Chapter 04 Current Electricity
Introduction
When we switch on the bulb of our rooms, it glows immediately. When we connect our room heater to the ‘pins’ placed on the walls, it produces heat. When we provide ‘cells’ to our torch and press its button, it glows and produces light. All of these phenomena are based on the subject matter of this chapter and the next.
Moving charges produce electric current. Electric current produces many effects like heating effect, chemical effect, magnetic effect, etc. This chapter provides us a strong base on how electric current is produced, which instruments are used for electrical measurement, the factors affecting electric current, etc. The chapter also deals with the factor which resists the flow of electric current known as ’electrical resistance’.
ELECTRIC CURRENT
We define the electric current, or simply the current, to be the net amount of charges passing per unit time across any section through a conductor. If the charges are positive, the current is assumed to flow in the direction of charges and if the charges are negative, the current is in a direction opposite to the charges.
If $\Delta Q$ is the amount of charge that passes through a particular area in a time interval $\Delta t$, the average current, $I _{u v}$, is equal to the ratio of the charge to the time interval.
$ I _{a v}=\frac{\Delta Q}{\Delta t} $
The $S I$ unit of current is ampere $(A)$, where $1 A=1 C / s$
That is, $1 A$ of current is equivalent to $1 C$ of charge passing through the surface in $1 s$. In practice, smaller units of current are often used, such as the milliampere $(1 mA=10^{-3} A)$ and the microampere $(1 \mu A=10^{-6} A)$.
In a conductor such as copper, the current is due to the motion of the negatively charged electrons. Therefore, when we speak of current in an ordinary conductor, such as a copper wire, the direction of the current will be opposite to the direction of flow of electrons.
If the charge on an electron is $e$ and $n$ electrons pass through a point in time $t$ then the total charge passing through that point will be $Q=n e$. Therefore, the current
$I=\frac{n e}{t} \quad \text{ where } e=1.6 \times 10^{-19} \text{ coulomb. }$
CONVENTIONAL FLOW NOTATION
When Benjamin Franklin made his conjecture regarding the direction of charge flow (from the smooth wax to the rough wool), he set a precedent for electrical notation that exists to this day, despite the fact that we know electrons are the constituent units of charge, and that they are displaced from the wool to the wax – not from the wax to the wool – when those two substances are rubbed together. This is why electrons are said to have a negative charge: because Franklin assumed electric charge moved in the opposite direction that it actually does, and so objects he called “negative” (representing a deficiency of charge) actually have a surplus of electrons.
By the time the true direction of electron flow was discovered, the nomenclature of “positive” and “negative” had already been so well established in the scientific community that no effort was made to change it, although calling electrons “positive” would make more sense in referring to “excess” charge. You see, the terms “positive” and “negative” are human inventions, and as such have no absolute meaning beyond our own conventions of language and scientific description. Franklin could have just as easily referred to a surplus of charge as “black” and a deficiency as “white,” in which case scientists would speak of electrons having a “white” charge (assuming the same incorrect conjecture of charge position between wax and wool).
However, because we tend to associate the word “positive” with “surpius” and “negative” with “deficiency,” the standard label for electron charge does seem backward. Because of this, many engineers decided to retain the old concept of electricity with “positive” referring to a surplus of charge, and label charge flow (current) accordingly. This is known as conventional flow notation.
Others chose to designate charge flow according to the actual motion of electrons in a circuit. This form of symbology is known as electron flow notation:
Electron flow notation
Fig. 4.2 (b)
In conventional flow notation, we show the motion of charge according to the (technically incorrect) labels of + and - . This way the labels make sense, but the direction of charge flow is incorrect. In electron flow notation, we follow the actual motion of electrons in the circuit, but the + and - labels seem backward. Does it matter, really, how we designate charge flow in a circuit? Not really, so long as we’re consistent in the use of our symbols. You may follow an imagined direction of current (conventional flow) or the actual (electron flow) with equal success insofar as circuit analysis is concerned.
Types of Current
(a) Direct Current : The current whose magnitude and direction does not vary with time is called direct current (dc). The various sources are cells, battery, dc dynamo etc.
(b) Alternating Current : The current whose magnitude continuously changes with time and periodically changes its direction is called alternating current. It has constant amplitude and has alternate positive and negative halves. It is produced by ac dynamo.
When an electric light switch is made an, the light cames on immediately. $a_n$ electric signal in a canductar travels at a speed of about $3 \times 10^{8} m / s$.
The current may be canstituted by mation of different type of charge carviers in different situations.
| S.Na. | Nature of material | Catse of current |
|---|---|---|
| 1. | Conductors | Motion of etectrons |
| 2. | Vacuum tubes | Mation of electrans |
| 3. | Semiconductors | Motion of hotes and electrons |
| 4. | Electrolytes | Mation of pasitue and negative ions |
| 5. | Discharge tube | Motion of pasitive and negative ions |
CHECK Po int
If a current of one-or two-tenths of an ampere were to flow into one of your hands and out the other, you would protably be electrocuted. But, if the same current were to flow into your hand and out the elbow above the same hand, you censld survive, even though the current might be large enough to burn your flesh. Explain.
Show Answer
SOLUTION
If a current of one-or-two-tenths of an ampere flows into one of our hands and out the other, we may be electrocuted, probably, because the electric circuit formed becomes closed. But when the same current flows into the hand and out the elbors above the same hand, the electric circuit is not completed, and we could survive.
ILLUSTRATION-4.1
If a current 3.2 ampere per second flows through a conducting wire, calculate the number of electrons passing through the wire.
Show Answer
SOLUTION:
Here, $I=3.2 A, t=1$ second, $n=?, e=1.6 \times 10^{-19}$ coulomb
$ \begin{aligned} & I=\frac{n e}{t} ; 3.2=\frac{n \times 1.6 \times 10^{-19}}{1} \\ & \text{ or } n \times 1.6 \times 10^{-19}=3.2 \quad \text{ or } \quad n=\frac{3.2}{1.6 \times 10^{-19}} \quad \therefore n=2 \times 10^{19} \text{ electrons } \end{aligned} $
ILLUSTRATION-4.2
If 25 coulomb charge flow in a conductor in 5 seconds then find out the value of electric current in it.
Show Answer
SOLUTION:
Electric current
$ \begin{aligned} & I=\frac{Q}{t} \quad[Q=25 \text{ coulomb, } t=5 \text{ second, } I=?] \\ & I=\frac{25}{5}=5 \text{ ampere } \end{aligned} $
ILLUSTRATION-4.3
If a light bulb uses 0.50 ampere of current, how much charge flows through it in 5 minutes ?
Show Answer
SOLUTION:
Five minute is 300 seconds, so from
$ \begin{aligned} & \Delta q=I \Delta t \\ & \Delta q=(0.50 A)(300 s)=150 C \end{aligned} $
RESISTANCE
Directly connecting the poles of a voltage source together with a single piece of wire is dangerous because the magnitude of electric current may be very large in such a short circuit, and the release of energy very dramatic (usually in the form of heat),you may have heard about firing due to short circuit. Usually, electric circuits are constructed in such a way as to make practical use of that released energy, in as safe a manner as possible. One practical and popular use of electric current is for the operation of electric lighting. The simplest form of electric lamp is a tiny metal “filament” inside of a clear glass bulb, which glows white-hot (incandesces) with heat energy when sufficient electric current passes through it. Like the battery, it has two conductive connection points, one for electrons to enter and the other for
something like this:
As the electrons work their way through the thin metal filament of the lamp, they encounter more opposition to motion than they typically would in a thick piece of wire. This opposition to electric current depends on the type of material, its cross-sectional area, and its temperature. It is technically known as resistance. (It can be said that conductors have low resistance and insulators have very high resistance.) This resistance serves to limit the amount of current through the circuit with a given amount of voltage supplied by the battery, as compared with the “short circuit” where we had nothing but a wire joining one end of the voltage source (battery) to the other.
When electrons move against the opposition of resistance, due to friction heat is generated. Just like mechanical friction, the friction produced by electrons flowing against a resistance manifests itself in the form of heat. The concentrated resistance of a lamp’s filament results in a relatively large amount of heat energy dissipated at that filament. This heat energy is enough to cause the filament to glow white-hot, producing light, whereas the wires connecting the lamp to the battery (which have much lower resistance) hardly even get warm while conducting the same amount of current. How to deal with resistance mathematically? You will find the answer in Ohm’s law section.
Knowledge Enhancer
The reciprocal of resistance is called conductance $G=1 / R$ Its SI unit is ohm-1 or mho or siemen (s). The substances which obey Ohm’s law are called Ohmic or linear conductors. The resistance of such conductors is independent of magnitude and polarity of applied potential difference. Here the graph between $I$ and $V$ is a straight line passing through the origin. The reciprocal of slope of straight line gives resistance
$R=\frac{V}{I}=\frac{1}{\tan \theta}=\text { constant. }$
Examples silver, copper, mercury, carbon, mica etc. The substances which do not obey Ohm’s law are called non-ohmic or non linear conductors. The $I-V$ curve is not a straight line. i.e. $p-n$ diode, transistors, thermionic valves, rectifiers etc.
FACTORS AFFECTING RESISTANCE
The resistance of a conductor depends on temperature, nature of material, length and area of cross-section. If the resistance of the wire at $0^{\circ} C$ be $R_0$ and at $t^{\circ} C$ be $R_t$, then the value of $R_t$ is obtained by the following formula
$ R_t=R_0(1+\alpha t) $
where $\alpha$ is a constant which is called the ’temperature coefficient of resistance’ of the material of the wire.
From eq. (1), we have $\alpha=\frac{R_t-R_0}{R_0 \times t}$ per $^{\circ} C$
If $R_0=1 ohm, t=1^{\circ} C$ then $\alpha=(R_t-R_0)=$ increase in resistance.
Therefore, if the resistance of a wire at $0^{\circ} C$ be $1 ohm$, then on raising its temperature to $1^{\circ} C$, the increase in its resistance will be equal to the “temperature coefficient of resistance’ of the material of that wire. For most of the metals, the value of $\alpha$ is nearly (1/273) per ${ }^{\circ} C$. Hence, by eq. (1), we have
$ R_t=R_0(1+\frac{t}{273})=R_0 \frac{(273+t)}{273}=R_0(\frac{T}{273}) $
where $T$ is the absolute temperature. Thus, $R_t \propto T$
Thus, the resistance of a pure metallic wire or the resistivity of a metal is directly proportional to its absolute temperature. The resistivity of alloys also increases with rise in temperature, but this increase is much smaller compared to pure metals. There are certain alloys such as manganin, constantan, nichrome, etc., whose resistivity is little affected by temperature, that is, their temperature coefficient of resistance is negligible. On account of their high resistivity and negligible temperature coefficient of resistance, these alloys are used to make wires for standard resistances, resistance boxes, etc.
170 Current Electricity $\mid$ Physics |
COLOUR CODE FOR CARBON RESISTORS
First Colour Code
| Colour | Strip A | Strip B | Strip C | Strip D Tolerance |
|---|---|---|---|---|
| Black | 0 | 0 | $10^{0}$ | |
| Brown | 1 | 1 | $10^{1}$ | |
| Red | 2 | 2 | $10^{2}$ | |
| Orange | 3 | 3 | $10^{3}$ | |
| Yellow | 4 | 4 | $10^{4}$ | |
| Green | 5 | 5 | $10^{5}$ | |
| Blue | 6 | 6 | $10^{6}$ | |
| Violet | 7 | 7 | $10^{7}$ | |
| Grey | 8 | 8 | $10^{8}$ | |
| White | 9 | 9 | $10^{9}$ | |
| Gold | - | - | $10^{-1}$ | $\pm 5 %$ |
| Silver | - | - | $10^{-2}$ | $\pm 10 %$ |
| NoColour | - | - | - | $\pm 20 %$ |
Aid to memory
BBROY of Great Britain has a Very Good Wife.
The first two rings give first two significant figures of resistance. The third ring indicates the decimal multiplier i.e. the number of zeros that will follow the two significant figures. The fourth ring shows tolerance or percentage accuracy.
Factors on which the resistance of a conductor depends
The resistance of ohmic circuit elements such as metal wires or carbon resistor depends on their geometries.
To find dependence, first we will keep the thickness of the object (like a copper wire) fixed and just increase or decrease the length of the wire. If we apply a potential difference across the ends of the wire and use current and potential difference measurements, we can determine its resistance as a function of length. We find that there is a proportionality between the length $L$ of the wire and the resistance of the wire $R$. Thus we can write
$ R=k L $
If instead we fix the length of the copper wire and vary the thickness or crosssectional area A then we find the resistance of the copper decreases as the cross-sectional area A of the wire increases. In fact we get an inverse relation-
ship so that $R=k^{\prime} \frac{1}{A}$
Combining these two expressions gives us: $R=k k^{\prime} \frac{L}{A}$
To simplify this expression, we can replace the product of the two constants with a single new constant and write $R=\rho \frac{L}{A}$
Fig. $4.8(a)$
$\longrightarrow 1 /$ A rea of cross section
Fig. 4.8 (b)
Where $\rho$ is a constant, which is called the specific resistance or resistivity of the conducting material. It does not depend on the area of cross-section. Resistivity depends only on the nature of material and temperature.
$ \rho=R \frac{L}{A} \quad \therefore \text{ unit of } \rho=\frac{\text{ ohm } \times m^{2}}{m}=ohm \times \text{ meter } $
Definition of resistivity
If $L=1$ meter, $A=1$ square meter then $\rho=R$ (numerically) i.e. resistance of a wire of unit length and unit area of cross section is numerically equal to the resistivity or specific resistance of the material of wire. Remember resistance is a property of an object. Resistivity is a property of a material.
Resistance of some metallic conductors rises with increase in temperature, e.g. copper, silver etc. Resistance of some alloys such as manganin and constantan changes very slowly with increase in temperature. Contrary to this, resistance of semiconductors such as germanium $(Ge)$, silicon $(Si)$ etc. decreases with increase in temperature.
On decreasing temperature of some metals, the resistance becomes zero at a certain temperature.
For example : Resistance of mercury becomes zero at $4.2 K$ temperature.
Activity
Complete an electric circuit consisting of a cell, an ammeter, a nichrome wire of length $l$ [say, marked (1)] and a plug key, as shown in Fig. 4.9
Fig. 4.9 : Electric circuit to study the factors on which the resistance of conducting wires depends.
Now, plug the key. Note the current in the ammeter.
Replace the nichrome wire by another nichrome wire of same thickness but twice the length, that is $2 l$ [marked (2) in the Fig]. Note the ammeter reading.
Now replace the wire by a thicker nichrome wire, of the same length $l$ [marked (3)]. A thicker wire has a larger cross-sectional area. Again note down the current through the circuit.
Instead of taking a nichrome wire, connect a copper wire [marked (4) in Fig. ] in the circuit. Let the wire be of the same length and same area of cross-section as that of the first nichrome wire [marked (1)]. Note the value of the current.
Notice the difference in the current in all cases.
Does the current depend on the length of the conductor?
Does the current depend on the area of cross-section of the wire used?
Knowledge Enhancer
If a wire is stretched such that length becomes $\mathrm{n}$ times, resistance becomes $n^2$ times. Due to stretching, the cross-sectional area decreases to $n$ times and volume (or mass) remains constant.
$\text { Volume }(V)=A \ell,\\ A \ell=A^{\prime} n \ell $
New cross-sectional arqea $A^{\prime}=\frac{A}{n}$ New resistance $R^{\prime}=\frac{\rho n \ell}{A / n}=\frac{n^2 \rho \ell}{A}=n^2 R$ The fractional change in resistance without change in volume or mass are : (a) When change in length is small ( $55 %$ ) fractional change in $R$ is $\frac{\Delta R}{R}=\frac{2 \Delta L}{L}$ (b) When change in radius is small ( $(5 %)$ fractional change in $R$ is $\frac{\Delta R}{R}=\frac{-4 \Delta r}{r}$ (c) When change in area is small ( $55 %$ ) fractional change in $R$ is $\frac{\Delta R}{R}=\frac{-2 \Delta A}{A}$
CHECK POINT :
Choose the correct alternative
(a) Alloys of metals usually have (greater/less) resistivity than that of their constituent metals.
(b) Alloys usually have much (lower/higher) temperature coefficients of resistance than puremetals.
(c) The resistance of graphite and most non-metals increases/decreases with increase in temperature.
(d) The resistivity of a semiconductor increases/decreases rapidly with increasing temperature.
(e) The resistivity of the alloy manganin is nearly independent of/increases rapidly with increases of temperature.
(f) The resistivity of a typical insulator (e.g., amber) is greater than that of a metal by a factor of the order of $(10^{22} / 10^{23})$.
Show Answer
Solution
(a) Alloys of metals usually have greater resistivity than that of their constituent metals.
(b) Alloys usually have much lower temperature coefficients of resistance than pure metals.
(c) The resistance of graphite and most non metals decreases with increase in temperature.
(d) The resistivity of a semiconductor decreases rapidly with increasing temperature.
(e) The resistivity of the alloy manganin is nearly independent of increasing temperature.
(f) The resistivity of a typical insulator (e.g. amber) is greater than that of a metal by factor of the order of $10^{22}$.
ILLUSTRATION-4.4
The resistivity of copper at room temperature is $1.72 \times 10^{-6} ohm$-centimeter. What is the resistance of a copper wire 35 meters long and 0.025 centimeter in cross section.
Show Answer
SOLUTION:
$R=\frac{\rho \ell}{A}=\frac{(1.72 \times 10^{-6} \Omega cm)(3500 cm)}{0.025 cm^{2}}=0.24 \Omega$
ILLUSTRATION-4.5
The resistance of a wire of length $100 cm$ and of uniform area of cross-section $0.020 cm^{2}$, is found to be $2.0 ohm$. Calculate sp. resistance of wire
Show Answer
SOLUTION:
$\ell=100 cm, A=0.020 cm^{2}, R=2.0 \Omega$
Sp. resistance $\rho=\frac{R A}{\ell}=\frac{2.0 \times 0.020}{100}=0.0004 \Omega-cm$
ILLUSTRATION-4.6
What length of resistance wire of resistivity $100 \times 10^{-8} \Omega m$, area of cross-section $2.5 \times 10^{-7} m^{2}$, would be needed to make a resistor of $57.6 \Omega$.
Show Answer
SOLUTION:
$R=57.6 \Omega, A=2.5 \times 10^{-7} m^{2}, l=?$
$\rho=100 \times 10^{-8} \Omega m, \quad R=\frac{\rho l}{A}$ so, $\quad l=\frac{R A}{\rho} \Rightarrow l=\frac{57.6 \times 2.5 \times 10^{-7}}{100 \times 10^{-8}} \frac{m^{2}}{m}=1.44 m$
ILLUSTRATION-4.7
Two wires $A$ and $B$ of equal masses and of same metal are taken. The diameter of the wire $A$ is half the diameter of the wire $B$. If the resistance of the wire $A$ be $24 ohm$, calculate the resistance of the wire $B$.
Show Answer
SOLUTION:
Let $r_A$ be the radius and $\ell_A$ the length of the wire $A$, and $r_B$ the radius and $\ell_B$ the length of the wire $B>$ Both the wires are of the same material (density $d$ ). Since the masses of the two wires are equal, we have
$ \begin{aligned} & \pi r_A^{2} \ell_A \times d=\pi r_B^{2} \ell_B \times d \\ \therefore \quad & \frac{\ell_A}{\ell_B}=\frac{r_B^{2}}{r_A^{2}} \end{aligned} $
Resistance of the wire $A$ is $R_A=\rho \frac{\ell_A}{\pi r_A^{2}}$, Resistance of the wire $B$ is $R_B=\rho \frac{\ell_B}{\pi r_B^{2}}$
where $\rho$ is the specific resistance of the material of the wires
$ \therefore \quad \frac{R_A}{R_B}=\frac{\ell_A}{\ell_B} \times \frac{r_B^{2}}{r_A^{2}}=\frac{r_B^{4}}{r_A^{4}} $
But $r_A=\frac{1}{2} r_B$ (given) $\therefore \quad \frac{R_A}{R_B}=16 \quad$ or $\quad R_B=\frac{1}{16} R_A=\frac{1}{16} \times 24 ohm=1.5 ohm$
ILLUSTRATION-4.8
Calculate the resistance of $1 km$ long copper wire of radius $1 mm$. (Resistivity of copper is $1.72 \times 10^{-8} \Omega m$ )
Show Answer
SOLUTION:
Given $\ell=1 km=1000 m, r=1 mm=10^{-3} m, A=\pi r^{2}=3.14 \times(10^{-3})^{2}=3.14 \times 10^{-6} m^{2}$
Resistivity $\rho=1.72 \times 10^{-8} \Omega m$
$ R=\rho \frac{l}{A}=\frac{1.72 \times 10^{-8} \times 1000}{3.14 \times 10^{-6}} \quad \text{ or } \quad R=5.5 \Omega \text{. } $
ELECTRIC CIRCUIT
To learn more on electricity we need to draw electric circuits with elements like batteries (combination of cells), bulbs, wire, switches, ammeter, voltmeter, Rheostat etc.
1. Cell
Direct current source of electromotive force.
Fig. 4.10 : Simple circuit showing battery and resistor.
2. Rheostat (variable resistance)
Wire of special type of alloy like manganin, eureka, nichrome etc. is wound on a hollow cylinder of china clay. It controls the current in the electric circuit.
3. Switch
It is used to close or open the electric circuit, controls the movement of electrons in a circuit.
4. Ammeter
Determines the value of current flowing in the circuit. The resistance of ammeter is small and it is used in series with the circuit.
5. Voltmeter
Determines the potential difference between two points in the circuit. Its resistance is high and it is used in parallel with the resistance wire.
CHECK POINT :
One example of a water system is a garden hose that waters a garden. Another is the cooling system of an automobile. Which of these exhibits behavior more analogous to that of an electric circuit? Why?
Show Answer
SOLUTION:
The garden hose is more analogous to an electric circuit as it contains a source of water in which there is a pressure difference like an electric circuit contains a source of electrical energy which has a potential difference. Also, the water is supplied through a pipe in the same way as a conducting path is provided for the flow of electrical energy in a circuit.
OHM’s LAW
According to Ohm’s law ‘The current passing through a conductor is directly proportional to the potential difference between its ends, provided the physical conditions of conductor remain unchanged.”
$ V \propto I \quad \text{ or } \quad V=R I $
where $R$ is a constant which is called resistance.
Unit of resistance
$ R=\frac{V}{I}=\frac{\text{ volt }}{\text{ ampere }}=ohm(\Omega) $
Fig. 4.15
The unit of resistance is ohm $(\Omega)$.
If $I=1$ ampere, $V=1$ volt then $R=1 \Omega$.
The resistance of a conductor is 1 ohm if a potential difference of 1 volt is produced across its ends when 1 ampere of the current flows through it.
The conductors, which obey the Ohm’s law are called the ohmic conductors or linear resistances. All metallic conductors (such as silver, aluminium, copper, iron, etc.) are the ohmic conductors.
The conductors, which do not obey the Ohm’s law are called the non-ohmic conductors or non-linear resistances. Examples are, diode valve, triode valve, transistors, electrolyte, etc.
It is sometime contended that $R=\Delta V / i($ or $\Delta V=i \mathcal{R})$ is a statement of Ohm’s law. Shat is not true! Ihis equation is the defining equation for resistance, and it applies ta all conducting devices, whether they abey Ohm’s law or not. If we measure the potential difference $\Delta V$ across and the current $i$ through the device, even a bulb or other nonahmic device, we can find its resistance at that value of $\Delta V$ as $\mathcal{R}=\Delta V /$. The essence of $\mathcal{O h m}$ ’s law, however, is a plat of $i$ versus $\Delta V$ that it is straight line, sa that the value of $\mathcal{R}$ is independent of the value of $\Delta V$.
Activity
Take a nichrome wire, a torch bulb, a $10 W$ bulb and an ammeter ( $0-5 A$ range), a plug key and some connecting wires.
Set up the circuit by connecting four dry cells of $1.5 V$ each in series with the ammeter leaving a gap $X Y$ in the circuit, as shown in Figure
Complete the circuit by connecting the nichrome wire in the gap XY. Plug the key. Note down the ammeter reading. Take out the key from the plug. Always take out the key from the plug after measuring the current through the circuit.
Replace the nichrome. wire with the torch bulb in the circuit and find the current through it by measuring the reading of the ammeter.
Now repeat the above step, with the $10 W$ bulb in the gap $X Y$.
Are the ammeter readings differ for different components connected in the gap $X Y$ ? What do the above observations indicate?
You may repeat this Activity by keeping any material component in the gap. Observe the ammeter readings in each case. Analyse the observations.
VERIFICATION OF OHM’s LAW
Set up a circuit as shown in figure with a battery switch $(S)$, a rheostat $(R h)$, ammeter $(a)$, a resistor $(R)$ and a voltmeter $(V)$. Close the circuit and adjust the rheostat such that the ammeter and the voltmeter show a measurable reading $I$ and $V$ respectively. Repeat the experiment for different values of $I$ and $V$ by adjusting the rheostat.
Tabulate your results as shown below :
| Trial No. | Potential difference $V$ (volt) | Current I (ampere) | $\frac{V}{I}=r($ ohm) |
|---|---|---|---|
| 1. | |||
| 2. | |||
| 3. |
Plot a graph of $V$ against $I$. If it is a straight line passing Fig. 4.18 through the origin we can conclude that P.D. ’ $V$ ’ is directly proportional to the current ’ $l$ ’ and Ohm’s law is verified.
Fig. 4.18
Limitation of Ohm’s Law
1. Ohm’s law does not apply to conductors such as diode, radio valves, metal rectifiers, electricity through gases.
2. Ohm’s law is applicable only when the physical conditions remain constant, for example the resistance of some conductors will vary if the wires are heated or placed under tension.
3. Ohm’s law is applicable only when the temperature of the conductor is constant. Generally the resistance of pure metals increases with temperature but the resistance of certain other conducting materials like carbon decreases with temperature. There are certain metal alloys of magnesium and copper whose resistance changes with temperature.
CHECK POINT :
Is Ohm’s law universally applicable for all conducting elements? If not, give examples of elements which do not obey Ohm’s law.
Show Answer
SOLUTION:
Ohm’s law is not a fundamental law in nature. It is not universally followed semiconductor diodes, Transistor, Thermistors, vacuum tubes do not follow ohm’s law.
ILLUSTRATION- 4.9
Calculate the potential difference required across ’ a conductor of resistance $5 \Omega$ to flow a current of $1.5 A$ through it.
Show Answer
SOLUTION:
Given $R=5 \Omega, I=1.5 A, V=$ ?
From Ohm’s, law, $V=I R=1.5 \times 5=7.5 V$
ILLUSTRATION- 4.10
How much electric current is produced on applying potential difference of 30 volts on a wire of resistance $20 ohm$ ?
Show Answer
SOLUTION:
$I=\frac{V}{R}=\frac{30}{20}=1.5$ ampere
ILLUSTRATION-4.11
If ampere current flowing through a resistance wire develops a potential difference of 50 volts across its ends, find the resistance of the resistance wire.
Show Answer
SOLUTION:
$R=\frac{V}{I}=\frac{50}{5}=10 ohm$
ILLUSTRATION-4.12
The potential difference between the terminals of an electric heater is $60 V$ when it draws a current of $4 A$ from the source. What current will the heater draw if the.potential difference is increased to $120 V$ ?
Show Answer
SOLUTION:
We are given, potential difference $V=60 V$, current $I=4 A$.
According to Ohm’s law, $R=\frac{V}{I}=\frac{60 V}{4 A}=15 \Omega$
When the potential difference is increased to $120 V$ the current is given by
$ \text{ current }=\frac{V}{R}=\frac{120 V}{15 \Omega}=8 A $
The current through the heater becomes $8 A$.
ILLUSTRATION- 4.13
A relay with a resistance of 12 ohms is in an electric circuit with one side at a potential of 85 volts and the other at 71 volts. How much current is in the relay?
Show Answer
SOLUTION:
The potential difference across the relay is $(85 V-71 V)=14 V$
$ I=\frac{\Delta V}{R}=\frac{14 V}{12 \Omega}=1.2 A $
COMBINATION OF RESISTORS
In many practical applications to have desired value of resistance two or more resistances are required to be combined. This can be done in two ways : (i) In series and (ii) In parallel. Sometimes resistances are to be combined in such a way that some resistances be in series and some in parallel. Such a combination is called mixed grouping.
If, in an electrical circuit, two or more resistances connected between two points are replaced by a single resistance such that there is no change in the current of the circuit and in the potential difference between those two points, then the single resistance is called the ’equivalent resistance’.
When the resistance of a circuit is to be increased, they are combined in series and when heavy current is to be passed, they are combined in parallel so as to decrease the total resistance.
SERIES COMBINATION
In this combination, the resistances are joined end to end. Thus the second end of each resistance is joined to the first end of the next resistance, and so on. The first end of the first resistance and the second end of the last resistance are connected to the cell. In this combination, the same current flows in all the resistances but the potential differences between their ends are different according to their resistances.
In figure, three resistances $AB, BC$ and $CD$ are connected in series. Suppose their resistances are $R_1, R_2$ and $R_3$ respectively. Let the ’equivalent resistance’ of these resistances be $R$.
Fig. 4.19
Suppose a current is flowing in all the three resistances. Suppose the potential differendes between the ends of the resistances $R_1, R_2$ and $R_3$ are $V_1, V_2$ and $V_3$ respectively. Then, according to Ohm’s law we have
$ V_1=i R_1, \quad V_2=i R_2 \text{ and } V_3=i R_3 \text{. } $
If the potential difference between points $A$ and $D$ be $V$, then
$ V=V_1+V_2+V_3=i R_1+i R_2+i R_3=i(R_1+R_2+R_3) $
The equivalent resistance between points $A$ and $D$ is $R$. Therefore
$ V=i R $
Comparing eq. (1) and (2), we get
$ \begin{aligned} & i R=i(R_1+R_2+R_3) \\ & R=R_1+R_2+R_3 \end{aligned} $
That is, the equivalent resistance of the resistances connected in series is equal to the sum of those resistances and is thus greater than any individual resistance.
Activity
Join three resistors of different values in series. Connect them with a battery, an ammeter and a plug key, as shown in Fig. You may se the resistors of values like $1 \Omega, 2 \Omega, 3 \Omega$ etc., and a battery of $6 V$ for performing this Activity.
Plug the key. Note the ammeter reading.
Change the position of ammeter to anywhere in between the resistors. Note the ammeter reading each time.
Do you find any change in the value of current through the ammeter ?
Activity
In previous Activity, insert a voltmeter across the ends $X$ and $Y$ of the series combination of three resistors, as shown in figure.
Plug the key in the circuit and note the voltmeter reading. It gives the potential difference across the series combination of resistors. Let it be $V$. Now, measure the potential difference across the two terminals of the battery. Compare the two values.
Take out the plug key and disconnect the voltmeter. Now, insert the voltmeter across the ends $X$ and $P$ of the first resistor, as shown in Figure.
Plug the key and measure the potential difference across the first resistor. Let it be $V_1$.
Similarly, measure the potential difference across the other two resistors, separately. Let these values be $V_2$ and $V_3$, respectively.
Deduce a relationship between $V, V_1, V_2$ and $V_3$.
PARALLEL COMBINATION
When two or more resistances are combined in such a way that their first ends are connected to one point and the second ends to another point then this combination is in parallel. resistances is same but the current in different resistances are different.
In Fig. 4.22, three resistances $R_1, R_2$ and $R_3$ are joined in parallel between the points $A$ and $B$. Suppose the current flowing from the cell is $i$. At the point $A$, this current is divided into three parts. Suppose $i_1, i_2$ and $i_3$ are the currents in $R_1, R_2$ and $R_3$ respectively. At the point $B$, these three currents meet and form the main current $i$. Thus, we have
$ i=i_1+i_2+i_3 $
Let the potential difference between the points $A$ and $B$ be $V$. Since each resistance is connected between $A$ and $B$, the potential difference between the ends of each will be $V$. Therefore,
Fig. 4.22
$ i_1=\frac{V}{R_1}, \quad i_2=\frac{V}{R_2} \text{ and } i_3=\frac{V}{R_3} $
Substituting these values in eq. (1) we get,
$ i=\frac{V}{R_1}+\frac{V}{R_2}+\frac{V}{R_3} $
If the equivalent resistance between the points $A$ and $B$ be $R$, then
$ i=V / R $
Comparing eq. (2) and (3), we get
$ \frac{V}{R}=\frac{V}{R_1}+\frac{V}{R_2}+\frac{V}{R_3} \quad \text{ or } \quad \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3} $
That is, the reciprocal of the equivalent resistance of the resistances connected in parallel is equal to the sum of the reciprocals of those resistances.
In parallel combination, the equivalent resistance $R$ is less than even the smallest resistance connected. For example, if we connect two resistors 1 $\Omega$ and 100 $\Omega$ in parallel, the equivalent resistance R will be less than 1 $\Omega$ as shown below:
$ \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{2}+\frac{1}{100}=\frac{101}{100} \quad \text{ or } \quad R=\frac{100}{101}=0.99 \Omega $
If resistance $R_1$ and $R_2$ are connected in parallel and $i$ is the total current drawn from cell then
$ \begin{aligned} & i=i_1+i_2 \\ & V=i_1 R_1=i_2 R_2 \end{aligned} $
Solving eq. (1) and (2)
$ i_2=\frac{i R_1}{R_1+R_2} \quad \text{ and } \quad i_2=\frac{i R_1}{R_1+R_2} $
Activity:
Make a parallel combination, $X Y$, of three resistors having resistances $R_1, R_2$, and $R_3$, respectively. Connect it with a battery, a plug key and an ammeter, as shown in Fig 4.24 (a). Also connect a voltmeter in parallel with the combination of resistors. a a Plug the key and note the ache potential difference across each resistor is also $V$. This can be checked by connecting the $V$, across the combination. The potential difference across each resistor is also V. This can be checked by connecting the voltmeter across each individual resistor [Fig. 4.24(b)]
Take out the plug from the key. Remove the ammeter and voltmeter from the circuit. Insert the ammeter in series with the resistor $R_1$, as shown in Fig. 4.24 (b). Note the ammeter reading, $I_1$. Similarly, measure the currents through $R_2$ and $R_3$. Let these be $I_2$ and $I_3$, respectively. What is the relationship between $I, I_1, I_2$ and $I_3$ ?
CHECK POINT :
Given $n$ resistors each of resistance $R$, how will you combine them to get the (i) maximum (ii) minimum effective resistance? What is the ratio of the maximum to minimum resistance?
Show Answer
SOLUTION:
(i) For maximum effective resistance, all the resistors should be joined in series.
$ R _{\max }=R+R+R+\ldots \ldots . n \text{ or } R _{\max }=n R $
(ii) For minimum effective resistance all the resistors should be joined in parallel
$\frac{1}{R_{\min }}=\frac{1}{R}+\frac{1}{R}+\frac{1}{R}+\ldots \ldots . . n$ or $\frac{1}{R_{\min }}=\frac{n}{R}$
So $R_{\min }=\frac{R}{n}$
Now, $\frac{R_{\max }}{R_{\min }}=n^2$
ILLUSTRATION-4.14
Three resistances of 2,3 and 5 ohms are connected in series. If the combination is connected to a battery of 4 volt then what will be (i) current in each resistance (ii) potential difference across each resistance?
Show Answer
SOLUTION:
Total resistance in series
$ \begin{aligned} & R=R_1+R_2+R_3 \\ & R=2+3+5=10 \Omega \end{aligned} $
(i) Current flowing through each resistance will be same
$ I=\frac{V}{R}=\frac{4}{10}=0.4 \text{ ampere } $
(ii) Potential difference across each resistance will be
determined by formula $V=I R$.
Potential difference across $2 \Omega$ resistance $\quad V _1=\mathbb R _1=0.4 \times 2=0.8$ volt
Potential difference across $3 \Omega$ resistance $\quad V_2=IR_2=0.4 \times 3=1.2$ volt
Potential difference across $5 \Omega$ resistance $V_3=IR_3=0.4 \times 5=2.0$ volt
ILLUSTRATION-4.15
Three resistors of $2 \Omega, 3 \Omega$ and $4 \Omega$ are connected in (i) series, (ii) parallel. Find the equivalent resistance in each case.
Show Answer
SOLUTION:
Given $R_1=2 \Omega, R_2=3 \Omega$ and $R_3=4 \Omega$
(i) In series, the equivalent resistance is
$ R=R_1+R_2+R_3=2+3+4=9 \Omega . $
(ii) In parallel, the equivalent resistance is $R$, then
$ \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3} \quad \text{ or } \quad \frac{1}{R}=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{13}{12} $
$ \frac{1}{R}=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{13}{12} $
ILLUSTRATION-4.16
Two resistors of $4 \Omega$ and $6 \Omega$ are connected in parallel. The combination is connected across a $6 V$ battery of negligible resistance. Calculate : (i) the current through the battery (ii) through each resistor. Draw the circuit diagram.
Show Answer
SOLUTION:
The circuit diagram is shown in figure.
(i) If resistance of the combination is $R$, then
$ \frac{1}{R}=\frac{1}{4}+\frac{1}{6}=\frac{3+2}{12}=\frac{5}{12} $
or $\quad R=\frac{12}{5} \Omega=2.4 \Omega$
Let current through the battery be $I$.
Fig. 4.26
$ I=\frac{V}{R}=\frac{6}{2.4}=2.5 A $
(ii) The potential difference across each resistor is $V=6$ volt (same as of battery) since they are in parallel.
Current through $4 \Omega$ resistor, $\quad I_1=\frac{V}{R_1}=\frac{6}{4}=1.5 A$
Current through $6 \Omega$ resistor, $\quad I_2=\frac{V}{R_2}=\frac{6}{6}=1 A$
ILLUSTRATION-4.17
$6 \Omega$ and $3 \Omega$ resistors are connected in parallel and an $8 \Omega$ resistor is connected in series with them. A current of $2 A$ passes through the $8 \Omega$ resistor. Find
(a) the combined resistance
(b) the potential difference across the combined resistor
(c) the current through the $3 \Omega$ resistor
Show Answer
SOLUTION:
Find the combined resistance $R$ of the resistors in parallel. $R=?, R_1=6 \Omega, R_2=3 \Omega$,
$ \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{6}+\frac{1}{3}=\frac{3}{6}=\frac{1}{2} $
$ R=2 \Omega $
$3 \Omega$
$ \text{ Fig. } 4.27(a) $
The equivalent circuit is shown in Fig. 4.27 (b).
(a) Let the combined resistance of all three resistors be $R$
Fig. 4.27 (b)
$ \begin{aligned} & R=?, R_1=8 \Omega, R_2=2 \Omega \\ & R=R_1+R_2=8 \Omega+8 \Omega=10 \Omega \end{aligned} $
(b) Let the potential difference across the combined resistors be $V$.
$ \begin{aligned} & V=?, I=2 A, R=10 \Omega \\ & V=I \times R=2 A \times 10 \Omega=20 V \end{aligned} $
(c) The combined resistance of the parallel network is $2 \Omega$. Hence potential difference $V$ across the parallel resistor is
$ V=I \times R=2 A \times 2 \Omega=40 V $
For the $3 \Omega$ resistor $V=4 V, I=?, R=3 \Omega$
$ I=\frac{V}{R}=\frac{4 V}{3 \Omega}=1 \frac{1}{3} A $
ILLUSTRATION-4.18
Three resistors are connected in series to 24 volt battery, and an ammeter in the circuit reads 0.50 amperes. The first resistor is rated at $22 ohms$, and the second at 8 ohms. Find (a) the total resistance (b) the resistance of the third resistor, and (c) the potential difference across the third resistor.
Show Answer
SOLUTION:
(a) The total resistance of the circuit is
$ R=\frac{\Delta V}{I}=\frac{24 V}{0.50 A}=48 \Omega $
(b) In a series circuit, resistance add, so $22 \Omega+8 \Omega+R_3=48 \Omega$ and $R_3=18 \Omega$.
(c) For any circuit element, $\Delta V=I R$ so $V_3=(0.50 A)(18 \Omega)=9 V$
ILLUSTRATION-4.19
In the circuit shown, find :
(a) the equivalent resistance between points $A$ and $B$.
(b) the current through the battery.
Show Answer
SOLUTION:
Given: $R_1=12 \Omega, R_2=12 \Omega, R_3=6.0 \Omega, V=12 V$
Find: (a) $R_p$
(b) $I$
(a) The three resistors are in a parallel combination.
$ \frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}=\frac{1}{12 \Omega}+\frac{1}{12 \Omega}+\frac{1}{6.0 \Omega}=\frac{1}{12 \Omega}+\frac{1}{12 \Omega}+\frac{2}{12 \Omega}=\frac{4}{12 \Omega}=\frac{1}{3.0 \Omega} $
So, $R_p=3.0 \Omega$.
Or using $R_p=\frac{R R}{R+R}=\frac{R}{2}$ for the two 12 ohm resistors in parallel, $R_p=\frac{12 \Omega}{2}=6 \Omega$
Now the two $6.0 \Omega$ resistances are in parallel and the equivalent resistance is $R_p=\frac{6.0 \Omega}{2}=3.0 \Omega$
(b) From Ohm’s law, $I=\frac{V}{R}=\frac{V}{R_p}=\frac{12 V}{3.0 \Omega}=4.0 A$
ELECTRICAL MEASURING INSTRUMENTS
Galvanometer
A galvanometer is an instrument which is used to measure current or voltage, as per requirement.
Ammeter
An ammeter is a low resistance galvanometer used to measure strength of current in an electrical circuit.
An ammeter is always connected in series in a circuit because when an ammeter is connected in series it does not appreciably change the resistance of circuit and hence the main current flowing through the circuit. An ideal ammeter has zero resistance. The reading of an ammeter is always less than actual current in the circuit because all practical ammeters have low finite resistance.
Smaller is the resistance of an ammeter more accurate will be the reading.
A galvanometer can be converted to an ammeter by connecting a low resistance shunt in parallel to coil of galvanometer.
Here $\quad I_g G=(I-I_g) S \quad$ or $\quad S=\frac{I_g}{I-I_g} G$
Here $G$ is resistance of galvanometer and $I_g$ is current required to produce full scale deflection of current.
Shunt (S)
Current Electricity
It is a low resistance connected in parallel to coil of galvanometer to convert it to ammeter. It protects a galvanometer from strong currents. It is also used to alter range of an ammeter.
The resistance of converted ammeter is $R_A=\frac{G . S}{G+S}$
The range of an ammeter is increased by reducing shunt resistance $S$.
If $I=N I_g \quad$ then $\quad S=\frac{I_g}{N I_g-I_g} G=\frac{G}{N-1}$
The range of an ammeter can be increased $N$ times by reducing shunt to $S=G / N-1$.
It is possible to increase the range of an ammeter because it lowers the resistance of ammeter further.
The length of the shunt required $\ell=\pi r^{2} S / \rho$ where $r$ is radius of shunt wire and $\rho$ is specific resistance of material of shunt wire.
Reducing the shunt resistance may increase range but it reduces the sensitivity.
Voltmeter
A voltmeter is a high resistance galvanometer used to measure potential difference.
A voltmeter is connected in parallel to a circuit element because when connected in parallel it draws least current from the main current. So it measures nearly accurate potential difference.
An ideal voltmeter has infinite resistance.
The reading of a voltmeter is always less than actual value because all practical voltmeter may have large but finite resistance. Greater is the resistance of voltmeter more accurate is its reading.
A galvanometer can be converted to a voltmeter by connecting a high resistance in series with coil of galvanometer.
$ V=I_g(R+G) \quad \text{ or } \quad R=\frac{V}{I_g}-G $
The resistance of converted voltmeter is $R _{\nu}=R+G$
The range of a voltmeter is increased by increasing the series resistance.
Fig. 4.30
$ \text{ If } \quad V=N V_g=N I_g G \text{ then } R=\frac{N I_g G-G}{I_g}=(N-1) G $
The value of resistance required to increase range $N$ times is $R=(N-1) G$
KIRCHHOFF’S LAW
This law is used to solve a problem regarding electrical circuit. The law has two parts.
(a) Kirchhoff’s first law or current law or junction rule
In any electrical network, the algebraic sum of currents meeting at a junction is always zero i.e. $\Sigma I=0$
$ I_1-I_2+I_3+I_4-I_5=0 \text{ or } I_1+I_3+I_4=I_2+I_5 $
The sum of currents flowing towards a junction is equal to sum of currents
leaving the junction.
By convention the current directed towards the junction is positive while those directed away from the junction is taken as negative.
The first law is in accordance with conservation of charge.
Fig. 4.31
The charges do not accumulate at a junction. The total charge entering a junction is equal to total charge leaving the junction.
(b) Kirchhoff’s second law or voltage law or loop rule
The algebraic sum of all the potential drops and emf’s along any closed path in a network is zero. i.e. $\Sigma V=0$
The second law is in accordance with conservation of energy.
According to second law, the electric energy given to the charge by a source of emf is lost in passing through resistance. The change in potential in covering a resistance in the direction of current is negative (-IR) while in opposite direction it is positive.
The potential falls along direction of current. The potential fall is taken as negative while potential rise is taken as positive.
The emf of source is taken as positive when it is traversed from negative to positive terminal while it is taken as negative when it is traversed from positive to negative irrespective of the direction of current.
If there are $n$ loops their will be $(n-1)$ equations according to loop rule. The algebraic sum of products of currents and resistances in a closed loop is equal to sum of emf’s applied in the circuit i.e., $\Sigma E=\Sigma I R$.
ILLUSTRATION-4.20
Determine the current through the dc source in the following network:
Show Answer
SOLUTION:
We assume the loop, or mush, currents $I_1, I_2$ and $I_3$ in the loops, or meshes, $A B D A, B C D B$ and $A D C E F A$, respectively, and arbitrarily select their directions clockwise, as shown, and traversal direction also clockwise. Applying the Kirchhoff’s voltage law to the three loops, we have respectively,
$ \begin{aligned} & +I_1 \times 10+(I_1-I_3) 5+(I_1-I_3) 5=0 \Rightarrow 4 I_1-I_2=I_3 \\ & +I_2 \times 5+(I_2-I_3) 10+(I_2-I_1) 5=0 \Rightarrow-I_1+4 I_2=2 I_3 \\ & +(I_3-I_1) 5+(I_3-I_2) 10+I_3 \times 10=10 \Rightarrow-I_1-2 I_2+5 I_3=2 \end{aligned} $
Solving the equations, (1) and (2), we get
$ I_1=\frac{2 I_3}{5}, I_2=\frac{3 I_3}{5} $
Then, substituting these values of $I_1$ and $I_2$ into the equation (3), we get the current through the de source, i.e.,
$ I_3=\frac{10}{17} A $
ILLUSTRATION-4.21
Calculate the current in each resistance in the given circuit.
Fig. 4.35
Show Answer
SOLUTION:
Let us assume the loop currents $I_1$ and $I_2$ in the clockwise direction in the two meshes $A B E F A$ and $B C D E B$, respectively. Let us also assume the traversal direction in the clockwise sense. Then, the equations for the two loops are, respectively,
$ (I_1-I_2) \times 1=+3-3 \Rightarrow I_1=I_2 $
The current in the resistance of $1 \Omega$ between $B$ and $E$ is
$ I _{B E}=I_1-I_2=0 $
Knowledge Enhancer
Wheatstone Bridge
It is an arrangement of four resistances devised by Charles Wheatstone which is used to measure an unknown resistance.
The Wheatstone bridge principle states that if four resistances $P, Q, R$ and $S$ are arranged to form a bridge with a cell and key between $A$ and $C$, and a galvanometer between $B$ and $D$ then bridge is said to be balanced when galvanometer shows a zero deflection.
In balanced condition $I_g=0 \quad$ so $V_B=V_D$ or $\frac{P}{Q}=\frac{R}{S}$. This is called condition of balance.
The condition of balance depends on resistance $P, Q, R$ and $S$. This is independent of emf of battery.
In state of balance the cell and galvanometer can be interchanged.
Fig. 4.36 While performing an experiment at start press the cell key $K_c$ first and then the galvanometer key $K_g$ and at end remove $K_g$ first and then $K_c$ to avoid induced effects.
$P$ and $Q$ are called ratio arms, $R$ is known as resistance arm and $S$ is the unknown arm. $B D$ and $A C$ are called conjugate arms.
The resistance of balanced Wheatstone bridge between $A$ and $C$ is $\frac{(P+Q)(R+S)}{P+Q+R+S}$. The sensitivity of bridge depends upon value of resistances. The sensitivity is maximum when $P, Q, R$ and $S$ are of same order. Unbalanced Wheatstone bridge
In unbalanced bridge $I_g \neq 0, V_B \neq V_D$ so $\frac{P}{Q} \neq \frac{R}{S}$.
If $\frac{P}{Q}<\frac{R}{S}$ then $V_B>V_D$ so current flows from $B$ to $D$.
If $\frac{P}{Q}>\frac{R}{S}$ then $V_B<V_D$ so current flows from $D$ to $B$.
METER BRIDGE
The metre bridge, shown, is a practical arrangement of the Wheatstone’s bridge. It consists of a $1 m$-long resistance wire $A C$ of the uniform cross-sectional area and uniform material, i.e., magnanin, or constantan, having the high electrical resistivity of low temperature coefficient.
A null situation is obtained by moving the jockey $J$ on the bridge wire $A C$ till no deflection is observed in the galvanometer. Now,
the balance condition gives
$ \begin{aligned} \frac{R}{S}=\frac{P}{Q} & \Rightarrow \frac{R}{S}=\frac{\rho L_1}{\rho L_2} \\ & \Rightarrow \frac{R}{S}=\frac{L_1}{L_2} \\ & \Rightarrow \frac{R}{S}=\frac{L_1}{(1-L_1)} \end{aligned} $
where $\rho$ is the resistance per unit length of the bridge wire.
ILLUSTRATION-4.22
If the current through the branch $B D$ of the circuit, shown, is zero.
Find the resistance $R$ and current through it.
Show Answer
SOLUTION:
The given circuit, when rearranged, is found to be a
Wheastone’s bridge circuit, as shown.
Since, the current through the branch $B D$ is zero, the bridge is balanced, i.e.,
$ \frac{P}{Q}=\frac{R}{S} \Rightarrow \frac{1.5}{1}=\frac{R}{4} $
$ \Rightarrow R=4 \times 1.5=6 \Omega $
The branch $B D$ is ineffective as it carries no current and, therefore, we have two parallel paths $A B C$ and $A D C$ of the resistances $2.5 \Omega$ and $10 \Omega$, respectively. The combined resistance between $A$ and $C$ becomes
$ R _{A C}=\frac{2.5 \times 10}{(2.5+10)}=\frac{25}{12.5}=2 \Omega $
Further, this resistance is connected in series with the resistance of $2 \Omega$ so that the total resistance becomes $(2+2)=4 \Omega$. The current from the source is
$ I=\frac{10}{4}=2.5 A $
Then, the current through the resistance $R$ is given by
$ \begin{aligned} & I_R \times(R+4)=I \times(R _{A C}) \\ & I_R \times(6+4)=2.5 \times 2 \\ & I_R=\frac{2.5 \times 2}{10}=0.5 A \end{aligned} $
Fig. 4.38
ILLUSTRATION-4.23
A metre bridge circuit is shown in the balanced state. The bridge wire $A C$ has a resistance per metre $\rho=100 \Omega / m$. Calculate the unknown resistance $S$ and the current drawn from the dc source of emf of $6 V$ and negligible internal resistance.
Show Answer
SOLUTION:
From the balance condition, we have
$ \begin{gathered} \frac{P}{Q}=\frac{R}{S} \Rightarrow \frac{0.6 \times 100}{0.4 \times 100}=\frac{3}{S} \\ \Rightarrow S=\frac{4 \times 3}{6}=2 \Omega \end{gathered} $
The equivalent resistance between $A$ and $C$ is given by
$ \begin{gathered} R _{A C}=\frac{(P+Q)(R+S)}{(P+Q)+(R+S)} \doteq \frac{(1 \times 100)(3+2)}{(1 \times 100)+(3+2)} \\ =\frac{100 \times 5}{100+5}=\frac{500}{105} \Omega \end{gathered} $
Then, the current drawn from the dc source is
$ I=\frac{6}{R _{A C}}=\frac{6 \times 105}{500}=\frac{630}{500}=1.26 A $
MISCELLANEOUS SOLVED EXAMPLES
1. If $3.0 \times 10^{15}$ electrons flow through a section of a wire of diameter $2.0 \mathrm{~mm}$ in $4.0 \mathrm{~s}$, what is the electric current in the wire?
Show Answer
SOLUTION:
Given $n=3.0 \times 10^{15}$ electrons, $t=4.0 s, d=2.0 mm$.
The charge in $3.0 \times 10^{15}$ electrons is $q=n e=(3.0 \times 10^{15}.$ electrons $)(1.60 \times 10^{-19}.$ C/electrons $)=4.8 \times 10^{-4}$
So, $\quad I=\frac{q}{t}=\frac{4.8 \times 10^{4} C}{4.0 s}=1.2 \times 10^{4} A=0.12 mA$
2. In the circuit shown, find : (a) the equivalent resistance between points $A$ and $B$, (b) the current through each resistor.
Show Answer
SOLUTION:
Given : $R_1=12 \Omega, R_2=3.0 \Omega, R_3=5.0 \Omega, R_4=4.0 \Omega, V=12 V$.
(a) All four resistors are in series combination, so
$ R_s=R_1+R_2+R_3+R_4=12 \Omega+3.0 \Omega+5.0 \Omega+4.0 \Omega=24 \Omega $
(b) The current through all resistors in series is the same
$ I=\frac{V}{R}=\frac{V}{R_s}=\frac{12 V}{24 \Omega}=0.50 A $
3. For the combination of resistors shown in figure, find the equivalent resistance between (i) $C$ and $D$ (ii) $A$ and $B$.
Show Answer
SOLUTION:
(i) The resistors $R_2, R_3, R_4$ are in series. They can be replaced by an equivalent resistance $R^{\prime}$, where $R^{\prime}=R_2+R_3+R_4=3+3+3=9 \Omega$.
The resistance $R_5(=3 \Omega)$ and $R^{\prime}(=9 \Omega)$ are in parallel, between the points $C$ and $D$. The equivalent resistance between $C$ and $D$ is $R^{\prime \prime}$, where $\quad \frac{1}{R^{\prime \prime}}=\frac{1}{3}+\frac{1}{9} \quad$ or $\quad \frac{1}{R^{\prime \prime}}=\frac{3+1}{9}=\frac{4}{9}$ or $\quad \frac{1}{R^{\prime \prime}}=\frac{4}{9}=2.25 \Omega$
Thus the equivalent resistance between $C$ and $D$ is $2.25 \Omega$ (ii) Now the resistor $R 1=(3 \Omega) \cdot R^{\prime \prime}(=2.25 \Omega)$ and $R_6(=3 \Omega)$ are in series between the points $A$ and $B$. The equivalent resistance between $A$ and $B$ is : $R=R_1+R^{\prime \prime}+R_6=3+2.25+3=8.25 \Omega$
4. Two copper wires $A$ and $B$ of length $30 m$ and $10 m$ have radius $2 mm$ and $1 mm$ respectively, Compare the resistance of the two wires. Which will have less resistance?
Show Answer
SOLUTION:
From the relation $R=\rho \frac{\ell}{A}$
Resistance of wire $A$ is $R_A=\rho \frac{30}{\pi(2 \times 10^{-3})^{2}}$
Resistance of wire $B$ is $\quad R_B=\rho \frac{10}{\pi(1 \times 10^{-3})^{2}}$
On dividing
$ \frac{R_A}{R_B}=\frac{30}{10}(\begin{matrix} 1 \times 10^{-3} \\ 2 \times 10^{-3} \end{matrix} )^{2}=\frac{3}{4} $
Thus the wire $A$ has less resistance.
5. How much energy is given to each coulomb of charge passing through a $6 V$ battery?
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SOLUTION:
Energy $=$ charge $\times$ potential difference
$ =1 \times 6=6 \text{ joule } $
6. (a) How much current will an electric bulb draw from a $220 V$ source, if the resistance of the bulb filament is $1200 \Omega$ ?
(b) How much current will an electric heater coil draw from a $220 V$ source, if the resistance of the heater coil is $100 \Omega$ ?
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SOLUTION:
(a) We are given $V=220 V ; R=1200 \Omega$.
From Eq. $R=V I I$, we have the current $I=220 V / 1200 \Omega=0.18 A$.
(b) We are given, $V=220 V, R=100 \Omega$.
From Eq. $R=V I I$, we have the current $I=220 V / 100 \Omega=2.2 A$.
The difference of current drawn by an electric bulb and electric heater from the same $220 V$ source.
7. Resistance of a metal wire of length $1 m$ is $26 \Omega$ at $20^{\circ} C$. If the diameter of the wire is $0.3 mm$, what will be the resistivity of the metal at that temperature?
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SOLUTION:
We are given the resistance $R$ of the wire $=26 \Omega$, the diameter $d=0.3 mm=3 \times 10^{-4} m$, and the length $l$ of the wire $=1 m$.
Therefore, from the resistivity of the given metallic wire is $\rho=(R A / l)=(R \pi d^{2} / 4 l)$
Substitution of values in this gives $\rho=1.84 \times 10^{-8} \Omega m$
The resistivity of the metal at $20^{\circ} C$ is $1.84 \times 10^{-8} \Omega m$. This is the resistivity of manganese.
8. An electric lamp of $100 \Omega$, a toaster of resistance $50 \Omega$, and a water filter of resistance $500 \Omega$ are connected in parallel to a $220 V$ source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?
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SOLUTION:
$R_l=100 \Omega, R_t=50 \Omega, R_f=500 \Omega, V=220$ volt,
$I$ drawn by, $\quad$ lamp $=I_l=\frac{V}{R_l}=\frac{220}{100}=2.2 A$
toaster $=I_I=\frac{220}{50}=4.4 A$, water filter $=I_I=\frac{220}{500}=0.44 A$
Total current drawn $=7.04 A$
Current to be drawn by iron $=7.04 A$
Resistance of electric iron $=R_1=\frac{V}{I}=\frac{220}{7.04}=31.25 \Omega$
9. What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance $4 \Omega$, $8 \Omega, 12 \Omega, 24 \Omega$ ?
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SOLUTION:
(a) Highest - By connecting the resistors in series. So $R=4+8+12+24=48 \Omega$
(b) Lowest - By connecting the resistors in parallel.
So, $\quad R=\frac{64}{6+3+2+1}=\frac{24}{12}=2 \Omega$
10. How many $176 \Omega$ resistors (in parallel) are required to carry $5 A$ on a $220 V$ line?
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SOLUTION:
When $N$ resistors each $R \Omega$ are in parallel.
$ R_p=\frac{R}{N} $
Current drawn from cell $=I=\frac{V}{R _{e q}}=\frac{V N}{R}$
$\therefore \quad N=\frac{I R}{V}=\frac{5 \times 176}{220}=4$
11. A nichrome wire has a resistance of $10 \Omega$. Find the resistance of another nichrome wire, whose length is three times and area of cross-section four times the first wire.
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SOLUTION:
Case $1: R=10 \Omega$. Let $\ell$ be length and ’ $a$ ’ the area of cross-section.
$\therefore \quad R=\rho \frac{\ell}{a}$ or $\quad 10=\rho \frac{\ell}{a}$
Case $2: R_1=$ ?, length $3 \ell$, area $=4 a$
$\therefore \quad R_1=\rho \frac{3 \ell}{4 a}$
Dividing (2) by (1)
$ \begin{aligned} & \quad \frac{R_1}{10}=\frac{\rho .3 \ell}{4 a} \times \frac{a}{\rho . \ell} \\ & \therefore \quad R_1=10 \times \frac{3}{4}=7.5 \Omega \end{aligned} $
12. Calculate the equivalent resistance between points $A$ and $C$, as shown in figure.
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SOLUTION:
Resistance of $2 \Omega$ and $6 \Omega$ are connected in parallel between points $B$ and $C$. Let $R_p$ be their equivalent resistance.
$ \frac{1}{R_p}=\frac{1}{2}+\frac{1}{6}=\frac{4}{6}=\frac{2}{3} \quad \text{ or } \quad R_p=\frac{3}{2}=1.5 \Omega $
Now, equivalent resistance $R_p$ is in series with $10 \Omega$ resistance. Therefore, equivalent resistance of circuit $(R_s)$ is
$ \therefore \quad R_s=10+1.5=11.5 \Omega \text{. } $
13. Three resistors of $2 \Omega, 4 \Omega$ and $6 \Omega$ are connected in series to a battery of emf $12 V$ and negligible internal resistance. Draw the circuit diagram and calculate (i) current drawn from cell (ii) p.d. at the ends of each resistor.
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SOLUTION:
Resistance of $2 \Omega, 4 \Omega$ and $6 \Omega$ in series.
$R=2+4+6=12 \Omega$
(i) $\therefore$ Current drawn from cell $(I)=\frac{V}{R}=\frac{12}{12}=1 A$
(ii) P.D. across $2 \Omega$ resistor $V_1=I . R=1 \times 2=2 V$
P.D. across $4 \Omega$ resistor $V_2=I . R=1 \times 4=4 V$
P.D. across $6 \Omega$ resistor $V_3=I . R=1 \times 6=6 V$
14. Two cells of $2 V$ each are connected in parallel. (figure) An external resistance of $0.5 \Omega$ is connected in series to the junction of two parallel resistors of $6 \Omega$ and $2 \Omega$ and then to the common terminal of the battery through each resistor. Calculate the
(i) total resistance of the circuit.
(ii) current flowing through $0.5 \Omega$ resistors
(iii) current flowing through $6 \Omega$ resistor.
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SOLUTION:
EMF of the battery $=2 V$
Effective resistance of the parallel resistors is given by $R_1$.
$ \frac{1}{R_1}=\frac{1}{6}+\frac{1}{2}=\frac{1+3}{6}=\frac{4}{6}=\frac{2}{3} \Rightarrow R_1=\frac{3}{2}=1.5 \Omega $
Total resistance of the circuit, $R=1.5+0.5=2 \Omega$
Main current $I=\frac{V}{R}=\frac{2}{2}=1 A$
(ii) Current flowing through $0.5 \Omega$ coil $=1 A$
P.D. across the junctions : $V_1=I R_1=1 \times 1.5=1.5 V$
(iii) Hence, current $I_1$, flowing through $6 \Omega$ resistor $I_1=\frac{V_1}{6}=\frac{1.5}{6}=0.25 A$
15. Four cells each of emf $2 V$ and internal resistance $0.1 \Omega$ are connected in series. The combination in series is joined to an ammeter of negligible resistance, a $1.6 \Omega$ resistor and an unknown resistor $R_1$. The current in the circuit is $2 A$. Draw a labelled diagram for the above arrangement and calculate : (i) the total resistance in the circult (ii) the total emf (iil) the value of $R_1$ (Iv) the potentlal difference across $R_1$.
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SOLUTION:
The circuit diagram is shown in figure.
(i) Total internal resistance $=0.1+0.1+0.1+0.1=0.4 \Omega$. Total resistance in the circuit $=1.6+R_1+0.4=(2.0+R_1) \Omega$.
(ii) Total emf $=2+2+2+2=8 V$
(iii) Current in circuit $=\frac{\text{ Total emf }}{\text{ Total resistance }}$
$ \begin{aligned} & \therefore \quad 2=\frac{8}{(2.0+R_1)} \text{ or } 4+2 R_1=8 \\ & 2 R_1=8-4=4 \text{ or } R_1=2 \Omega . \end{aligned} $
(iv) Potential difference across $R_1$ is $V=I R_1=2 \times 2=4 V$
1 EXERCISE
Fill In The Blanks
DIRECTIONS : Complete the following statements with an appropriate word / term to be filled in the blank space(s).
1. The colour coding of wires is ………….. for earth ………….. for live and ………….. for neutral.
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Answer: Green, red, black2. The rate of flow of electric charge is called …………..
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Answer: current3. Current is measured with an instrument called a (an). …………..
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Answer: ammeter4. If there is no current, a voltmeter connected across a resistor will register. ………….. voltage.
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Answer: zero5. Between any two points in a circuit, the sum of all ………….. is the same through any pathway.
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Answer: potential difference6. Combined resistance is the sum of separate resistances provided that the various conductors are connected in …………..
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Answer: series7. In a parallel circuit, each circuit element has the same…………..
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Answer: potential difference8. Copper is a preferred material for making wire because of its low. …………..
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Answer: resistivity9. The SI unit of electric current is …………..
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Answer: ampere10. ………….. is a property that resists the flow of electrons in a conductor.
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Answer: Resistance11. The SI unit of resistance is …………..
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Answer: ohm (2)12. The potential difference across the ends of a resistor is
………….. to the current through it, provided its ………….. remains the same.
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Answer: directly proportional, temperature13. The resistance of a conductor depends directly on its
………….. inversely on its ………….. , and also on the ………….. of the conductor.
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Answer: length, area of cross-section, material14. The slope of voltage (along $y$-axis) and current (along $\mathrm{x}$-axis) is called …………..
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Answer: resistance15. An ammeter should have very, ………….. resistance.
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Answer: low16. Current is measured by an instrument called, …………..
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Answer: ammeter17. The phenomenon of losing of electrical resistance by a conductor on cooling it to an extremely low temperature is known as …………..
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Answer: superconductivityTrue / False:
DIRECTIONS : Read the following statements and write your answer as true or false.
1. A copper wire of length $L$ and cross-sectional area $A$ carries a current $I$. If the specific resistance of copper is $S$, then electric field in the wire is $I S / A$.
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Answer: True2. The drift velocity of electrons in a metallic wire will decrease if the temperature of the wire is increased.
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Answer: True3. Two wires, each of radius $r$, but of different materials are connected together end to end. If the densities of charge carriers in the two wires are in the ratio $1: 4$, the drift velocity of electrons in the two wires will be in the ratio $1: 2$.
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Answer: False4. The drift velocity of an electron is doubled if the applied electric field across the conductor is doubled.
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Answer: True5. The temperature coefficient of resistance of a wire is $0.00125 /{ }^{\circ} C$. The resistance of the wire is $1 ohm$ at $300 K$. The resistance will be $2 ohm$ at $1100 K$.
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Answer: False6. The quantity of charge flowing past a point multiplied by time is a current.
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Answer: False7. The resistivity of all pure metals increases with the rise in temperature.
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Answer: True8. Ohm’s law is a relation between the power used in a circuit to the current and the potential difference.
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Answer: False9. Direction of current is taken opposite to the direction of flow of electrons.
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Answer: True10. A cell generates a potential difference across its terminals
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Answer: True11. The equivalent resistance of several resistors in series is equal to the sum of their individual resistances.
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Answer: True12. In parallel combination, the reciprocal of equivalent resistance is the sum of the reciprocal of individual resistance.
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Answer: True13. A voltmeter is always connected in series in the circuit to measure the potential difference between two points.
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Answer: False14. The resistivity of an alloy is generally lower than that of pure metals which form the alloy.
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Answer: False15. The series arrangement is used for domestic circuits.
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Answer: FalseMatch the following
DIRECTIONS : Each question contains statements given in two columns which have to be matched. Statements $(A, B, C, D)$ in column I have to be matched with statements $(p, q, r, s)$ in column II.
1. $ \begin{array}{ll} \text{Column I} & \text{Column II} \\ \text{(A) Ohm} & \text{(p) $\frac{\rho L}{A}$} \\ \text{(B) Resistance} & \text{(q) $\frac{1 \text{ volt }}{1 \text{ ampere }}$} \\ \text{(C) Resistivity} & \text{(r) zero resistance} \\ \text{(D) Super conducțor} & \text{(s) ohm-meter} \\ \end{array} $
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Answer: (a) $\to q$;(b) $\to p$;(c) $\to s$;$(d) \to r$2. In the figure shown, the value of each resistance is R. match the entries of column I with the entries of column II.
$ \begin{array}{ll} \text{Column I} & \text{Column II} \\ \text{(A) Resistance between a and b} & \text{ (p) $\frac{R}{2} $ } \\ \text{(B) Resistance between a and c} & \text{(q) $\frac{5}{8} R$ } \\ \text{(C) Resistance between b and d } & \text{(r) R} \\ & \text{(s) None} \\ \end{array} $
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Answer: (a) $\to$ q;(b) $\to p$; (c) $\to r$;(d) $\to s$3. Column II gives order of resistivity for materials in column I Column I
$ \begin{array}{ll} \text{Column I} & \text{Column II} \\ \text{(A) Semi-conductor} & \text{ (p) $3 \times 10^{3} \Omega-m$} \\ \text{(B) Conductor} & \text{(r) $10^{16} \Omega-m$ } \\ \text{(C) Insulator } & \text{(r) R} \\ \text{(D) Super conductor} & \text{(s) $1 \Omega-m$} \\ \end{array} $
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Answer: (a) $\to$ q;(b) $\to r$; (c) $\to p$;$(d) \to q$Very Short Answer Questions:
DIRECTIONS : Give answer in one word or one sentence.
1. What are the factors on which resistance depends?
2. Which combination have maximum value of equivalent resistance?
3. If one coulomb charge flows in a circuit for one second then what will be the value of current in the circuit.
4. Define one ohm resistance.
5. Name the instrument used to measure electric current.
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Answer: Ammeter6. Write the unit of electrical resistance.
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Answer: Ohm7. Name the best conductor of electricity.
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Answer: Silver8. In domestic wiring do we connect various distribution circuits in series?
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Answer: No, various distribution circuits are connected in parallel.9. Why is an ammeter likely to be burnt out if you connect it in parallel?
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Answer: If an ammeter is connected in parallel, the resultant resistance of the circuit decreases and more current passes through the instrument. Hence, the ammeter is likely to be burnt out.10. What is the resistance of an air gap?
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Answer: It is very-very large, say almost infinity.11. Manganin is used for making standard resistors. Why?
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Answer: Manganin is an alloy whose temperature coefficient of resistance is almost negligible. So its resistance does not change with change in temperature, that is why it is used in making standard resistors.12. A wire of resistivity $\rho$ is stretched to double its length. What will be its new resistivity?
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Answer: Resistivity will not change as it depends on the material not on the length.13. A copper wire of resistivity $\rho$ is stretched to reduce its diameter to half of its previous value. What will be its new resistivity?
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Answer: Resistivity will not change as it depends on the material not on the diamater of the wire.14. Why is a voltmeter always connected in parallel with a circuit element across which voltage is to be measured?
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Answer: Voltage is measured across any two points of a conductor by a voltmeter connected in parallel because potential difference across parallel combination of elements remains same.15. Which physical quantity does the voltage vs current graph for a metallic conductor depict?
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Answer: Slope of voltage vs. current graph gives resistance of a metallic conductor. S.I. unit is ohm.16. A carbon resistor of $74 k \Omega$ is to be marked with rings of different colours for its identification. Write the sequence of colours.
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Answer: $74 k \Omega=74 \times 10^{3} \Omega$. Violet, yellow and orange.17. Specific resistance of copper, silver and constant an are $1.18 \times 10^{-6} \Omega cm .1 \times 10^{-6} \Omega cm$ and $48 \times 10^{-6} \Omega cm$ respectively. Which is the best electrical conductor and why?
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Answer: The specific resistance of silver is least. So it is the best conductor as the conductivity is the reciprocal of specific resistance.18. Sketch a graph showing variation of resistivity of carbon with temperature.
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Answer:
19. Is a current-carrying wire electrically charged?
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Answer: A current carrying wire consists of equal number of electrons and protons. Therefore, the wire is not electrically charged.20. A wire has the non-uniform cross-sectional area. A steady current flows through the wire in the direction of increasing cross-sectional area. How does the drift speed of charge carrier change?
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Answer: $I=q n A v_d \Rightarrow v_d \propto \frac{1}{A}$
i.e, as the area of cross-section of wire increases in the direction of current $I$, the drift speed of charge carriers decreases in that direction.
21. V-I graph for a metallic wire at two different temperatures $T_1$ and $T_2$ is as shown in the following figures. Which of the two temperature is higher and why?
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Answer: Resistance increases with increase in temperature. Resistance of metallic wire is the inverse of slope of the graph $R$ is more at $T_2 \therefore T_2>T_1$.22. What is the effect on current through a circuit of steady resistance when both voltage and resistance are doubled?
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Answer: If both voltage and resistance are doubled, there will be no change in the current.23. Which has the greater electrical resistance, wet skin or dry skin?
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Answer: Dry skin has greater resistance than wet skin. There is usually a layer of salt remaining on our skin from respiration. When the skin is wet the salt lowers its resistance.24. What is the source of electrons that makes a shock when you touch a charged conductor?
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Answer: The source of electrons that makes a shock when we touch a charged conductor is our body itself.25. A carbon resistor is marked in coloured bands of red, black, organge and silver. What is the resistance and tolerance value of the resistor?
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Answer: Red -2 , Black -0 , Orange -3 , Silver $- \pm 10 %=$ tolerance SHORT ANSWER GUESTIONS :Short Answer Questions:
DIRECTIONS : Give answer in 2-3 sentences.
1. Write Ohm’s law. Explain it by giving diagram of an electric circuit.
2. Write the expression for the equivalent resistance $R$ of three resistors $R_1, R_2$ and $R_3$ joined in (i) parallel, (ii) series.
3. Define the term current and state its SI unit.
4. Define the term resistance.
5. A lamp draws a current of $0.5 A$ when it is connected to a 230 V supply. What is the resistance of the filament of the lamp?
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Answer: $I=0.5 A, V=230$ volt
$R=\frac{V}{l}=\frac{230}{0.5}=460 \Omega$
6. Join three resistances of $2 ohm$ each such that the total resistance of the circuit is $3 ohm$.
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Answer: Two resistance in parallel, and the third in series with them.7. Judge the equivalent resistance when the following are connected in parallel (a) $1 \Omega$ and $10^{6} \Omega$ (b) $1 \Omega$ and $10^{3} \Omega$ and $10^{6} \Omega$.
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Answer: (a) When a number of resistances are connected in parallel then their combined resistance is less than the smallest individual resistance. Therefore, equivalent resistance will be less than 1 .
(b) In this case, also the equivalent resistance will be less than 1 .
8. Resistance of a metal wire of length $1 m$ is $26 \Omega$ at $20^{\circ} C$. If the diameter of the wire is $0.3 mm$, what will be the resistivity of the metal at that temperature.
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Answer: We are given the resistance $R$ of the wire $=26 \Omega$, the diameter $d=0.3 mm=3 \times 10^{-4} m$ and the length $\ell$ of the wire $=1 m$ We know that the resistivity of the given metallic wire is
$\rho=(\frac{RA}{\ell})=(\frac{R \pi d^{2}}{4 \ell})$
$=\frac{26 \times 22 \times 3 \times 10^{-4} \times 3 \times 10^{-4}}{7 \times 4 \times 1}=183.857 \times 10^{-8}$
$=1.84 \times 10^{-8} \Omega m$
The resistivity of the metal at $20^{\circ} C$ is $1.84 \times 10^{-8} \Omega m$
9. The equivalent resistance of circuit diagram is $4 \Omega$. Calculate
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Answer: Equivalent resistance of circuit $R=4 \Omega$.
Equivalent resistance of $2 \Omega$ and $x$ in series
$ R_1=2 \Omega+x $
Equivalent resistance of $7 \Omega$ and $5 \Omega$ in series
$ R_2=7+5=12 \Omega $
As $R_1$ and $R_2$ are in parallel
$\therefore \frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2} \Rightarrow \frac{1}{4}=\frac{1}{R_I}+\frac{1}{12}$
$\therefore \frac{1}{R_1}=\frac{1}{4}-\frac{1}{12}=\frac{2}{12}=\frac{1}{6}$
Putting value of $R_1$ in (a)
$ 6=2+x $
$\therefore x=6-2=4 \Omega$.
10. The $I-V$ characteristics of a resistor are observed to deviate from a straight line for higher values of current as shown below. Why?
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Answer: For higher current there are higher heat or energy losses as $V=I^{2} R$11. A potential difference $V$ is applied across a copper wire of diameter $d$ and length $l$. When $d$ is doubled, how does the drift velocity change?
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Answer: $I=q n A v_d \Rightarrow \frac{V}{R}=q n A v_d \Rightarrow \frac{V A}{\rho l}=q n A v_d \Rightarrow v_d=\frac{V}{\rho l q n}$ i.e., the drift speed of charge carriers does not depend on the diameter of wire.12. Why do electrons, rather than protons, make up the flow of charge in a metal wire?
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Answer: In conductors, one or more electrons from each atom are free to move throughout the atomic lattice. The movement of electrons produce electric current in the conductor. Protons, on the other hand, are bound to the nuclei of atoms that are more or less locked in a fixed positions. Therefore, protons can’t move and hence can’t contribute to the current.13. Does electric charge flow across a circuit or through a circuit? Does voltage flow across a circuit or is it impressed across a circuit? Explain.
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Answer: Electric charges flow through a circuit just like the molecules of water flow through a pipe if there is a pressure difference between its ends.
On the other hand, voltage doesn’t flow across a circuit but it is just impressed across a circuit. Voltage doesn’t go anywhere in a circuit just like pressure doesn’t go through the pipe in the example above.
14. Two resistors of $15 \Omega$ and $30 \Omega$ are connected in parallel. What resistance should be connected in series of the combination to get an equivalent resistance of $20 \Omega$ ?
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Answer: $10 \Omega$15. The equivalent resistance between the point $A$ and $B$ in the adjoining circuit is $2 \Omega$. Determine the value of $R$.
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Answer: $6 \Omega$Long Answer Questions:
DIRECTIONS : Give answer in four to five sentences.
1. How does the resistance of a metallic wire depend on its temperature? Explain the reason to your answer.
2. A circuit consists of a $1 \Omega$ wire in series with a parallel arrangement of $6 \Omega$ and $3 \Omega$ with a P.D. of $12 V$ is connected across the whole circuit. Draw the circuit diagram and calculate the main current in the circuit.
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Answer: Effective resistance in the parallel circuit is
$ \begin{aligned} & \therefore \quad \frac{1}{R_1}=\frac{1}{6}+\frac{1}{3}=\frac{1+2}{6}=\frac{3}{6}=\frac{1}{2} \\ & \Rightarrow \quad R_1=2 \Omega \end{aligned} $
Total resistances $R=1+2=3 \Omega$
Current $I=\frac{V}{R}=\frac{12}{3}=4 A$
3. A copper wire has diameter $0.5 mm$ and resistivity of $1.6 \times$ $10^{-8} \Omega m$. What will be the length of this wire to make its resistance $10 \Omega$ ? How much does the resistance change if the diameter is doubled?
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Answer: $122.7 m, 1 / 4$ minutes.4. Three resistors of $20 ohm, 30 ohms$ and $50 ohm$ resistance are joined in series. Across this combination a source of 150 volts is connected then determine the current in each resistor and potential difference across their ends.
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Answer: 1.5 ampere, 30 volts, 45 volts.5. When a potential difference of 2 volt is applied across the ends of a wire of $5 m$ length, a current of $1 A$ is found to flow through it. Calculate :
(i) the resistance per unit length of the wire
(ii) the resistance of $2 m$ length of this wire
(iii) the resistance across the ends of the wire it is doubled on itself.
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Answer: (i) $0.4 \Omega m^{-1}$ (ii) $0.8 \Omega$ (iii) 16. How can three resistors of resistance $2 \Omega, 3 \Omega$ and $6 \Omega$ be connected to give a total resistance of (a) $4 \Omega$ (b) $1 \Omega$ ?
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Answer: (a) As the total resistance (equivalent resistance) is $4 \Omega$, the $6 \Omega$ resistor cannot be in series. So, it must be in parallel with some other resisters.
In parallel connection, the equivalent resistance $(4 \Omega)$ has to be less than all the resistances. So, the resistors of $2 \Omega$ and $3 \Omega$ cannot be in parallel at one time with $6 \Omega$.
So, the resistors have to be in a mixed combination. Let us consider the combination show in the figure,
The equivalent resistance between $B$ and $C$ (which are in
parallel) $=\frac{3 \Omega \times 6 \Omega}{3 \Omega+6 \Omega}=\frac{18 \Omega}{9 \Omega}=2 \Omega$
The resistance between $A$ and $D=2 \Omega+2 \Omega=4 \Omega$.
So, the combination shown in the ligure is true.
7. A battery of $9 V$ is connected in series with resistors of $0.2 \Omega$, $0.3 \Omega, 0.4 \Omega, 0.5 \Omega$ and $12 \Omega$ respectively. How much current would flow through the $12 \Omega$ resistor?
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Answer: Resistors are connected in series.
So, equivalent resistance
$ R=0.2 \Omega+0.3 \Omega+0.4 \Omega+0.5 \Omega+12 \Omega=13.4 \Omega $
Potential difference, $V=9 V$
Current, through the circuit, $I=\frac{V}{R}=\frac{9}{13.4}=0.67 A$
8. What is drift velocity of electrons and relaxation time of free electrons in a metallic conductor carrying a current? Establish a relation between them?
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Answer: Drift velocity is defined as the average velocity with which the free electrons get drilled towards the positive end of the conductor under the influence of an external electric ficld.
Relaxation time is the average time that has elapsed since each electron suffered its last collision with the atoms or ions of conductor.
Consider a metalic conductor of length $l$, and $V$ be the potential difference applied across the ends. Then the
magnitude of electric field, $E=\frac{V}{\ell}$.
Since the charge on electron is $-e$, each electron experiences a force, $\vec{F}=-e \vec{E}$
If $m$ is the mass of an electric, the acceleration of each
electron is, $\vec{a}=\frac{-e \vec{E}}{m}$
Due to this acceleration, apart from its thermal velocity, acquires additional velocity component in a direction opposite to the direction of electric field. At any instant of time, the velocity acquired by electron having thermal
velocity $ u _1 $ will be $ \overrightarrow v _1=\overrightarrow u _1+\overrightarrow{a} \tau _1 $ and so on.
Where $\tau_1$ is the time elapsed after it’s last collision.
$\therefore \quad$ The average velocity of all the electrons in the conductor (i.e., the drift velocity)
$ \vec{v} _d=\frac{ \vec{v} _1+ \vec{v} _2+\ldots .+ \vec{v} _n}{n}=\frac{\vec{a}(\tau_1+\tau_2+\ldots+\tau_n)}{n}$ (Since the average thermal velocity of electrons is zero).
$=\vec{a} \tau$ where $\tau=\frac{\tau_1+\tau_2+\ldots+\tau_n}{n}$ is the relaxation time.
Putting the value of $\vec{a}$ from (2) drift velocity speed $ \vec{v} _d=\frac{e \vec{E} \tau}{m}$.
$\therefore \quad$ Average drift speed $v_d=\frac{e E \tau}{m}$
9. Define resistivity of material. Staie its S.I. units and discuss its variation with temperature in case of (i) metals (ii) semiconductors and (iii) insulators.
Show Answer
Answer: The resistivity of the material of a conductor is defined as the resistance of unit length and unit area of cross-section of the conductor. The S.I. unit of resistivity is ohm metre $(\Omega m)$.
Resistivity of a material,
$\rho=\frac{m}{n e^{2} \tau}$ or $\rho \alpha \frac{1}{n \tau}$ Where $m$ is the mass of electron, $n$ is the number density of electron and $\tau$ is the average relaxation time. This shows that the resistivity is related to two parameters of the material namely $n$ and $\tau$.
The variation of resistivity with temperature is different in different materials. (a) For metallic conductors
The temperature dependence of resistivity of a metal is given by the relation,
$ \rho=\rho_0[1+\alpha(T-T_0)] $
Where $\rho$ and $\rho_0$ are the resistivity at temperature $T$ and $T_0$ respectively and $\alpha$ is called temperature coefficient of resistivity.
$ \therefore \quad \alpha=\frac{\rho-\rho_0}{\rho_0(T-T_0)}=\frac{d \rho}{\rho_0} \cdot \frac{1}{d T} $
The resistivity of a conductor increases with increase in temperature since $\alpha$ is positive. The variation of resistivity of copper with temperature is as shown.
(b) For Semiconductors
For Semiconductors, the resistivity decreases as temperature increases since the value of $\alpha$ is negative.
(c) For insulators
The resistivity increases exponentially with decrease in temperature. It becomes infinitely large at temperatures near absolute zero.
10. Explain the various types of grouping of cells and find the condition for the maximum current in the external resistor connected to the combination of cells (i) in series (ii) in parallel and (iii) in mixed grouping.
Show Answer
Answer: (a) When cells are arranged in series
Consider three cells of e.m.f $E_1, E_2, E_3$ of internal resistances $r_1, r_2, r_3$ are connected in series through an external resistor $R$. The same current $I$ flows through the cells.
The equivalent internal resistance of this combination is $r _{e q}=r_1+r_2+r_3$.
If $n$ cells each of e.m.f $E$ and internal resistance $r$ are connected to the external resistor $R$, then, the equivalent e.m.f of this combination, $E _{e q}=n E$. Equivalent internal resistance of $n$ cells in series $r _{e q}=n r$. $\therefore$ Total resistance of the circuit $=n r+R$.
$\therefore$ Current in the circuit $I=\frac{n E}{n r+R}$.
(b) When cells are arranged in parallel
In this combination of cells, the total current is the sum of the currents due to individual cells. Total emf of this combination is the same $(E)$.
The equivalent internal resistance $\frac{1}{r _{e q}}=\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}$
If $m$ cells are connected in parallel, each of emf $E$ and internal resistance $r$ to the external resistor $R$, then the current,
$I=\frac{E}{R+r / m}=\frac{m E}{m R+r}$
(c) When the cells are in mixed grouping
If there be $n$ cells in series in one row and $m$ rows of cells in parallel. Suppose the cells are identical with each cell be of e.m.f. $E$ and internal resistance $r$.
Total resistance in the circuit $=\frac{R+n r}{m .}$
Total emf of all the cells $=n E$
$\therefore$ Current in the external resistance $R$ is
$I=\frac{n E}{R+\frac{n r}{m}}=\frac{m n E}{m R+n r}$
2 EXERCISE
Multiple Choice Questions:
DIRECTIONS : This section contains multiple choice questions. Each question has 4 choices (a), (b), (c) and (d) out of which ONLY ONE is correct.
1. A cylindrical rod is reformed to twice its length with no change in its volume. If the resistance of the rod was $R$, the new resistance will be
(a) $R$
(b) $2 R$
(c) $4 R$
(d) $8 R$
Show Answer
Answer: (c)2. What is the current through a $5.0 ohm$ resistor if the voltage across it is $10 V$
(a) zero
(b) $0.50 A$
(c) $2.0 A$
(d) $5.0 A$
Show Answer
Answer: (c)3. A wire carries a steady current of $1.0 A$ over a period of $20 s$. What total charge passes through the wire in this time interval
(a) $200 C$
(b) $20 C$
(c) $2.0 C$
(d) $0.20 C$
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Answer: (b)4. The length of a wire is doubled and the radius is doubled. By what factor does the resistance change
(a) 4 times as large
(b) twice as large
(c) unchanged
(d) half as large
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Answer: (d)5. A circular conductor is made of a uniform wire of resistance $2 \times 10^{-3} ohm / metre$ and the diameter of this circular conductor is 2 metres. Then the resistance measured between the ends of the diameter is (in ohms)
(a) $\pi \times 10^{-3}$
(c) $4 \pi \times 10^{-3}$
(b) $2 \pi \times 10^{-3}$
(d) $4 \times 10^{-3}$
Show Answer
Answer: (c)6. A $24 V$ potential difference is applied across a parallel combination of four $6 ohm$ resistor. The current in each resistor is
(a) $1 A$
(b) $4 A$
(c) $16 A$
(d) $36 A$
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Answer: (b)7. Three resistances of 2,3 and $5 \Omega$ are connected in parallel to a $10 V$ battery of negligible internal resistance. The potential difference across the $3 \Omega$ resistance will be
(a) $2 V$
(b) $3 V$
(c) $5 V$
(d) $10 V$
Show Answer
Answer: (d)8. Two unequal resistances are connected in parallel. Which of the following statement is true
(a) current in same in both
(b) current is larger in higher resistance
(c) voltage-drop is same across both
(d) voltage-drop is lower in lower resistance
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Answer: (c)9. You are given $n$ identical wires, each of resistance R. When these are connected in parallel, the equivalent resistance is $X$. When these will be connected in series, then the equivalent resistance will be
(a) $X / n^{2}$
(b) $n^{2} X$
(c) $X / n$
(d) $nX$
Show Answer
Answer: (b)10. A piece of wire of resistance $R$ is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is $R^{\prime}$, then the ratio $R / R^{\prime}$ is
(a) $1 / 25$
(b) $1 / 5$
(c) 5
(d) 25
Show Answer
Answer: (d)11. 2 ampere current is flowing through a conductor from a 10 volt emf source then resistance of conductor is
(a) $20 \Omega$
(b) $5 \Omega$
(c) $12 \Omega$
(d) $8 \Omega$
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Answer: (b)12. Charge on an electron is $1.6 \times 10^{-19}$ coulomb. Number of electrons passing through the wire per second on flowing of 1 ampere current through the wire will be
(a) $0.625 \times 10^{-19}$
(b) $1.6 \times 10^{-19}$
(c) $1.6 \times 10^{-19}$
(d) $0.625 \times 10^{19}$
Show Answer
Answer: (d)13. 20 coulomb charge is flowing in 0.5 second from a point in an electric circuit then value of electric current in amperes will be
(a) 10
(b) 40
(c) 0.005
(d) 0.05
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Answer: (b)14. Three resistors of $4.0 \Omega, 6.0 \Omega$ and $10.0 \Omega$ are connected in series. What is their equivalent resistance
(a) $20 \Omega$
(b) $7.3 \Omega$
(c) $6.0 \Omega$
(d) $4.0 \Omega$
Show Answer
Answer: (c)15. A letter ’ $A$ ’ is constructed of a uniform wire of resistance 1 ohm per $cm$. The sides of the letter are $20 cm$ and the cross piece in the middle is $10 cm$ long. The resistance between the ends of the legs will be
(a) $32.4 ohm$
(b) $28.7 ohm$
(c) $26.7 ohm$
(d) $24.7 ohm$
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Answer: (c)16. A wire of resistance $R$ is cut into ten equal parts which are then joined in parallel. The new resistance is
(a) $0.01 R$
(b) $0.1 R$
(c) $10 R$
(d) $100 R$
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Answer: (a)17. According to international convention of colour coding in a wire
(a) live is red, neutral is black and earth is green
(b) live is red, neutral is green and earth is black
(c) live is brown, neutral is blue and earth is black
(d) live is red, neutral is black and earth is green
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Answer: (d)18. If a wire is stretched to make its length three times, its resistance will become
(a) three times
(b) one-third
(c) nine times
(d) one-ninth
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Answer: (c)19. The resistivity of a wire depends on
(a) length
(b) area of cross-section
(c) material
(d) all the above three factors
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Answer: (c)20. Ampere-second stands for the unit of
(a) power
(c) e.m.f.
(b) charge
(d) energy
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Answer: (d)21. In the circuits shown below the ammeter $A$ reads $4 amp$. and the voltmeter $V$ reads 20 volts. The value of the resistance $R$ is
(a) slightly more than 5 ohms
(b) slightly less than 5 ohms
(c) exactly 5 ohms
(d) None of the above
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Answer: (a)22. Three resistors are connected to form the sides of a triangle $ABC$ as shown below.
The resistance of side $A B$ is 40 ohms, of side $B C 60$ ohms and of side CA $100 ohms$. The effective resistance between the point $A$ and $B$ in ohms is
(a) 50
(b) 64
(c) 32
(d) 100
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Answer: (c)23. If one micro-amp. current is flowing in a wire, the number of electrons which pass from one end of the wire to the other end in one second is
(a) $6.25 \times 10^{12}$
(c) $6.25 \times 10^{18}$
(b) $6.25 \times 10^{15}$
(d) $6.25 \times 10^{19}$
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Answer: (a)24. The unit for specific resistance is
(a) ohm $\times$ second
(b) ohm $\times cm$
(c) ohm
(d) $ohm / cm$
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Answer: (b)25. If the temperature of a conductor is increased, its resistance will
(a) not increase
(b) increase
(c) decrease
(d) change according to the whether
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Answer: (b)26. The unit for electric conductivity is
(a) ohm per cm
(b) ohm $\times cm$
(c) ohm per second
(d) mho
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Answer: (c)27. Two wires of resistance $R_1$ and $R_2$ are joined in parallel. The equivalent resistance of the combination is
(a) $R_1 R_2 / R_1+R_2$
(b) $R_1+R_2$
(c) $R_1 \times R_2$
(d) $R_1 / R_2$
Show Answer
Answer: (a)28. Gases are good conductors of electricity at
(a) high pressure
(b) low pressure
(c) low temperature
(d) high temperature
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Answer: (b)29. The resistance of a thin wire in comparison of a thick wire of the same material
(a) is low
(b) is equal
(c) depends upon the metal of the wire
(d) is high
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Answer: (d)30. If the specific resistance of a wire of length $l$ and radius $r$ is $k$ then resistance is
(a) $k \pi r^{2} / l$
(b) $\pi r^{2} / l k$
(c) $k l / \pi r^{2}$
(d) $k / l r^{2}$
Show Answer
Answer: (c)31. If a charge of $1.6 \times 10^{-19}$ coulomb flows per second through any cross section of any conductor, the current constitute will be
(a) $2.56 \times 10^{-19} A$
(c) $.1 .6 \times 10^{-19} A$
(b) $6.25 \times 10^{-19} A$
(d) $3.2 \times 10^{-19} A$
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Answer: (c)32. The number of electrons flowing per second through any cross section of wire; if it carries a current of one ampere, will be
(a) $2.5 \times 10^{18}$
(c) $12.5 \times 10^{18}$
(b) $6.25 \times 10^{18}$
(d) $5 \times 10^{18}$
Show Answer
Answer: (b)33. The number of electron passing through a heater wire in one minute, if it carries a current of 8 ampere, will be
(a) $2 \times 10^{20}$
(c) $3 \times 10^{20}$
(b) $2 \times 10^{21}$
(d) $3 \times 10^{21}$
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Answer: (d)34. The total momentum of electrons in a straight wire of length $\ell=1000 m$ carrying a current $I=70 A$, will be (in N. s)
(a) $0.40 \times 10^{-6}$
(c) $0.80 \times 10^{-6}$
(b) $0.20 \times 10^{-6}$
(d) $0.16 \times 10^{-6}$
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Answer: (a)35. In a wire of length $4 m$ and diameter $6 mm$, a current of 120 ampere is passed. The potential difference across the wire is found to be 18 volt. The resistance of wire will be
(a) $0.15 ohm$
(b) $0.25 ohm$
(c) $6.660 ohm$
(d) none of the above
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Answer: (a)36. The resistance between two rectangular faces of a block of dimensions $4 cm \times 4 cm \times 10 cm$ of maganin $(r=48 \times 10^{-8} ohm, m)$, will be
(a) $4.8 \mu \Omega$
(b) $3.8 \mu \Omega$
(c) $30 \mu \Omega$
(d) $3 \mu \Omega$
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Answer: (a)37. If resistance of a wire formed by $1 . c c$ of copper be $2.46 \Omega$. The diameter of wire is $0.32 mm$, then the specific resistance of wire will be
(a) $1.59 \times 10^{-6}$ ohm. cm
(b) $2.32 \times 10^{-6}$ ohm. cm
(c) $3.59 \times 10^{-6}$ ohm. cm
(d) $1.59 \times 10^{-8}$ ohm. cm
Show Answer
Answer: (a)38. A given piece of wire length $\ell$, cross sectional area $A$ and resistance $R$ is stretched uniformly to a wire of length $2 \ell$. The new resistance will be
(a) $2 R$
(b) $4 R$
(c) $R 2$
(d) Remains unchanged
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Answer: (b)39. A given piece of wire of length $\ell$, radius $r$ and resistance $R$ is stretched uniformly to a wire of radius $(r / 2)$. The new resistances will be
(a) $2 R$
(b) $4 R$
(c) $8 R$
(d) $16 R$
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Answer: (d)40. There are two wires of the same length and of the same material and radial $r$ and $2 r$. The ratio of their specific resistance is
(a) $1: 2$
(b) $1: 1$
(c) $1: 4$
(d) $4: 1$
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Answer: (b)41. Specific resistance of a wire depends on the
(a) length of the wire
(b) area of cross-section of the wire
(c) resistance of the wire
(d) material of the wire
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Answer: (d)42. The resistance of some substances become zero at very low temperature, then these substances are called
(a) good conductors
(b) super conductors
(c) bad conductors
(d) semi conductors
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Answer: (b)43. The resistance of wire is $20 \Omega$. The wire is stretched to three time its length. Then the resistance will now be
(a) $6.67 \Omega$
(b) $60 \Omega$
(c) $120 \Omega$
(d) $180 \Omega$
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Answer: (d)44. When the resistance of copper wire is $0.1 \Omega$ and the radius is $1 mm$, then the length of the wire is (specific resistance of copper is $3.14 \times 10^{-8} ohm \times m$ )
(a) $10 cm$
(b) $10 m$
(c) $100 m$
(d) $100 cm$
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Answer: (b)45. When the resistance wire is passed through a die the crosssection area decreases by $1 %$, the change in resistance of the wire is
(a) $1 %$ decrease
(b) $1 %$ increase
(c) $2 %$ decrease
(d) $2 %$ increase
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Answer: (d)46. Two wires of the same material having lengths in the ratio of $1: 2$ and diameters in the ratio $2: 1$ are connected with a cell of 6 volt and internal resistance $1 \Omega$. The ratio of the potential difference across the two wires will be
(a) $1: 2$
(b) $2: 1$
(c) $1: 8$
(d) $8: 1$
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Answer: (c)47. The resistance $4 R, 16 R, 64 R, \ldots . . . . . \infty$ are connected in series, their resultant will be
(a) 0
(b) $\infty$
(c) $4 / 3 R$
(d) $3 / 4 R$
Show Answer
Answer: (b)48. Resistance $R, 2 R, 4 R, 8 R$…………. $\infty$ are connected in parallel. Their resultant resistance will be
(a) $R$
(b) $R / 2$
(c) 0
(d) $\infty$
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Answer: (b)49. The equivalent resistance between points $X$ & $Y$
(a) $R$
(b) $2 R$
(c) $R 2$
(d) $4 R$
Show Answer
Answer: (b)50. The equivalent resistance between points $X$ & Y
(a) $5 \Omega$
(b) $10 \Omega$
(c) $15 \Omega$
(d) $60 \Omega$
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Answer: (c)51. The equivalent resistance between points $X$ & Y
(a) $4 \Omega$
(b) $4.5 \Omega$
(c) $2 \Omega$
(d) $20 \Omega$
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Answer: (b)52. Three resistances $4 \Omega$ each of are connected in the form of an equilateral triangle. The effective resistance between two corners is
(a) $8 \Omega$
(b) $12 \Omega$
(c) $3 / 8 \Omega$
(d) $8 / 3 \Omega$
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Answer: $\text { (d) } R_{A B}=\frac{(4+4) \times 4}{4+4+4}=\frac{8}{3} \Omega$
53. Two wires of same metal have the same length but their cross-sections area in the ratio $3: 1$. They are joined in series. The resistance of the thicker wire is $10 \Omega$. The total resistance of the combination will be .
(a) $40 \Omega$
(b) $40 / 3 \Omega$
(c) $5 / 2 \Omega$
(d) $100 \Omega$
Show Answer
Answer: (a) Same metal means same sip. resistance
$ \begin{aligned} \frac{R_1}{R_2} & =\frac{A_2}{A_1}=\frac{1}{3} \Rightarrow R_2=3 R_1=3 \times 10=30 \Omega \\ R & =R_1+R_2=40 \Omega \end{aligned} $
54. A certain piece of silver of given mass is to be made like a wire. Which of the following combination of length $(L)$ and the area of cross-sectional (a) will lead to the smallest resistance
(a) L and A
(b) $2 L$ and $A / 2$
(c) $L 2$ and $2 A$
(d) Any of the above, because volume of silver remains same
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Answer: (c) $R \propto \frac{\ell}{A}$. Hence minimum for option (c)55. A certain wire has a resistance $R$. The resistance of another wire identical with the first except having twice its diameter is
(a) $2 R$
(c) $4 R$
(b) $0.25 R$
(d) $0.5 R$
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Answer: (b) $R \propto \frac{1}{A} \propto \frac{1}{r^{2}}$. Hence $\frac{1}4^{\text{th }}$56. What length of the wire of specific resistance $48 \times 10^{-8} \Omega$ $m$ is needed to make a resistance of $4.2 \Omega$ (diameter of wire $=0.4 mm$ )
(a) $4.1 m$
(c) $2.1 m$
(b) $3.1 m$
(d) $1.1 m$
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Answer: (d) $R=\frac{\rho \ell}{A} ; \quad \ell= _{\rho}^{R A}=\frac{4.2 \times \pi(\frac{0.4}{2} \times 10^{-3})^{2}}{4.8 \times 10^{-8}}=1.1 m$57. The resistance of an ideal voltmeter is
(a) zero
(c) very large
(b) very low
(d) Infinite
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Answer: (c) Ideal voltmeter should not draw any current flow source hence its resistance $=\infty$,
Practically $\infty$ resistance is not possible, but ideal voltmeter is possible with the help of potentiometer that you will learn in higher classes.
58. Masses of 3 wires of same metal are in the ratio $1: 2: 3$ and their lengths are in the ratio $3: 2: 1$. The electrical resistances are in ratio
(a) $1: 4: 9$
(c) $1: 2: 3$
(b) $9: 4: 1$
(d) $27: 6: 1$
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Answer: (d) $R=\frac{\rho \ell}{A}=\frac{d \rho \ell^{2}}{m} ; R \propto \frac{\ell^{2}}{m}$
$ [V=A \ell, d=\frac{m}{v}=\frac{m}{A \ell} \Rightarrow A= m \\ d \ell] \\ \cdot \\ R_1: R_2: R_3 \equiv \frac{9}{1}: \frac{4}{2}: \frac{1}{3} \equiv 9: 2: \frac{1}{3}=27: 6: 1 $
59. When a $12 \Omega$ resistor is connected in series with a moving coil galvanometer then its deflection reduces from 50 divisions to 10 divisions. The resistance of the galvanometer is
(a) $24 ohm$
(c) 3 ohm
(b) $36 ohm$
(d) $60 ohm$
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Answer: (c) $i \propto$ division deflected. $$ \mathrm{V}=50 \mathrm{G}, \hat{\wedge} 50 \mathrm{G}=10(G+12) ; G=3 \Omega $$60. In a Wheatstone’s bridge all the four arms have equal resistance $R$. If the resistance of the galvanometer arm is also $R$, the equivalent resistance of the combination across the battery is
(a) $R / 2$
(b) $R$
(c) $2 R$
(d) $R / 4$
Show Answer
Answer: 
As Wheatstone bridge is in balanced condition potential difference across $A B$ is zero hence resistance across $A B$ is ineffective circuit reduce to simple circuit as shown resistance $R$ in upper branch are in series and in parallel with lower branch.
$R _{e q}=R$.
61. We have two wires $A$ and $B$ of same mass and same material. The diameter of the wire $A$ is half of that $B$. If the resistance of wire $A$ is $24 ohm$ then the resistance of wire $B$ will be
(a) $12 ohm$
(b) $3.0 ohm$
(c) $1.5 ohm$
(d) None of the above
Show Answer
Answer: (c) Same material $\to$ same density, specific resistance as they are material property.
$ \begin{aligned} & R=\frac{\rho \ell}{A}=\frac{\rho V}{A^{2}}=\frac{\rho m}{d A^{2}} \\ & R \propto \frac{1}{A^{2}} \propto \frac{1}{r^{4}} \\ & \frac{R_A}{R_B}=\frac{r_B^{4}}{r_A^{4}}=2^{4}=16 \Rightarrow R_B=\frac{24}{16}=1.5 \Omega \end{aligned} $
62. The electric resistance of a certain wire of iron is $R$. If its length and radius are both doubled, then
(a) The resistance will be doubled and the specific resistance will be halved
(b) The resistance will be halved and the specific resistance will remain unchanged
(c) The resistance will be halved and the specific resistance will be doubled
(d) The resistance and the specific resistance, will both remain unchanged
Show Answer
Answer: $ \text{ (b) } \begin{aligned} R & =\frac{\rho \ell}{A}=\frac{\rho \ell}{\pi r^{2}} \\ R^{\prime} & =\frac{\rho 2 \ell}{\pi(2 r)^{2}}=\frac{R}{4} \end{aligned} $
Specific resistance will remain same as it is a material property but remember it depends on temperature.
63. When a wire of uniform cross-section $a$, length $\ell$ and resistance $R$ is bent into a complete circle, resistance between any two of diametrically opposite points will be
(a) $R / 4$
(b) $R / 8$
(c) $4 R$
(d) $R / 2$
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Answer: (a) Two resistance ( $R / 2$ ) will be in parallel, hence $R _{e q}=R / 4$64. A solenoid is at potential difference $60 V$ and current flows through it is 15 ampere, then the resistance of coil will be
(a) $4 \Omega$
(b) $8 \Omega$
(c) $0.25 \Omega$
(d) $2 \Omega$
Show Answer
Answer: (a) $V=i \times R ; R=60 / 15=4 \Omega$65. The resistance of a discharge tube is
(a) Ohmic
(b) Non-ohmic
(c) Both
(a) and
(d) Zero
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Answer: (b)66. Conductivity increases in the order of
(a) $Al, Ag, Cu$
(b) $Al, Cu, Ag$
(c) $Cu, Al, Ag$
(d) $Ag, Cu, Al$
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Answer: (b) Silver is the best conductor of electricity.67. By increasing the temperature, the specific resistance of a conductor and a semiconductor
(a) Increases for both
(b) Decreases for both
(c) Increases, decreases
(d) Decreases, increases
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Answer: (c)68. A strip of copper and another of germanium are cooled from. room temperature to $80 K$. The resistance of
(a) Each of these increases
(b) Each of these decreases
(c) Copper strip increases and that of germanium decreases
(d) Copper strip decreases and that of germanium increases
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Answer: (d) Copper is a conductor while germanium is a semiconductor. Resistance of temperature decreases with temperature while that of semi-conductor increases hence resistance of copper strip decreases and that of germanium increases.69. In the circuit shown in the figure, the current through
(a) the $3 \Omega$ resistor is $0.50 A$
(b) the $3 \Omega$ resistor is $0.25 A$
(c) the $4 \Omega$ resistor is $0.50 A$
(d) the $4 \Omega$ resistor is $0.25 A$
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Answer: (d) $2 \Omega, 4 \Omega, 2 \Omega$ on right side are in series resultant parallel to $8 \Omega$ then in series with $2 \Omega, 2 \Omega$ then in parallel with $8 \Omega$, then in series with $3 \Omega, 2 \Omega$. Thus, $R _{e q}=9 ohm$. $i=9 / 9=1 amp$ flow from battery.
Passing through $3 \Omega$ it will divide into equal parts ( $1 / 2$ amp) in $8 \Omega$ (near to cell) and remaining section then again divide into equal parts ( $1 / 4 amp$ ) in $8 \Omega$ (middle one) and remaining section hence 1/4 amp. passes through $4 \Omega$.
70. Which of the following has a negative temperature coefficient?
(a) $C$
(b) $Fe$
(c) $Mn$
(d) $Ag$
Show Answer
Answer: (a)More than One Correct:
DIRECTIONS: This section contains multiple choice questions. Each question has 4 choices (a), (b), (c) and (d) out of which ONE OR MORE may be correct.
1. A current passes through a wire of non-uniform crosssection. Which of the following quantities are independent cross-section?
(a) The charge crossing in a given time interval
(b) Drift speed
(c) Current density
(d) Free-electron density
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Answer: (a,d)2. Two conductors made of the same material have lengths $L$ and $2 L$, but have equal resistances. The two are connected in series in a circuit in which current is flowing. Which of the following is/are correct?
(a) The potential difference across the two conductors is the same
(b) The electron drift velocity is larger in the conductor of length $2 L$
(c) The electric field in the first conductor is twice than that in the second
(d) The electric field in the second conductor is twice than that in the first
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Answer: (a,b,c)3. When a potential difference is applied across the ends of a linear-metallic conductor:
(a) the free electrons are set in motion from their position of rest
(b) the free electrons are accelerated continuously from the lower potential end to the higher potential end of the conductor
(c) the free electrons are accelerated continuously from the higher potential end to the lower potential end of the conductor
(d) the free electrons acquire a constant drift velocity from the lower potential end to the higher potential end of the conductor
(e) the vibrating atomic ions in the conductor start vibrating more vigorously
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Answer: (d,e)4. Electric current arises from the flow of charged particles. Now:
(a) in metals it is dominant due to flow of electrons
(b) in semiconductors it is jointly due to flow of holes and electrons
(c) in electrolytes it is due to flow of negative ions only
(d) in discharge tubes containing gases at low pressure it is due to flow of positive ions only
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Answer: (a,d)5. The potential difference between the points $A$ and $B$ in the circuit shown here is 16 volts. Then:
(a) the current through the $2 \Omega$ resistance is $3.5 amp$
(b) the current through the $4 \Omega$ resistance is $2.5 amp$
(c) the current through the $3 \Omega$ resistance is $1.5 amp$
(d) the potential difference between the terminals of the $9 V$ battery is $7 V$
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Answer: (a,c,d)6. Electric current is due to flow of charge carriers in the conductors. Which of the following is/are correct?
(a) The drift speed of charge carriers is a very small fraction of the mean thermal agitation speed of the same charge carriers
(b) The number of charge carriers per unit volume is always the same as the number of atoms of the conductor per unit volume
(c) The drift velocity is proportional to the electric field applied ordinarily
(d) In an intrinsic semiconductor, the charge carriers are either electrons only or holes only; both of them may not participate in conduction
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Answer: (a,c)7. When some potential difference is maintained between $A$ and $B$, current $I$ enters the network at $A$ and leaves at $B$ :
(a) the equivalent resistance between $A$ and $B$ is $8 \Omega$
(b) $C$ and $D$ are at the same potential
(c) no current flows between $C$ and $D$
(d) current (3//5) flows from $D$ to $C$
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Answer: (a,b,d)
As $C$ and $D$ are joined, they must be at the same potential, and may be treated as the same point. This gives the equivalent resistance as $8 \Omega$. If we distribute current in the network, using symmetry $$ \begin{array}{ll} & V_A-V_D=V_A-V_C \ \text { or } & 20 I_1=5\left(I-I_1\right) \text { or } I_1=\frac{I}{5} \ \therefore \quad & I-2 I_1=1-\frac{2}{5}=\frac{3 I}{5} \ & =\text { Current flowing from } D \text { to } C \end{array} $$
8. In the given circuit:
(a) the current through the battery is $5.0 amp$
(b) $P$ and $Q$ are at the same potential
(c) $P$ is $\overline{2} .5 V$ higher than $Q$
(d) $Q$ is $2.5 V$ higher than $P$
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Answer: (a,d)9. Three voltmeters, all having different resistances, are joined as shown in the figure. When some potential difference is applied across $A$ and $B$, their readings are $V_1, V_2, V_3$ :
(a) $V_1=V_2$
(b) $V_1 \neq V_2$
(c) $V_1+V_2=V_3$
(d) $V_1+V_2>V_3$
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Answer: (b,c)10. In the circuit, the battery is ideal. A voltmeter of resistance $600 \Omega$ is connected in turn across $R_1$ and $R_2$, giving readings $V_1$ and $V_2$ respectively:
$ E=120 V \frac{\xi _{+}}{T^{-}} R_1=600 \Omega $
(a) $V_1=80 V$
(b) $V_1=60 V$
(c) $V_2=30 V$
(d) $V_2=40 V$
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Answer: (b, c) Treat all voltmeters and ammeters as resistances. Draw the circuit and find the currents and potential differences for each section.11. Three ammeters $A, B$ and $C$ of resistances $R_A, R_B$ and $R_C$ respectively are joined as shown in the figure. When some potential difference is applied across the terminals $T_1$ and $T_2$, their readings are $I_A, I_B$ and $I_C$ respectively:
(a) $I_A=I_B$
(b) $I_A R_A+I_B R_B=I_C R_C$
(c) $\frac{I_A}{I_C}=\frac{R_C}{R_A}$
(d) $I_B=\frac{R_C}{I_C+R_B}$
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Answer: (a,b,d)Fill in the Passage.
DIRECTIONS : Complete the following passage(s) with an appropriate word/term to be filled in the blank spaces.
I.
resistance $\quad $ conductor $\quad $ free electrons $\quad $ high potential
low potential $\quad $ positive ions $\quad $ obstructions
Conductors conduct electricity owing to the presence of a large number of …… 1….. When a potential difference is applied across a conductor these electrons begin to drift from the end at …… 2…… to the end at …… 3……. While drifting they collide with other electrons as well as fixed …… 4…… . The fixed ions are formed from the atoms which given free electrons. Thus, even as conductor conducts electricity, at the same time it offers some …… 5……. to the flow of charges.
The obstruction offered by a …… 6….. to the flow of electric current is called its …… 7…….
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Answer:
- free electrons
2 . low potential
-
high potential
-
positive ions
-
obstructions
6 . conductor
- resistance
II.
$\begin{matrix} \text{ flowing } \\ \text{ conductor }\end{matrix} $ $\begin{matrix} \text{ amount } \\ \text{ directly }\end{matrix} $ $\begin{matrix} \text{ unit \qquad time \quad surface } \\ \text{ potential difference. }\end{matrix} $
The electric current is defined as the …………… of charge for …………… that crosses the ………….., in much the same sense that a river current is the amount of water ………….. per unit time. When physical conditions remaining same, then the electric current flowing through a ………………. is …………… proportional to the ……………… across the two ends of the conductor.
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Answer: 1. amount
- unit time
3 . surface
-
flowing
-
conductor,
-
directly
-
potential difference.
III.
Starting point $\quad $ end point $\quad $ sum difference $\quad $ individual
$\begin{matrix} R_1-R_2 & R_1+R_2 \quad \text{ resistances \quad potential difference }\end{matrix} $ series.
In a series combination of resistors, the ……………. of one resistor is connected to the …………… of second resistor. Then the effective resistance of the combination is equal to the …………… of the resistances of …………… resistors. If $R_1$ and $R_2$ are the resistances of two resistors connected in series combination, their effective or net resistance is given by $R _{\text{net }}=\ldots \ldots . .5$……. This is true for ’ $n$ ’ number of resistors connected in series combination. Thus, we have
$R _{n e t}=R_1+R_2+R_3 \ldots \ldots \ldots . R_n$, where $R_1, R_2, R_3+\ldots \ldots \ldots, R_n$ are the resistances of ’ $n$ ’ resistors connected in …………… arrangement.
Thus, the net resistance of the combination obtained is greater than the largest value of resistance in the combination. In this type of combination, the potential difference across the combination of resistors is divided among the resistors according to the values of their ……..7……. Hence, the resistors in this type of combination act as dividers of ……………. .
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Answer:
- end point,
- starting point,
- sum difference,
- individual
- $R_1-R_2, R_1+R_2$
- series
- resistances
- potential difference,
Passage Based Questions:
DIRECTIONS : Study the given paragraph(s) and answer the following questions.
Passage -I
Answer the following questions based on the given circuit.
1. The potential drop across the $3 \Omega$ resistor is
(a) $IV$
(b) $15 V$
(c) $2 V$
(d) $3 V$
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Answer: (a) Equivalent resistance of $3 \Omega$ and $6 \Omega=\frac{3 \times 6}{3+6}=2 \Omega$ as they are in parallel they have same p.d. $$ i=\frac{3}{6}=\frac{1}{2} $$ P.D. across $2 \Omega=\frac{1}{2} \times 2=1$ volt2. The equivalent between points $A$ and $B$ is
(a) $7 \Omega$
(b) $6 \Omega$
(c) $13 \Omega$
(d) $5 \Omega$
3. The current flowing through in the given circuit is
(a) $0.5 A$
(b) $1.5 A$
(c) $6 A$
(d) $3 A$
Passage-II
Answer the following questions based on the given circuit.
1. The equivalent resistance between points $A$ and $B$, is
(a) $12 \Omega$
(b) $36 \Omega$
(c) $32 \Omega$
(d) $24 \Omega$
2. The current through each resistor is (a) $1 A$
(b) $2.3 A$
(c) $0.5 A$
(d) $0.75 A$
3. The potential drop across the $12 \Omega$ resistor is
(a) $12 V$
(b) $6 V$
(c) $8 V$
(d) $0.5 V$
Passage-III
Answer the following questions based on the given circuit.
1. The equivalent resistance between points $A$ and $B$
(a) $6.2 \Omega$
(b) $5.1 \Omega$
(c) $13.33 \Omega$
(d) $1.33 \Omega$
2. The current through the battery is
(a) $2.33 A$
(b) $3.12 A$
(c) $4.16 A$
(d) $5.19 A$
3. The current through the $4.0 ohm$ resistor is
(a) $5.6 A$
(b) $0.98 A$
(c) $0.35 A$
(d) $0.68 A$
Passage -IV
Answer the following questions based on the given circuit.
1. The total resistance of the circuit is
(a) $2 \Omega$
(b) $4 \Omega$
(c) $1.5 \Omega$
(d) $0.5 \Omega$
2. The current flowing through $0.5 \Omega$ resistor is
(a) $1 A$
(b) $1.5 A$
(c) $3 A$
(d) $2.5 A$
3. The current flowing through $6 \Omega$ resistor is
(a) $0.50 A$
(b) $0.75 A$
(c). $0.80 A$
(d) $0.25 A$
Assertion & Reason:
DIRECTIONS : Each of these questions contains an Assertion followed by reason. Read them carefully and answer the question on the basis of following options. You have to select the one that best describes the two statements.
(a) If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
(b) If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
(c) If Assertion is correct but Reason is incorrect.
(d) If Assertion is incorrect but Reason is correct.
1. Assertion : When a battery is short-circuited, the terminal voltage is zero.
Reason : In the situation of a short-circuit, the current is zero
2. Assertion : The equation $V=R i$ does not apply to those conducting devices which do not obey Ohm’s law.
Reason : $V=Ri$ is a statement of Ohm’s law.
3. Assertion : All electric devices shown in the circuit are ideal: The reading of each of ammeter (a) and voltmeter (V) is zero.
Reason : An ideal voltmeter draws almost no current due to very large resistance, and hence $(V)$ and (a) will read zero.
4. Assertion : If $\rho_1$ and $\rho_2$ be the resistivities of the materials of two resistors of resistances $R_1$ and $R_2$ respectively and $R_1>R_2$, then $\rho_1>\rho_2$.
Reason : The resistance $R=\rho \frac{\ell}{A} \Rightarrow \rho_1>\rho_2$ if $R_1>R_2$
5. Assertion : The product of resistivity and conductivity of a conductor depends on the material of the conductor. Reason : Because each of resistivity and conductivity depends on the material of the conductor.
6. Assertion : Insulators do not allow flow of current through themselves.
Reason : They have no free-charge carriers.
7. Assertion : Positive charge inside the cell always goes from positive terminal to the negative terminal.
Reason : Positive charge inside the cell may go from negative terminal to the positive terminal.
8. Assertion : Wire $A$ is thin in comparison to wire. $B$ of same material same length then resistance of wire $A$ is greater than resistance of wire $B$.
Reason : Resistivity of wire $A$ is greater than resistance of wire B.
9. Assertion : Resistivity of material may change with temperature.
Reason : Resistivity is a material property & independent on temperature.
Multiple Matching Questions:
DIRECTIONS : Following question has four statements $(A, B$, $C$ and $D$ ) given in Column I and four statements ( $p, q, r$ and $s$ ) in Column II. Any given statement in Column I can have correct matching with one or more statement(s) given in Column II. Match the entries in column I with entries in column II.
1.
| Column I | Column II |
|---|---|
| (A) Ohm’s Law | (p) Direct proportional to area |
| (B) Resistivity | (q) Voltage $\alpha$ current |
| (C) For Ohmic-conductor | (r) $\frac{\text{ charge }}{\text{ time }}$ |
| (D) Electric current | (s) $V=I R$ |
2.
| Column I | Column II | |
|---|---|---|
| (A) |  |
(p) $R _{eq}=1 \Omega, I=12 A$ |
| (B) |  |
(q) $\alpha$ Length |
| (C) | Resistance | (r) $R _{\text{eq }}=1 \Omega, I=6 A$ |
| (D) |  |
(s) $\alpha \frac{1}{\text{ Area }}$ |
3.
| Column I | Column II |
|---|---|
| (A) Ohm | (p) 1 Volt/ 1 Amp |
| (B) Current | (q) Depends on matter of conductor |
| (C) Resistivity | (r) $\frac{\text{ charge }}{\text{ time }}$ |
| (D) Super conductor | (s) $\frac{\text{ Resistance }}{\text{ Zero }}$ |
HOTS Subjective Questions:
1. In the circuit shown, if a wire is connected between points A and B. How much current will flow through that wire?
2. A potential difference $\mathbf{V}$ is applied across a conductor of length $L$ and diameter $D$. How the electric field $E$ and resistance $R$ effected when
(i) $V$ is halved
(ii) $L$ is halved
(iii) $D$ is doubled
Justify your answer in each case.
3. What is the number of electrons that would flow per second through the cross-section of a wire when 1 A current flows in it?
4. $0.7 C$ charge passes through a cross-section of a conductor in $7 s$. Find the current.
5. A current of $1.2 A$ flows through a conductor for $3.0 s$. What amount of charge passes through the conductor?
6. Name a metal which offers higher resistance to the passage of electricity other than copper.
7. Define the term resistivity of a material.
8. Calculate the resistance of a conductor, if the current flowing through it is $0.2 A$ when the applied potential difference is 0.8 volt.
9. A cylinder of a material is $10 cm$ long and has a cross-section of $2 cm^{2}$. If its resistance along the length be $20 \Omega$, what will its resistivity value be in number and units?
10. Two wires, one of manganin and the other of copper have equal lengths and resistances. Which one of these wires will be thicker?
11. A student obtains resistances $3,4,12$ and $16 \Omega$ using only two metallic resistance wires are either separately or joined together. What is the value of resistance of each of these wires?
