Current Electricity Chapter-Wise Test 1

Correct answer Carries: 4.

Wrong Answer Carries: -1.

In a circuit with a battery, why does the internal resistance of the battery affect the maximum power delivered to an external load?

Maximum power transfer occurs when the external resistance equals the internal resistance (\( R = r \)), as \( P = I^2 R = (\varepsilon / (R + r))^2 R \). Internal resistance limits current, influencing the power distribution.

Increases emf
Reduces voltage drop
Increases total resistance
Limits current through the circuit
4

A copper wire carries \( 2.72 \, \text{A} \) with a drift speed of \( 8 \times 10^{-5} \, \text{m/s} \). If \( n = 8.5 \times 10^{28} \, \text{m}^{-3} \) and \( e = 1.6 \times 10^{-19} \, \text{C} \), what is the cross-sectional area?

Drift speed: \( v_d = \frac{I}{n e A} \).

Rearrange: \( A = \frac{I}{n e v_d} \).

Substitute: \( A = \frac{2.72}{8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times 8 \times 10^{-5}} \).

Calculate: \( A = \frac{2.72}{1.088 \times 10^5} \approx 2.5 \times 10^{-5} \, \text{m}^2 \).

\( 2.2 \times 10^{-5} \, \text{m}^2 \)
\( 2.4 \times 10^{-5} \, \text{m}^2 \)
\( 2.45 \times 10^{-5} \, \text{m}^2 \)
\( 2.5 \times 10^{-5} \, \text{m}^2 \)
4

A circuit has a \( 10 \, \text{V} \) battery with \( 1 \, \Omega \) internal resistance and two resistors \( 5 \, \Omega \) and \( 10 \, \Omega \) in parallel. What is the total current?

Parallel resistance: \( \frac{1}{R_p} = \frac{1}{5} + \frac{1}{10} = \frac{2 + 1}{10} = \frac{3}{10} \Rightarrow R_p = \frac{10}{3} \approx 3.33 \, \Omega \).

Total resistance: \( R_{\text{total}} = 1 + 3.33 = 4.33 \, \Omega \).

Current: \( I = \frac{\varepsilon}{R_{\text{total}}} = \frac{10}{4.33} \approx 2.31 \, \text{A} \).

\( 2.0 \, \text{A} \)
\( 2.2 \, \text{A} \)
\( 2.31 \, \text{A} \)
\( 2.5 \, \text{A} \)
3

A \( 15 \, \text{V} \) battery with \( 1 \, \Omega \) internal resistance delivers a current of \( 2.5 \, \text{A} \) to a resistor. What is the resistance of the resistor?

Terminal voltage: \( V = \varepsilon - I r = 15 - 2.5 \times 1 = 12.5 \, \text{V} \).

Resistance: \( R = \frac{V}{I} = \frac{12.5}{2.5} = 5 \, \Omega \).

\( 4.0 \, \Omega \)
\( 4.5 \, \Omega \)
\( 5.0 \, \Omega \)
\( 5.5 \, \Omega \)
3

A copper wire carries \( 2 \, \text{A} \) with a drift speed of \( 8 \times 10^{-5} \, \text{m/s} \). If \( n = 8.5 \times 10^{28} \, \text{m}^{-3} \) and \( e = 1.6 \times 10^{-19} \, \text{C} \), what is the cross-sectional area?

Drift speed: \( v_d = \frac{I}{n e A} \).

Rearrange: \( A = \frac{I}{n e v_d} \).

Substitute: \( A = \frac{2}{8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times 8 \times 10^{-5}} \).

Calculate: \( A = \frac{2}{1.088 \times 10^5} \approx 1.84 \times 10^{-5} \, \text{m}^2 \).

\( 1.5 \times 10^{-5} \, \text{m}^2 \)
\( 1.7 \times 10^{-5} \, \text{m}^2 \)
\( 1.8 \times 10^{-5} \, \text{m}^2 \)
\( 1.84 \times 10^{-5} \, \text{m}^2 \)
4

What is the physical significance of the temperature coefficient of resistivity being positive for metals?

A positive temperature coefficient (\( \alpha \)) means resistivity (\( \rho_t = \rho_0 [1 + \alpha (T - T_0)] \)) increases with temperature, as increased lattice vibrations reduce the mean free time between collisions, raising resistance.

Resistance decreases with temperature
Conductivity increases
Resistance increases with temperature
Charge carriers multiply
3

A wire of length \( 6 \, \text{m} \) and cross-sectional area \( 5 \times 10^{-6} \, \text{m}^2 \) has a resistance of \( 12 \, \Omega \). What is the resistivity of the material?

Resistance: \( R = \frac{\rho l}{A} \).

Rearrange: \( \rho = \frac{R A}{l} \).

Substitute: \( \rho = \frac{12 \times 5 \times 10^{-6}}{6} = 10 \times 10^{-6} = 1.0 \times 10^{-5} \, \Omega \text{m} \).

\( 1.0 \times 10^{-5} \, \Omega \text{m} \)
\( 1.2 \times 10^{-5} \, \Omega \text{m} \)
\( 1.5 \times 10^{-5} \, \Omega \text{m} \)
\( 1.8 \times 10^{-5} \, \Omega \text{m} \)
1

A conductor has a resistivity of \( 8 \times 10^{-8} \, \Omega \text{m} \) and \( \alpha = 4 \times 10^{-3} \, ^\circ\text{C}^{-1} \) at \( 20^\circ \text{C} \). What is its resistivity at \( 70^\circ \text{C} \)?

Use: \( \rho_t = \rho_0 [1 + \alpha (T - T_0)] \).

Substitute: \( \rho_t = 8 \times 10^{-8} [1 + 4 \times 10^{-3} (70 - 20)] \).

Calculate: \( \rho_t = 8 \times 10^{-8} [1 + 0.2] = 8 \times 10^{-8} \times 1.2 = 9.6 \times 10^{-8} \, \Omega \text{m} \).

\( 9.0 \times 10^{-8} \, \Omega \text{m} \)
\( 9.2 \times 10^{-8} \, \Omega \text{m} \)
\( 9.6 \times 10^{-8} \, \Omega \text{m} \)
\( 10.0 \times 10^{-8} \, \Omega \text{m} \)
3

A Wheatstone bridge has \( R_1 = 18 \, \Omega \), \( R_2 = 36 \, \Omega \), \( R_3 = 12 \, \Omega \). What is \( R_4 \) for balance?

Balance condition: \( \frac{R_1}{R_2} = \frac{R_3}{R_4} \).

Substitute: \( \frac{18}{36} = \frac{12}{R_4} \).

Solve: \( 0.5 = \frac{12}{R_4} \Rightarrow R_4 = \frac{12}{0.5} = 24 \, \Omega \).

\( 20 \, \Omega \)
\( 24 \, \Omega \)
\( 28 \, \Omega \)
\( 32 \, \Omega \)
2

Why does the power dissipated in a resistor increase quadratically with current?

Power \( P = I^2 R \). Since power depends on the square of the current (\( I^2 \)), doubling the current quadruples the power, assuming resistance remains constant.

Voltage increases linearly
Power depends on current squared
Resistance decreases
Energy loss cancels out
2

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