Current Electricity Chapter-Wise Test 7

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A wire of length \( 2 \, \text{m} \) and resistance \( 5 \, \Omega \) is stretched to \( 4 \, \text{m} \). What is the new resistance?

Volume constant: \( l A = l' A' \Rightarrow A' = \frac{A}{2} \).

New resistance: \( R' = \frac{\rho l'}{A'} = \frac{\rho (2l)}{\frac{A}{2}} = 4 \frac{\rho l}{A} = 4R = 4 \times 5 = 20 \, \Omega \).

\( 15 \, \Omega \)
\( 18 \, \Omega \)
\( 20 \, \Omega \)
\( 25 \, \Omega \)
3

Two cells in parallel have emf \( 12 \, \text{V} \) and \( 3 \, \text{V} \) with internal resistances \( 4 \, \Omega \) and \( 1 \, \Omega \). What is the equivalent emf?

For parallel: \( \varepsilon_{\text{eq}} = \frac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1 + r_2} \).

Substitute: \( \varepsilon_{\text{eq}} = \frac{12 \times 1 + 3 \times 4}{4 + 1} = \frac{12 + 12}{5} = \frac{24}{5} = 4.8 \, \text{V} \).

\( 4.5 \, \text{V} \)
\( 4.8 \, \text{V} \)
\( 5.0 \, \text{V} \)
\( 5.5 \, \text{V} \)
2

A \( 10 \, \text{V} \) battery with negligible internal resistance is connected to a cubical network of 12 resistors, each \( 1.5 \, \Omega \). What is the current through one edge from a corner?

Equivalent resistance: \( R_{\text{eq}} = \frac{5}{6} R = \frac{5}{6} \times 1.5 = 1.25 \, \Omega \).

Total current: \( I_{\text{total}} = \frac{V}{R_{\text{eq}}} = \frac{10}{1.25} = 8 \, \text{A} \).

Corner current: \( I = \frac{I_{\text{total}}}{3} = \frac{8}{3} \approx 2.67 \, \text{A} \).

\( 2.0 \, \text{A} \)
\( 2.5 \, \text{A} \)
\( 2.67 \, \text{A} \)
\( 3.0 \, \text{A} \)
3

A copper wire of cross-sectional area \( 8 \times 10^{-7} \, \text{m}^2 \) carries a current of \( 2 \, \text{A} \). If \( n = 8.5 \times 10^{28} \, \text{m}^{-3} \) and \( e = 1.6 \times 10^{-19} \, \text{C} \), what is the drift speed?

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

Substitute: \( v_d = \frac{2}{8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times 8 \times 10^{-7}} \).

Calculate: \( v_d = \frac{2}{1.088 \times 10^4} \approx 1.84 \times 10^{-4} \, \text{m/s} \).

\( 1.6 \times 10^{-4} \, \text{m/s} \)
\( 1.84 \times 10^{-4} \, \text{m/s} \)
\( 2.0 \times 10^{-4} \, \text{m/s} \)
\( 2.2 \times 10^{-4} \, \text{m/s} \)
2

A wire has a resistance of \( 24 \, \Omega \) at \( 25^\circ \text{C} \) and \( 25.8 \, \Omega \) at \( 75^\circ \text{C} \). What is the temperature coefficient of resistivity?

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

Substitute: \( 25.8 = 24 [1 + \alpha (75 - 25)] \).

Solve: \( 25.8 = 24 + 1200\alpha \Rightarrow 1200\alpha = 1.8 \Rightarrow \alpha = \frac{1.8}{1200} = 1.5 \times 10^{-3} \, ^\circ\text{C}^{-1} \).

\( 1.5 \times 10^{-3} \, ^\circ\text{C}^{-1} \)
\( 1.8 \times 10^{-3} \, ^\circ\text{C}^{-1} \)
\( 2.0 \times 10^{-3} \, ^\circ\text{C}^{-1} \)
\( 2.2 \times 10^{-3} \, ^\circ\text{C}^{-1} \)
1

A nichrome wire has a resistance of \( 60 \, \Omega \) at \( 25^\circ \text{C} \) and \( \alpha = 1.7 \times 10^{-4} \, ^\circ\text{C}^{-1} \). What is its resistance at \( 225^\circ \text{C} \)?

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

Substitute: \( R_t = 60 [1 + 1.7 \times 10^{-4} (225 - 25)] \).

Calculate: \( R_t = 60 [1 + 1.7 \times 10^{-4} \times 200] = 60 [1 + 0.034] = 60 \times 1.034 = 62.04 \, \Omega \).

\( 61 \, \Omega \)
\( 62.04 \, \Omega \)
\( 63 \, \Omega \)
\( 64 \, \Omega \)
2

A conductor has a resistivity of \( 1.2 \times 10^{-7} \, \Omega \text{m} \) and \( \alpha = 4 \times 10^{-3} \, ^\circ\text{C}^{-1} \) at \( 20^\circ \text{C} \). What is its resistivity at \( 80^\circ \text{C} \)?

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

Substitute: \( \rho_t = 1.2 \times 10^{-7} [1 + 4 \times 10^{-3} (80 - 20)] \).

Calculate: \( \rho_t = 1.2 \times 10^{-7} [1 + 0.24] = 1.2 \times 10^{-7} \times 1.24 = 1.488 \times 10^{-7} \, \Omega \text{m} \).

\( 1.4 \times 10^{-7} \, \Omega \text{m} \)
\( 1.45 \times 10^{-7} \, \Omega \text{m} \)
\( 1.488 \times 10^{-7} \, \Omega \text{m} \)
\( 1.5 \times 10^{-7} \, \Omega \text{m} \)
3

A \( 18 \, \text{V} \) battery with \( 3 \, \Omega \) internal resistance delivers a current of \( 2 \, \text{A} \) to a resistor. What is the resistance of the resistor?

Terminal voltage: \( V = \varepsilon - I r = 18 - 2 \times 3 = 12 \, \text{V} \).

Resistance: \( R = \frac{V}{I} = \frac{12}{2} = 6 \, \Omega \).

\( 5.0 \, \Omega \)
\( 5.5 \, \Omega \)
\( 6.0 \, \Omega \)
\( 6.5 \, \Omega \)
3

A cell of emf \( 10 \, \text{V} \) and internal resistance \( 1 \, \Omega \) is connected to a \( 9 \, \Omega \) resistor. What is the terminal voltage?

Total resistance: \( R_{\text{total}} = 9 + 1 = 10 \, \Omega \).

Current: \( I = \frac{\varepsilon}{R_{\text{total}}} = \frac{10}{10} = 1 \, \text{A} \).

Terminal voltage: \( V = \varepsilon - I r = 10 - 1 \times 1 = 9 \, \text{V} \).

\( 9.0 \, \text{V} \)
\( 9.5 \, \text{V} \)
\( 10.0 \, \text{V} \)
\( 11.0 \, \text{V} \)
1

A cell of emf \( 6 \, \text{V} \) and internal resistance \( 1.5 \, \Omega \) is connected to a \( 4.5 \, \Omega \) resistor. What is the terminal voltage?

Total resistance: \( R_{\text{total}} = 4.5 + 1.5 = 6 \, \Omega \).

Current: \( I = \frac{\varepsilon}{R_{\text{total}}} = \frac{6}{6} = 1 \, \text{A} \).

Terminal voltage: \( V = \varepsilon - I r = 6 - 1 \times 1.5 = 4.5 \, \text{V} \).

\( 4.5 \, \text{V} \)
\( 5.0 \, \text{V} \)
\( 5.5 \, \text{V} \)
\( 6.0 \, \text{V} \)
1

Performance Summary

Score:

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