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 $I_{max} = \epsilon / r$ .

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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

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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

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