1. When a current $I$ flows through a wire of radius $r$ , the drift velocity is $v_d$ . If the same current flows through a wire of same material but radius $2r$ , the new drift velocity will be:

(a) $v_d/4$
(b) $v_d/2$
(c) $2v_d$
(d) $4v_d$

Answer: (a)
$I = neAv_d \Rightarrow v_d = I / (ne \pi r^2)$ . Since $I$ and $material$ are constant, $v_d \propto 1/r^2$ . Increasing $r$ to $2r$ makes the new velocity $1/4$ of original.

2. The resistivity of a alloy like Manganin or Nichrome is:

(a) Nearly independent of temperature
(b) Increases rapidly with temperature
(c) Decreases with temperature
(d) Becomes zero at absolute zero

Answer: (a)
Alloys like Manganin have a very small temperature coefficient of resistance, making them ideal for standard resistors.

3. Kirchhoff’s First Law ( $\sum I = 0$ ) and Second Law ( $\sum V = 0$ ) are based on which conservation laws respectively?

(a) Energy, Charge
(b) Charge, Momentum
(c) Charge, Energy
(d) Energy, Momentum

Answer: (c)
The Junction rule is based on conservation of charge, and the Loop rule is based on conservation of energy.

4. In a Wheatstone bridge, the four resistances are $10 \Omega, 10 \Omega, 10 \Omega$ and $X \Omega$ . If the bridge is balanced, the value of $X$ is:

(a) $10 \Omega$
(b) $20 \Omega$
(c) $5 \Omega$
(d) Zero

Answer: (a)
For a balanced bridge: $R_1/R_2 = R_3/R_4$ . Here, $10/10 = 10/X \Rightarrow X = 10 \Omega$ .

5. Terminal potential difference $V$ of a cell of EMF $\epsilon$ and internal resistance $r$ while it is being charged is:

(a) $V = \epsilon - Ir$
(b) $V = \epsilon + Ir$
(c) $V = \epsilon$
(d) $V = Ir$

Answer: (b)
During charging, current enters the positive terminal, so $V = \epsilon + Ir$ . (During discharging, $V = \epsilon - Ir$ ).

6. A wire of resistance $R$ is stretched to twice its original length. Its new resistance will be:

(a) $2R$
(b) $4R$
(c) $R/2$
(d) $R/4$

Answer: (b)
When stretched, volume is constant ( $A \times l = A' \times l'$ ). If $l' = 2l$ , then $A' = A/2$ . New $R' = \rho (2l) / (A/2) = 4 \rho (l/A) = 4R$ .

7. The current density in a wire is $\vec{j}$ . The relation between $\vec{j}$ and the electric field $\vec{E}$ is given by:

(a) $\vec{j} = \sigma \vec{E}$
(b) $\vec{j} = \rho \vec{E}$
(c) $\vec{E} = \sigma \vec{j}$
(d) $\vec{j} \cdot \vec{E} = 0$

Answer: (a)
This is the microscopic or vector form of Ohm’s Law, where $\sigma$ is conductivity.

8. Two cells of EMFs $1.5V$ and $2.0V$ with internal resistances $0.2\Omega$ and $0.3\Omega$ are connected in parallel. The equivalent EMF is:

(a) $3.5V$
(b) $1.7V$
(c) $0.5V$
(d) $1.75V$

Answer: (b)
$\epsilon_{eq} = (\epsilon_1 r_2 + \epsilon_2 r_1) / (r_1 + r_2) = (1.5 \times 0.3 + 2.0 \times 0.2) / (0.2 + 0.3) = (0.45 + 0.40) / 0.5 = 0.85 / 0.5 = 1.7V$ .

9. The relaxation time in a conductor:

(a) Increases with increase in temperature
(b) Decreases with increase in temperature
(c) Is independent of temperature
(d) Increases with increase in electric field

Answer: (b)
With higher temperature, ions vibrate more vigorously, leading to more frequent collisions and thus a smaller relaxation time ( $\tau$ ).

10. Current is a scalar quantity because:

(a) It does not have direction
(b) It does not obey laws of vector addition
(c) Its magnitude is very small
(d) It flows in a wire

Answer: (b)
Current has magnitude and direction (sense of flow), but it is added algebraically, not vectorially at a junction.

Part 2: Assertion-Reason Questions

(A) Both A & R are true, R explains A.
(B) Both A & R are true, R does NOT explain A.
(C) A is true, R is false.
(D) A is false, R is true.

1. Assertion (A): The drift velocity of electrons is very small (mm/s), yet the bulb glows instantly when switched on.
Reason (R): The electric field is established throughout the circuit with the speed of light.

Answer: (A)
Electrons everywhere in the wire start drifting almost simultaneously because the electromagnetic signal travels at speed of light.

2. Assertion (A): Bending a wire does not affect its electrical resistance.
Reason (R): Resistance depends on length, area, and resistivity, which remain unchanged by bending.

Answer: (A)
As long as the cross-section and length aren’t distorted significantly, bending doesn’t change $R$ .

3. Assertion (A): Alloys like Constantan are used to make standard resistors.
Reason (R): Alloys have a high temperature coefficient of resistance.

Answer: (C)
Assertion is True. Reason is False—they are used because they have a very low temperature coefficient.

4. Assertion (A): When cells are connected in series, the total internal resistance increases.
Reason (R): In series, equivalent resistance is the sum of individual resistances.

Answer: (A)
$r_{eq} = r_1 + r_2 + ... + r_n$ .

5. Assertion (A): A Wheatstone bridge is most sensitive when all four resistances are of the same order.
Reason (R): In a balanced bridge, the galvanometer shows zero deflection.

Answer: (B)
Both are true, but the definition of balance (R) does not explain why sensitivity is max at equal orders (A).

6. Assertion (A): A cell is a device that maintains a constant potential difference.
Reason (R): A cell converts chemical energy into electrical energy.

Answer: (D)
Assertion is False—a cell maintain a current, but terminal PD ( $V$ ) changes with current. Reason is True.

7. Assertion (A): For a conductor, the $V-I$ graph is a straight line passing through the origin at constant temperature.
Reason (R): Resistance of a conductor is independent of the current flowing through it.

Answer: (A)
This is Ohm’s Law.

8. Assertion (A): Resistance of a semiconductor decreases with increase in temperature.
Reason (R): In semiconductors, more charge carriers (electrons and holes) become available as temperature rises.

Answer: (A)
The increase in carrier density ( $n$ ) dominates over the decrease in relaxation time.

9. Assertion (A): Connecting wires are made of copper.
Reason (R): Copper has high resistivity.

Answer: (C)
Copper is used because it has low resistivity.

10. Assertion (A): Power dissipated in a resistor is $V^2/R$ .
Reason (R): According to Ohm’s Law, $V = IR$ .

Answer: (A)
Starting from $P = VI$ , substituting $I = V/R$ gives $P = V^2/R$ .

Part 3: Important Derivations & Theory

1. Define drift velocity. Derive the relation between current ( $I$ ) and drift velocity ( $v_d$ ).

Definition: The average velocity with which free electrons in a conductor get drifted in a direction opposite to the applied electric field.
Derivation:
1. Distance traveled in $\Delta t$ is $v_d \Delta t$ .
2. Volume $= A \times v_d \Delta t$ .
3. No. of electrons $= n \times A v_d \Delta t$ .
4. Total Charge $\Delta Q = (n A v_d \Delta t) \times e$ .
5. Current $I = \Delta Q / \Delta t = \mathbf{neAv_d}$ .

2. State Kirchhoff’s Laws for an electrical network. Explain the sign conventions used.

1. Junction Rule: Algebraic sum of currents at any junction is zero ( $\sum I = 0$ ).
2. Loop Rule: Algebraic sum of changes in potential around any closed loop is zero ( $\sum \Delta V = 0$ ).
Sign Conventions:
– Potential drop across resistor in current direction is $-IR$ .
– Potential rise across battery from negative to positive is $+\epsilon$ .

3. Derive the balanced condition for a Wheatstone bridge using Kirchhoff’s laws.

1. For balance, current through galvanometer $I_g = 0$ .
2. Loop 1 (Left): $-I_1 R_1 + I_2 R_2 = 0 \Rightarrow I_1 R_1 = I_2 R_2$ .
3. Loop 2 (Right): $-I_1 R_3 + I_2 R_4 = 0 \Rightarrow I_1 R_3 = I_2 R_4$ .
4. Divide equations: $R_1 / R_3 = R_2 / R_4$ or $\mathbf{R_1 / R_2 = R_3 / R_4}$ .

Part 4: Numericals

1. The resistance of a platinum wire of a thermometer at the ice point is $5 \Omega$ and at steam point is $5.39 \Omega$ . When the thermometer is inserted in a hot bath, the resistance is $5.795 \Omega$ . Calculate the temperature of the bath.

Formula: $t = \frac{R_t - R_0}{R_{100} - R_0} \times 100$ .
$t = \frac{5.795 - 5}{5.39 - 5} \times 100 = \frac{0.795}{0.39} \times 100$ .
$t \approx 203.85 ^\circ C$ .

2. A network of resistors is connected to a $16V$ battery with internal resistance of $1\Omega$ . (Assume the network is a $4\Omega$ equivalent resistance). Calculate (a) the total current in the circuit and (b) the terminal voltage of the battery.

(a) $I = \epsilon / (R + r) = 16 / (4 + 1) = 16/5 = 3.2 A$ .
(b) $V = \epsilon - Ir = 16 - (3.2 \times 1) = 16 - 3.2 = 12.8 V$ .

3. Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area $1.0 \times 10^{-7} m^2$ carrying a current of $1.5 A$ . Assume $n = 8.5 \times 10^{28} m^{-3}$ .

$v_d = I / (neA)$ .
$v_d = 1.5 / (8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times 1.0 \times 10^{-7})$ .
$v_d \approx 1.1 \times 10^{-3} m/s = 1.1 mm/s$ .

Part 5: Case Study

Case Study: Internal Resistance and Battery Performance
Every real cell has an internal resistance due to the electrolyte and electrodes. This resistance is responsible for the ‘voltage drop’ when a circuit is closed. As a battery gets old, its internal resistance increases significantly, which is why an old battery might show the full EMF on a voltmeter but fail to start a car engine (which requires high current). The terminal voltage $V$ is always less than $\epsilon$ during discharge because some energy is wasted as heat inside the cell itself.

Under what condition is the terminal potential difference equal to the EMF of the cell?
Why does a car battery need to have a very low internal resistance?
What happens to the terminal voltage if the external resistance connected to the cell is increased?
Define ‘Short Circuit’ of a cell and find the maximum current it can draw.

1. When the circuit is **open** ( $I = 0$ ), then $V = \epsilon$ .
2. To provide a **high starting current** ( $I = \epsilon/r$ when $R \approx 0$ ) required for the starter motor.
3. It **increases**. As $R$ increases, $I$ decreases, so the drop $Ir$ decreases, making $V$ closer to $\epsilon$ .
4. Short circuit is when $R = 0$ . Maximum current $I_{max} = \epsilon / r$ .

← Back to Chapter Notes Go to CBSE Hub →

Current Electricity: Question Bank

Current Electricity: Question Bank

Part 1: Multiple Choice Questions (1 Mark)

Part 2: Assertion-Reason Questions

Part 3: Important Derivations & Theory

Part 4: Numericals

Part 5: Case Study

Editorial Team

Leave a Reply Cancel reply

Part 1: Multiple Choice Questions (1 Mark)

Part 2: Assertion-Reason Questions

Part 3: Important Derivations & Theory

Part 4: Numericals

Part 5: Case Study

Editorial Team

You might also like

10 Solved Numericals on Current Electricity

10 Solved Numericals on Moving Charges and Magnetism

10 Solved Numericals on Electrostatic Potential and Capacitance

Leave a Reply Cancel reply