}{\mathrm{~L} _{2}}\right) \mathrm{R}\right)$

(iii) Measurements of current.

(iv) Measurement of high emf.

(v) Measurement of a very small emf.

19. If a current of I ampere traverses a conductor between the ends of which there is a potential difference equal to $\mathrm{V}$, the work done during a time $\mathrm{t}$ is $\mathrm{V}$ I $t$. If the whole of energy liberated is converted into heat, the heat generated is given by

$\mathrm{H}=\mathrm{I}^{2} \mathrm{R} \mathrm{t}$

Average

Combination of Cells

1. N identical cells, each of emf $E$, and internal resistance $r$, are connected in parallel. The variation of the net emf, $E$ of the combination, with the number of cells $N$, is expressed corretly by the graph labelled as:

$~$

Show Answer

Correct Answer: (2)

Solution:

Ifn cells, of e.m.f. $\mathrm{E} _{1}, \mathrm{E} _{2} \ldots . \ldots . \mathrm{E} _{\mathrm{n}}$ and internal resistance $\mathrm{r} _{1}, \mathrm{r} _{2} \ldots . \ldots . \mathrm{r} _{\mathrm{n}}$, are connected in parallel, then we have,

$ \frac{1}{\mathrm{r} _{\mathrm{eq}}}=\frac{1}{\mathrm{r} _{1}}+\frac{1}{\mathrm{r} _{2}}+\ldots . \ldots \frac{1}{\mathrm{r} _{\mathrm{n}}} $

and $\frac{E _{\text {eq }}}{r _{e q}}=\frac{E _{1}}{r _{1}}+\frac{E _{2}}{r _{2}}+\ldots . \frac{E _{n}}{r _{n}}$

From these equations we have,

$ \frac{1}{r _{e q}}=\frac{1}{r}+\frac{1}{r}+\ldots . \frac{1}{r}=\frac{N}{r} $

and $\frac{E _{e q}}{r _{e q}}=\frac{E}{r}+\frac{E}{r}+\ldots . . \frac{E}{r}=\frac{\frac{n E}{r}}{\frac{n}{r}}=E$

Hence option (2) is correct.

Average

Terminal p.d

2. A cell having an emf $E$ and internal resistance $r$ is connected across a variable resistance $R$. When the resistance $R$ is increasing and the potential difference $V$ across $R$ is measured, the graph between increases value of $R$ and the corresponding value of $V$ is represented by the graph, labelled as graph.

$~$

Show Answer

Correct Answer: (3)

Solution:

The circuit is a shown.

The current in this circuit is $i=\frac{E}{R+r}$

Potential V across $\mathrm{R}$ is

$ V=i R=\left(\frac{E}{R+r}\right) R=\frac{E}{1+\frac{r}{R}} $

It is easy to see that

$~$

When $\mathrm{R} \rightarrow 0 \mathrm{~V}=0$ and

When $\mathrm{R} \rightarrow \infty \mathrm{V} \rightarrow \mathrm{E}$

Hence option (3) is correct.

Average

Cells in Series

3. N identical cells, each of emf $E$ and internal resistance $r$, are joined in series. Out of these n cells are incorrectly connected, i.e., these terminals are connected in reverse of that required for series connection $\left(\mathrm{n}<\frac{\mathrm{N}}{3}\right) . \mathrm{E} _{0}$ and $\mathrm{r} _{0}$ represent the resulting emf and internal resistance of the combination.

Then $\mathrm{E} _{0}$ and $\mathrm{r} _{0}$ are given by:

(1) $\quad(N-2 n) E, r _{0}=\mathrm{Nr}$

(2) $\left(\frac{N-n}{2}\right)$ E, $r _{0}=\left(\frac{N-n}{2}\right) r$

(3) $(n-n) E, r _{0}=\left(\frac{N}{n}\right) r$

(4) $(2 \mathrm{~N}-\mathrm{n}) \mathrm{E}, \mathrm{r} _{0}=(\mathrm{N}-2 \mathrm{n}) \mathrm{r}$

Show Answer

Correct Answer: (1)

Solution:

$~$

$\mathrm{F} _{0}=(\mathrm{N}-\mathrm{n}) \mathrm{E}-\mathrm{nE}=(\mathrm{N}-2 \mathrm{n}) \mathrm{E}$

$\mathrm{r} _{\mathrm{n}}=\mathrm{r}+\mathrm{r}+ \ldots . \ldots \mathrm{n}$ times $=\mathrm{Nr}$

Hence the option (1) is correct.

Average

Cell Combinations

4. A battary of 24 cells each of emf $1.5 \mathrm{~V}$ and internal resisatnce $2 \Omega$ is connected so as to send maximum current through a 125 resistance. In order to achieve this we must connect.

(1) 6 cells in series and connect 4 such rows in parallel

(2) all the 24 cells in series

(3) all the 24 cells in parallel

(4) 12 cells in series and connect 2 such rows in parallel

Show Answer

Correct Answer: (4)

Solution:

Let us connect cells in mixed grouping where $\mathrm{n}$ cells are connected in series in one row and $\mathrm{m}$ such rows are connected in parallel. Let $\mathrm{N}=\mathrm{Nr}=24$

Then for maximum current in external resistance.

$$ \begin{aligned} & \mathrm{r}=\sqrt{\frac{\mathrm{NR}}{\mathrm{r}}}=\sqrt{\frac{24 \times 12}{2}}=12 \\ & \mathrm{~m}=\frac{\mathrm{N}}{\mathrm{n}}=2 \end{aligned} $$

Hence option (4) is correct.

Difficult

Cell Combinations

5. (MN) identical cells, of e.m.f. $E$ and internal resistance $r$, connected in $\mathbf{N}$ parallel rows where each row contains $M$ cells in series. This combination of cells is connected across an external resistance $R$. What should be arrangment of the cells for getting maximum current through $R$ if the total number of cells is fixed?

(1) Number of cells in a row $=\frac{\mathrm{RN}}{\mathrm{r}}$ and number of rows $=\frac{\mathrm{Nr}}{\mathrm{R}}$

(2) Number of cells in a row $=\sqrt{\frac{\mathrm{RN}}{\mathrm{r}}}$ and number of rows $=\sqrt{\frac{\mathrm{Nr}}{\mathrm{R}}}$

(3) Number of cells in a row $=\sqrt{\frac{\mathrm{MR}}{\mathrm{r}}}$ and number of rows $=\sqrt{\frac{\mathrm{MR}}{\mathrm{R}}}$

(4) Number of cells in a row $=\sqrt{\frac{\mathrm{Nr}}{\mathrm{R}}}$ and number of rows $=\sqrt{\frac{\mathrm{nR}}{\mathrm{r}}}$

Show Answer

Correct Answer: (2)

Solution:

The simplified version of the given circuit is shown in diagram (2) we have,

$$ \mathrm{i}=\frac{\mathrm{mE}}{\mathrm{R}+\frac{\mathrm{mr}}{\mathrm{n}}}=\frac{\mathrm{mnE}}{\mathrm{mr}+\mathrm{nR}} $$

$~$

If the total number of cell is fixed, than $m$ is fixed. Then we need to find the value of the $m$ and $n$ so that the current is maximum. We can write

$$ i=\frac{m n E}{(\sqrt{m r}-\sqrt{n r})^{2}+2 \sqrt{m n R}} $$

As $\mathrm{mn}$ is fixed. I is maximum when denominator is minimum

$\therefore(\sqrt{\mathrm{mr}}-\sqrt{\mathrm{nr}})^{2}$

or $\quad \frac{m}{n}=\frac{R}{r}$

If $\mathrm{N}$ is the total number of cells then we can get $\mathrm{mn}=\mathrm{N}$ and $\frac{\mathrm{m}}{\mathrm{n}}=\mathrm{R} / \mathrm{r}$

Hence for the maximum current. Number of cell in a row $=\sqrt{\frac{\mathrm{RN}}{\mathrm{r}}}$

and number of rows $=\sqrt{\frac{\mathrm{Nr}}{\mathrm{R}}}$

Average

Circuit Analysis

6. You are provided with a test resistance $R$, a high resistance $R _{1}$ and a small resistance $R _{2}$, two identical galvanometer $G _{1}$ and $G _{2}$ and a variable voltage source $V$. You are required to set up a circuit for the verification of ohm’s law. Which of the following would be a correct circuit?

$~$

Show Answer

Correct Answer: (3)

Solution:

Here we have to use galvanometer $\mathrm{G} _{1}$, with a high resistance in series, in place of the voltmeter. Similarly galvanometer $\mathrm{G} _{2}$, with a parallel small resist once,takes the place of the ammeter. Hence the circuit diagram (3) is the correct circuit for the verification of ohm’s law.

Average

Drift Velocity

7: A conductor contains $8 \times 10^{-28}$ free electrons per cubic meter. The conductor carries a current of $1 \mathrm{~A}$ and has a length of $10 \mathrm{Cm}$. The area of cross section of the conductor is $7.81 \times 10^{-6} \mathrm{~m}^{2}$. The time taken by an electron to move from one end of the conductor to the other end is:

(1) $10^{-3} \mathrm{~s}$

(2) $10^{-4} \mathrm{~s}$

(3) $10^{5} \mathrm{~s}$

(4) $10^{6} \mathrm{~s}$

Show Answer

Correct Answer: (3)

Solution:

We known that drift velocity $\mathrm{v} _{\mathrm{d}}$ is given by the relation.

$$ v _{d}=\frac{I}{n e A} $$

where the symbols have their usual meanings.

Now it is easy to see that $\mathrm{v} _{\mathrm{d}}$ can be written as $\mathrm{v} _{\mathrm{d}}=\frac{\mathrm{L}}{\mathrm{t}}$

where $L$ is the distance between one end of a conductor to the other end and t is the time taken by the free electrons to travel from one one to the other end of the conductor.

Hence, $\frac{\mathrm{L}}{\mathrm{t}}=\frac{\mathrm{I}}{\mathrm{neA}}$

or $\frac{\mathrm{t}}{\mathrm{L}}=\frac{\mathrm{neA}}{\mathrm{I}}$

or $\quad \mathrm{t}=\frac{\text { neA }}{\mathrm{I}} \mathrm{L}$

Substituting we given values we get, $\mathrm{t}=10^{-3} \mathrm{~s}$

Hence option (1) is correct.

Easy

Mean Free Path

8: The mean free path of electron in a metal is $4 \times 10^{-8} \mathrm{~m}$. The electric field which can give on average $2 \mathrm{eV}$ energy to an electron in the metal will be

(1) $10 \times 10^{7} \frac{\mathrm{V}}{\mathrm{m}}$

(2) $5 \times 10^{-7} \frac{\mathrm{V}}{\mathrm{m}}$

(3) $5 \times 10^{7} \frac{\mathrm{V}}{\mathrm{m}}$

(4) $2.5 \times 10^{7} \quad \frac{\mathrm{V}}{\mathrm{m}}$

Show Answer

Correct Answer: (3)

Solution:

It is easy to see that the work done is (qE) $(\lambda)$.

Hence,

$2 \mathrm{eV}=\mathrm{qE} \times 4 \times 10^{-8}$

or $\quad E=\frac{1}{4 \times 10^{-8}} \frac{V}{m}=5 \times 10^{7} \frac{V}{m}$

Easy

Resistivity

9. The length of a wire is doubled and current through it is increased by a factor of $\frac{3}{2}$ without any charge of temperature. The resistivity of the material of the wire

(1) remains same

(2) increase by a factor of 3

(3) decrease by a factor of 3

(4) increase by a factor of 2

Show Answer

Correct Answer: (1)

Solution:

The resistivity of a wire depends only on the nature of wire. Changing factors like length, area of cross section does not charge the resistivity of the material of the wire.

Average

Combination of Resistance

10. A wire of uniform resistance $\mathrm{z} \mathrm{m}^{-1}$ is sent into a circle of radius $r$. The same wire is connected between point $A$ and $B$ as shown. The equivalent resistance between $A$ and $B$ is:

$~$

(1) $\frac{\mathrm{zr}}{(3 \pi+16)}$

(2) $\frac{6 \pi(z / r)}{(3 \pi+16)}$

(3) $\frac{6 \pi \mathrm{zr}}{(3 \pi+16)}$

(4) $\left(\frac{\mathrm{z}}{\mathrm{r}}\right)(3 \pi+16)$

Show Answer

Correct Answer: (3)

Solution:

Let $R _{1}$ be the resistance of one quadrant and $R _{2}$ be the resistancr of three quadrants of circle.

$\mathrm{R} _{1}=\left(\frac{2 \pi \mathrm{r}}{4}\right) \mathrm{z}=\frac{\pi \mathrm{zr}}{2}$

$\mathrm{R} _{2}=\frac{3}{2} \pi \mathrm{zr}$

If the resistance of straight wires $\mathrm{AO}$ and $\mathrm{OB}$ taken together is $\mathrm{R} _{3}$. Then

$\mathrm{R} _{3}=2 \mathrm{rz}$

$\mathrm{R} _{1}, \mathrm{R} _{2}$ and $\mathrm{R} _{3}$ are connected in parallel

$\frac{1}{\mathrm{R}}=\frac{2}{\pi \mathrm{zr}}+\frac{2}{3 \pi \mathrm{zr}}+\frac{1}{2 \mathrm{zr}}=\frac{3 \pi+16}{6 \pi \mathrm{zr}}$

$\mathrm{R}=\frac{6 \pi \mathrm{zR}}{3 \pi+16}$

Hence optino (3) is correct.

Easy

Non Ohmic Devices

11. The relation between $V$ and $I$ for a device not obeying $O h m ’ s$ is shown in the figure where there is more than one value for current for a given interval of $V$. This device can be

(1) a junction diode

(2) a super conductor

(3) a light emitting diode

(4) a transmiter

Show Answer

Correct Answer: (3)

Solution:

Informative (The material of the light emitting diode is GaAs).

$~$

Average

Combination of Resistances

12. Sixteen resistance, each of $16 \Omega$ are connected as shown in figure. The net resistance, between $A$ and B, is.

(1) $4 \Omega$

(2) $6 \Omega$

(3) $8 \Omega$

(4) $3 \Omega$

$~$

Show Answer

Correct Answer: (4)

Solution:

In each of the four arms of network of resistance four resistance are converted in parallel. The effective resistance of each arm of network is

$\frac{\mathrm{R}}{4}=\frac{16}{4}=4 \Omega$

The total resistance is then given by:

$\mathrm{R} _{\mathrm{T}}=\frac{(4+4+4) \times 4}{(4+4+4)+4}=3 \Omega$

Hence option (4) is correct.

Easy

Resistance of Wire

13. A copper wire is stretched to make it $0 . z \%$ longer. Then the percentage charge in its resistance is

(1) $0 .\left(\frac{1}{2}\right) \times \mathrm{z} \%$

(2) No charge

(3) $0.4(\mathrm{z} \%)$

(4) $2 \times(0 . z \%)$

Show Answer

Correct Answer: (4)

Solution:

If A is the area of cross sectino of the wire of length $L$ and density $d$ its mass is given by:

$$ \mathrm{m}=\mathrm{ALd} \quad \text { or } \quad \mathrm{A}=\frac{\mathrm{m}}{\mathrm{Ld}} $$

The resistance, $R$ of the wire of, resistively $\rho$ is given by

$\mathrm{R}=\frac{\rho \mathrm{L}}{\mathrm{A}}=\frac{\rho \mathrm{L}^{2} \mathrm{~d}}{\mathrm{~m}}$

Differentating we get,

$\frac{\mathrm{dR}}{\mathrm{R}}=\frac{2 \mathrm{dL}}{\mathrm{L}}$

Hence option (4) is correct.

Difficult

Resistance Combinations

14. $x$ and $y$ are two points on a uniform ring of resistance $R$. The angle $x z y=\theta$ where $z$ is the centre of ring. Then the equivalent resistance between $x$ and $y$ is:

(1) $\frac{\mathrm{R}}{4 \pi^{2}}(2 \pi-\theta) \theta$

(2) $\frac{\mathrm{R}}{4 \pi^{2}} \theta$

(3) $\frac{\mathrm{R}}{2 \pi} \theta(2 \pi-\theta)$

(4) $\frac{\mathrm{R}}{(2 \pi-\theta)} \times 4 \pi^{2} \theta$

$~$

Show Answer

Correct Answer: (1)

Solution:

$$ \mathrm{R} _{x y}=\frac{\rho \ell _{x y}}{\mathrm{~A}} $$

Also, $\mathrm{R}=\frac{\rho(2 \pi \mathrm{a})}{\mathrm{A}} \quad$ and $\quad \ell _{x y}=\theta \mathrm{A}$

$$ \mathrm{R} _{x y}(\text { Smaller portion })=\frac{\mathrm{R}}{2 \pi \mathrm{a}}(\theta \mathrm{a})=\frac{\mathrm{R} \theta}{2 \pi} $$

Similarly, $\mathrm{R} _{x y}($ Larger portion $)=\frac{\mathrm{R}}{2 \pi}=(2 \pi-\theta)$

$$ R _{\text {eq }}=\frac{\frac{R \theta}{2 \pi} \times \frac{R}{2 \pi}(2 \pi-\theta)}{\frac{R \theta}{2 \pi}+\frac{R}{2 \pi}(2 \pi-\theta)}$$

$$ =\frac{\operatorname{R\theta }(2 \pi-\theta)}{4 \pi^{2}} $$

Hence the option (1) is correct.

Difficult

Resistance Combinations

15. A cylinder of radius a, lengths $L$ is made of a material of resistivity $\rho _{1}$. It is electroplated with a material, of resistivity $\rho _{2}$, till it radius increases to $b$. The resistance of the elctroplated cylinder will be given by the relation.

(1) $ \left[\frac{1}{\frac{a^{2}}{\rho _{1}}+\left(\frac{b^{2}-a^{2}}{\rho _{2}}\right)}\right]$

(2) $\frac{1}{\pi}\left[\frac{1}{\frac{a^{2}}{\rho _{1}}+\left(\frac{b^{2}-a^{2}}{\rho _{2}}\right)}\right]$

(3) $\frac{\mathrm{a}^{2}}{\rho _{2}}+\left(\frac{\mathrm{b}^{2}-\mathrm{a}^{2}}{\rho _{1}}\right)$

(4) $\frac{1}{2 \pi}\left[\frac{a _{1} \rho _{1}+b \rho _{2}}{\frac{a^{2}}{\rho _{1}}+\frac{b^{2}}{\rho _{2}}}\right]$

Show Answer

Correct Answer: (2)

Solution:

Let $R _{1}$ be resistance of inner cylinder of radius a and resistivity $\rho$. Then $R _{1}=\rho _{1} \frac{L}{\pi a^{2}}$

The area of cross section of electroplated part is $\pi\left(b^{2}-a^{2}\right)$. The resistance $R _{2}$ of electroplated part of cylinder is

$$ \mathrm{R} _{2}=\rho _{2} \frac{\mathrm{L}}{\pi\left(\mathrm{b}^{2}-\mathrm{a}^{2}\right)} $$

Resistance $\mathrm{R} _{1}$ and $\mathrm{R} _{2}$ are connected in parallel. The equivalent resistance $\mathrm{R}$ is

$R=\frac{R _{1} R _{2}}{R _{1}+R _{2}}=\frac{\rho _{1} \rho _{2} L^{2} /\left[\pi^{2} a^{2}\left(b^{2}-a^{2}\right)\right]}{\frac{1}{\pi}\left[\frac{\rho _{1}}{a^{2}}+\frac{\rho _{2}}{\left(b^{2}-a^{2}\right)}\right]}$

$$ =\frac{\rho \rho _{2} \mathrm{~L}}{\pi\left[\rho _{1}\left(b^{2}-a^{2}\right)+\rho _{2} a^{2}\right]}=\frac{1}{\pi}\left[\frac{1}{a^{2} / \rho _{1}+\frac{\left(b^{2}-a^{2}\right)}{\rho^{2}}}\right] $$

Hence option (2) is correct.

Difficult

Continuous Variation of Resistance

16. A cylinderical wire of radius $R$ and is of uniform area of cross section. The current density varies with radial distance $r$ as $J=\alpha r^{2}$. Then, the current, through the outer portion of the wire, between radial distances, $\frac{R}{2}$ and $R$, is given by

(1) $\pi \mathrm{a}^{2} \mathrm{R}^{4}$

(2) $\frac{15}{32} \pi \mathrm{aR}^{4}$

(3) $\frac{15}{32} \pi \mathrm{aR}^{3}$

(4) $\frac{15}{32}(\pi \mathrm{aR})^{3}$

$~$

Show Answer

Correct Answer: (2)

Solution:

We consider an element of the cylinderical wire of radius $r$ and thikness $d r$ as shown. Then the area of cross section of the element is $\mathrm{dA}=2 \pi \mathrm{rdr}$.

It can be seen that the current is along the wire and the elementary area $\mathrm{dA}$ is perpendicular to the cross section of the wire and both have the same direction.

Hence the current through the element of the wire is

$$ \mathrm{dI}=\mathrm{dA}=\alpha \mathrm{r}^{2} \times 2 \pi \mathrm{rdr}=2 \pi \alpha \mathrm{r}^{3} \mathrm{dr} $$

Hence the current through the outer portion of the wire is $\mathrm{I}=\int _{\mathrm{R} / 2}^{\mathrm{R}} 2 \pi \alpha \mathrm{R}^{3} \mathrm{dr}=2 \pi \alpha\left[\frac{\mathrm{r}^{4}}{4}\right] _{\mathrm{R} / 2}^{\mathrm{R}}=\frac{15}{32} \pi \alpha \mathrm{R}^{4}$

Hence option (2) is correct.

Average

V-I Graphs

17. The voltage current graph for a platinum wire conductor at temperature $T _{1}$ and $T _{2}$ is as shown in the figure.

$~$

(Given temperature coefficient of platinum is $\alpha$ and $\mathbf{R} _{0}$ to the resistance of platinum wire at $\left.0^{\circ} \mathrm{C}\right)$. The value of $\left(\mathrm{T} _{2}-\mathrm{T} _{1}\right)$ is

(1) $\frac{2}{\alpha \mathrm{R} _{0}} \tan \theta$

(2) $\frac{\mathrm{R} _{0} \alpha}{2} \cos 2 \theta$

(3) $\frac{2}{\alpha \mathrm{R} _{0}} \sin 2 \theta$

(4) $\frac{2}{\alpha \mathrm{R} _{0}} \cot 2 \theta$

Show Answer

Correct Answer: (4)

Solution:

$$ \begin{aligned} & \mathrm{R} _{1}=\tan \theta=\mathrm{R} _{0}\left(1+\alpha \mathrm{T} _{1}\right) \\ & \mathrm{R} _{2}=\cot \theta=\mathrm{R} _{0}\left(1+\alpha \mathrm{T} _{2}\right) \\ & \cot \theta-\tan \theta=\mathrm{R} _{0}\left(1+\alpha \mathrm{T} _{2}\right)-\mathrm{R} _{0}\left(1+\alpha \mathrm{T} _{1}\right) \\ & \text { or } \quad \mathrm{T} _{2}-\mathrm{T} _{1}=\frac{1}{\alpha \mathrm{R} _{0}}(\cot \theta-\tan \theta) \\ & \mathrm{T} _{2}-\mathrm{T} _{1}=\frac{1}{\alpha \mathrm{R} _{0}}\left(\frac{\cos \theta}{\sin \theta}-\frac{\sin \theta}{\cos \theta}\right) \\ & \quad=\frac{2 \cos 2 \theta}{\alpha \mathrm{R} _{0} \sin 2 \theta}=\frac{2}{\alpha \mathrm{R} _{0}} \cot 2 \theta \end{aligned} $$

Easy

Dependence of Resistivity on Temperature

18. Which of the following statement is correct?

(1) For metals, resistivity increases linearly with temperature, at all temperatures.

(2) For insulators resistivity decreases expontentially with decreasing temperatures.

(3) For semi conductors resistivity becomes very large at low temperatures tending to infinitely large values as $\mathrm{T} \rightarrow 0$

(4) Resistivity increases with temperatures for metals, insulators as well as semi conductors.

Show Answer

Correct Answer: (3)

Solution:

For semi conductor the resistivity increases exponentially with decreasing temperature, the variation being expressed by the relation of the form

$$ \rho(T)=\rho _{0}(T) \exp \left(\frac{E _{g}}{k T}\right) $$

Here $\rho _{0}(T)$ is a factor that varies slightly will temperature, $k$ is the Boletzman’s constant and $\mathrm{E} _{\mathrm{g}}$ is the positive energy known as gap energy.

Because of expontatial dependance the resistance of semi conductors becomes very large at low temperature tending to infinite large values as $\mathrm{T} \rightarrow 0$.

Hence option (3) is correct.

Average

Dependance of Resistance on Temperature

19. Which of the following statements is incorrect?

(1) The resistance of a thermistor charges very rapidly with charge in temperature.

(2) The temperature co-efficient of resistivity can be both positive and negative.

(3) A thermistor can easily measure a charge in temperature of $10^{-3} \mathrm{C}$.

(4) The temperature co-efficient of resistivily of a thermistor is normally very low.

Show Answer

Correct Answer: (4)

Solution:

The thermistor is a semi conducting device. The resistivity of a semi conductor increases expontentially with decrease in temperature. The resistivity of a semi conductor increases very large with decrease in temperature tending to inifinity as $\mathrm{T} \rightarrow 0$. The option 1,2 and 3 are known characterstics of a thermistor. Hence the option (4) is not correct.

Easy

Non-Ohmic Devices

20. Which of the following graphs represent current voltage relationship for a water voltamtere?

$~$

Show Answer

Correct Answer: (1)

Solution:

For certain electrolytes like platinum and water, we find that the current flowing through is extremely small until the potential difference increase to a value $\mathrm{V}^{\prime}$. After that the current increases linearly, with $\mathrm{V}$, as shown in option (1).

Hence option (1) is correct.

Average

Circuit Analysis

21. Find the steady state voltage drop across the capacitor $C$ in the diagram given below.

(1) V

(2) $\frac{\mathrm{VR} _{1}}{\left[\mathrm{R} _{2}+\left(\frac{\mathrm{R} _{1} \mathrm{R} _{3}}{\mathrm{R} _{1}+\mathrm{R} _{3}}\right)\right]}$

(3) $\frac{\mathrm{VR} _{2}}{\mathrm{R} _{1}+\mathrm{R} _{2}}$

(4) $\frac{\mathrm{VR} _{2}}{\mathrm{R} _{1}+\mathrm{R} _{3}}$

$~$

Show Answer

Correct Answer: (3)

Solution:

Since no d.c. current flows through a capacitor the steady state voltage drop across capacitor is same as across resistance $\mathrm{R} _{2}$. The current is given by

$$ I=\frac{V}{R _{1}+R _{2}} $$

Here the voltage drop across $R _{2}=\mathrm{IR} _{2}=\frac{\mathrm{VR} _{2}}{\mathrm{R} _{1}+\mathrm{R} _{2}}$

Hence option (3) is correct.

Easy

Meter Bridge

22. Figure shows a meter bridge circuit for the determination of an unknown resistance. When we use different values of $R$, (say $R _{1}, R _{2}$ and $R _{3}$, respectively) we obtain ‘balance points’ at A, B and C, respectively. For which resistor, the value of $\boldsymbol{x}$ will be determined more accurately?

$~$

(1) $\mathrm{R} _{1}$

(2) $\mathrm{R} _{2}$

(3) $\mathrm{R} _{3}$

(4) The accuracy of determination will be practically the same whether we use $R _{1}$, or $R _{2}$, or $R _{3}$.

Show Answer

Correct Answer: (2)

Solution:

Meter bridge is based on the principle of Wheatstone bridge. The balance condition $\frac{\mathrm{R} _{1}}{\mathrm{R} _{2}}=\frac{\mathrm{R} _{3}}{\mathrm{R} _{4}}$ in a Wheatstone bridge is an ideal one since it is not achieveable in practice. This is because current in the galvanoemeter can never be made equal to zero. Hence the value of variable resistance $\left(R _{5}\right)$ in a Wheatstone bridge will always be a little different from ideal value $R _{3}$ for balance. $\left(R _{3}^{\prime}-R _{3}\right)$ is known as want of balance. It can be shown that, for optimum sensitivity of a wheatstone bridge for a given want of balance is obtained when the resistance in its all four arms are of the same order of magnitude. The balance point B, using resistance $\mathrm{R} _{2}$, is close to the middle of the wire. This satisfies the condition for optimum sansitivity. Hence option (2) is correct.

Average

Meter Bridge

23. Figure shows a meter bridge circuit for measurements of resistance. However the meter bridge wire has end corrections $\alpha$ and $\beta$ (in terms of the length of meter bridge wire).

Taking known $P$ and $Q$, we find a balance point is obtained at a length $\ell _{1}$ of meter bridge wire. If we inter change, the position of $P$ and $Q$, the balance point is obtained at a length $\ell _{2}$ of meter bridge wire; then the value of $\alpha$ and $\beta$ are

$~$

(1) $\alpha=\frac{\mathrm{Q} \ell _{2}-\mathrm{P} \ell _{1}}{\mathrm{P}-\mathrm{Q}}, \beta=\frac{\mathrm{Q} \ell _{1}-\mathrm{P} \ell _{2}}{\mathrm{Q}-\mathrm{P}}$

(2) $\alpha=\frac{\mathrm{P} \ell _{2}-\mathrm{Q} \ell _{1}}{\mathrm{Q}-\mathrm{P}}, \beta=\frac{\mathrm{P} \ell _{1}-\mathrm{Q} \ell _{2}}{\mathrm{P}-\mathrm{Q}}$

(3) $\alpha=\frac{\ell _{1}-\ell _{2}}{\mathrm{P}}, \beta=\frac{\ell _{2}-\ell _{1}}{\mathrm{Q}}$

(4) $\alpha=\frac{\mathrm{Q} \ell _{1}-\mathrm{P} \ell _{2}}{\mathrm{P}-\mathrm{Q}}, \beta=\frac{\mathrm{P} \ell _{1}-\mathrm{Q} \ell _{2}}{\mathrm{P}-\mathrm{Q}}$

Show Answer

Correct Answer: (4)

Solution:

We can write

$ \frac{P}{Q}=\frac{\ell _{1}+\alpha}{100-\ell _{1}+\beta} $

and $\frac{Q}{P}=\frac{\ell _{2}+\alpha}{\left(100-\ell _{2}+\beta\right)}$

Solving these two equations for $\alpha$ and $\beta$ we get the values of $\alpha$ and $\beta$ as given in option (4).

Average

Wheatstone Bridge

24. To obtain the optimum sensitivity of a Wheatstone bridge, for a given ‘want of balance’, and with a freedom of choice of arrangment, we should.

(1) Connect the battery so that resistance in series with the resistance to be measured is greater than the resistance in parallel will it.

(2) Connect the battary so that the resistance in series with the resistance to be measured is less than the resistance in parallel will it.

(3) Interchange the position of the battery and the galvanometer.

(4) Use a galvanometer with a reasonable high resistance.

Show Answer

Correct Answer: (1)

Solution:

We can obtain an ideal relation for the sensitivity of a Wheatstone bridge for a given want of balance. However this ideal relation cannot be satisfied in actual working conditions. Hence we look for a rule for optimum senstivity of a wheatstone bridge. The option (1) is a working rule given by Callender for optimum senstivity of a Wheatstone bridge.

Average

Combination of Resistance

25. An infinite ladder network of resistance is constructed with $1 \Omega$ and $2 \Omega$ resistances as shown. The $6 \mathrm{~V}$ battery has negligible resistance. Then the effective resistance between $\mathrm{A}$ and $B$ is

(1) $1 \Omega$

(2) $2 \Omega$

(3) $3 \Omega$

(4) Infinite

$~$

Show Answer

Correct Answer: (2)

Solution:

Let $x$ be the effective resistance between points $A$ and $B$. Since the chain is an infininte one this may also be taken as the effective resistance of the chain beyond the points $\mathrm{C}$ and $\mathrm{D}$. Hence the may replace the given chain by the equivalent diagram as shown:

We this have,

$x=1+\frac{1}{(1 / 2+1 / x)}, x=1+\frac{2 x}{2+x}$

or $(x-1)(2+x)=2 x$

Simplifying we have, $x^{2}-x-2=0$

or $\quad x=1 \pm \sqrt{\frac{1+8}{2}}=2$ or -1

$~$

Since the equivalent resistance cannot be negative we have,

$$ x=2 \Omega $$

Hence option (2) is correct.

Average

Combination of Resistance

26. Figure shows a network of 5 resistances each of value $4 \Omega$ connected as shown. The value of the resistnace between terminals $P$ and $Q$ is

(1) $2 \Omega$

(2) $4 \Omega$

(3) $8 \Omega$

(4) $0.4 \Omega$

Correct answer: (2)

$~$

Show Answer

Solution:

The given network forms a balanced wheatstone bridge. Hence the resistance between $A B$ can be removed. All the resistance have value of $4 \Omega$. Hence the resistance between points $\mathrm{P}$ and $\mathrm{Q}$ is also $4 \Omega$.

Hence option (2) is correct.

Average

Combination of Resistances

27. Six equal resistance each of value $x \mathrm{ohm}$ are connected to form a network as shown in figure. Then the resistance between $A$ and $B$ is

(1) $2 x$

(2) $6 x$

(3) $x / 2$

(4) $x / 6$

Correct answer: (3)

$~$

Show Answer

Solution:

The upper five resistance between $A$ and $B$ are forming a balanced Wheatstone bridge. Hence there will be no current in arm CO. Hence CO are can be removed. Also resistance of arm ACB, AOB and AB are all in pararllel.

The effective resistance $\mathrm{R}$ is given by

$\frac{1}{\mathrm{R}}=\frac{1}{(x+x)}+\frac{1}{(x+x)}+\frac{1}{x}$

or $\quad \mathrm{R}=\frac{x}{2}$

Hence option (3) is correct.

Difficult

Combination of Resistances

28. Figure shows the connections of six resistances each of value $x$ between the points A, B and $C$. Then the net resistance will be maximum between.

$~$

(1) A and B

(2) B and C

(3) A and C

(4) The net value of the resistance between $\mathrm{A}$ and $\mathrm{B}$; $\mathrm{B}$ and $\mathrm{C}$ and $\mathrm{A}$ and $\mathrm{C}$ will be the same.

Show Answer

Correct Answer: (1)

Solution:

In the branch $\mathrm{AC}$ the effective resistance is $\frac{x}{3}$. In the branch $\mathrm{BC}$ the effective resistance is $\frac{x}{2}$

$\therefore$ Resistance in the branch $\mathrm{ACB}=\frac{x}{3}+\frac{x}{2}=\frac{5 x}{6}$

The resistance of branch ACB is in parallel to resistance of branch AB. This effective resistance between $\mathrm{A}$ and $\mathrm{B}=\frac{x \times 5 x / 6}{x+5 x / 6}=\frac{5}{11} x$.

Similarly the resistance between B and C is $\frac{4}{11} x$ and the resistance between A and C is $\frac{3}{11} x$.

Hence option (1) is correct.

Difficult

Circuit Analysis

29. A battery, of internal resistance $4 \Omega$, is connected to the network of resistance as shown in the following figure.

$~$

To deliver the maximum power to the network the value of, $R$, should be

(1) 2

(2) $4 / 9$

(3) $2 / 9$

(4) $2 / 3$

Show Answer

Correct Answer: (1)

Solution:

The given network can be reduced to a balanced Wheatstone bridge as shown. Hence no current flows through side $\mathrm{BD}$ of $6 \mathrm{R}$ resistance.

The given circuit forther reduces to.

$~$

To obtain maximum power, the net external resistance must be equal to total internal resistance.

$\therefore 2 \mathrm{R}=4$

or $\mathrm{R}=2 \Omega$

Hence option (1) is correct.

Average

Potentiometer

30. A potentiometer circuit is shown in figure. $x$ and $y$ being the terminal of an unknown e.m.f. It is observed that when the jockey $J$ is at $A$ there is a small deflection in the galvanometer. But when the jockey is moved to $B$ the deflection remains in the same direction but increases. This happens because

(1) The main battery is connected in an incorrect manner.

(2) The e.m.f. of the main battery is less than unknown e.m.f.

(3) The unknown e.m.f. is less than that of the main battery.

(4) That is a broken connection in the galvanomter circuit.

Show Answer

Correct Answer: (1)

Solution:

$~$

In a potentiometer we balance an unknown e.m.f. against potential drop a portion of length of the potentiometer wire. We find that the deflaction is in the same direction and increases as we move from one end A to another and B of the potentiometer wire it is clear that external e.m.f. is superimposing on drop of potential across AJ. This indicates the external e.m.f. between points $x$ y is wrongly connected.

Hence option (1) is correct.

Average

Potentiometer

31. A potentiometer circuit is shown in figure $x$ and $y$ being the tarminals of an unknown e.m.f. It is observed that when the jockey $J$ is at $A$, there is a small deflection in the galvanometer and the deflection remains $m$ the same direction but decreases when the jockey is moved to $B$. This is because

$~$

(1) The unknown e.m.f. $x y$ is connected incorrectly.

(2) The unknown e.m.f. $x y$ is just less than $\mathrm{E}$.

(3) The unknown e.m.f. is much less than E.

(4) The total drop of potential e.m.f. across the pontentiometer wire is less than the unknown.

Show Answer

Correct Answer: (4)

Solution:

When we find that deflation decreases when we move form end A to the other end B, the unknown emf $x y$ is correctly connected. We all not able to obtain a balance point since the total drop of potential across the total length of potential meter is less than unknown e.m.f. Hence to obtain the balance we must use source battery $E$ of higher value of emf.

Average

Potentiometer

32. Figuqre shows a potentiometer with a cell of $20 \mathrm{~V}$ and internal resistance of $0.04 \Omega$ maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf $1.02 \mathrm{~V}$ gives a balance point at 67.3 length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of $600 \mathrm{k} \Omega$ is put in series with it which is shorted close to the balance point. The standard cell is replaced by a cell of unknown emf $E$ and the balance point found

$~$

similarly turn out to be $82.3 \mathrm{~cm}$ length of the wire. The value of $E$ from the above given data is found to be 1.25 volt. We wish to determine the emf of, say, a themocouple (or the order of a few millivolt.

(1) To aclieve this measurement we should replace the high resistance of $600 \Omega$ with an infinite resistance.

(2) Increase the potential gradient of the potentiometer wire by a factor of at least 100.

(3) Decrease the potential gradient of the potentiometer wire by at least a factor of 100 by connecting suitable low resistance in series with the potentiometer wire.

(4) Decrease in potential gradient of potentiometer wire by a factor of more than 100 by connecting a suitable high resistance in series with the potentiometer wire.

Show Answer

Correct Answer: (4)

Solution:

The potential gradient of given wire is

$$ \rho=\frac{2 \mathrm{R}}{(\mathrm{R}+0.5) \mathrm{L}} \frac{\mathrm{V}}{\mathrm{cm}} $$

Usually R is of the order of a few ohm while $L$ is of the order of $400 \mathrm{~cm}$ to $1000 \mathrm{~cm}$.

Hence $\rho=\frac{2.0 \times 10}{10.5 \times 10^{3}}$

It follows that the balance length for a small e.m.f. would be only a few $\mathrm{cm}$. We would need to reduce the value of $s$ by a factor of at least hundred or so. This can be achieved by adding a suitable high resistance in series with the potentiometer wire.

Hence option (4) is correct.

Potentiometer

33. Which of the following statements is correct?

(1) A potentiometer can only be used for comparing two e.m.f.’s

(2) A potentiometer cannot be used to measure a high e.m.f.

(3) A potentiometer cannot be used for measuring thermo e.m.f.

(4) A potentiometer can be used for measuring high emf’s as well as very small e.m.f.’s

Show Answer

Correct Answer: (4)

Solution:

A potentiometer can be used for measuring a high e.m.f. The circuit diagram is shown in figures. Here the large e.m.f. to be measured is connected with two resistances $R _{1}$ and $R _{2}$

$\left(\mathrm{R} _{1} \sim 100 \Omega \quad \mathrm{R} _{2} \sim 9900 \Omega\right)$

We first connect 1 and 3 and 2’ and 3’. The potential drop produced across $\mathrm{R} _{1}$ due to $\mathrm{H}$. T is balanced across a length $\mathrm{L} _{1}$ of the potentiometer wire. Then potential drop across $\mathrm{R} _{1} \propto \mathrm{L} _{1}$. Since $\mathrm{R} _{1}+\mathrm{R} _{2}=10000 \Omega$ we can assume that the total potential drop across $R _{1}$ and $R _{2}$ is equal to e.m.f. of high e.m.f. using ohm’s law, we can write.

Potential drop across $R _{1}=\frac{E}{R _{1}+R _{2}} R _{1}$

$\frac{100}{10000} \mathrm{E}=\frac{\mathrm{E}}{100}$

or $\frac{\mathrm{E}}{100} \propto \mathrm{L} _{1}$

Drop across potentiometer wire is $\mathrm{E} _{\mathrm{S}}\left(\frac{\mathrm{r}}{\mathrm{R} _{2}}\right)$

$~$

The potential gradient therefore is $\mathrm{E} _{\mathrm{S}} \frac{\mathrm{r}}{\mathrm{R}} \cdot \frac{1}{\mathrm{~L}}$

The small e.m.f. to be measured is now brought into circuit by making contact between 2 and K. If L’ is the balance length, the $\mathrm{e}=$ small emf $=\left(\mathrm{E} _{\mathrm{S}} \frac{\mathrm{r}}{\mathrm{R} _{2}} \frac{1}{\mathrm{~L}}\right) \mathrm{L}^{\prime}$

The circuit shown is used for the measurement of low resistance. Here we set the total potential drop across the potentioneles small.

The small potential drop cross the potentiometer is measured accurately using a standard cell let $r$ be the resistence per unit length of potentiometer wire and $\mathrm{L}$ is its total length then the current is given by

$$ I=\frac{E}{\left(R _{1}+R _{2}+r\right)} $$

Total potential drop across potentiometer wire

$$ I^{r}=\frac{\mathrm{Er}}{\mathrm{R} _{1}+\mathrm{R} _{2}+\mathrm{r}} $$

$~$

$\therefore$ Potential Gradient $\rho=\frac{\mathrm{Er}}{\left(\mathrm{R} _{1}+\mathrm{R} _{2}+\mathrm{r}\right) \mathrm{L}}$

We adjust the value of $\left(R+R _{2}\right)$ so that $\rho$ is of the order of magnitude required to determine $\rho$ very accurately use is made of standard cell $\mathrm{E} _{\mathrm{s}}$. With the help of two way key contact is made between 1 and $R$. The values of $R _{1}$ and $R _{2}$ are altered keeping their sum constant till the balance point is obtained of the end of potentiometer wire. Now the potential drop across $R _{2}$ is exactly equal to e.m.f. of standard cell. The current $I$ therefore is $\mathrm{E} _{5} / R _{2}$. The potential drop across potentiometer wire is $E _{s}\left(r / R _{2}\right)$.

The potential gradient therefore is $E _{S} \frac{r}{R} \cdot \frac{1}{L}$.

The small emf to be measured is now brought into circuit by making connection between 2 and $\mathrm{k}$. If L’ is the balancing length. Then

e $=$ small e.m.f. $=\left(\mathrm{E} _{\mathrm{s}} \mathrm{r} / \mathrm{R} _{2} \cdot \frac{1}{\mathrm{~L}}\right) \mathrm{L}^{\prime}$

Hence option (4) is correct.

Average

Heating Effect of Current

34. An electric tea kettle has two heating coils. When one of coils is switched on, the kettle begins to boil in 6 minutes. When the other is switched on the kettle begin to boil in 8 minutes. It boil the coils are switched on simultaneously and connected in parallel then the kettle will begin to boil in.

(1) 14 minutes

(2) 205.7 seconds

(3) $7 / 24$ minutes

(4) 48 minutes

Show Answer

Solution:

Let $\mathrm{t}$ be the time the kettle take to begin boiling when they are connected in parallel.

Now, $H=\frac{V^{2}}{R _{1}} t _{1}=\frac{V^{2}}{R _{2}} t=\frac{V^{2}}{R _{1} R _{2}}\left(R _{1}+R _{2}\right) t$

Thus, $\mathrm{R} _{1}=\frac{\mathrm{V}^{2} \mathrm{t} _{1}}{\mathrm{H}}$ and $\mathrm{R} _{2}=\frac{\mathrm{V}^{2} \mathrm{t} _{2}}{\mathrm{H}}$

$H=\frac{V^{2}\left(R _{1}+R _{2}\right)}{R _{1} R _{2}} t, H=V^{2} t\left(\frac{H}{V^{2} t _{1}}+\frac{H}{V^{2} t _{2}}\right)$

or $\quad t=\frac{t _{1} t _{2}}{t _{1}+t _{2}}=\frac{6 \times 8}{6+8}=\frac{24}{7}$ minute

$=205.7 \mathrm{~s}$

Hence option (2) is correct.

Easy

Heating Effect of Current

35. Lamps A and $B$ have filaments of carbon and tangsten respectively. At room temperature the resistance of two bulbs is same. Both are connected in parallel to a $22 \mathrm{~V}$ supply. Which of the following statements is correct?

(1) Both bulbs have same intensily

(2) Intensity of bulb $A$ is more than that of $B$

(3) Intensity of bulb $A$ is less than thab $B$

(4) Initially both have same intensity but, after a sufficiently long time, intensity of bulb $A$ is less than that of B

Show Answer

Correct Answer: (1)

Solution:

Since both carbon and tengsten have same resistance (say $x$ ) and are connected in parallel the potential drop across each of then is $220 \mathrm{~V}$.

$\mathrm{Q} _{1}=$ heat produced in carbon filament bulb $=\frac{(220)^{2}}{x}$

$\mathrm{Q} _{2}=$ heat produced in tungsten filament bulb $=\frac{(220)^{2}}{x}$

$ \therefore \quad \frac{\mathrm{Q} _{1}}{\mathrm{Q} _{2}}=1 $

$~$

Hence both bulbs have the same intensity.

Hence option (1) is correct.

Average

Power Consumption in a Circuit

36. Six Diwali tree lights are arranged in a parallel circuit as shown in figure. Each bulb dissipates $10 \mathrm{~W}$ when operated at $120 \mathrm{~V}$. The total power consumption, of the array, would be (nearly)

$~$

(1) 600 Watt

(2) 0.6 Watt

(3) 120 Watt

(4) 60 Watt

Show Answer

Correct Answer: (4)

Solution:

Resistance of the each buls $=\mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{P}}=\frac{120 \times 120}{10}=1440 \Omega$

The given 6 resistance are connected in parallel, therefore equivalent resistance is given by:

$\frac{1}{\mathrm{R} _{\mathrm{eq}}}=\frac{1}{\mathrm{R}}+\frac{1}{\mathrm{R}}+\frac{1}{\mathrm{R}}+\frac{1}{\mathrm{R}}+\frac{1}{\mathrm{R}}+\frac{1}{\mathrm{R}}=\frac{6}{\mathrm{R}}$

$R _{\text {eq }}=\frac{R}{6}=\frac{1440}{6}=240$

Hence the power consumed by the given arrangment is given by

$\mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}=\frac{120 \times 120}{240}=60 \mathrm{~W}$

Hence option (4) is correct.

Average

Heating Effect of Currents

37. A copper wire, having cross sectional area of $0.5 \mathrm{~mm}^{2}$ and a length of $0.1 \mathrm{~m}$, is initially at $25^{\circ} \mathrm{C}$ and thermally insulted from the surroundings a current of $1.0 \mathrm{~A}$ is set up in this wire. Neglecting the change of resistance with the temperature, the time in which wire will start melting is given by :

(1) $(555) \mathrm{s}$

(2) $\quad(55.5) \mathrm{s}$

(3) $\quad(5.55) \mathrm{s}$

(4) $\quad(0.55) \mathrm{s}$

Melting point of copper $=1050^{\circ} \mathrm{C}$

Spefic resistance of copper $=1.6 \times 10^{-8} \Omega$

Density of copper $=9 \times 10^{3} \mathrm{Kg} / \mathrm{m}^{3}$

Specific heat capacity of copper $=9 \times 10^{-2} \mathrm{Cal} / \mathrm{kg}^{0} \mathrm{C}, 1 \mathrm{Cal}=4.2 \mathrm{~J}$

Show Answer

Correct Answer: (2)

Solution:

We can easily see that $\mathrm{I}^{2} \mathrm{Rt}=\mathrm{ms}(\Delta \theta)$

or $\quad I^{2}\left(\rho _{1} \frac{\ell}{\mathrm{A}}\right) \mathrm{t}=\mathrm{A} \ell \mathrm{d}(\mathrm{s} \Delta \theta)$

In above relation $\rho _{1}$ is the specific resistance and d the density of the copper.

$\therefore \mathrm{t}=\frac{\mathrm{A}^{2} \mathrm{~d}(\mathrm{~s} \Delta \theta)}{\mathrm{I}^{2} \rho _{1}}$

Substituting the values we get

$$ \begin{aligned} & \mathrm{t}= \frac{\left(0.5 \times 10^{-6}\right)^{2}\left(9 \times 10^{3}\right)\left(9 \times 10^{-2} \times 4.18 \times 1050\right)}{(10)^{2}\left(1.6 \times 10^{-8}\right)} \\ &=55.5 \mathrm{~S} \end{aligned} $$

Hence option (2) is correct.

Average

Heating Effect of Currents

38. Two wires, made of tinned copper, having identical cross section and different lengths, are used as fuses. Then

(1) Two wires fuses will melt at different values of the current.

(2) Two wire fuses will melt at the same value of current

(3) Two wire fuses will only melt at the same value of the current only if they we of equal lenghts

(4) Two wires fuses will only melt at the same value of current if they have same resistivity

Show Answer

Correct Answer: (2)

Solution:

If a current I flows into a fuse wire, the rate of loss of heat from the fuse wire is given by the relation.

$ \mathrm{I}^{2}=2 \pi^{2} \frac{\mathrm{Qr}{ }^{3}}{\rho} $

or $\quad \mathrm{Q}=\frac{\mathrm{I}^{2} \rho}{2 \pi^{2} \mathrm{r}^{3}}$

or $\quad Q \propto \frac{\rho}{r^{3}}$ for the same value of the current.

Since the area of cross section of bulb wires is identical, so the radius of both the wires will be same. Since $\mathrm{Q}$ is independent of $\ell$, both the wires of different lengths will melt at the same value of the current. Hence option (2) is correct.

Average

Heating Effect of Currents

39. A fuse, with a radius of $0.1 \mathrm{~mm}$ blows at $10 \mathrm{~A}$. What should be the radius of fuse wire made of the same material which melts at $20 \mathrm{~A}$ ?

(1) $0.5 \mathrm{~mm} \quad$

(2) $0.05 \mathrm{~mm} \quad$

(3) $0.159 \mathrm{~mm} \quad$

(4) $0.318 \mathrm{~mm} \quad$

Show Answer

Correct Answer: (3)

Solution:

The relation between the heat produced and current in a fuse wire of radius $r$ and having same material is given by

$ \mathrm{I}^{2}=\frac{2 \pi^{2} \mathrm{Hr}^{3}}{\rho} $

Hence we have,

$(10)^{2}=\frac{2 \pi^{2} \mathrm{H}(0.1)^{3}}{\rho}$

$(20)^{2}=\frac{2 \pi^{2} \mathrm{H} x^{3}}{\rho}$

$\frac{(20)^{2}}{(10)}=\frac{x^{3}}{(0.1)^{3}}$

Solving for $x$ we get

$$ x=0.159 \mathrm{~mm} $$

Hence option (3) is correct.

Average

Heating Effect of Currents

40. Three equal resistance each of value $x \mathrm{ohm}$ are connected as shown in figure. A battery of e.m.f. $E$ and internal resistance $r$ is connected across the circuit. To generate maximum heat in the circuit, the value of $x$ should be

$~$

(1) $3 \mathrm{R} _{\mathrm{e}}\left(\mathrm{R} _{\mathrm{e}}\right.$ being the equivalent resistance of resistances $x$ )

(2) $3 \mathrm{r}$

(3) $\frac{3}{x} \mathrm{r}^{2}$

(4) $\frac{3 x^{2}}{\mathrm{r}}$

Show Answer

Correct Answer: (2)

Solution:

If $\mathrm{H}$ is the heat produced, then

$\mathrm{H}=\mathrm{I}^{2} \mathrm{R} _{e} \mathrm{t}$ where $\mathrm{I}$ is the current in crcuit and $\mathrm{R} _{\mathrm{e}}$ is the equivalent resistance.

Now, $I=\frac{E}{\left(R _{e}+r\right)}$ where $r$ is the internal resistance of the cell.

Also, $\frac{1}{\mathrm{R} _{\mathrm{e}}}=\frac{1}{x}+\frac{1}{x}+\frac{1}{x}=\frac{3}{x}$

$\therefore \mathrm{R} _{\mathrm{e}}=\mathrm{x} / \mathrm{x}$

Hence, $H=\frac{E^{2}}{\left(R _{e}+r\right)}\left(R _{e} t\right)$

For maximum heat to be produced $\frac{\mathrm{dH}}{\mathrm{d} x}$ should be zero.

$\therefore \frac{E^{2} t}{\left(r+R _{e}\right)^{2}}-\frac{3 E^{2} R _{e} t}{\left(r+R _{e}\right)^{3}}=0$

or $\quad I=\frac{2 R _{e}}{r+R _{e}} \quad$ or $\quad r=R _{e}$

But, $\mathrm{R} _{\mathrm{e}}=(x / 3)$

$\therefore(x / 3)$ or $\quad x=3 \mathrm{r}$

Hence option (2) is correct.

Average

Resistance Variation with Length, Area & Nature of Material

41. Following figure showns cross section through three long conductor of same length and material, with circular cross sections as shown in figure. Conductor B will fit exactly within conductor $A$, and conductor $C$ will fit exactly withing conductor $B$, relationship between their end to end resistance is

$~$

(1) $R _{A}=R _{B}=R _{C}$

(2) $R _{A}>R _{B}>R _{C}$

(3) $\mathrm{R} _{\mathrm{A}}<\mathrm{R} _{\mathrm{B}}<\mathrm{R} _{\mathrm{C}}$

(4) Given data not sufficient

Show Answer

Correct Answer: (1)

Solution:

As the length of all conductors is equal and ’ $\rho$ ’ reistivity is also same as they are made up of same material.

Now, area of cross section of conductor $\mathrm{A}=\pi\left[(\sqrt{3 \mathrm{a}})^{2}-(\sqrt{2 \mathrm{a}})^{2}\right]=\pi \mathrm{a}^{2}$

and of conductor $\mathrm{B}=\pi\left[(\sqrt{2 \mathrm{a}})^{2}-(\sqrt{\mathrm{a}})^{2}\right]=\pi \mathrm{a}^{2}$

Area of cross section of conductor $\mathrm{C}=\pi \mathrm{a}^{2}$

$\therefore$ Areas of cross section of the conductors $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are also equal.

As resistance of a conductor is given by $R=\rho \frac{\ell}{A}$

$\therefore \mathrm{R} _{\mathrm{A}}=\mathrm{R} _{\mathrm{B}}=\mathrm{R} _{\mathrm{C}}$.

Hence answer (1) is correct.

Difficult

Grouping of Resistances

42. The equivalent resistance of the network between points $A$ and $B$ in the given figure.

$~$

(1) $\frac{14}{9} \Omega$

(2) $\frac{21}{5} \Omega$

(3) $\frac{18}{7} \Omega$

(4) $\frac{11}{2} \Omega$

Show Answer

Correct Answer: (1)

Solution:

Current distribution at point $\mathrm{A}$ and $\mathrm{B}$ will be similar

Therefore, circuit can be redrawn like as shown

Applying $\mathrm{KCl}$ at junection Q.

$\mathrm{i} _{1}+\mathrm{i}^{\prime}-\mathrm{i} _{2}=0 \quad \mathrm{i}^{\prime}=\mathrm{i} _{2}-\mathrm{i} _{1}$

$\mathrm{V} _{\mathrm{P}}-\mathrm{V} _{\mathrm{R}}=6 \mathrm{i} _{1}+3 \mathrm{i} _{2}$

$\mathrm{V} _{\mathrm{P}}-\mathrm{V} _{\mathrm{R}}=6 \mathrm{i} _{1}-3\left(\mathrm{i} _{2}-\mathrm{i} _{1}\right)+6 \mathrm{i} _{1}=15 \mathrm{i} _{1}-3 \mathrm{i} _{2}$

$\Rightarrow 6 \mathrm{i} _{1}+3 \mathrm{i} _{2}=15 \mathrm{i} _{1}-3 \mathrm{i} _{2}$

$6 i _{2}=9 i _{1} \Rightarrow i _{2}=(3 / 2) i _{1}$

But, $i=i _{1}+i _{2}=i _{1}+\frac{3}{2} i _{1}=(5 / 2) i _{1}$

$\mathrm{i} _{1}=\frac{2}{5} \mathrm{i} \quad \quad \mathrm{i} _{2}=\frac{3}{2} \times \frac{2}{5} \mathrm{i}=\frac{3}{5} \mathrm{i}$

$\mathrm{V} _{\mathrm{P}}-\mathrm{V} _{\mathrm{Q}}=\mathrm{i} \times \mathrm{R}^{\prime}$

$6 \mathrm{i} _{1}+3 \mathrm{i} _{2}=\mathrm{iR}^{\prime}$

$6 \times \frac{2}{5} \mathrm{i}+3 \times \frac{3}{5} \mathrm{i}=\mathrm{iR}^{\prime}$

$~$

$\frac{12 \mathrm{i}}{5}+\frac{9}{5} \mathrm{i}=\mathrm{iR}^{\prime}$ (Where $\mathrm{R}^{\prime}$ is the equivalent resistance of Wheatstone bridge)

$\frac{21}{5} \mathrm{i}=\mathrm{iR}^{\prime}$

$\mathrm{R}^{\prime}=\frac{21}{5} \Omega$

Hence effective resistance between ponits A& B

$\frac{1}{\mathrm{R} _{\text {eq }}}=\frac{1}{\mathrm{R}^{\prime}}+\frac{1}{6}=\frac{1}{\mathrm{R} _{\text {eq }}}=\frac{5}{21}+\frac{1}{6}=\frac{30+21}{21 \times 6}=\frac{51}{21 \times 6}=\frac{27}{21 \times 2}=\frac{9}{14} \Omega$

$$ \mathrm{R} _{\mathrm{eq}}=\frac{14}{9} \Omega $$

Alternative (Short cut)

$\mathrm{R}^{\prime}=$ equivalent of Wheatstone bridge

$\mathrm{R}^{\prime}=\frac{(\mathrm{P}+\mathrm{Q}) \mathrm{RS}+(\mathrm{R}+\mathrm{S}) \mathrm{PQ}+\mathrm{G}(\mathrm{P}+\mathrm{Q})(\mathrm{R}+\mathrm{S})}{\mathrm{G}(\mathrm{P}+\mathrm{Q}+\mathrm{R}+\mathrm{S})+\mathrm{G}(\mathrm{P}+\mathrm{R})(\mathrm{Q}+\mathrm{S})}$

Here, $\mathrm{P}=6 \Omega, \mathrm{Q}=3 \Omega, \mathrm{R}=3 \Omega, \mathrm{S}=6 \Omega, \mathrm{G}=3 \Omega$

$\mathrm{R}^{\prime}=\frac{18 \times 9+9 \times 18+3 \times 9 \times 9}{3(18)+9 \times 9}=\frac{12}{5} \Omega$

Hence effective resistance between point A&B

$$ \frac{1}{\mathrm{R} _{\mathrm{eq}}}=\frac{5}{12}+\frac{\mathrm{i}}{6}=\frac{14}{9} \Omega $$

Hence option(1) is correct

Difficult

Ohm’s Law and Current Density

43. The given figure shows a conductor of length ’ $\ell$ ’ and cross sectional area ’ $A$ ‘, its conductively varies as $\sigma(x)=\frac{\sigma _{0} \ell}{\sqrt{x}}$ where ’ $x$ ’ is the distance along the axis of the cylinder from one end of it. The electric field at each point in the cylinder will be; when is the potential difference across its ends is $V _{0}$.

(1) $\frac{2 \sqrt{\ell} V _{0}}{3 A \sigma _{0}}$

(2) $\frac{3 \mathrm{~V} _{0} \sqrt{x}}{2 \ell 3 / 2}$

(3) $\frac{2}{3} \frac{V _{0} \sqrt{x}}{\ell 3 / 2}$

(4) $\frac{3}{2} \mathrm{~V} _{0} \sqrt{\frac{x}{\ell}}$

$~$

Show Answer

Correct Answer: (2)

Solution:

$\mathrm{dR}=\frac{1}{\sigma(x)} \frac{\mathrm{d} x}{\mathrm{~A}}=\frac{\sqrt{x} \mathrm{~d} x}{\sigma _{0} \ell \mathrm{A}}$

$~$

$\mathrm{R}=\int _{0}^{\ell} \frac{\sqrt{x} \mathrm{~d} x}{\sigma _{0} \ell \mathrm{A}}=\frac{1}{\mathrm{~A} \sigma _{0} \ell} \cdot \frac{2}{3}(\ell)^{3 / 2}=\frac{2 \sqrt{\ell}}{3 \mathrm{~A} \sigma _{0}}$

Current flowing $\mathrm{i}=\frac{\mathrm{V} _{0}}{\mathrm{R}}=\frac{3 A \sigma _{0} \mathrm{~V} _{0}}{2 \sqrt{\ell}}$

Current density $\mathrm{j}=\frac{\mathrm{i}}{\mathrm{A}}=\frac{3 \sigma _{0} \mathrm{~V} _{0}}{2 \sqrt{\ell}}$

Electric field $(\mathrm{E})=\frac{\mathrm{j}}{\sigma}=\frac{2 \mathrm{~V} _{0} \sqrt{x}}{2 \ell 3 / 2}$

Hence option (2) is correct.

Average

Discharging of Cell

44. Given graph shows the variation of power dissipated in external resistnace (Load) $R$, delivered by a cell of emf ’ $E$ ‘, internal resistance ’ $r$ ‘, when supplying current ’ $i$ ’ to the load. The value of load corresponding to point $A$ in the given graph is

(1) $\mathrm{r}$

(2) $2 \mathrm{r}$

(3) $\frac{r}{2}$

(4) $\sqrt{\mathrm{r}}$

Correct answer: (1)

$~$

Show Answer

Solution:

Current given by the cell of the resistance $r=\frac{E}{R+r}$

Power dissipated in external resistance (load) $P=i^{2} R=\frac{E^{2}}{(R+r)^{2}} R$

Power will be maximum when $\frac{\mathrm{dP}}{\mathrm{dR}}=0$

$\frac{d P}{d R}=\frac{d}{d R}\left[E^{2} R(R+r)^{-2}\right]$

$=E^{2}\left[R \times(R+r)^{-3} \times(-2)+(R+r)^{-2} \times 1\right]=E^{2}\left[\frac{-2 R}{(R+r)^{3}}+\frac{1}{(R+r)^{2}}\right]$

For $\frac{\mathrm{dP}}{\mathrm{dR}}=0$,

$\frac{2 \mathrm{R}}{(\mathrm{R}+\mathrm{r})^{3}}=\frac{1}{(\mathrm{R}+\mathrm{r})^{2}} \Rightarrow \mathrm{R}=\mathrm{r}$

Hence option(1) is correct

Difficult

Potentiometer

45. In the given figure $A B$ is a potentiometer wire of length $600 \mathrm{~cm}$, the current flowing through the galvanometer, if jockey is touched at $560 \mathrm{~cm}$ lengths of wire from end $A$.

(1) $\frac{4 \mathrm{E}}{22 \mathrm{r}}$

(2) $\frac{3 \mathrm{E}}{22 \mathrm{r}}$

(3) $\frac{7 \mathrm{E}}{22} \mathrm{r}$

(4) $\frac{E}{22} \mathrm{r}$

$~$

Show Answer

Correct Answer: (2)

Solution:

Resistance of potentiometer wire $=15 \mathrm{r}$

Resistance of the length $560 \mathrm{~cm}$

$$ =\frac{15 \mathrm{r}}{600} \times 560=14 \mathrm{r} $$

Applying KVL in loop APMA

$-14 r i _{1}+\left(i-i _{1}\right) r+\frac{E}{2}=0$

$-15 r i _{1}+i r+\frac{E}{2}=0$

$\Rightarrow 15 \mathrm{i} _{1} \mathrm{r}-\mathrm{ir}=\frac{\mathrm{E}}{2} \quad \qquad \ldots . \ldots (1) $

$~$

Applying KVL to loop AEBA

$-\mathrm{E}+\mathrm{ir}+\mathrm{ir}+\mathrm{i} _{1} 14 \mathrm{r}=0$

$14 i _{1} r+2 i r=E$

$7 \mathrm{i} _{1} \mathrm{r}+\mathrm{ir}=\frac{\mathrm{E}}{2} \quad \qquad \ldots . \ldots (2)$

Solving (1) & (2)

$\mathrm{i} _{1}=\frac{\mathrm{E}}{22 \mathrm{r}}$

Putting the value of $\mathrm{i} _{1}$ in equations (2)

$\frac{7 \mathrm{E}}{22}+\mathrm{ir}=\frac{\mathrm{E}}{2}$

ir $=\frac{E}{2}-\frac{7 \mathrm{E}}{22}=\frac{4 \mathrm{E}}{22}$

$\mathrm{i}=\frac{4 \mathrm{E}}{22} \mathrm{r}$

Hence current through galvanometer $i-i _{1}=\frac{4 E}{22 r}-\frac{E}{22 r}=\frac{3 E}{22 r}$

Hence option (2) is correct.

Difficult

Meter Bridge

46. In the given experimental setup of meter bridge the length of the wire is one meter but its radius is increasing linearly from end $A$ to end $B$ which becomes double at end $B$ with respect to end $A$. The null point will be obtained at?

$~$

(1) $66.67 \mathrm{~cm}$

(2) $20 \mathrm{~cm}$

(3) $33.33 \mathrm{~cm}$

(4) $80 \mathrm{~cm}$

Show Answer

Correct Answer: (2)

Solution:

Let ’ $d x$ ’ be a small element of length at distance $x$ from end $A$, and radius of the wire at that position will be $=\left(\mathrm{r}+\frac{\mathrm{r}}{1} x\right)=\mathrm{r}(1+x)$ where $\mathrm{r}$ is the radius at end $\mathrm{A}$.

‘dR’ = Resistance of this element $\mathrm{dR}=\frac{\rho \mathrm{d} x}{\pi \mathrm{r}^{2}(1+x)^{2}}$

$\mathrm{R} _{1}=\int _{0}^{x} \mathrm{dR}=\int _{0}^{x} \frac{\rho \mathrm{d} x}{\pi \mathrm{r}^{2}(1+x)^{2}}=\frac{\rho}{\pi \mathrm{r}^{2}} \int _{0}^{x}(1+x)^{-2} \mathrm{~d} x$

$\mathrm{R} _{1}=\frac{\rho}{\pi \mathrm{r}^{2}}\left(1-\frac{1}{1+x}\right)=\frac{\rho x}{\pi \mathrm{r}^{2}(1+x)}$

Similarly resistance of the remaining wire i.e. from $\mathrm{J}$ to B.

$$ \begin{gathered} \mathrm{R} _{2}=\frac{\rho}{\pi \mathrm{r}^{2}} \int _{x}^{1}(1+x)^{-2} \mathrm{~d} x=\frac{\rho}{\pi \mathrm{r}^{2}}\left(-\frac{1}{2}+\frac{1}{1+x}\right) \\ \mathrm{R} _{2}=\frac{\rho(1-x)}{2 \pi \mathrm{r}^{2}(1+x)} \end{gathered} $$

As the bridge is balanced at distance $x$ from end A. Hence,

$\frac{\mathrm{R} _{1}}{\mathrm{R} _{2}}=\frac{8}{16}$ or $\frac{2 x}{(1-x)}=\frac{1}{2}$

$\Rightarrow 4 x=1-x$ or $5 x=1$

$$ x=\frac{1}{5} \mathrm{~m}=20 \mathrm{~cm} $$

Hence option (2) is correct.

Difficult

Grouping of Resistors

47. In the given circuit, each resistance is of $12 \Omega$, the effective resistance between points $A$ and $B$ will be

(1) $7 \Omega$

(2) $9 \Omega$

(3) $12 \Omega$

(4) $10 \Omega$

Correct answer: (1)

$~$

Show Answer

Solution:

As the paths $\mathrm{AD}$ and $\mathrm{AH}$ are equivalent, hence they carry equal currents i.e. $\mathrm{I} _{1}$, therefor current through $\mathrm{AB}$ is $\mathrm{I}-2 \mathrm{I} _{1}$

The potential difference $\mathrm{V}$ between points $\mathrm{A}$ and $\mathrm{B}$ may be written from the path $\mathrm{AB}$, AHEB, AHGFCB.

$\mathrm{V}=\left(\mathrm{I}-2 \mathrm{I} _{1}\right) 12$

$$ \begin{aligned} \mathrm{V} & =12 \mathrm{I} _{1}+12 \mathrm{I} _{2}+12 \mathrm{I} _{1}=\left(2 \mathrm{I} _{1}+\mathrm{I} _{2}\right) 12 \\ \mathrm{V} & = 12 \mathrm{I} _{1}+12\left(\mathrm{I} _{1}-\mathrm{I} _{2}\right)+24\left(\mathrm{I} _{1}-\mathrm{I} _{2}\right)+12\left(\mathrm{I} _{1}-\mathrm{I} _{2}\right)+12 \mathrm{I} _{1} \\ & =\left(6 \mathrm{I} _{1}-4 \mathrm{I} _{2}\right) 12 \end{aligned} $$

Hence on elimination of $\mathrm{I} _{1}$ and $\mathrm{I} _{2}$

$\mathrm{V}=7 \Omega$

Hence option (1) is correct.

$~$

Difficult

Combination of Resistances

48. The equivalent resistance between points $a$ and $c$ in the given figure will be when each resistance is of $8 \Omega$

(1) $\frac{20}{3} \Omega$

(2) $8 \Omega$

(3) $6 \Omega$

(4) $32 \Omega$

Correct answer: (3)

$~$

Show Answer

Solution:

Current distribution in the given figure will be as follows, (due to symmetry at points a and c).

$~$

$\frac{1}{\mathrm{R} _{\mathrm{ac}}}=\frac{1}{8}+\frac{1}{24}=\frac{3+1}{24}=\frac{4}{24} \Rightarrow \mathrm{R} _{\mathrm{ac}}=6 \Omega$

Hence option (3) is correct.

Difficult

Power Delivered to a Resistor (Max. Power Transfer Theorem)

49. In the given figure, $R$ is a variable resistor and its value is $1 \Omega$ at the time $t=0$, i.e. when switch is closed. It is increasing continuously by an automatic regulator at the rate of $4.0 \times 10^{-2} \mathrm{r} / \mathrm{s}$ and regulator is switched of when maximum power is delivered (dissipated) in the resistor $(R)$, the time for which regulator was on

(1) 500 second

(2) 400 second

(3) 100 second

(4) 150 second

Correct answer: (3)

$~$

Show Answer

Solution:

In loop 1: $\quad-7 \mathrm{I} _{1}-\mathrm{R}\left(\mathrm{I} _{1}-\mathrm{I} _{2}\right)+5-3 \mathrm{I} _{1}+10=0$

$$ \begin{equation*} \mathrm{I} _{1}(\mathrm{R}+10)=15+\mathrm{RI} _{2} \quad \qquad \ldots . \ldots {(i)} \end{equation*} $$

In loop 2: $\quad-5-4 \mathrm{I} _{2}-6 \mathrm{I} _{2}+\mathrm{R}\left(\mathrm{I} _{1}-\mathrm{I} _{2}\right)=0$

$$ \begin{align*} & I _{2}(10+R)=I _{1} R-5 \\ & \Rightarrow I _{1}=\frac{(10+R) I _{2}+5}{R} \quad \qquad \ldots . \ldots {(ii)}\\ & I _{2}=\frac{I _{1} R-5}{10+R} \end{align*} $$

Putting the value of $\mathrm{I} _{2}$ in equation(i)

$$ \begin{aligned} & I _{1}(R+10)=15+R\left(\frac{I _{1} R-5}{10+R}\right) \Rightarrow I _{1}=\frac{R+15}{2 R+10} \\ & \therefore I _{2}=\frac{\left[\frac{R+15}{2 R+10}\right] \cdot R-5}{10+R}=\frac{R-5}{2 R+10} \end{aligned} $$

$\therefore$ Current flowing through variable resistor $\mathrm{R},=\mathrm{I} _{1}-\mathrm{I} _{2}$

$$ =\frac{R+15}{2 R+10}-\frac{(R-5)}{(2 R+10)}=\frac{10}{R+5} $$

Hence power dissipated $\mathrm{P}=\left(\mathrm{I} _{1}-\mathrm{I} _{2}\right)^{2} \mathrm{R}=\frac{100 \mathrm{R}}{(\mathrm{R}+5)^{2}}$

For power to be maximum $\frac{\mathrm{dP}}{\mathrm{dR}}=0$

$=100\left[-2 \mathrm{R} \times(\mathrm{R}+5)^{-3}+(\mathrm{R}+5)^{-2} \times 1\right]$

$=100\left[-\frac{2 \mathrm{R}}{(\mathrm{R}+5)^{3}}+\frac{1}{(\mathrm{R}+5)^{2}}\right]$

$\Rightarrow \frac{2 \mathrm{R}}{(\mathrm{R}+5)^{3}}=\frac{1}{(\mathrm{R}+5)^{2}} \Rightarrow 2 \mathrm{R}=\mathrm{R}+5 \Rightarrow \mathrm{R}=5 \Omega$

Since rate of increase of resistance is $4.0 \times 10^{-2} \Omega / \mathrm{s}$.

$\therefore$ Time required to attain value $(5 \Omega)$ from $1(\Omega)$ will be 100 seconds.

i.e. $\quad R _{t}=R _{\text {initial }}+\alpha($ time $)$

$$ \begin{aligned} & 5=1+4 \times 10^{-2} \times t \\ & \frac{4.0}{4.0 \times 10^{-2}}=\mathrm{t} \Rightarrow 100 \text { second } \end{aligned} $$

Hence optino (3) is correct.

Difficult

Heating Effect

50. In the given figure, the charge supplied by the source varies as a function of time as $Q=3 t-5 t^{2}$, total heat produced in a resistance of $4 \Omega$ will be

$~$

(1) $2.025 \mathrm{~J}$

(2) $1.350 \mathrm{~J}$

(3) $4.050 \mathrm{~J}$

(4) $0.750 \mathrm{~J}$

Show Answer

Correct Answer: (1)

Solution:

Current $\mathrm{i}=\frac{\mathrm{dQ}}{\mathrm{dt}}=3-10 \mathrm{t}$

Hence current in $4 \Omega$

$\mathrm{i} _{2}=\frac{\mathrm{i} \times 12}{12+4}=\frac{12 \mathrm{i}}{16}=\frac{3 \mathrm{i}}{4}$

$\mathrm{i} _{2}=\frac{3}{4}(3-10 \mathrm{t})$

$~$

Charge flowing from the source will be zero at $\mathrm{t}=0$ and will be maximum when $\frac{\mathrm{dQ}}{\mathrm{dt}}=0$ i.e. at $\mathrm{t}=\frac{3}{10}$.

Hence total heat produced in $4 \Omega$ resistor.

$$ \begin{aligned} & \mathrm{H}=\int 1^{2} \mathrm{Rdt}=\int _{0}^{3 / 10}\left[\frac{3}{4}(3-10 \mathrm{t})\right]^{2} 4 \mathrm{dt} \\ & \mathrm{H}=\frac{9}{4}\left[\int _{0}^{3 / 10} 9 \mathrm{dt}+\int _{0}^{3} 100 \mathrm{tdt}-\int _{0}^{3} 60+\mathrm{dt}\right]=\frac{9}{4}\left[9(\mathrm{t}) _{0}^{3 / 10}+100\left(\frac{\mathrm{t}^{3}}{3}\right) _{0}^{3 / 10}-60\left(\frac{\mathrm{t}^{2}}{2}\right) _{0}^{3 / 10}\right] \\ & =\frac{9}{4}\left[9\left(\frac{3}{10}-0\right)+\frac{100}{3}\left(\frac{27}{1000}-0\right)-\frac{60}{2}\left(\frac{9}{100}-0\right)\right] \\ & \quad=\frac{9}{4}\left[\frac{27}{10}+\frac{9}{10}-\frac{27}{10}\right]=\frac{81}{40} \mathrm{~J}=2.025 \mathrm{~J} \end{aligned} $$

Hence option (1) is correct.

Difficult

Current Density

51. The current density in a cylindrical wire of radius $R=3 \mathrm{~cm}$ varies with radial distance ’ $r$ ’ as $\mathrm{mr}^{3}$, where ’ $\mathrm{m}$ ’ is a constant having value $2.0 \times 10^{7} \mathrm{amp} / \mathrm{m}^{5}$ and distance ’ $r$ ’ is in meter, the current flowing through the outer portion of the wire between $R / 2$ and $R$ is

(1) $\quad 0.591 \mathrm{~A}$

(2) $\quad 0.047 \mathrm{~A}$

(3) $1.52 \mathrm{~A}$

(4) $1.47 \mathrm{~A}$

$~$

Show Answer

Correct Answer: (1)

Solution:

Area dA $=2 \pi \mathrm{rdr}$

$I($ current $)=\oint \mathbf{j} \cdot \mathbf{d A}$

$\oint \mathbf{j} \cdot \mathbf{d A}=\mathrm{jdA} \cos \theta=\mathrm{jdA} \quad($ as current flows normal to area $)$

$~$

Hence, $I=\oint j d A=\int _{R / 2}^{R} m^{3} 2 \pi d r$

$$ \begin{aligned} & =\mathrm{m} \times 2 \pi \int _{\mathrm{R} / 2}^{\mathrm{R}} \mathrm{r}^{4} \mathrm{dr}=2 \pi \mathrm{m}\left[\frac{\mathrm{r}^{5}}{5}\right] _{\mathrm{R} / 2}^{\mathrm{R}} \\ & =\frac{2 \pi \mathrm{m}}{5}\left[\mathrm{R}^{5}-\frac{\mathrm{R}^{5}}{32}\right]=\frac{2 \pi \mathrm{m}}{5} \times \frac{31 \mathrm{R}^{5}}{32} \\ & =\frac{2 \times 3.14 \times 2.0 \times 10^{7}}{5} \times \frac{31}{32} \times\left(3 \times 10^{-2}\right)^{5}=591.34 \times 10^{-3}=0.591 \mathrm{~A} \end{aligned} $$

Hence option (1) is correct.

Average

Current

52. The given graph shows the variation of current with time in a circuit for $\mathbf{0}$ second to $\mathbf{1 0}$ second. It increases upto 12A linearly in 5 second. Then decreases to zero in next five second. The charge flowing in the circuit between 2 second and 6 seconds will be

(1) 60 Coulomb

(2) 30 Coulomb

(3) 24 Coulomb

(4) 36 Coulomb

$~$

Show Answer

Correct Answer: (4)

Solution:

The rate of growth of current upto 5 seconds.

$$ =\frac{12-0}{5}=2.4 \mathrm{amp} / \mathrm{s} $$

$~$

$\therefore$ Current at the end of 2 second $=2.4 \times 2=4.8 \mathrm{amp}$

The rate of decay of current from $A$ to $B=-2.4 \mathrm{amp} / \mathrm{s}$

$\therefore$ Current at sixth second $=12-2.4=9.8$ amp.

Hence chrage between 2 second and 6 second

$$ \begin{aligned} & =\text { area of trapezium PQAR }+ \text { Area of trapezium RAST } \\ & =\frac{1}{2}(4.8+12) \times 3+\frac{1}{2}(12+9.6) \times 1 \\ & =(25.2+10.8)=36 \text { Coulomb } \end{aligned} $$

Hence option (4) is correct.

Difficult

Unbalanced Wheatstone Bridge

53. The equivalent resistance between points $A$ and $B$ will be

(1) $\frac{45}{14} \Omega$

(2) $\frac{115}{14} \Omega$

(3) $\frac{175}{71} \Omega$

(4) $2.5 \Omega$

$~$

Show Answer

Correct Answer: (3)

Solution:

$~$

$$ \begin{aligned} & \mathrm{R} _{\mathrm{eq}}=\frac{\mathrm{PQ}(\mathrm{R}+\mathrm{S})+(\mathrm{P}+\mathrm{Q}) \mathrm{RS}+\mathrm{G}(\mathrm{P}+\mathrm{Q})(\mathrm{R}+\mathrm{S})}{\mathrm{G}(\mathrm{P}+\mathrm{Q}+\mathrm{R}+\mathrm{S})+(\mathrm{P}+\mathrm{R})(\mathrm{Q}+\mathrm{S})} \\ & \therefore \mathrm{R} _{\mathrm{eq}}=\frac{2 \times 3(1+4)+(2+3)(1 \times 4)+5(2+3)(1+4)}{5(2+3+1+4)+(2+1)(3+4)} \\ & \quad=\frac{6 \times 5+5 \times 4+5 \times 5 \times 5}{5 \times 10+3 \times 7}=\frac{30+20+125}{50+21}=\frac{175}{71} \Omega \end{aligned} $$

Hence option (3) is correct.

Alternative : can be determined using Kirchoff’s laws.

Average

Current Distribution

54. In the network points $A B$ and $C$ are at potentials of $70 \mathrm{~V}, 0 \mathrm{~V}$ and $10 \mathrm{~V}$ respectively. Then

$~$

(1) Potential at point $\mathrm{D}$ is $40 \mathrm{~V}$

(2) The currents in the section $\mathrm{AD}, \mathrm{DB}$ and $\mathrm{DC}$ are in the ratio 1:2:3

(3) The network draws a totoal power of $360 \mathrm{~W}$.

(4) All the above

Show Answer

Correct Answer: (1)

Solution:

$\mathrm{V} _{\mathrm{A}}-\mathrm{V} _{\mathrm{D}}=70-\mathrm{V} _{\mathrm{D}}=10 \mathrm{i}$

$\mathrm{V} _{\mathrm{D}}-\mathrm{V} _{\mathrm{B}}=\mathrm{V} _{\mathrm{D}}-0=20 \mathrm{i} _{1}$

$\mathrm{V} _{\mathrm{D}}-\mathrm{V} _{\mathrm{C}}=\mathrm{V} _{\mathrm{D}}-10=30\left(\mathrm{i}-\mathrm{i} _{1}\right)$

$\Rightarrow 70-\mathrm{V} _{\mathrm{D}}=10 \mathrm{i}$ or $70-20 \mathrm{i} _{1}=10 \mathrm{i} _{\mathrm{i}}$

$70=10\left(2 \mathrm{i} _{1}+\mathrm{i}\right)$ or $70=20\left(\mathrm{i} _{1}+10 \mathrm{i}\right) \quad \qquad \ldots . \ldots (1)$

$20 \mathrm{i} _{1}-10=30\left(\mathrm{i}-\mathrm{i} _{1}\right)$

$10=-30 \mathrm{i}+50 \mathrm{i} _{1} \quad \qquad \quad \qquad \ldots . \ldots (2)$

From (1) & (2)

$\left(70=20 \mathrm{i} _{1}+10 \mathrm{i}\right) \times 3$, or $10=50 \mathrm{i} _{1}-30 \mathrm{i}$ $220=110 \mathrm{i} _{1}$ $\mathrm{i} _{1}=\frac{220}{110}=2 \mathrm{~A}$

Hence $V _{D}=20 i _{1}=20 \times 2=40 \mathrm{~V}$

Hence option (1) is correct

Aeverage

Combination of Resistances

55. The equivalent resistance between point $A \& B$ in the given circuit will be

(1) $\frac{8}{7} \mathrm{r}$

(2) $\frac{10}{3} \mathrm{r}$

(3) $\frac{4}{5} \mathrm{r}$

(4) $\frac{22}{3} \mathrm{r}$

$~$

Show Answer

Correct Answer: (3)

Solution:

Due tosymmetry of the figure. It can be reduced.

$~$

$\frac{1}{\mathrm{R} _{\mathrm{eq}}}=\frac{3}{8 \mathrm{r}}+\frac{1}{2 \mathrm{r}}+\frac{3}{8 \mathrm{r}}$

$\frac{1}{\mathrm{R} _{\text {eq }}}=\frac{3+4+3}{8 \mathrm{r}}=\frac{10}{8 \mathrm{r}}$

$\Rightarrow \mathrm{R} _{\mathrm{eq}}=\frac{8 \mathrm{r}}{10}=\frac{4 \mathrm{r}}{5}$

Hence option (3) is correct

Average

Calculation of Resistance

56. A material of resistivity ’ $\rho$ ’ is filed in space between two concentric spheres of radius a and ’ $b$ ’ uniformly $(a<b)$, the resistance of this material will be:

(1) $\frac{\rho}{4 \pi(b+a)}$

(2) $\frac{\rho}{4 \pi}\left(\frac{1}{\mathrm{~b}^{2}}-\frac{1}{\mathrm{a}^{2}}\right)$

(3) $\frac{\rho}{4 \pi}\left(\frac{1}{b}-\frac{1}{a}\right)$

(4) $\frac{\rho}{4 \pi}\left(\frac{1}{\mathrm{a}}-\frac{1}{\mathrm{~b}}\right)$

Show Answer

Correct Answer: (4)

Solution:

Let the resistance of the small element $\mathrm{d} x$ of radius $x$ is $\mathrm{dR}$

$~$

$\mathrm{dR}=\frac{\rho \mathrm{d} x}{4 \pi x^{2}}$

$\therefore \mathrm{R}=\frac{\rho}{4 \pi} \int _{\mathrm{a}}^{\mathrm{b}} \frac{\mathrm{d} x}{x^{2}}=\frac{\rho}{4 \pi}\left[-\frac{1}{x}\right] _{\mathrm{a}}^{\mathrm{b}}$

$=\frac{\rho}{4 \pi}\left[\frac{1}{\mathrm{a}}-\frac{1}{\mathrm{~b}}\right]$

Hence option (4) is correct.

Average

Potentiometer

57. In an arrangement, for the determination of internal resistance of a battery using potentiometer, the null point is obatined at $300 \mathrm{~cm}$ of the wire when the experimental cell is shunted with resistance of $5 \Omega$, when another resistance of $20 \Omega$ is connected in parallel to this shunt, galvanometer shows no deflection at $270 \mathrm{~cm}$ of wire, the internal resistance of the cell and the balancing length for the cell without any shunt will be

(1) $4.0 \Omega$ and $540 \mathrm{~cm}$

(2) $0.55 \Omega$ and $300 \mathrm{~cm}$

(3) $0.45 \Omega$ and $270 \mathrm{~cm}$

(4) $7.78 \Omega$ and $435 \mathrm{~cm}$

Show Answer

Correct Answer: (1)

Solution:

$\mathrm{r}=\left(\frac{\ell}{300}-1\right) 5$, where $\ell$ being balancing length without shunting.

When a resistance of $20 \Omega$ is joined in parallel to $5 \Omega$ the effective shunt $=\frac{5 \times 20}{5+25}=4 \Omega$

$$ \begin{aligned} \therefore \mathrm{r} & =\left(\frac{\ell}{270}-1\right) 4 \\ & \Rightarrow\left(\frac{\ell}{300}-1\right) 5=\left(\frac{\ell}{270}-1\right) 4 \\ & \Rightarrow\left(\frac{\ell-300}{6}\right)=\left(\frac{\ell-270}{27}\right) 4=9(\ell-300)=8(\ell-270) \quad \ell=2700-2160 \text { or } \ell=540 \mathrm{~cm} \end{aligned} $$

Hence internal resistance $\mathrm{r}=\left(\frac{540}{300}-1\right) 5=\frac{240}{300} \times 5$

$ \mathrm{r}=4 \Omega $

Hence option (1) is correct.

Average

Heating Effect of Current

58. A nichrome Filament heater consumes $100 \mathrm{~W}$ power at $27^{\circ} \mathrm{C}$, when it is used its temperature rises to $3027^{\circ} \mathrm{C}$, the power consumed by it at the moment when switch is, put on, voltage remaining same. Given the temperature coefficient of resistively of nichrome is $\mathbf{4} \times 10^{-4} \mathrm{~K}^{-1}$.

(1) $90.9 \mathrm{~W}$

(2) $220 \mathrm{~W}$

(3) $1300 \mathrm{~W}$

(4) $7.96 \mathrm{~W}$

Show Answer

Solution:

$\mathrm{P} _{0}=\frac{\mathrm{V}^{2}}{\mathrm{R} _{0}}, \Rightarrow \mathrm{R} _{0}=\frac{\mathrm{V}^{2}}{\mathrm{P} _{0}}$

$P _{t}=\frac{V^{2}}{R _{t}} \Rightarrow R _{t}=\frac{V^{2}}{P _{t}}$

As, $\mathrm{R} _{\mathrm{t}}=\mathrm{R} _{0}(1+\alpha \Delta \mathrm{t})$

$\frac{\mathrm{V}^{2}}{\mathrm{P} _{\mathrm{t}}}=\frac{\mathrm{V}^{2}}{\mathrm{P} _{0}}(1+\alpha \Delta \mathrm{t})$

$\Rightarrow \mathrm{P} _{0}=\mathrm{P} _{\mathrm{t}}(1+\alpha \Delta \mathrm{t})$

$=100\left[1+(3027-27) \times 4 \times 10^{-4}\right]$

$=100\left[1+3000 \times 4 \times 10^{-4}\right]=100 \times 2.2=200 \mathrm{~W}$

Hence correct option is (2).

Average

Grouping of Cells

59. Two cells $A$ and $B$ of emf $1.5 \mathrm{~V}$ and $1.7 \mathrm{~V}$, internal resistances $r _{1}$ and $r _{2}$ respecting are arranged as shown in figure, the voltmeter reads $1.6 \mathrm{~V}$ (the voltmeter is ideal) the relationship between $r _{1}$ and $r _{2}$ will be.

$~$

(1) $\mathrm{r} _{1}=3 \mathrm{r} _{2}$

(2) $\mathrm{r} _{1}=-3 \mathrm{r} _{2}$

(3) $\mathrm{r} _{2}=2 \mathrm{r} _{2}$

(4) $\mathrm{r} _{2}=2 \mathrm{r} _{1}$

Show Answer

Correct Answer: (1)

Solution:

$1.65=1.5+\mathrm{I} _{1} \mathrm{r} _{1} \Rightarrow \mathrm{I} _{1} \mathrm{r} _{1}=1.65-1.5=0.15$

Also, $1.65=1.7-\mathrm{I} _{1} \mathrm{r} _{2} \Rightarrow \mathrm{I} _{1} \mathrm{r} _{2}=1.70-1.65=.05$

Ideal Voltmeter

Dividing $\frac{\mathrm{r} _{1}}{\mathrm{r} _{2}}=\frac{0.15}{.05}=3$

$$ \Rightarrow \mathrm{r} _{1}=3 \mathrm{r} _{1} $$

Hence option (1) is correct.

Average

Potentiometer

60. The given figure shows a potentiometer wire of length $400 \mathrm{~cm}$ resistance ’ $r$ ’ and uniform area of cross section, A driver cell of $5.0 \mathrm{~V}$ along with a variable resistance ’ $R$ ’ is connected across it, to supply the current. During an experiment the null point is obtained at the length $160 \mathrm{~cm}$, when $R=0$ and at $320 \mathrm{~cm}$ for certain value of $R$ from end $A$, the emf of the experimental cell $(E)$ and the value of variable resistance $R$ will be

(1) $2 \mathrm{~V}$ and $1.25 \mathrm{r}$

(2) $1.0 \mathrm{~V}$ and $3 \mathrm{r}$

(3) $2.0 \mathrm{~V}$ and $3.5 \mathrm{r}$

(4) $1.0 \mathrm{~V}$ and $2.5 \mathrm{r}$

$~$

Show Answer

Correct Answer: (1)

Solution:

From figure, when the value of resistance $\mathrm{R}=0$

then current in the potentiometer wire $\mathrm{i} _{1}=\frac{5.0}{\mathrm{r}} \mathrm{A}$

Hence potential drop across the wire $=5.0 \mathrm{~V}$

$\therefore$ Potential gradient $\mathrm{k}=\frac{5.0}{4.0}=\frac{5}{4}$ volt $/ \mathrm{m}$

Since cell of emf ’ $E$ ’ is balanced at $160 \mathrm{~cm}$ length of the wire, therefore

$\mathrm{E}=\mathrm{k} \ell=\frac{5}{4} \times 1.6=2.0 \mathrm{~V}$

Now current becomes $\mathrm{i} _{2}=\frac{5.0}{\mathrm{R}+\mathrm{r}}$ when resistance $\mathrm{R}$ has certain value. $\therefore$ Potential drop across the whole length of the wire $=\mathrm{i} _{2} \mathrm{r}=\frac{5 \mathrm{r}}{\mathrm{R}+\mathrm{r}}$

Now potential gradient $k^{\prime}=\frac{5 r}{4(R+r)}$ and null point is obatined at $360 \mathrm{~cm}$

$\therefore \mathrm{E}=\mathrm{k}^{\prime} \ell^{\prime} \Rightarrow 2.0=\frac{5 \mathrm{r}}{4(\mathrm{R}+\mathrm{r})} \times 3.6$

$2.0=\frac{4.5 \mathrm{r}}{\mathrm{R}+\mathrm{r}}$

2.0 or $+2.0 \mathrm{r}=4.5 \mathrm{r}$

$2.5 \mathrm{r}=2.0 \mathrm{R}$

$\mathrm{R}=\frac{2.5 \mathrm{r}}{2.0}=1.25 \mathrm{r}$

Hence option (1) is correct.

Average

Current Distribution

61. In the given figure, ideal voltmeter reads $30 \mathrm{~V}$ when connected across $15 \Omega$ resistance.

Then potential difference across $B$ and $C$ and the resistance $R$ is given by

$~$

(1) $15 \mathrm{~V}$ and $42.5 \Omega$ respectively

(2) $10 \mathrm{~V}$ and $40 \Omega$ respectively

(3) $-15 \mathrm{~V}$ and $42.5 \Omega$ respectively

(4) $-10 \mathrm{~V}$ and $70 \Omega$ respectively

Show Answer

Correct Answer: (1)

Solution:

The current flowing in branch BD is $\frac{30}{15}=2 \mathrm{~A}$

$\therefore$ Current in branch $\mathrm{BC}=3 \mathrm{~A}$

Hence potential difference between points $B \& C=3 \times 5=15 \mathrm{~V}$

Applying KVL to loop BDCB

$-15 \times 2-\mathrm{IR}+100+5 \times 3=0$

$\mathrm{IR}=-30+15+100$

$ 2 \times \mathrm{R}=115-30=85 \mathrm{~V} $

$\mathrm{R}=\frac{85}{2}=42.5 \Omega$

Hence option (1) is correct.

Average

Combination of Resistances

62. In the given figure, each resistance is of $6 \Omega$, the effective resistance between points $A$ and B will be

(1) $3.5 \Omega$

(2) $5 \Omega$

(3) $7.2 \Omega$

(4) $4.5 \Omega$

$~$

Show Answer

Correct Answer: (2)

Solution:

Let current I enters at point A, and divide into three branches equally i.e. $\frac{I}{3}$. Due to symmetry same current will leave at end B.

Hence $\mathrm{R} _{\mathrm{AB}}=5 \Omega$

$~$

Option (2) is correct.

Average

Resistance of conductors

63. Two conductos, one solid and other hollow are made up of same material, the length of solid conductor is twice the length of hollow conductor and diameter is $D$, while the outer diameter of hollow conductor is $2 D$ and inner is $D$, the ratio of resistance between the ends of two conductor will be

(1) $2: 3$

(2) $1: 6$

(3) $3: 2$

(4) $3: 1$

Show Answer

Correct Answer: (3)

Solution:

$\mathrm{R} _{1}=\rho \frac{2 \ell}{\pi \frac{\mathrm{D}^{2}}{4}}=\frac{8 \rho \ell}{\pi \mathrm{D}^{2}}$

$\mathrm{R} _{2}=\rho \frac{\ell}{\left(\frac{\pi \mathrm{D}^{2}}{4}-\frac{\pi \mathrm{D}^{2}}{16}\right)}=\frac{16 \rho \ell}{3 \pi \mathrm{D}^{2}}$

$\frac{\mathrm{R} _{1}}{\mathrm{R} _{2}}=\frac{\frac{8 \rho \ell}{\pi \mathrm{D}^{2}}}{\frac{16 \rho \ell}{3 \pi \mathrm{D}^{2}}}=\frac{8 \times 3}{16}=3: 2$

Hence option (3) is correct.

Average

Potentiometer

64. In the following circuit, the potential difference between the points $C$ and $D$ is balanced against $40 \mathrm{~cm}$ length of the potentiometer wire, in order to balance the potential difference between the points $B$ and $C$, where should jockey on the wire be pressed?

(1) $10 \mathrm{~cm}$

(2) $40 \mathrm{~cm}$

(3) $60 \mathrm{~cm}$

(4) $20 \mathrm{~cm}$

$~$

Show Answer

Correct Answer: (4)

Solution:

Current in loop AKBCDA

$$ \begin{gathered} -5.0+\mathrm{I} \times 0.5+\mathrm{I} \times 1.5+3 \mathrm{I}=0 \\ 5 \mathrm{I}=5 \quad \mathrm{I}=1 \mathrm{~A} \end{gathered} $$

Hence potential difference between points $\mathrm{D}$ and $\mathrm{C}=3 \times 1=3 \mathrm{~V}$

and between points $\mathrm{C}$ and $\mathrm{B}=1.5 \times 1=1.5 \mathrm{~V}$

According to given figure

$ \mathrm{V} _{\mathrm{DC}}=\mathrm{K} \times 40 \Rightarrow 3=\mathrm{K} \times 40 \Rightarrow \mathrm{K}=\frac{3}{40} \text { volt } / \mathrm{cm} $

Now, $\mathrm{V} _{\mathrm{CB}}=\mathrm{K} \times \ell \Rightarrow \ell=\frac{1.5 \times 40}{3}=20 \mathrm{~cm}$

Hence option (4) is correct.

Average

Current Distribution

65. In the given circuit, the battery of emf $2.0 \mathrm{~V}$ and internal resistance $1.0 \Omega$ is connected to an external resistance $50 \Omega$, the resistance of voltmeter and ammeter are $200 \Omega$ and $2.0 \Omega$ respectively. The readings of voltmeter and ammeter will be, when switch ’ $S$ ’ is connected to point $A$.

(1) $3.72 \mathrm{~V} ~~ \& ~~ 0.047 \mathrm{~A}$

(2) $1.86 \mathrm{~V} ~~ \& ~~ 0.093 \mathrm{~A}$

(3) $3.72 \mathrm{~V} ~~ \& ~~ 0.093 \mathrm{~A}$

(4) $1.86 \mathrm{~V} ~~ \& ~~ 0.047 \mathrm{~A}$

$~$

Show Answer

Correct Answer: (4)

Solution:

Net resistance of the combination.

$\mathrm{R}=\left(\frac{50 \times 200}{200+50}\right)+2 \Omega+1 \Omega=43 \Omega$

Hence current idrawn from the cell.

$\mathrm{i}=\frac{2}{43} \mathrm{~A}$

In loop PQRS

$-50 \mathrm{i} _{1}+200\left(\mathrm{i}-\mathrm{i} _{1}\right)=0$

$200 \mathrm{i}=250 \mathrm{i} _{1}$ or $4 \mathrm{i}=5 \mathrm{i} _{1}$

$\Rightarrow \mathrm{i} _{1}=\frac{4}{5} \times \mathrm{i}=\frac{4}{5} \times \frac{2}{43} \mathrm{~A}=\frac{8}{215} \mathrm{~A}$

$~$

Hence current through voltmeter $=i-i _{1}=\frac{2}{43}-\frac{4}{5} \times \frac{2}{43}=\frac{1}{5} \times \frac{4}{43}=\frac{4}{215} \mathrm{~A}$

Hence voltmeter reading $=200 \times \frac{4}{215}=3.72$

Ammeter readind $=\frac{2}{43}=0.047 \mathrm{~A}$

Easy

Calculation of Current

66. The electron in an hydrogen atom moves in a circular orbit of radius $5 \times 10^{-11} \mathrm{~m}$ with a speed of $0.6 \pi \times 10^{6} \mathrm{~m} / \mathrm{s}$, then

(1) The frequency of the electron is $3 \times 10^{15}$ revolutions

(2) The electron carries $3.2 \times 10^{-19} \mathrm{C}$ charge around the loop

(3) The current in the orbit is $0.96 \mathrm{~mA}$

(4) The current flows in the same direction in which electrons move

Show Answer

Correct Answer: (3)

Solution:

$I=\frac{q}{T}=\frac{q}{\left(\frac{2 \pi}{\omega}\right)}=\frac{q \omega}{2 \pi}$

Also, $\omega=\frac{v}{\mathrm{r}}$

$\Rightarrow \mathrm{I}=\frac{\mathrm{q} v}{2 \pi \mathrm{r}}=\frac{1.6 \times 10^{-19} \times 0.6 \times 1{ } _{0}^{6} \pi}{2 \times 5 \times 10^{-11}}=0.96 \mathrm{~mA}$

Hence option (3) is correct.

Average

Potential Differnece

67. In the circuit shown, the cell is ideal, with emf $10 \mathrm{~V}$, each resistance is of $2 \Omega$, the potential difference across the capacitor is

(1) $12 \mathrm{~V}$

(2) $10 \mathrm{~V}$

(3) $8 \mathrm{~V}$

(4) zero

$~$

Show Answer

Correct Answer: (3)

Solution:

No current will flow through the branch containing capacitor.

$\therefore \mathrm{R} _{\text {eff }}=\left(\frac{4 \times 2}{4+2}\right)+2=\frac{10}{3} \Omega$

Hence current drawn from the cell $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R} _{\text {eff }}}=\frac{10}{10 / 3}=3 \mathrm{~A}$

$\therefore$ Current in branch $\mathrm{ABC}=1 \mathrm{~A}$

Hence p.d. between points B & D

$=\left(\mathrm{V} _{\mathrm{B}}-\mathrm{V} _{\mathrm{C}}\right)+\left(\mathrm{V} _{\mathrm{C}}-\mathrm{V} _{\mathrm{D}}\right)$

$=2 \times 1+2 \times 3=2+6=8 \mathrm{~V}$

Hence option (3) is correct.

Easy (Knowledge Based)

Current Density

68. Two bars of radius $r$ and $2 r$ are kept in contact as shown in figure, an electric current $i$ is passed through bars, which one of the following statement is cuuret;

$~$

(1) Heat produced in bar $\mathrm{BC}$ is 4 time the heat produced in bar $\mathrm{AB}$

(2) Electric field in both bars have the same value

(3) Current density across $\mathrm{AB}$ is double that of across $\mathrm{BC}$

(4) Potential difference across $\mathrm{AB}$ is 4 time that of across $\mathrm{BC}$

Show Answer

Solution:

As the resistance of Bar $\mathrm{BC}$ will be four time of resistance of $\mathrm{Bar} \mathrm{AB}$.

(Area) $\mathrm{A} _{\mathrm{BC}}=\frac{1}{4} \mathrm{~A} _{\mathrm{AB}}$

$\mathrm{R} _{\mathrm{BC}}=4$ times of $\mathrm{R} _{\mathrm{AB}}$

Heat produced $\mathrm{H}=\mathrm{i}^{2} \mathrm{Rt}$

For same current heat produced will be four times.

Hence option (1) is correct.

Average

Effect on Resistance on Stretching

69. A wire of resistance $16 \Omega$ is cut into four equal parts, each part is stretched to the original length of the wire, then each streteched part first connected in series than in parallel, the ratio of resistance will be.

(1) $16: 1$

(2) $1: 16$

(3) $4: 1$

(4) $1: 4$

Show Answer

Correct Answer: (1)

Solution:

Resistance of each part on cutting $=4 \Omega$

After stretching resistance of each part will be 16 times $=64 \Omega$

Effective resistance in series $=64 \times 4=256 \Omega$

Effective resistance in parallel $\frac{64}{4}=16 \Omega$

$\therefore$ Ratio of resistances in series to parallel $=256: 16$

$ =16: 1 $

Hence option (1) is correct.

Average

Combination of Resistance and Division of Current in Parallel Combination

70. The current of 3A flows in the combination of resistance as shown in figure, the potential difference between points $A$ and $B$. Will be

(1) $+0.75 \mathrm{~V}$

(2) zero

(3) $8.25 \mathrm{~V}$

(4) $-0.75 \mathrm{~V}$

$~$

Show Answer

Correct Answer: (4)

Solution:

The given circuit can be reduced to the circuit as shown here

Current in $10 \Omega$ resistance

$\mathrm{I} _{1}=\frac{3 \times 6}{10+6}=\frac{18}{16}=\frac{9}{8} \mathrm{~A}$

Current in $6 \Omega$ resistance $\frac{3 \times 10}{10+6}=\frac{30}{16}=\frac{15}{8} \mathrm{~A}$

$~$

From given figure $V _{C}-V _{A}=\frac{9}{8} \times 4=\frac{9}{2} \mathrm{~V}=4.5 \mathrm{~V}$

$\mathrm{V} _{\mathrm{C}}-\mathrm{V} _{\mathrm{B}}=\frac{15}{8} \times 2=\frac{15}{4} \mathrm{~V}=3.75 \mathrm{~V}$

Hence, $\left(\mathrm{V} _{\mathrm{C}}-\mathrm{V} _{\mathrm{B}}\right)-\left(\mathrm{V} _{\mathrm{C}}-\mathrm{V} _{\mathrm{A}}\right)=3.75-4.5=-0.75 \mathrm{~V}$

Hence option(4) is correct

Average

Cell and Distribution of Current

71. In the given figure, the emf of the battery will be

$~$

(1) $12 \mathrm{~V}$

(2) $13 \mathrm{~V}$

(3) $16 \mathrm{~V}$

(4) $18 \mathrm{~V}$

Show Answer

Correct Answer: (2)

Solution:

Current entering at point $\mathrm{Q}$ is $2 \mathrm{~A}$ as it will divide equally.

Hence current through $\mathrm{PQ}=2 \mathrm{~A}$

and current in branch $\mathrm{PQRS}=2 \mathrm{~A}$

$\therefore$ Current in branch PS $=\frac{6}{4} \mathrm{~A}=1.5 \mathrm{~A}$

$\therefore$ Therefore current drawn from cell $=(2.0+1.5) \mathrm{A}=3.5 \mathrm{~A}$

Total resistance in the circuit $=2 \Omega+\frac{12}{7} \Omega=\frac{26}{7} \Omega$

Hence $\operatorname{emf} E=I R=3.5 \times \frac{26}{7}=13 \mathrm{~V}$

Hence option (2) is correct.

Average

Distribution of Current

72. In the given circuit diagram, the current flowing through branch ab and bd will be:

(1) $2.5 \mathrm{~A}$ and $10 \mathrm{~A}$

(2) $0 \mathrm{~A}$ and $10 \mathrm{~A}$

(3) $0 \mathrm{~A}$ and $2.5 \mathrm{~A}$

(4) $2.5 \mathrm{~A}$ and $0 \mathrm{~A}$

$~$

Show Answer

Correct Answer: (4)

Solution:

The resistance between the points $\mathrm{b}$ and $\mathrm{d}$ becomes ineffective in the circuit, because points $\mathrm{b}$ and $\mathrm{d}$ are equal potential points. hence current in branch bd $=0 \mathrm{~A}$

$~$

Current in branch $\mathrm{ab}=\frac{10}{4}=2.5 \mathrm{~A}$

Hence correct option is (4).

Average

Distribution of Current

73. In the given circuit, current through branch de and the ammeter reading will be

$~$

(1) $0 \mathrm{~A}$ and $\frac{9}{13} \mathrm{~A}$

(2) 0 A and $\frac{9}{40} \mathrm{~A}$

(3) 0 A and $1 \mathrm{~A}$

(4) $0.5 \mathrm{~A}$ and $0.5 \mathrm{~A}$

Show Answer

Correct Answer: (3)

Solution:

The effective resistance between points $a$ and $f$.

$~$

Hence current drawn from battery $\mathrm{i}=\frac{9}{9}=1 \mathrm{~A}$

Therefore current through branch $\mathrm{de}=0$ ( $\mathrm{d}$ and e being equal potential points, and ammeter reading $1 \mathrm{~A})$. Hence option (3) is correct.

Easy

Combination of Resistances

74. Two resistances $\mathbf{R} _{1}$ and $\mathbf{R} _{2}$ provides series to parallel equivalents as $\left(\frac{\mathrm{n}}{1}\right)$, the correct relationship between $\mathbf{R} _{1}, \mathbf{R} _{2}$ and is

(1) $\left(\frac{\mathrm{R} _{1}}{\mathrm{R} _{2}}\right)^{2}+\left(\frac{\mathrm{R} _{2}}{\mathrm{R} _{1}}\right)^{2}=\mathrm{n}^{2}$

(2) $\left(\frac{\mathrm{R} _{1}}{\mathrm{R} _{2}}\right)^{3 / 2}+\left(\frac{\mathrm{R} _{2}}{\mathrm{R} _{1}}\right)^{3 / 2}=\mathrm{n}^{3 / 2}$

(3) $\frac{\mathrm{R} _{1}}{\mathrm{R} _{2}}+\frac{\mathrm{R} _{2}}{\mathrm{R} _{1}}=\mathrm{n}$

(4) $\left(\frac{\mathrm{R} _{1}}{\mathrm{R} _{2}}\right)^{1 / 2}+\left(\frac{\mathrm{R} _{2}}{\mathrm{R} _{1}}\right)^{1 / 2}=\sqrt{\mathrm{n}}$

Show Answer

Correct Answer: (4)

Solution:

In series $R _{S}=R _{1}+R _{2}$

In parallel $R _{P}=\frac{R _{1} R _{2}}{R _{1}+R _{2}}$

Given, $\frac{\mathrm{R} _{\mathrm{S}}}{\mathrm{R} _{\mathrm{P}}}=\frac{\mathrm{n}}{1}$

$\Rightarrow \frac{\left(\mathrm{R} _{1}+\mathrm{R} _{2}\right)^{2}}{\mathrm{R} _{1} \mathrm{R} _{2}}=\frac{\mathrm{n}}{1} \Rightarrow \frac{\mathrm{R} _{1}+\mathrm{R} _{2}}{\sqrt{\mathrm{R} _{1} \times \mathrm{R} _{2}}}=\sqrt{\mathrm{n}}$ $\Rightarrow \frac{\sqrt{\mathrm{R} _{1}^{2}}}{\sqrt{\mathrm{R} _{1} \mathrm{R} _{2}}}+\frac{\sqrt{\mathrm{R} _{2}^{2}}}{\sqrt{\mathrm{R} _{1} \mathrm{R} _{2}}}=\sqrt{\mathrm{n}}$

$\Rightarrow \frac{\sqrt{\mathrm{R} _{1}}}{\sqrt{\mathrm{R} _{2}}}+\frac{\sqrt{\mathrm{R} _{2}}}{\sqrt{\mathrm{R} _{1}}}=\sqrt{\mathrm{n}}$

$\Rightarrow\left(\frac{\mathrm{R} _{1}}{\mathrm{R} _{2}}\right)^{1 / 2}+\left(\frac{\mathrm{R} _{2}}{\mathrm{R} _{1}}\right)^{1 / 2}=\sqrt{\mathrm{n}}$

Hence alternative (4) is correct.

Average

Resistance

75. A hollow cylinder of length ’ $\ell$ ’ and radii $a$ and $b$ is filled with a material of resistivity ’ $\rho$ ’ and is connected to a battery of emf ’ $E$ ’ through an ammeter, the reading of ammeter will be

(1) $\frac{\mathrm{E} 2 \pi \ell}{\rho \log _{\mathrm{e}}(\mathrm{b} / \mathrm{a})}$

(2) $\frac{\mathrm{E} 2 \pi \ell}{\rho \log _{e}(\mathrm{a} / \mathrm{b})}$

(3) infinite

(4) zero

$~$

Show Answer

Correct Answer: (1)

Solution:

Resistance of an imaginary cylinder of radius $x$, thickness $\mathrm{d} x, \mathrm{dR}=\frac{\rho \mathrm{d} x}{2 \pi x \ell}$

So the resistance of given cylinder $(\mathrm{R})=\int _{\mathrm{a}}^{\mathrm{b}} \frac{\rho \mathrm{d} x}{2 \pi x \ell}=\frac{\rho}{2 \pi \ell} \int _{\mathrm{a}}^{\mathrm{b}} \frac{1}{x} \mathrm{~d} x=\frac{\rho}{2 \pi \ell} \log _{\mathrm{e}}(\mathrm{b} / \mathrm{a})$

Hence reading of ammeter $I=\frac{E}{R}=\frac{2 \pi \ell E}{\rho \log _{e}(b / a)}$

Hence option (1) is correct.

Easy

Temperature Dependence of Resistance

76. A piece of silver and another of germanium cooled from room temperature to $85 \mathrm{~K}$. Then the resistance of

(1) Each of then increases

(2) Each of then decreases

(3) That of silver increases and germanium decreases

(4) That of silver decreases and that of germanium decreases

Show Answer

Correct Answer: (4)

Solution:

Silver is a metal and germanium is a semi conductor. Resistance of metal decrease and that of a semi conductor increases with decreases in temperature.

Hence option (4) is correct

Average

Using Ohm’s Law

77. In a discharge tube the number of hydrogen ions drifting across a cross section per second is $n _{1}$ while the number of electrons drifting in the opposite direction across another cross section is $n _{2}$. If the supply voltage is $V$, the effective resistance, of the tube is

(1) $\frac{V}{\left(n _{1}-n _{2}\right) e}$

(2) $\frac{V}{\left(n _{1}+n _{2}\right) e}$

(3) $\frac{\mathrm{eV}}{\mathrm{n} _{1}-\mathrm{n} _{2}}$

(4) $\frac{\mathrm{eV}}{\mathrm{n} _{1}+\mathrm{n} _{2}}$

Show Answer

Correct Answer: (2)

Solution:

The effective current in the tube will be obtained by adding the current due to flow of protons in one direction and the flow of electrons in opposite direction. Hence the total current through the tube is $\left(n _{1}+n _{e}\right)|e|$. Now from Ohm’s law, the resistance of the tube is given by

$$ \mathrm{R}=\frac{\mathrm{V}}{\left(\mathrm{n} _{1}+\mathrm{n} _{2}\right) \mathrm{e}} $$

Average

Driff Velocity

78. A silver wire has a radius of $0.1 \mathrm{~cm}$ and carries a current of $2 \mathrm{~A}$. The drift velocity of the electron in the silver wire is:

We are given:

Auogaidro’s number $=6.023 \times 10^{23}$

Atomic weight of silver $=108$

Density of silver $10.5 \times 10^{3} \mathrm{Kg} / \mathrm{m}^{3}$

Charge on the electron $=1.6 \times 10^{-19} \mathrm{C}$

(1) $7 \times 10^{-4} \mathrm{~m} / \mathrm{s}$

(2) $7 \times 10^{-6} \mathrm{~m} / \mathrm{s}$

(3) $7 \times 10^{-3} \mathrm{~m} / \mathrm{s}$

(4) 0.7

Show Answer

Correct Answer: (1)

Solution:

Since the valence of silver is one so each atom of silver can be assumed to contribute one electron.

By Avogadro’s law, we have $1 \mathrm{~g}$ mole of substance $=6.023 \times 10^{23}$ atoms

Hence number of electrons per unit volume will be

$\mathrm{n}=\frac{6.023 \times 10^{23}}{10^{8} \times 10^{-3}} \times 10.5 \times 10^{3}=5.86 \times 10^{28} / \mathrm{m}^{3}$

Cross sectional area $=\pi \mathrm{r}^{2}=3.14 \times 10^{-6} \mathrm{~m}^{2}$

Now we know I = nevA where $v$ is the drift velocity.

$\therefore \mathrm{v}=\frac{\mathrm{I}}{\mathrm{neA}}=\frac{2}{5.86 \times 10^{28} \times 1.6 \times 10^{-19} \times 3.14 \times 10^{-6}}$

or $\quad \mathrm{v}=7 \times 10^{-4} \mathrm{~m} / \mathrm{s}$

Hence option (1) is correct.

Average

Factors Affecting Resistance

79. A one meter long metallic wire is broken into two unequal parts $P$ and $Q$. $P$ part of the wire is uniformly extended into another wire $R$. Length of $R$ is twice the length of $P$ and resistance of $R$ is equal to that of $Q$. The ratio of the resistance of $P$ and $R$, is

(1) $\frac{1}{2}$

(2) $\frac{1}{4}$

(3) $\frac{1}{8}$

(4) $\frac{1}{16}$

Show Answer

Correct Answer: (2)

Solution:

Let $\ell$ be the length of part $\mathrm{P}$ of the wire.

The length of wire $\mathrm{R}=2 \ell$

Length of wire $\mathrm{Q}=1-\ell$

Mass of $\mathrm{P}=$ Mass of $\mathrm{R}$

$\pi \mathrm{r} _{\mathrm{P}}^{2} \ell \mathrm{d}=\pi \mathrm{r} _{\mathrm{R}}^{2} 2 \ell \mathrm{d}$

$r _{P}^{2}=2 r _{R}^{2} \quad$ or $\quad r _{p}=\sqrt{2} r _{R}$

Than, $\mathrm{R} _{\mathrm{P}}=\frac{\rho \ell}{\pi \mathrm{r} _{\mathrm{P}}^{2}}, \quad \mathrm{R} _{\mathrm{Q}}=\frac{\rho(1-\ell)}{\pi \mathrm{r} _{\mathrm{Q}}^{2}}$

and $\mathrm{R} _{\mathrm{R}}=\rho \frac{2 \ell}{\pi \mathrm{r} _{\mathrm{R}}^{2}}$

Hence, $\frac{\mathrm{R} _{\mathrm{P}}}{\mathrm{R} _{\mathrm{R}}}=\frac{\mathrm{r} _{\mathrm{R}}^{2}}{2 \mathrm{r} _{\mathrm{P}}^{2}}=\frac{\mathrm{r} _{\mathrm{R}}^{2}}{2\left(\sqrt{2} \mathrm{r} _{\mathrm{R}}\right)^{2}}=\frac{1}{4}$

Hence option (2) is correct.

Average

Heating Effects of Currents

80. A wire, of resistance $r$, connects points $A$ and $B$; two points in a circuit; the resistance of the remainder of which is $R$. Without any other charge being made $A$ and $B$ are also connected by ( $n-1)$ other wires, the resistance of each of which is $r$. Then the heat produced in the $n$ wires null be greater than that produced in the first wire; if

(1) $r>(\sqrt{\mathrm{n}}) \mathrm{R}$

(2) $r=(\sqrt{\mathrm{n}}) \mathrm{R}$

(3) $\mathrm{r}<\sqrt{\mathrm{n}} \mathrm{R}$

(4) $\mathrm{r}>(\sqrt{\mathrm{n}-1}) \mathrm{R}$

$~$

Show Answer

Correct Answer: (3)

Solution:

In the first case have,

$\mathrm{H} _{1}=$ heat produced per second in the wire $\mathrm{r}=\mathrm{i} _{1}^{2} \mathrm{r}$

or $\mathrm{H} _{1}=\frac{\mathrm{E}^{2}}{(\mathrm{R}+\mathrm{r})^{2}} \mathrm{r}$

$~$

In the second case the $\mathrm{n}$ wire are are connected in parallel having are equivalent resistance $\frac{\mathrm{r}}{\mathrm{n}}$

$\therefore \mathrm{H} _{2}=\mathrm{i} _{2}^{2} \frac{\mathrm{r}}{\mathrm{n}}=\frac{\mathrm{n}^{2} \mathrm{E}^{2}}{(\mathrm{nR}+\mathrm{r})^{2}} \frac{\mathrm{r}}{\mathrm{n}} \quad\left[\because \mathrm{i} _{2}=\frac{\mathrm{E}}{\mathrm{R}+\frac{\mathrm{r}}{\mathrm{n}}}=\frac{\mathrm{nE}}{\mathrm{nR}+\mathrm{r}}\right]$

or $\mathrm{H} _{2}=\frac{\mathrm{nE}^{2}}{(\mathrm{nR}+\mathrm{r})^{2}} \mathrm{r}$

$=\frac{\mathrm{H} _{2}}{\mathrm{H} _{1}}=\frac{\mathrm{n}(\mathrm{R}+\mathrm{r})^{2}}{(\mathrm{nR}+\mathrm{r})^{2}}$

If $\mathrm{H} _{2}>\mathrm{H} _{1}$ then we must have,

$\mathrm{n}(\mathrm{R}+\mathrm{r})^{2}>(\mathrm{nR}+\mathrm{r})^{2}$

or $n\left[R^{2}+2 R r+r^{2}\right]>n^{2} R^{2}+2 n R r+r^{2}$

or $\quad \mathrm{nR}^{2}-\mathrm{n}^{2} \mathrm{R}^{2}>\mathrm{r}^{2}-\mathrm{nr}^{2}$

$n R^{2}\left(1-n^{2}\right)>r^{2}(1-n)$

$\mathrm{nR}>\mathrm{r}^{2}$

$\mathrm{R}>\frac{\mathrm{r}}{\sqrt{\mathrm{n}}}$

or $\mathrm{r}<(\sqrt{\mathrm{n}}) \mathrm{R}$

Hence option (3) is correct.

81. The supply voltage to a room is $120 \mathrm{~V}$. The resistance of the lead wires is $6 \Omega$. $60 \mathrm{~W}$ bulb is already switched on. What is the decrease of voltage across the bulb, when a $240 \mathrm{~W}$ heater is switched on in parallel to the bulb?

(1) 2.9 Volt

(2) 13.3 Volt

(3) 10.04 Volt

(4) zero Volt

Show Answer

Correct Answer: (3)

Solution:

’ $\mathrm{R}$ ’ for bulb $=\frac{120 \times 120}{60} \Omega=240 \Omega$

‘R’ for Heater $=\frac{120 \times 120}{240} \Omega=60 \Omega$\

$~$

Voltage, across the bulb,

(i) Before heater is switched on $\mathrm{V} _{1}=\frac{120}{246} \times 240 \mathrm{~V}$

(ii) After heat is switched on $\mathrm{V} _{2}=\frac{120}{54} \times 48 \mathrm{~V}$

$\therefore$ Decrease in voltage $=\mathrm{V} _{1}-\mathrm{V} _{2}=120\left[\frac{240}{240}-\frac{48}{54}\right] \mathrm{V}$

$$ \simeq 10.04 \mathrm{~V} $$

Hence option (3) is correct.

82. In a large building, there are 15 bulbs of $40 \mathrm{~W}, 5$ bulbs of $100 \mathrm{~W}, 5$ fans of $80 \mathrm{~W}$ and 1 heater of $1 \mathrm{~kW}$. The voltage of the electric mains is $220 \mathrm{~V}$. The minimum capacity of the main fuse of the building will be:

(1) $12 \mathrm{~A}$

(2) $14 \mathrm{~A}$

(3) $8 \mathrm{~A}$

(4) $10 \mathrm{~A}$

Show Answer

Correct Answer: (1)

Solution:

Total power of all the devices

$=[15 \times 40 \times 5 \times 100+5 \times 80+1 \times 1000] \mathrm{W}=2500 \mathrm{~W}$

$\therefore$ Current $=\frac{2500 \mathrm{~W}}{200 \mathrm{~V}}=11.36 \mathrm{~A} \approx 12 \mathrm{~A}$

$\therefore$ Minimum capacity of the main fuse should be $12 \mathrm{~A}$.

Hence option (1) is correct.

83. When $5 \mathrm{~V}$ potential difference is applied across a wire of length $0.1 \mathrm{~m}$, the drift speed of electrons $2.5 \times 10^{-4} \mathrm{~ms}^{-1}$. If the electron density of the wire is $8 \times 10^{28} \mathrm{~m}^{-3}$, the resistivity of the materials is close to

(1) $1.6 \times 10^{-6} \Omega \mathrm{m}$

(2) $1.6 \times 10^{-5} \Omega \mathrm{m}$

(3) $1.6 \times 10^{-8} \Omega \mathrm{m}$

(4) $1.6 \times 10^{-7} \Omega \mathrm{m}$

Correct answer: (2)

(IIT 2015)

Show Answer

Solution:

$$ \begin{aligned} & \mathrm{R}=\frac{\rho \mathrm{L}}{\mathrm{A}}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{\mathrm{V}}{\mathrm{neAv} _{\mathrm{d}}} \\ & \begin{aligned} \therefore & =\frac{\mathrm{V}}{\mathrm{neV}} \cdot \frac{1}{\mathrm{~L}} \\ & \simeq 1.6 \times 10^{-5} \Omega-\mathrm{m} \end{aligned} \end{aligned} $$

Hence option (2) is correct.

84. In the circuit shown, the current in the $1 \Omega$ resistor is:

(1) $0.13 \mathrm{~A}$, from $\mathrm{Q}$ to $\mathrm{P}$

(2) $0.13 \mathrm{~A}$, from $\mathrm{P}$ to $\mathrm{Q}$

(3) $1.3 \mathrm{~A}$, from $\mathrm{P}$ to $\mathrm{Q}$

(4) $0 \mathrm{~A}$

Correct answer: (1)

$~$

Show Answer

Solution:

$~$

By using Kirchoff’s laws, we find the current in the $1 \Omega$ resistor, as per option (1).

85. The temperature dependence of resisatnces of $\mathrm{Cu}$ and undoped $\mathrm{Si}$ in the temperature range $300-400 \mathrm{~K}$, is best described by:

(1) Linear increase for $\mathrm{Cu}$, exponential increase for $\mathrm{Si}$

(2) Linear increase for $\mathrm{Cu}$, exponential decrease for $\mathrm{Si}$

(3) Linear decrease for $\mathrm{Cu}$, linear decrease for $\mathrm{Si}$

(4) Linear increase for $\mathrm{Cu}$, linear increase for $\mathrm{Si}$

Show Answer

Correct Answer: (2)

Solution:

For conductors

Resistance increases linearly (for not to large increase in temperature) with temperature.

For semi-conductors

Resistance decreases non-linearly is an (almost) exponential way.

Hence option (2) is correct.

UNIT 12 CURRENT ELECTRICITY

Learning Objectives

Electric Current

The Direction of a Current

Ohm’s Law

Electrical Resistivity

Colour Code for Resistances

Physics of the Oh’m Law

Effect of Temperature on Resistance

Super Conductors

Resistance in Series and Parallel

Electric Cell

e.m.f. of a Cell

Terminal Potential Differnece

Grouping of Cells

Heating Effect of Curretn (Joule Heat)

Electrical Power

Kirchoff’s Laws

First Law

Second Law

Metre Bridge

Potentiometer

Comparision of e.m.f. of Two Cells

Determination of Internal Resistance of a Cell

PROBLEMS FOR PRACTICE

Question Bank

Key Learning Points

Table of Contents

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Mentorship

NCERT Exercise Video Solution

UNIT 12 CURRENT ELECTRICITY

Learning Objectives

Electric Current

The Direction of a Current

Ohm’s Law

Electrical Resistivity

Colour Code for Resistances

Physics of the Oh’m Law

Effect of Temperature on Resistance

Super Conductors

Resistance in Series and Parallel

Electric Cell

e.m.f. of a Cell

Terminal Potential Differnece

Grouping of Cells

Heating Effect of Curretn (Joule Heat)

Electrical Power

Kirchoff’s Laws

First Law

Second Law

Metre Bridge

Potentiometer

Comparision of e.m.f. of Two Cells

Determination of Internal Resistance of a Cell

PROBLEMS FOR PRACTICE

Question Bank

Key Learning Points

Table of Contents

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Mentorship

NCERT Exercise Video Solution

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