NCERT Solutions For Class 12 Physics Chapter 3 Current Electricity

NCERT Solutions For Class 12 Physics Chapter 3 gives us an introduction to the motion of electrons, and current electricity, solving the problems at the end of the chapter. Class 12 Physics NCERT Solutions Chapter 3 is provided in a detailed manner, so the students can easily understand and practice the solutions. This chapter introduces the concept of motion of electrons and the effects that arise due to it, current electricity. This chapter is a continuation of the topics and concepts that are introduced in the previous chapter. They can read NCERT Class 12 Physics Solutions for more solutions regarding other chapters. They can also check this NCERT Solutions For Class 12 Physics Chapter 3 post for reference.

Class 12 Physics NCERT Solutions Chapter 3 Current Electricity:

Topic-wise overview of NCERT Solutions For Class 12 Physics Chapter 3 is given in the following table.

Section Name Topic Name
3 Current Electricity
3.1 Introduction
3.2 Electric Current
3.3 Electric Currents in Conductors
3.4 Ohm’s law
3.5 Drift of Electrons and the Origin of Resistivity
3.6 Limitations of Ohm’s Law
3.7 Resistivity of Various Materials
3.8 Temperature Dependence of Resistivity
3.9 Electrical Energy, Power
3.10 Combination of Resistors — Series and Parallel
3.11 Cells, EMF, Internal Resistance
3.12 Cells in Series and in Parallel
3.13 Kirchhoff’s Rules
3.14 Wheatstone Bridge
3.15 Meter Bridge
3.16 Potentiometer

Class 12 Physics NCERT Solutions Chapter 3 For Problems Given at the End of the Lesson:

Question 3.1:

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

ADVERTISEMENT

Solution:
EMF of the battery, E = 12 V
The internal resistance of the battery, r = 0.4 Ω
Maximum current drawn from the battery = I
According to Ohm’s law,
V = IR
⇒ I = V/R
= 12/0.4 = 30 A
The maximum current drawn from the given battery is 30 A.


Question 3.2:

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?

ADVERTISEMENT

Solution:

EMF of the battery, E = 10 V
The internal resistance of the battery, r = 3 Ω
Current in the circuit, I = 0.5 A
Resistance of the resistor = R
The relation for current using Ohm’s law is,
NCERT-Solutions-For-Class-12-Physics-Chapter-3-Current-Electricity-Formulae_1

The terminal voltage of the resistor = V

According to Ohm’s law,

V = IR

= 0.5 × 17

= 8.5 V

Therefore, the resistance of the resistor is 17 Ω and the terminal voltage is

8.5 V.


Question 3.3:

(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.

Solution:

a) Three resistors of resistances 1 Ω, 2 Ω, and 3 Ω are combined in series. The total resistance of the combination is given by the algebraic sum of individual resistances.

Total resistance = 1 + 2 + 3 = 6 Ω

(b) Current flowing through the circuit = I

EMF of the battery, E = 12 V

The total resistance of the circuit, = 6 Ω

The relation for current using Ohm’s law is,

I = V/R

⇒ 12/6 = 2A

The potential drop across 1 Ω resistor = V1

From Ohm’s law, the value of Vcan be obtained as

V1 = 2 × 1= 2 V … (i)

Potential drop across 2 Ω resistor = V2

Again, from Ohm’s law, the value of Vcan be obtained as

V2 = 2 × 2= 4 V … (ii)

Potential drop across 3 Ω resistor = V3

Again, from Ohm’s law, the value of Vcan be obtained as

V3 = 2 × 3= 6 V … (iii)

Therefore, the potential drop across 1 Ω, 2 Ω, and 3 Ω resistors are 2 V, 4 V, and 6 V respectively.


Question 3.4:

(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.

Solution:

(a) There are three resistors of resistances,

R1 = 2 Ω, R2 = 4 Ω, and R3 = 5 Ω

They are connected in parallel. Hence, the total resistance (R) of the combination is given by,
NCERT-Solutions-For-Class-12-Physics-Chapter-3-Current-Electricity-Formulae_2
Therefore, the total resistance of the combination is 20/19 ohms.

(b) EMF of the battery, V = 20 V

Current (I1) flowing through resistor R1 is given by,
NCERT-Solutions-For-Class-12-Physics-Chapter-3-Current-Electricity-Formulae_3

Current (I2) flowing through resistor R2 is given by,
NCERT-Solutions-For-Class-12-Physics-Chapter-3-Current-Electricity-Formulae_4
Current (I3) flowing through resistor R3 is given by,
NCERT-Solutions-For-Class-12-Physics-Chapter-3-Current-Electricity-Formulae_5
Total current, I = I1 + I2 + I3 = 10 + 5 + 4 = 19 A

Therefore, the current through each resister is 10 A, 5 A, and 4 A respectively and the total current is 19 A.


Question 3.5:

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 coefficient of the material of the resistor is 1.70 x 10-4 C-1

Solution:

Room temperature, T = 27°C

Resistance of the heating element at TR = 100 Ω

Let T1 is the increased temperature of the filament.

Resistance of the heating element at T1R1 = 117 Ω

Temperature co-efficient of the material of the filament,
NCERT-Solutions-For-Class-12-Physics-Chapter-3-Current-Electricity-Formulae_6
Therefore, at 1027°C, the resistance of the element is 117Ω.


Question 3.6:

A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6.0 × 10−7 m2, and its resistance is measured to be 5.0 Ω. What is the resistivity of the material at the temperature of the experiment?

Solution:

Length of the wire, l =15 m

Area of the cross-section of the wire, a = 6.0 × 10−7 m2

Resistance of the material of the wire, R = 5.0 Ω

The resistivity of the material of the wire = ρ

Resistance is related to the resistivity as

NCERT-Solutions-For-Class-12-Physics-Chapter-3-Current-Electricity-Formulae_7

Therefore, the resistivity of the material is 2 × 10−7 Ω m.


Question 3.7:

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 coefficient of resistivity of silver.

Solution:

Temperature, T1 = 27.5°C

Resistance of the silver wire at T1R1 = 2.1 Ω

Temperature, T2 = 100°C

Resistance of the silver wire at T2R2 = 2.7 Ω

Temperature coefficient of silver = α

It is related with temperature and resistance as
NCERT-Solutions-For-Class-12-Physics-Chapter-3-Current-Electricity-Formulae_8
Therefore, the temperature coefficient of silver is 0.0039°C−1.


Question 3.8:

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? The temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.70 × 10−4 °C −1.

Solution:

Supply voltage, V = 230 V

The initial current drawn, I1 = 3.2 A

Initial resistance = R1, which is given by the relation,
NCERT-Solutions-For-Class-12-Physics-Chapter-3-Current-Electricity-Formulae_9

Steady-state value of the current, I2 = 2.8 A
Resistance at the steady-state = R2, which is given as
NCERT-Solutions-For-Class-12-Physics-Chapter-3-Current-Electricity-Formulae_10

Temperature co-efficient of nichrome, α = 1.70 × 10−4 °C −1

The initial temperature of nichrome, T1= 27.0°C

Study state temperature reached by nichrome = T2

Tcan be obtained by the relation for α,
NCERT-Solutions-For-Class-12-Physics-Chapter-3-Current-Electricity-Formulae_11
Therefore, the steady temperature of the heating element is 867.5°C


Question 3.9:

Determine the current in each branch of the network shown in fig 3.30:
NCERT-Solutions-For-Class-12-Physics-Chapter-3-Current-Electricity-Formulae_12
Solution:

Current flowing through various branches of the circuit is represented in the given figure.
NCERT-Solutions-For-Class-12-Physics-Chapter-3-Current-Electricity-Formulae_13

I1 = Current flowing through the outer circuit

I2 = Current flowing through branch AB

I3 = Current flowing through branch AD

I2 − I4 = Current flowing through branch BC

I3 + I4 = Current flowing through branch CD

I4 = Current flowing through branch BD

For the closed circuit ABDA, potential is zero i.e.,

10I2 + 5I4 − 5I3 = 0

2I2 + I4 −I3 = 0

I3 = 2I2 + I4 … (1)

For the closed circuit BCDB, potential is zero i.e.,

5(I2 − I4) − 10(I3 + I4) − 5I= 0

5I2 + 5I4 − 10I3 − 10I− 5I= 0

5I2 − 10I3 − 20I= 0

I2 = 2I3 + 4I… (2)

For the closed circuit ABCFEA, potential is zero i.e.,

−10 + 10 (I1) + 10(I2) + 5(I2 − I4) = 0

10 = 15I2 + 10I− 5I4

3I2 + 2I− I4 = 2 … (3)

From equations (1) and (2), we obtain

I3 = 2(2I+ 4I4) + I4

I3 = 4I+ 8I4 + I4

− 3I3 = 9I4

− 3I4 = + I… (4)

Putting equation (4) in equation (1), we obtain

I3 = 2II4

− 4I4 = 2I2

I2 = − 2I4 … (5)

It is evident from the given figure that,

I1 = II2 … (6)

Putting equation (6) in equation (1), we obtain

3I2 +2(II2) − I4 = 2

5I2 + 2I− I4 = 2 … (7)

Putting equations (4) and (5) in equation (7), we obtain

5(−2 I4) + 2(− 3 I4) − I4 = 2

− 10I4 − 6I− I4 = 2

17I4 = − 2
I4 = -2∕17 A

Equation (4) reduces to

I3 = − 3(I4)
NCERT-Solutions-For-Class-12-Physics-Chapter-3-Current-Electricity-Formulae_14
Therefore, current in branch AB = 4/17 A

In branch BC = 6/17 A

In Branch CD = -4/17 A

In Branch AD = (6/17) A

In Branch BD = (-2/17) A

Total Current = NCERT-Solutions-For-Class-12-Physics-Chapter-3-Current-Electricity-Formulae_15


Question 3.10:

(a) In a meter bridge [Fig. 3.27], the balance point is found to be at 39.5 cm from the end A, when the resistor 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 and 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?

Solution:

A meter bridge with resistors X and Y is represented in the given figure.

NCERT-Solutions-For-Class-12-Physics-Chapter-3-Current-Electricity-Formulae_16

(a) Balance point from end A, l1 = 39.5 cm

Resistance of the resistor Y = 12.5 Ω

The condition for the balance is given as,

NCERT-Solutions-For-Class-12-Physics-Chapter-3-Current-Electricity-Formulae_17

Therefore, the resistance of resistor X is 8.2 Ω.

The connection between resistors in a Wheatstone or metre bridge is made of thick copper strips to minimize the resistance, which is not taken into consideration in the bridge formula.

(b) If and Y are interchanged, then l1 and 100−l1 get interchanged.

The balance point of the bridge will be 100−l1 from A.

100−l= 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 deflection. Hence, no current would flow through the galvanometer.


Question 3.11:

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?

Solution:

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 = V1

is connected to the storage battery in series. Hence, it can be written as

V1 = V − E

V= 120 − 8 = 112 V

Current flowing in the circuit = I, which is given by the relation,

NCERT-Solutions-For-Class-12-Physics-Chapter-3-Current-Electricity-Formulae_18

Voltage across resistor R given by the product, IR = 7 × 15.5 = 108.5 V

DC supply voltage = Terminal voltage of battery + Voltage drop across R

Terminal voltage of battery = 120 − 108.5 = 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.


Question 3.12:

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?

Solution:

EMF of the cell, E1 = 1.25 V

Balance point of the potentiometer, l1= 35 cm

The cell is replaced by another cell of EMF E2.

New balance point of the potentiometer, l2 = 63 cm

The balance condition is given by the relation,

NCERT-Solutions-For-Class-12-Physics-Chapter-3-Current-Electricity-Formulae_19

Therefore, EMF of the second cell is 2.25V.


Question 3.13:

The number density of free electrons in a copper conductor estimated in Example 3.1 is 8.5 × 1028 m−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 × 10−6 m2 and it is carrying a current of 3.0 A.

Solution:

Number density of free electrons in a copper conductor, n = 8.5 × 1028 m−3 Length of the copper wire, l = 3.0 m

Area of cross-section of the wire, A = 2.0 × 10−6 m2

Current carried by the wire, I = 3.0 A, which is given by the relation,

nAeVd

Where,

e = Electric charge = 1.6 × 10−19 C

Vd = Drift velocity = Length of the wire (l) / Time taken to cover l (t)

NCERT-Solutions-For-Class-12-Physics-Chapter-3-Current-Electricity-Formulae_20
Therefore, the time taken by an electron to drift from one end of the wire to the other is 2.7 × 104 s.


Question 3.14:

The earth’s surface has a negative surface charge density of 10−9 C m−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 field, how much time (roughly) would be required to neutralize 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 × 106 m.)

Solution:

Surface charge density of the earth, σ = 10−9 C m−2

Current over the entire globe, I = 1800 A

Radius of the earth, r = 6.37 × 106 m

Surface area of the earth,

A = 4πr2

= 4π × (6.37 × 106)2

= 5.09 × 1014 m2

Charge on the earth surface,

σ × A

= 10−9 × 5.09 × 1014

= 5.09 × 10C

Time taken to neutralize the earth’s surface = t

Current, I = q/t

⇒ t = q / i

NCERT-Solutions-For-Class-12-Physics-Chapter-3-Current-Electricity-Formulae_21

Therefore, the time taken to neutralize the earth’s surface is 282.77 s.


Question 3.15:

(a) Six lead-acid types 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 is 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?

Solution:

(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,

NCERT-Solutions-For-Class-12-Physics-Chapter-3-Current-Electricity-Formulae_22

Terminal voltage, V = IR = 1.39 × 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
Hence, maximum currentNCERT-Solutions-For-Class-12-Physics-Chapter-3-Current-Electricity-Formulae_24
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.


Question 3.16:

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

Solution:

Resistivity of aluminum, ρAl = 2.63 × 10−8 Ω m

Relative density of aluminum, d1 = 2.7

Let l1 be the length of aluminum wire and m1 be its mass.

Resistance of the aluminum wire = R1

Area of cross-section of the aluminum wire = A1

Resistivity of copper, ρCu = 1.72 × 10−8 Ω m

Relative density of copper, d2 = 8.9

Let l2 be the length of copper wire and m2 be its mass.

Resistance of the copper wire = R2

Area of cross-section of the copper wire = A2

The two relations can be written as

NCERT-Solutions-For-Class-12-Physics-Chapter-3-Current-Electricity-Formulae_25

It is given that,

NCERT-Solutions-For-Class-12-Physics-Chapter-3-Current-Electricity-Formulae_26
And,
l1 = l2
NCERT-Solutions-For-Class-12-Physics-Chapter-3-Current-Electricity-Formulae_27

Mass of the aluminum wire,

m1 = Volume × Density

A1l1 × dA1 l1d1 … (3)

Mass of the copper wire,

m2 = Volume × Density

A2l2 × dA2 l2d2 … (4)

Dividing equation (3) by equation (4), we obtain

NCERT-Solutions-For-Class-12-Physics-Chapter-3-Current-Electricity-Formulae_29

It can be inferred from this ratio that m1 is less than m2. Hence, aluminum is lighter than copper.

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


Question 3.17:

What conclusion can you draw from the following observations on a resistor made of alloy Manganin?

Current (A) Voltage (V) Current (A) Voltage (V)
0.2 3.94 3.0 59.2
0.4 7.87 4.0 78.8
0.6 11.8 5.0 98.6
0.8 15.7 6.0 118.5
1.0 19.7 7.0 138.2
2.0 39.4 8.0 158.0

 

Solution:

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.


Question 3.18:

(a) A steady current flows in a metallic conductor of non-uniform cross-section. Which of these quantities is constant along the conductor: current, current density, electric field, 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?

Solution:

(a) When a steady current flows in a metallic conductor of a non-uniform cross-section, the current flowing through the conductor is constant. Current density, electric field, and drift speed are inversely proportional to the area of the cross-section. Therefore, they are not constant.

(b) No, Ohm’s law is not universally applicable for all conducting elements. Vacuum diode semi-conductor 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 = IR

Voltage (V) is directly proportional to current (I).

R is the internal resistance of the source.

I = V/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 very 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.


Question 3.19:

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 coefficients 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 (1022/103).

Solution:

(a) Alloys of metals usually have greater resistivity than that of their constituent metals.

(b) Alloys usually have lower temperature coefficients 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 1022.


Question 3.20:

(a) Given 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. 3.31.
NCERT-Solutions-For-Class-12-Physics-Chapter-3-Current-Electricity-Formulae_30.1NCERT-Solutions-For-Class-12-Physics-Chapter-3-Current-Electricity-Formulae_30.2

Solution:

(a) Total number of resistors = n

Resistance of each resistor = R

(i) When n resistors are connected in series, effective resistance R1is the maximum, given by the product nR.

Hence, maximum resistance of the combination, RnR

(ii) When n resistors are connected in parallel, the effective resistance (R2) is the minimum, given by the ratio R/n

Hence, minimum resistance of the combination, R2 = R/n

(iii) The ratio of the maximum to the minimum resistance is,

NCERT-Solutions-For-Class-12-Physics-Chapter-3-Current-Electricity-Formulae_32

(b) The resistance of the given resistors is,

R1 = 1 Ω, R2 = 2 Ω, R3 = 3 Ω2

  1. Equivalent resistance, R′ = 11/3 ohms

Consider the following combination of the resistors.

NCERT-Solutions-For-Class-12-Physics-Chapter-3-Current-Electricity-Formulae_33

Equivalent resistance of the circuit is given by,

NCERT-Solutions-For-Class-12-Physics-Chapter-3-Current-Electricity-Formulae_34
2. Equivalent resistance, R’ = 11/5 ohms

Consider the following combination of the resistors.

NCERT-Solutions-For-Class-12-Physics-Chapter-3-Current-Electricity-Formulae_35

Equivalent resistance of the circuit is given by,

NCERT-Solutions-For-Class-12-Physics-Chapter-3-Current-Electricity-Formulae_36

(iii) Equivalent resistance, R = 6 Ω

Consider the series combination of the resistors, as shown in the given circuit.
NCERT-Solutions-For-Class-12-Physics-Chapter-3-Current-Electricity-Formulae_37

Equivalent resistance of the circuit is given by the sum,

R = 1 + 2 + 3 = 6 Ω
(iv) Equivalent resistance, R’ = 6/11 Ω

Consider the series combination of the resistors, as shown in the given circuit.
NCERT-Solutions-For-Class-12-Physics-Chapter-3-Current-Electricity-Formulae_38.
Equivalent resistance of the circuit is given by,

NCERT-Solutions-For-Class-12-Physics-Chapter-3-Current-Electricity-Formulae_39

(c) (a) It can be observed from the given circuit that in the first small loop, two resistors of resistance 1 Ω each are connected in series.

Hence, their equivalent resistance = (1+1) = 2 Ω

It can also be observed that two resistors of resistance 2 Ω each are connected in series.

Hence, their equivalent resistance = (2 + 2) = 4 Ω.

Therefore, the circuit can be redrawn as
NCERT-Solutions-For-Class-12-Physics-Chapter-3-Current-Electricity-Formulae_40

It can be observed that 2 Ω and 4 Ω resistors are connected in parallel in all the four loops. Hence, equivalent resistance (R) of each loop is given by,

NCERT-Solutions-For-Class-12-Physics-Chapter-3-Current-Electricity-Formulae_42

The circuit reduces to,
NCERT-Solutions-For-Class-12-Physics-Chapter-3-Current-Electricity-Formulae_43

All the four resistors are connected in series.

Hence, equivalent resistance of the given circuit is NCERT-Solutions-For-Class-12-Physics-Chapter-3-Current-Electricity-Formulae_44

(b) It can be observed from the given circuit that five resistors of resistance R each are connected in series.

Hence, equivalent resistance of the circuit = R + R + R + R

= 5 R


Question 3.21:

Determine the current drawn from a 12 V supply with internal resistance 0.5 Ω by the infinite network shown in Fig. 3.32. Each resistor has 1 Ω resistance.

NCERT-Solutions-For-Class-12-Physics-Chapter-3-Current-Electricity-Formulae_45

Solution:

The resistance of each resistor connected in the given circuit, R = 1 Ω

Equivalent resistance of the given circuit = R

The network is infinite. Hence, equivalent resistance is given by the relation,
NCERT-Solutions-For-Class-12-Physics-Chapter-3-Current-Electricity-Formulae_46

Negative value of R’ cannot be accepted. Hence, equivalent resistance,

R’ = (1+√3) = 1+1.73 = 2.73 Ω

Internal resistance of the circuit, r = 0.5 Ω

Hence, total resistance of the given circuit = 2.73 + 0.5 = 3.23 Ω

Supply voltage, V = 12 V

According to Ohm’s Law, current drawn from the source is given by the ratio, (12/3.23) = 3.72 A


Question 3.22:

Figure 3.33 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 up to 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.

NCERT-Solutions-For-Class-12-Physics-Chapter-3-Current-Electricity-Formulae_47

(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 1.0 V instead of 2.0 V?

(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?

Solution:

(a) Constant EMF of the given standard cell, E1 = 1.02 V

Balance point on the wire, l= 67.3 cm

A cell of unknown EMF, ε,replaced the standard cell. Therefore, new balance point on the wire, l = 82.3 cm

The relation connecting EMF and balance point is,

NCERT-Solutions-For-Class-12-Physics-Chapter-3-Current-Electricity-Formulae_48the value of unknown EMF is 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 modified if a series resistance is connected with the wire AB. The potential drop across AB is slightly greater than the EMF is measured. The percentage error would be small.


Question 3.23:

Figure 3.34 shows a potentiometer circuit for comparison of two resistances. The balance point with a standard resistor R = 10.0 Ω is found to be 58.3 cm, while that with the unknown resistance is 68.5 cm. Determine the value of X. What might you do if you failed to find a balance point with the given cell of EMF ε?

NCERT-Solutions-For-Class-12-Physics-Chapter-3-Current-Electricity-Formulae_50

Solution:

Resistance of the standard resistor, R = 10.0 Ω

Balance point for this resistance, l1 = 58.3 cm

Current in the potentiometer wire = i

Hence, potential drop across RE1 = iR

Resistance of the unknown resistor = X

Balance point for this resistor, l2 = 68.5 cm

Hence, potential drop across XE2 = iX

The relation connecting emf and balance point is,

NCERT-Solutions-For-Class-12-Physics-Chapter-3-Current-Electricity-Formulae_50
= (68.5/58.3) x 10 = 11.749 Ω

Therefore, the value of the unknown resistance, X, is 11.75 Ω.

If we fail to find a balance point with the given cell of emf, ε, then the potential drop across R and X must be reduced by putting a resistance in series with it. Only if the potential drop across R or X is smaller than the potential drop across the potentiometer wire AB, a balance point is obtained.


Question 3.24:

Figure 3.35 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.

NCERT-Solutions-For-Class-12-Physics-Chapter-3-Current-Electricity-Formulae_51

Solution:

Internal resistance of the cell = r

Balance point of the cell in open circuit, l1 = 76.3 cm

An external resistance (R) is connected to the circuit with R = 9.5 Ω

New balance point of the circuit, l= 64.8 cm

Current flowing through the circuit = I

The relation connecting resistance and emf is,

NCERT-Solutions-For-Class-12-Physics-Chapter-3-Current-Electricity-Formulae_52

Therefore, the internal resistance of the cell is 1.68Ω.