Current Electricity Chapter-Wise Test 21

Correct answer Carries: 4.

Wrong Answer Carries: -1.

Why does the equivalent resistance of two resistors in parallel always lie between zero and the smallest individual resistance?

For parallel resistors, \( 1/R_{\text{eq}} = 1/R_1 + 1/R_2 \), so \( R_{\text{eq}} = R_1 R_2 / (R_1 + R_2) \). This value is less than the smaller resistance (e.g., if \( R_1 < R_2 \), \( R_{\text{eq}} < R_1 \)) but greater than zero, as additional paths reduce resistance.

Voltage cancels out
Additional paths reduce total resistance
Current doubles
Resistances add inversely
2

Why does a superconductor exhibit zero resistance below its critical temperature?

Below the critical temperature, electrons in a superconductor form Cooper pairs, which move without scattering off lattice ions, eliminating resistance as there’s no energy loss to collisions.

Electrons stop moving
No collisions with lattice
Voltage becomes infinite
Charge carriers vanish
2

A conductor of length \( 1 \, \text{m} \) and cross-sectional area \( 1 \times 10^{-6} \, \text{m}^2 \) has a resistance of \( 2 \, \Omega \). What is the resistivity of the material?

Resistance is given by \( R = \frac{\rho l}{A} \).

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

Given: \( R = 2 \, \Omega \), \( A = 1 \times 10^{-6} \, \text{m}^2 \), \( l = 1 \, \text{m} \).

Substitute: \( \rho = \frac{2 \times 1 \times 10^{-6}}{1} = 2 \times 10^{-6} \, \Omega \text{m} \).

\( 1.5 \times 10^{-6} \, \Omega \text{m} \)
\( 2.0 \times 10^{-6} \, \Omega \text{m} \)
\( 2.5 \times 10^{-6} \, \Omega \text{m} \)
\( 3.0 \times 10^{-6} \, \Omega \text{m} \)
2

A \( 5 \, \Omega \) resistor carries a current of \( 3 \, \text{A} \) for \( 20 \, \text{s} \). What is the energy dissipated?

Energy: \( W = I^2 R t \).

Substitute: \( W = 3^2 \times 5 \times 20 = 9 \times 100 = 900 \, \text{J} \).

\( 600 \, \text{J} \)
\( 750 \, \text{J} \)
\( 800 \, \text{J} \)
\( 900 \, \text{J} \)
4

A \( 12 \, \text{V} \) battery with \( 2 \, \Omega \) internal resistance delivers a current of \( 3 \, \text{A} \) to a resistor. What is the resistance of the resistor?

Terminal voltage: \( V = \varepsilon - I r = 12 - 3 \times 2 = 6 \, \text{V} \).

Resistance: \( R = \frac{V}{I} = \frac{6}{3} = 2 \, \Omega \).

\( 1.5 \, \Omega \)
\( 1.8 \, \Omega \)
\( 2.0 \, \Omega \)
\( 2.5 \, \Omega \)
3

A copper wire of length \( 2 \, \text{m} \) and cross-sectional area \( 2 \times 10^{-6} \, \text{m}^2 \) carries a current of \( 2 \, \text{A} \). If the number density of free electrons in copper is \( 8.5 \times 10^{28} \, \text{m}^{-3} \) and \( e = 1.6 \times 10^{-19} \, \text{C} \), what is the drift speed of the electrons?

Drift speed is given by \( v_d = \frac{I}{n e A} \).

Given: \( I = 2 \, \text{A} \), \( n = 8.5 \times 10^{28} \, \text{m}^{-3} \), \( e = 1.6 \times 10^{-19} \, \text{C} \), \( A = 2 \times 10^{-6} \, \text{m}^2 \).

Substitute: \( v_d = \frac{2}{8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times 2 \times 10^{-6}} \).

Calculate: \( v_d = \frac{2}{2.72 \times 10^4} = 7.35 \times 10^{-5} \, \text{m/s} \).

\( 5.0 \times 10^{-5} \, \text{m/s} \)
\( 7.35 \times 10^{-5} \, \text{m/s} \)
\( 8.0 \times 10^{-5} \, \text{m/s} \)
\( 9.5 \times 10^{-5} \, \text{m/s} \)
2

A copper wire carries \( 4 \, \text{A} \) with a drift speed of \( 1.0 \times 10^{-4} \, \text{m/s} \). 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: \( A = \frac{4}{8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times 1.0 \times 10^{-4}} \).

Calculate: \( A = \frac{4}{1.36 \times 10^5} \approx 2.94 \times 10^{-5} \, \text{m}^2 \).

\( 2.5 \times 10^{-5} \, \text{m}^2 \)
\( 2.7 \times 10^{-5} \, \text{m}^2 \)
\( 2.9 \times 10^{-5} \, \text{m}^2 \)
\( 2.94 \times 10^{-5} \, \text{m}^2 \)
4

Why does the resistivity of an insulator decrease dramatically when impurities are added?

Impurities in insulators (e.g., semiconductors) introduce additional charge carriers by creating energy levels within the bandgap, allowing more electrons or holes to conduct, reducing resistivity.

Temperature increases
Charge carrier density increases
Electron mass decreases
Lattice structure collapses
2

A nichrome wire has a resistance of \( 50 \, \Omega \) at \( 25^\circ \text{C} \) and \( 55 \, \Omega \) at a higher temperature. If \( \alpha = 1.7 \times 10^{-4} \, ^\circ\text{C}^{-1} \), what is the temperature?

Use: \( R_t = R_0 [1 + \alpha (T - T_0)] \).

Substitute: \( 55 = 50 [1 + 1.7 \times 10^{-4} (T - 25)] \).

Solve: \( 55 = 50 + 50 \times 1.7 \times 10^{-4} (T - 25) \Rightarrow 5 = 8.5 \times 10^{-3} (T - 25) \).

\( T - 25 = \frac{5}{8.5 \times 10^{-3}} \approx 588 \Rightarrow T \approx 613^\circ \text{C} \).

\( 550^\circ \text{C} \)
\( 613^\circ \text{C} \)
\( 650^\circ \text{C} \)
\( 700^\circ \text{C} \)
2

A \( 7 \, \Omega \) resistor carries a current of \( 2 \, \text{A} \) for \( 25 \, \text{s} \). What is the energy dissipated?

Energy: \( W = I^2 R t \).

Substitute: \( W = 2^2 \times 7 \times 25 = 4 \times 175 = 700 \, \text{J} \).

\( 600 \, \text{J} \)
\( 650 \, \text{J} \)
\( 675 \, \text{J} \)
\( 700 \, \text{J} \)
4

Performance Summary

Score:

Category Details
Total Attempts:
Total Skipped:
Total Wrong Answers:
Total Correct Answers:
Time Taken:
Average Time Taken per Question:
Accuracy:
0