Current Electricity Chapter-Wise Test 10

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A wire has a resistance of \( 25 \, \Omega \) at \( 20^\circ \text{C} \) and \( 27.5 \, \Omega \) at \( 90^\circ \text{C} \). What is the temperature coefficient of resistivity?

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

Substitute: \( 27.5 = 25 [1 + \alpha (90 - 20)] \).

Solve: \( 27.5 = 25 + 1750\alpha \Rightarrow 1750\alpha = 2.5 \Rightarrow \alpha = \frac{2.5}{1750} \approx 1.43 \times 10^{-3} \, ^\circ\text{C}^{-1} \).

\( 1.43 \times 10^{-3} \, ^\circ\text{C}^{-1} \)
\( 1.5 \times 10^{-3} \, ^\circ\text{C}^{-1} \)
\( 1.6 \times 10^{-3} \, ^\circ\text{C}^{-1} \)
\( 1.8 \times 10^{-3} \, ^\circ\text{C}^{-1} \)
1

Two cells in parallel have emf \( 12 \, \text{V} \) and \( 6 \, \text{V} \) with internal resistances \( 4 \, \Omega \) and \( 2 \, \Omega \). What is the equivalent emf?

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

Substitute: \( \varepsilon_{\text{eq}} = \frac{12 \times 2 + 6 \times 4}{4 + 2} = \frac{24 + 24}{6} = \frac{48}{6} = 8 \, \text{V} \).

\( 7.5 \, \text{V} \)
\( 8.0 \, \text{V} \)
\( 8.5 \, \text{V} \)
\( 9.0 \, \text{V} \)
2

A \( 12 \, \text{V} \) battery with \( 1 \, \Omega \) internal resistance delivers a current of \( 2 \, \text{A} \) to a resistor. What is the resistance of the resistor?

Terminal voltage: \( V = \varepsilon - I r = 12 - 2 \times 1 = 10 \, \text{V} \).

Resistance: \( R = \frac{V}{I} = \frac{10}{2} = 5 \, \Omega \).

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

A Wheatstone bridge has \( R_1 = 30 \, \Omega \), \( R_2 = 60 \, \Omega \), \( R_3 = 25 \, \Omega \). What is \( R_4 \) for balance?

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

Substitute: \( \frac{30}{60} = \frac{25}{R_4} \).

Solve: \( 0.5 = \frac{25}{R_4} \Rightarrow R_4 = \frac{25}{0.5} = 50 \, \Omega \).

\( 45 \, \Omega \)
\( 50 \, \Omega \)
\( 55 \, \Omega \)
\( 60 \, \Omega \)
2

A copper wire carries \( 3.4 \, \text{A} \) with a drift speed of \( 1.25 \times 10^{-4} \, \text{m/s} \). 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: \( A = \frac{3.4}{8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times 1.25 \times 10^{-4}} \).

Calculate: \( A = \frac{3.4}{1.7 \times 10^5} = 2.0 \times 10^{-5} \, \text{m}^2 \).

\( 1.8 \times 10^{-5} \, \text{m}^2 \)
\( 1.9 \times 10^{-5} \, \text{m}^2 \)
\( 1.95 \times 10^{-5} \, \text{m}^2 \)
\( 2.0 \times 10^{-5} \, \text{m}^2 \)
4

A conductor has a resistivity of \( 6 \times 10^{-8} \, \Omega \text{m} \) and \( \alpha = 4 \times 10^{-3} \, ^\circ\text{C}^{-1} \) at \( 20^\circ \text{C} \). What is its resistivity at \( 90^\circ \text{C} \)?

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

Substitute: \( \rho_t = 6 \times 10^{-8} [1 + 4 \times 10^{-3} (90 - 20)] \).

Calculate: \( \rho_t = 6 \times 10^{-8} [1 + 0.28] = 6 \times 10^{-8} \times 1.28 = 7.68 \times 10^{-8} \, \Omega \text{m} \).

\( 7.2 \times 10^{-8} \, \Omega \text{m} \)
\( 7.5 \times 10^{-8} \, \Omega \text{m} \)
\( 7.68 \times 10^{-8} \, \Omega \text{m} \)
\( 8.0 \times 10^{-8} \, \Omega \text{m} \)
3

A circuit has a \( 30 \, \text{V} \) battery with \( 3 \, \Omega \) internal resistance and three resistors \( 3 \, \Omega \), \( 6 \, \Omega \), \( 12 \, \Omega \) in parallel. What is the current through the \( 6 \, \Omega \) resistor?

Parallel resistance: \( \frac{1}{R_p} = \frac{1}{3} + \frac{1}{6} + \frac{1}{12} = \frac{4 + 2 + 1}{12} = \frac{7}{12} \Rightarrow R_p = \frac{12}{7} \approx 1.71 \, \Omega \).

Total resistance: \( R_{\text{total}} = 3 + 1.71 = 4.71 \, \Omega \).

Total current: \( I = \frac{\varepsilon}{R_{\text{total}}} = \frac{30}{4.71} \approx 6.37 \, \text{A} \).

Voltage across parallel: \( V = I R_p = 6.37 \times 1.71 \approx 10.89 \, \text{V} \).

Current through \( 6 \, \Omega \): \( I_6 = \frac{V}{6} = \frac{10.89}{6} \approx 1.82 \, \text{A} \).

\( 1.6 \, \text{A} \)
\( 1.7 \, \text{A} \)
\( 1.8 \, \text{A} \)
\( 1.82 \, \text{A} \)
4

A \( 5 \, \Omega \) resistor carries a current of \( 4 \, \text{A} \) for \( 15 \, \text{s} \). What is the energy dissipated?

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

Substitute: \( W = 4^2 \times 5 \times 15 = 16 \times 75 = 1200 \, \text{J} \).

\( 1000 \, \text{J} \)
\( 1100 \, \text{J} \)
\( 1150 \, \text{J} \)
\( 1200 \, \text{J} \)
4

A circuit has a \( 6 \, \text{V} \) battery with \( 1 \, \Omega \) internal resistance and two resistors \( 2 \, \Omega \) and \( 3 \, \Omega \) in parallel. What is the total current?

Parallel resistance: \( \frac{1}{R_p} = \frac{1}{2} + \frac{1}{3} = \frac{3 + 2}{6} = \frac{5}{6} \Rightarrow R_p = \frac{6}{5} = 1.2 \, \Omega \).

Total resistance: \( R_{\text{total}} = 1 + 1.2 = 2.2 \, \Omega \).

Current: \( I = \frac{\varepsilon}{R_{\text{total}}} = \frac{6}{2.2} \approx 2.73 \, \text{A} \).

\( 2.0 \, \text{A} \)
\( 2.5 \, \text{A} \)
\( 2.73 \, \text{A} \)
\( 3.0 \, \text{A} \)
3

What happens to the total resistance of a circuit when an additional resistor is added in parallel to an existing resistor?

For resistors in parallel, \( 1/R_{\text{total}} = 1/R_1 + 1/R_2 \). Adding another resistor increases the sum of reciprocals, reducing \( R_{\text{total}} \) below the smallest individual resistance.

It decreases
It increases
It remains unchanged
It doubles
1

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