Chapter 3 Current Electricity Class 12 Physics

Question 3.1

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

Concept

Current $I = \frac{E}{R + r}$. For maximum current, external resistance $R = 0$.

$$I_{max} = \frac{E}{r} = \frac{12}{0.4} = 30 \text{ A}$$

Result: 30 Amperes

Question 3.2

A battery of emf 10 V and internal resistance $3 \Omega$ 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?

Part 1: Resistance (R)

$$0.5 = \frac{10}{R + 3} \implies R + 3 = 20 \implies R = 17 \Omega$$

Part 2: Terminal Voltage (V)

$$V = E – Ir = 10 – (0.5 \times 3) = 8.5 \text{ V}$$

Result: $17 \Omega$, 8.5 V

Question 3.3

At room temperature ($27.0^\circ \text{C}$) the resistance of a heating element is $100 \Omega$. What is the temperature of the element if the resistance is found to be $117 \Omega$, given $\alpha = 1.70 \times 10^{-4} \text{ }^\circ\text{C}^{-1}$?

Formula

$R = R_0 [1 + \alpha(T – T_0)] \implies T – T_0 = \frac{R – R_0}{R_0 \alpha}$

$$T – 27 = \frac{117 – 100}{100 \times 1.7 \times 10^{-4}} = \frac{17}{0.017} = 1000$$ $$T = 1000 + 27 = 1027^\circ \text{C}$$

Result: $1027^\circ \text{C}$

Question 3.4

A negligibly small current is passed through a wire of length 15 m and uniform cross-section $6.0 \times 10^{-7} \text{ m}^2$, and its resistance is measured to be $5.0 \Omega$. What is the resistivity of the material?

Calculation

Resistivity $\rho = \frac{RA}{L}$.

$$\rho = \frac{5.0 \times 6.0 \times 10^{-7}}{15} = 2.0 \times 10^{-7} \Omega \text{ m}$$

Result: $2.0 \times 10^{-7} \Omega \text{ m}$

Question 3.5

A silver wire has a resistance of $2.1 \Omega$ at $27.5^\circ \text{C}$, and a resistance of $2.7 \Omega$ at $100^\circ \text{C}$. Determine the temperature coefficient of resistivity of silver.

Formula

$\alpha = \frac{R_2 – R_1}{R_1(T_2 – T_1)}$.

$$\alpha = \frac{2.7 – 2.1}{2.1(100 – 27.5)} = \frac{0.6}{2.1 \times 72.5} = \frac{0.6}{152.25}$$ $$\alpha \approx 0.0039 \text{ }^\circ\text{C}^{-1}$$

Result: $0.0039 \text{ }^\circ\text{C}^{-1}$

Question 3.6

A heating element connected to a 230 V supply draws an initial current of 3.2 A which settles to 2.8 A. Room temp is $27.0^\circ \text{C}$. Find steady temperature. ($\alpha = 1.70 \times 10^{-4} \text{ }^\circ\text{C}^{-1}$).

1. Resistances

$R_1 = \frac{230}{3.2} = 71.87 \Omega$ (at $27^\circ \text{C}$).
$R_2 = \frac{230}{2.8} = 82.14 \Omega$ (at steady temp).

2. Temperature

$$T_2 – 27 = \frac{82.14 – 71.87}{71.87 \times 1.7 \times 10^{-4}} = \frac{10.27}{0.0122} \approx 840^\circ \text{C}$$ $$T_2 = 840 + 27 = 867^\circ \text{C}$$

Result: $867^\circ \text{C}$

Question 3.7

Determine the current in each branch of the network shown (Wheatstone Bridge Network: AB=10, BC=5, AD=5, DC=10, BD=5, Source=10V).

Kirchhoff’s Laws

Let currents be $I_1$ (AB), $I_2$ (AD), and $I_g$ (BD).
Loop ABDA: $10I_1 + 5I_g – 5I_2 = 0 \implies 2I_1 + I_g – I_2 = 0$ (1)
Loop BCDB: $5(I_1 – I_g) – 10(I_2 + I_g) – 5I_g = 0 \implies I_1 – 4I_g – 2I_2 = 0$ (2)

Solution

Solving these with the source equation ($10V$):

$$I_{AB} = \frac{4}{17} \text{ A}, \quad I_{AD} = \frac{6}{17} \text{ A}, \quad I_{BD} = \frac{-2}{17} \text{ A}$$

Total Current $I = \frac{10}{17} \text{ A}$.

Question 3.8

A storage battery of emf 8.0 V and internal resistance $0.5 \Omega$ is being charged by a 120 V DC supply using a series resistor of $15.5 \Omega$. What is the terminal voltage of the battery during charging?

1. Current

Net EMF = $120 – 8 = 112 \text{ V}$. Total Resistance = $15.5 + 0.5 = 16 \Omega$.

$$I = \frac{112}{16} = 7 \text{ A}$$

2. Terminal Voltage

During charging, $V = E + Ir$.

$$V = 8 + (7 \times 0.5) = 8 + 3.5 = 11.5 \text{ V}$$

Result: 11.5 V

Question 3.9

The number density of free electrons in a copper conductor is $8.5 \times 10^{28} \text{ m}^{-3}$. How long does an electron take to drift from one end of a wire 3.0 m long to the other? Area $2.0 \times 10^{-6} \text{ m}^2$, Current 3.0 A.

Calculation

Drift velocity $v_d = \frac{I}{nAe}$. Time $t = \frac{L}{v_d} = \frac{L n A e}{I}$.

$$t = \frac{3.0 \times 8.5 \times 10^{28} \times 2.0 \times 10^{-6} \times 1.6 \times 10^{-19}}{3.0}$$ $$t = 2.7 \times 10^4 \text{ s} \approx 7.5 \text{ hours}$$

Result: $2.7 \times 10^4 \text{ s}$

NCERT Solutions (2025-26)