NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity

NCERT
SOLUTIONS
CLASS - 12th

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Class : 12th
Subject : Physics
Chapter : 3
Chapter Name : Current Electricity

Q3.1 The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4 Ω, what
is the maximum current that can be drawn from the battery?

Answer. Emf of the battery, E = 12V
Internal resistance of the battery = r = 0.4Ω
Maximum current drawn from the battery = I
According to ohm’s law,
E = Ir
E
I= r
12
= 0.4 = 30A
The maximum current drawn from the given battery is 30 A.

Page : 127 , Block Name : Exercise

Q3.2 A battery of emf 10 V and internal resistance 3 Ω is connected to a resistor. If the current in the
circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the
circuit is closed?

Answer. Emf of the battery , E = 10V
Internal resistance of the battery, r = 3Ω
Current in the circuit , I = 0.5A
Resistance of the resistor = R
The relation of the current using ohm’s law is,
E
I = R+r
E
R+r= I
10
= 0.5 = 20Ω
∴ R = 20 − 3 = 17Ω
Terminal voltage of the resistor = V
According to Ohm’s law,
V=IR
= 0.5 x 17
= 8.5 V
Therefore , the resistance of the resistor is 17 Omega and the terminal voltage is 8.5 V.

Page : 127 , Block Name : Exercise

Q3.3 (a) Three resistors 1 Ω, 2 Ω, and 3 Ω are combined in series. What is the total resistance of the
combination?
(b) If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the
potential drop across each resistor.

Answer. (a) Three resistor of resistance 1 Ω , 2 Ω and 3 Ω are combined in series. Total resistance of the
combination is given by the algebraic sum of individual resistances. Total resistance = 1+2+3 =6 Ω.
(b) Current owing through the circuit = I
Emf of the battery , E = 12V
Total resistance of the circuit, R = 6Omega .
The relation of the current using Ohm’s law is ,
E
I= R
12
= 6 = 2A
Potential drop across 1Ω resistor = V 1
From Ohm’s law , the value of V 1 can be obtained as
V 1 = 2 x 1 = 2V …..(i)
Potential drop across 2Ω resistor = V 2
Again from Ohm’s law , the value of V 2 can be obtained as
V 2 = 2 x 2 = 4V …..(ii)
Potential drop across 3Ω resistor = V 3
Again from Ohm’s law , the value of V 3 can be obtained as
V 3 = 2 x 3 = 6V …..(iii)
Therefore, the potential drop across 1Ω, 2Ω and 3 Ω
Resistor are 2V, 4V and 6V respectively.

Page : 127 , Block Name : Exercise

Q3.4 (a) Three resistors 2 Ω, 4 Ω and 5 Ω are combined in parallel. What is the total resistance of the
combination?
(b) If the combination is connected to a battery of emf 20 V and negligible internal resistance, determine
the current through each resistor, and the total current drawn from the battery.

Answer. (a) There are three resistors of resistance, R 1 = 2Ω, R 2 = 4Ω, and R 3 = 5Ω
They are connected in parallel. Hence , total resistance ® of the combination is given by,
1 1 1 1
R
= R + R + R
1 2 3
1 1 1 10 + 5 + 4 19
= 2 + 4 + 5 = 20
= 20
20
∴ R = 19 Ω
20
Therefore, total resistance of the combination si 19 Ω.
(b) Emf of the battery, V = 20V
Current I 1 ( ) owing through resistor R1 is given by,
V
I1 = R
1

20
= 2 = 10A

( )
Current I 2 Flowing through resistor R 2 is given by,
V
I2 = R
2
20
= 4 = 5A

( )
Current I 3 Flowing through resistor R 3 is given by,
V
I3 = R
3
20
= 5 = 4A
Total current I = I 1 + I 2 + I 3 = 10 + 5 + 4 = 19A
Therefore, the current through each resistor is 10A, 5A, and 4A respectivly and the total current is 19A.

Page : 127 , Block Name : Exercise

Q3.5 At room temperature (27.0 °C) the resistance of a heating element is 100 Ω. What is the temperature
of the element if the resistance is found to be 117 ,Ω given that the temperature coef cient of the material
of the resistor is 1.70 × 10 − 4C − 1

Answer. Room temperature, T = 27.0 °C
Resistance of the heating element at T,R = 100 Ω
Let T 1 is the increased temperature of the lament.
Resistance of the heating element at T 1,R 1 = 117 Ω
Temperature coef cient of the material of the lament,
α = 1.70 × 10 − 4C − 1
alpha is given by relation,
R1 − R
α=
(
R T1 − T )
R1 − R
T1 − T = Rα
117 − 100
T 1 − 27 =
(
100 1.7 × 10 − 4 )
T 1 − 27 = 1000
T 1 = 1027 ∘ C
Therefore, at 1027 °C, the resistance of the element is 117 Ω.

Page : 127 , Block Name : Exercise

Q3.6 A negligibly small current is passed through a wire of length 15 m and uniform cross-section
6.0 × 10 − 7m 2 , and its resistance is measured to be 5.0 Ω. What is the resistivity of the material at the
temperature of the experiment?

Answer. Length of the wire, L = 15 m
Area of the cross-section of the wire, a= 6.0 × 10 − 7m 2
Resistance of the material of the wire,R = 5.0 Ω

Resistivity of the material of the wire = ρ
Resistance is related to resistivity as
l
R = ρA
RA
ρ= l
5 × 6 × 10 − 7
= 15
= 2 × 10 − 7Ωm
therefore , the resistivity of the material is 2 × 10 − 7Ωm.

Page : 127 , Block Name : Exercise

Q3.7 A silver wire has a resistance of 2.1 Ω at 27.5 °C, and a resistance of 2.7 Ω at 100 °C. Determine the
temperature coef cient of resistivity of silver.

Answer. Temperature, T 1 = 27.5 ∘ C
Resistance of the silver wire at T 1, R 1 = 2.1Ω
Temperature , T 2 = 100 ∘ C
Resistance of the silver wire at T 2, R 2 = 2.7Ω
Temperature coef cient of silver = alpha
It is related with temperature and resistance as
R2 − R1
α=
(
R1 T2 − T1 )
2.7 − 2.1
= 2.1 ( 100 − 27.5 ) = 0.0039 ∘ C − 1
Therefore, the temperature coef cient of silver is 0.0039 ∘ C − 1

Page : 127 , Block Name : Exercise

Q3.8 A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A
which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heating
element if the room temperature is 27.0 °C? Temperature coef cient of resistance of nichrome averaged
over the temperature range involved is 1.70 × 10 − 4 ∘ C − 1.

Answer. Supply voltage, V = 230V
Initial current drawn, I 1 = 3.2A
Initial resistance = R 1, which is given by the relation,
V
R1 = I
230
= 3.2 = 71.87Ω
Steady state value of the current, I 2 = 2.8A
Resistance at the steady state = R 2 which is given as
230
R 2 = 2.8 = 82.14Ω
Temperature coef cient of nichrome, a = 1.70 × 10 − 4 ∘ C − 1
Initial temperature of nichrome , T 1 = 27.0 ∘ C
T 2 Can be obtained by the relation for α,

R2 − R1
α=
(
R1 T2 − T1 )
82.14 − 71.87
T 2 − 27 ∘ C = = 840.5
71.87 × 1.7 × 10 − 4
∘
T 2 = 840.5 + 27 = 867.5 C
Therefore, the steady temperature of the heating element is 867.5 °C

Page : 127 , Block Name : Exercise

Q3.9 Determine the current in each branch of the network shown in Fig

Answer. Current through various branches of the circuit is represented in the given gure.

I 1 = current owing through the outer circuit
I 2 = current owing through branch AB
I 3 = current owing through branch AD
I 4 = current owing through branch BD
I 2- I 4 = current owing through branch BC
I 3 + I 4 = current owing through branch CD
For the closed circuit ABDA, potential is zero i.e.,
10I 2 + 5I 4 − 5I 3 = 0
2I 2 + I 4 − I 3 = 0
I 3 = 2I 2 + I 4…(1)
For second closed circuit BCDB, potential is zero i.e.,

( ) ( )
5 I 2 − I 4 − 10 I 3 + I 4 − 5I 4 = 0
5I 2 + 5I 4 − 10I 3 − 10I 4 − 5I 4 = 0
5I 2 − 10I 3 − 20I 4 = 0
I 2 = 2I 3 + 4I 4…(2)
For second closed circuit ABCFEA, potential is zero i.e.,

( ) ( ) (
− 10 + 10 I 1 + 10 I 2 + 5 I 2 − I 4 = 0 )
10 = 15I 2 + 10I 1 − 5I 4
3I 2 + 2I 1 − I 4 = 2…. (3)
From equation (1) and (2), we obtain

(
I 3 = 2 2I 3 + 4I 4 + I 4 )
I 3 = 4I 3 + 8I 4 + I 4
− 3I 3 = 9I 4
− 3I 4 = + I 3…(4)
Putting the equation (4) ij equation (1), we obtain
I 3 = 2I 2 + I 4
− 4I 4 = 2I 2
I 2 = − 2I 4…(5)
It is evident from the given gure that ,
I 1 = I 3 + I 2…(6)
Putting the equation (6) ij equation (1), we obtain

( )
3I 2 + 2 I 3 + I 2 − I 4 = 2
5I 2 + 2I 3 − I 4 = 2…(7)
Putting the equation (4) and (5) in equation (7), we obtain

( ) (
5 − 2I 4 + 2 − 3I 4 − I 4 = 2 )
− 10I 4 − 6I 4 − I 4 = 2
17I 4 = − 2
−2
I 4 = 17 A
Equation (4) reduce to
I3 = − 3 I4 ( )
= − 3 17 ( ) −2 6
= 17 A

I2 = − 2 I4 ( )
= − 2 17 ( ) −2 4
= 17 A

I 2 − I 4 = 17 −
4
( ) −2
17
6
= 17 A

I 3 + I 4 = 17 +
6
( ) −2
17
4
= 17 A

I1 = I3 + I2
6 4 10
= 17 + 17 = 17 A
4
Therefore the current in branch AB = 17 A
6
In branch BC = 17 A
−4
In branch CD = 17 A
6
In branch AD = 17 A

In branch BD =
( )
4
−2
17
A
6 −4 6 −2 10
Total current = 17 + 17 + 17 + 17 + 17 = 17 A

Page : 128 , Block Name : Exercise

Q3.10 (a) In a meter bridge, the balance point is found to be at 39.5 cm from the end A, when the resistor Y
is of 12.5 Ω. Determine the resistance of X. Why are the connections between resistors in a Wheatstone or
meter bridge made of thick copper strips?
(b) Determine the balance point of the bridge above if X and Y are interchanged.
(c) What happens if the galvanometer and cell are interchanged at the balance point of the bridge? Would
the galvanometer show any current?

Answer. (a)A meter bridge with resistor X and Y si represented in given gure.

Balance point from end A , I 1 = 39.5cm
Resistance of the resistor Y = 12.5 Ω
Condition for the balance is given as,
X 100 − l 1
Y = l1
100 − 39.5
X= 39.5
× 12.5 = 8.2Ω
Therefore, the resistance of resistor X is 8.2 Ω .
The connection between resistors in a wheatstone or meter bridge is made of thick copper strips to
minimize the resistance, which is not taken into consideration in the bridge formula.

(b) if X and Y are interchanged, then l 1 and 100 - l 1 get interchanged.
The balance point of the bridge will be 100 - l 1 from A.
100 - l 1 = 100 - 39.5 = 60.5 cm
Therefore, the balance point is 60.5 cm from A.

(c) When the galvanometer and cell are interchanged at the balance point of the bridge, the galvanometer
will show no de ection. Hence , no current would ow through the galvanometer.

Page : 128 , Block Name : Exercise

Q3.11 A storage battery of emf 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V dc supply
using a series resistor of 15.5 Ω. What is the terminal voltage of the battery during charging? What is the
purpose of having a series resistor in the charging circuit?

Answer. Emf of the storage battery, E = 8.0 V
Internal resistance of the battery, r = 0.5 Ω
DC supply voltage, V = 120 V
Resistance of the resistor, R = 15.5 Ω
Effective voltage in the circuit = V 1
R is connected in the storage battery in series. Hence, it can be written as
V1 = V − E
V ′ = 120 − 8 = 112V
Current owing in the circuit = I, which is given by the relation,
V1
I = R+r
112 112
= 15.5 + 5 = 16 = 7A
Voltage across resistor R given by the product, IR = 7 x 15.5 = 108.5 V
DC supply voltage = Terminal voltage of battery + voltage drop across R
Terminal voltage of the battery = 120 - 108.65 = 11.5 V
A series resistor in a charging circuit limits the current drawn from the external source. The current will be
extremely high in its absence. This is very dangerous.

Page : 128 , Block Name : Exercise

Q3.12 In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length of the
wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm, what is the emf of the
second cell?

Answer. Emf of cell, E 1 = 1.25V
Balance point of the potentiometer, l 1 = 35cm
The cell is replaced by another cell of emf E 2.
New balance point of the potentiometer, l 2 = 63cm
The balance condition is given by the relation,
E1 l1
E2
= l
2
l2
E2 = E1 × l
1

63
= 1.25 × 35 = 2.25V
Therefore , emf of the second cell is 2.25 V.

Page : 128 , Block Name : Exercise

Q3.13 The number density of free electrons in a copper conductor estimated in Example 3.1 is
8.5 × 10 28m − 3 . How long does an electron take to drift from one end of a wire 3.0 m long to its other end?
The area of cross-section of the wire is 2.0 × 10 − 6m 2 and it is carrying a current of 3.0 A.

Answer. Number density of free electrons in a copper conductor , n = 8.5 × 10 28m − 3 Length of the copper
wire, I = 3.0m
Area of cross-section of the wire , A = 2.0 × 10 − 6m 2
Current carried by the wire, I = 3.0A A which gives by the relation,
I = nAeV d
where
e = Electric charge = 1.6 × 10 − 19C
Length of the wire ( l )
= Time taken to cover l ( t )
V d = Drift velocity
l
I = nAe t
nAel
t= I
3 × 8.5 × 10 28 × 2 × 10 − 6 × 1.6 × 10 − 19
= 3.0
4
= 2.7 × 10 s
therefore , the time taken by an electron to drift from one end of the wire to the other is 2.7 × 10 4s.

Page : 128 , Block Name : Exercise

Q3.14 The earth's surface has a negative surface charge density of 10 − 9Cm − 2. The potential difference of
400 kV between the top of the atmosphere and the surface results (due to the low conductivity of the lower
atmosphere) in a current of only 1800 A over the entire globe. If there were no mechanism of sustaining
atmospheric electric eld, how much time (roughly) would be required to neutralise the earth's surface?
(This never happens in practice because there is a mechanism to replenish electric charges, namely the
continual thunderstorms and lightning in different parts of the globe). (Radius of earth = 6.37 × 10 6m)

Answer. Surface charge density of the earth, σ = 10 − 9Cm − 2
Current over the entire globe, I = 1800 A
Radius of the earth, r = 6.37 × 10 6m
Surface area of the earth,
A = 4nr 2

(
= 4π × 6.37 × 10 6 ) 2

= 5.09 × 10 14m 2
Charge on the earth surface,

q=σ×A
= 10 − 9 × 5.09 × 10 14
= 5.09 × 10 5C
Time taken to neutralize the earth's surface = t
q
I= t
Current,
q
t=
I
5.09 × 10 5
= = 282.77s
1800
Therefore, the time taken to neutralize the earth's surface is 282.77 s.

Page : 128 , Block Name : Additional Exercise

Q3.15 (a) Six lead-acid type of secondary cells each of emf 2.0 V and internal resistance 0.015 Ω are joined
in series to provide a supply to a resistance of 8.5 Ω. What are the current drawn from the supply and its
terminal voltage?
(b) A secondary cell after long use has an emf of 1.9 V and a large internal resistance of 380 Ω. What
maximum current can be drawn from the cell? Could the cell drive the starting motor of a car?

Answer. (a) Number of secondary cells, n = 6
Emf of each secondary cell, E = 2.0 V
Internal resistance of each cell, r = 0.015 Ω
series resistor is connected to the combination of cells.
Resistance of the resistor, R = 8.5 Ω
Current drawn from the supply = I, which is given by the relation,
nE
I=
R + nr
6×2
=
8.5 + 6 × 0.015
12
= = 1.39A
8.59
Terminal voltage, V = IR = 1.39 x 8.5 = 11.87 A
Therefore, the current drawn from the supply is 1.39 A and terminal voltage is 11.87 A.
(b) After a long use, emf of the secondary cell, E = 1.9 V
Internal resistance of the cell, r = 380 Ω
E 1.9
Hence, maximum current = r = 380 = 0.005A
Therefore, the maximum current drawn from the cell is 0.005 A. Since a large current is required to start
the motor of a car, the cell cannot be used to start a motor.

Page : 129 , Block Name : Additional Exercise

Q3.16 Two wires of equal length, one of aluminium and the other of copper have the same resistance.
Which of the two wires is lighter? Hence explain why aluminium wires are preferred for overhead power
cables. (ρ Al = 2.63 × 10 − 8Ωm, ρ Cu = 1.72 × 10 − 8Ωm Relative density of Al = 2.7, of Cu = 8.9)

Answer. Resistivity of aluminium,ρ Al = 2.63 × 10 − 8Ωm
Relative density of aluminium, d 1 = 2.7
Let l 1 be the length of aluminium wire and m 1 be its mass.
Resistance of the aluminium wire = = R 1
Area of cross-section of the aluminium wire = = A 1
Resistivity of copper, ρ Cu = 1.72 × 10 − 8Ωm
Relative density of copper, d 2 = 8.9
Let l 2 be the length of copper wire and m 2 be its mass.
Resistance of the copper wire = = R 2
Area of cross-section of the copper wire = = A 2
The two relations can be written as
l1
R 1 = ρ 1 A ……….(1)
1
l2
R 2 = ρ 2 A ……..(2)
2
It is given that,
R1 = R2
l1 i2
A A = ρ2 A
1 3
And,
l1 = l2
ρ1 ρ2
∴ A = A
1 2
A1 ρ1
=
A2 ρ2
2.63 × 10 − 8
2.63
= −8
=
1.72 × 10 1.72
Mass of the aluminium wire,
m 1 = Volume x Density
= A 1 / 1 × d 1 = A 1 / 1d 1…(3)
Mass of the copper wire,
m 2 = Volume x Density
= A 2l 2 × d 2 = A 2 / 2d 2…....(4)
Dividing equation (3) by equation (4), we obtain
m1 A 1l 1d 1
m 2 = A 2l 2d 2

For l 1 = l 2
m1 A 1d 1
m 2 = A 2d 2
A1 2.63
For A = 1.72
2
m1 2.63 2.7
m 2 = 1.72 × 8.9 = 0.46
It can be inferred from this ratio that m 1 is less than m 2. Hence, aluminium is lighter than copper Since

aluminium is lighter, it is preferred for overhead power cables over copper.

Page : 129 , Block Name : Additional Exercise

Q3.17 What conclusion can you draw from the following observations on a resistor made of alloy
manganin?

Answer. It can be inferred from the given table that the ratio of voltage with current is a constant, which is
equal to 19.7. Hence, manganin is an ohmic conductor i.e., the alloy obeys Ohm's law. According to Ohm's
law, the ratio of voltage with current is the resistance of the conductor. Hence, the resistance of manganin
is 19.7 Ω.

Page : 129 , Block Name : Additional Exercise

Q3.18 Answer the following questions:
(a) A steady current ows in a metallic conductor of non-uniform cross-section. Which of these quantities
is constant along the conductor: current, current density, electric eld, drift speed?
(b) Is Ohm’s law universally applicable for all conducting elements? If not, give examples of elements
which do not obey Ohm’s law.
(c) A low voltage supply from which one needs high currents must have very low internal resistance. Why?
(d) A high tension (HT) supply of, say, 6 kV must have a very large internal resistance. Why?

Answer. (a) When a steady current ows in a metallic conductor of non-uniform cross-section, the current
owing through the conductor is constant. Current density, electric eld, and drift speed are inversely
proportional to the area of cross-section. Therefore, they are not constant.
(b) No, Ohm's law is not universally applicable for all conducting elements. Vacuum diode semiconductor
is a non-ohmic conductor. Ohm's law is not valid for it.
(c) According to Ohm's law, the relation for the potential is V = IR Voltage (V) is directly proportional to
V
current (I). R is the internal resistance of the source. I = R
If V is low, then R must be very low, so that high current can be drawn from the source.
(d) In order to prohibit the current from exceeding the safety limit, a high tension supply must have a large
internal resistance. If the internal resistance is not large, then the current drawn can exceed the safety
limits in case of a short circuit.

Page : 129 , Block Name : Additional Exercise

Q3.19 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 coef cients of resistance than pure metals.
(c) The resistivity of the alloy manganin is nearly independent of/ increases rapidly with increase of
temperature.
(d) 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 3).

Answer. (a) Alloys of metals usually have greater resistivity than that of their constituent metals.
(b) Alloys usually have lower temperature coef cients of resistance than pure metals
(c) The resistivity of the alloy, manganin, is nearly independent of increase of temperature.
(d) The resistivity of a typical insulator is greater than that of a metal by a factor of the order of 10 22.

Page : 129 , Block Name : Additional Exercise

Q3.20 (a) 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?
(b) Given the resistances of 1 Ω, 2 Ω, 3 Ω, how will be combine them to get an equivalent resistance of (i)
(11/3) Ω (ii) (11/5) Ω, (iii) 6 Ω, (iv) (6/11) Ω?
(c) Determine the equivalent resistance of networks shown in Fig.

Answer. (a) Total number of resistors = n
Resistance of each resistor = R
(i) When n resistors are connected in series, effective resistance R 1 is the maximum, given by the product
nR.
Hence, maximum resistance of the combination, R 1= nR
(ii) When n resistors are connected in parallel, the effective resistance R 2 is the
R
minimum, given by the ratio n .
R
Hence, minimum resistance of the combination, R 2 = n
R1 nR
(iii) The ratio of the maximum to the minimum resistance is, R = R = n2
2
n

(b) The resistance of the given resistors is,R 1 = 1Ω, R 2 = 2Ω tR 3 = 3Ω2
11
(i) Equivalent resistance,R ∗ = 3 Ω

Consider the following combination of the resistors.

Equivalent resistance of the circuit is given by,
2×1 2 11
R′ = 2 + 1 + 3 = 3 + 3 = 3 Ω
11
(ii) Equivalent resistance,R ′ = 5 Ω
Consider the following combination of the resistors.

Equivalent resistance of the circuit is given by,
2×3 6 11
R′ = 2 + 3 + 1 = 5 + 1 = 5 Ω
(iii) Equivalent resistance, R = 6Ω
Consider the series combination of the resistors, as shown in the given circuit

Equivalent resistance of the circuit is given by the sum,
R ′ = 1 + 2 + 3 = 6Ω
6
(iv) Equivalent resistance,R ′ = 11 Ω
Consider the series combination of the resistors, as shown in the given circuit.

Equivalent resistance of the circuit is given by,
1×2×3 6
R ′ = 1 × + 2 × 3 + 3 × 1 = 11 Ω

(c)(a) It can be observed from the given circuit that in the rst small loop, two resistors of resistance 1 Ω
each are connected in series.
Hence, their equivalent resistance = (1+1) = 2Ω
It can also be observed that two resistors of resistance 2Ω each are connected in series.
Hence, their equivalent resistance = (2 + 2) = 4Ω.
Therefore, the circuit can be redrawn as

It can be observed that 2Ω and 4Ω resistors are connected in parallel in all the four loops. Hence,

( )
equivalent resistance R ′ of each loop is given by,
2×4 8 4
R′ = 2 + 4 = 6 = 3 Ω
The circuit reduces to,

All the four resistors are connected in series.
4 16
Hence, equivalent resistance of the given circuit is 3 × 4 = 3 Ω
(b) It can be observed from the given circuit that ve resistors of resistance Reach are connected in series.
Hence, equivalent resistance of the circuit = R+R+R+R+R = 5 R

Page : 129 , Block Name : Additional Exercise

Q3.21 Determine the current drawn from a 12V supply with internal resistance 0.5Ω by the in nite
network shown in Fig. Each resistor has 1Ω resistance.

Answer. The resistance of each resistor connected in the given circuit, R = 1Ω
Equivalent resistance of the given circuit =R ′
The network is in nite. Hence, equivalent resistance is given by the relation,
R
∴ R′ = 2 + ( R + 1 )

(R ) − 2R − 2 = 0
′ 2 ′

2 ± √4 + 8
R=
2
2 ± √12
= = 1 ± √3
2
Negative value of R cannot be accepted. Hence, equivalent resistance,
R ′ = (1 + √3) = 1 + 1.73 = 2.73Ω
Internal resistance of the circuit, r = 0.5 Ω
Hence, total resistance of the given circuit = 2.73 +0.5 = 3.23Ω
Supply voltage, V = 12 V
12
According to Ohm's Law, current drawn from the source is given by the ratio 3.23 = 3.72 A.

Page : 130 , Block Name : Additional Exercise

Q3.22 Figure shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Ω maintaining a
potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for
very moderate currents upto a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very
low currents drawn from the standard cell, a very high resistance of 600 kΩ is put in series with it, which is
shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf ε and the
balance point found similarly, turns out to be at 82.3 cm length of the wire.

(a) What is the value ε?
(b) What purpose does the high resistance of 600 kΩ have?
(c) Is the balance point affected by this high resistance?
(d) Is the balance point affected by the internal resistance of the driver cell ?
(e) Would the method work in the above situation if the driver cell of the potentiometer had an emf of 1.0V
instead of 2.0V?
(f) Would the circuit work well for determining an extremely small emf, say of the order of a few mV (such
as the typical emf of a thermo-couple)? If not, how will you modify the circuit?

Answer. (a) Constant emf of the given standard cell, E 1 = 1.02 V
Balance point on the wire, l 1 = 67.3 cm
A cell of unknown emf, ε replaced the standard cell. Therefore, new balance point on the wire, = 82.3 cm
The relation connecting emf and balance point is,
E1 ε
=
l1 l
l
ε= × E1
l1
82.3
= × 1.02 = 1.247V
67.3
The value of unknown em s 1.247 V.
(b) The purpose of using the high resistance of 600 kΩ is to reduce the current through the galvanometer
when the movable contact is far from the balance point.
(c) The balance point is not affected by the presence of high resistance.
(d) The point is not affected by the internal resistance of the driver cell
(e) The method would not work if the driver cell of the potentiometer had an emf of 1.0 V instead of 2.0 V.
This is because if the emf of the driver cell of the potentiometer is less than the emf of the other cell, then
there would be no balance point on the wire.
(f) The circuit would not work well for determining an extremely small emf. As the circuit would be
unstable, the balance point would be close to end A. Hence, there would be a large percentage of error. The
given circuit can be modi ed if a series resistance is connected with the wire AB. The potential drop across
AB is slightly greater than the emf measured. The percentage error would be small.

Page : 130 , Block Name : Additional Exercise

Q3.23 Figure shows a 2.0 V potentiometer used for the determination of internal resistance of a 1.5 V cell.
The balance point of the cell in open circuit is 76.3 cm. When a resistor of 9.5 Ω is used in the external
circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. Determine the
internal resistance of the cell.

Answer. Internal resistance of the cell = r
Balance point of the cell in open circuit, l 1 = 76.3 cm
An external resistance (R) is connected to the circuit with R = 9.5 Ω
New balance point of the circuit, l 2 = 64.8 cm
Current owing through the circuit = I
The relation connecting resistance and emf is,

r=
( )
l1 − l2
l2
R

76.3 − 64.8
= × 9.5 = 1.68Ω
64.8
Therefore, the internal resistance of the cell is 1.68Ω.

Page : 131 , Block Name : Additional Exercise