Current Electricity Chapter-Wise Test 18

Correct answer Carries: 4.

Wrong Answer Carries: -1.

Two cells of emf \( 5 \, \text{V} \) and \( 9 \, \text{V} \) with internal resistances \( 1 \, \Omega \) and \( 2 \, \Omega \) are connected in series with a \( 7 \, \Omega \) resistor. What is the current through the circuit?

Equivalent emf: \( \varepsilon_{\text{eq}} = 5 + 9 = 14 \, \text{V} \).

Total resistance: \( R_{\text{total}} = 1 + 2 + 7 = 10 \, \Omega \).

Current: \( I = \frac{\varepsilon_{\text{eq}}}{R_{\text{total}}} = \frac{14}{10} = 1.4 \, \text{A} \).

\( 1.4 \, \text{A} \)
\( 1.5 \, \text{A} \)
\( 1.6 \, \text{A} \)
\( 1.8 \, \text{A} \)
1

Two cells of emf \( 5 \, \text{V} \) and \( 7 \, \text{V} \) with internal resistances \( 2 \, \Omega \) and \( 4 \, \Omega \) are connected in series with a \( 6 \, \Omega \) resistor. What is the current through the circuit?

Equivalent emf: \( \varepsilon_{\text{eq}} = 5 + 7 = 12 \, \text{V} \).

Total resistance: \( R_{\text{total}} = 2 + 4 + 6 = 12 \, \Omega \).

Current: \( I = \frac{\varepsilon_{\text{eq}}}{R_{\text{total}}} = \frac{12}{12} = 1 \, \text{A} \).

\( 1.0 \, \text{A} \)
\( 1.2 \, \text{A} \)
\( 1.5 \, \text{A} \)
\( 2.0 \, \text{A} \)
1

A \( 8 \, \Omega \) resistor carries a current of \( 2.5 \, \text{A} \) for \( 10 \, \text{s} \). What is the energy dissipated?

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

Substitute: \( W = (2.5)^2 \times 8 \times 10 = 6.25 \times 80 = 500 \, \text{J} \).

\( 400 \, \text{J} \)
\( 450 \, \text{J} \)
\( 475 \, \text{J} \)
\( 500 \, \text{J} \)
4

A circuit has a \( 8 \, \text{V} \) battery with \( 2 \, \Omega \) internal resistance and two resistors \( 4 \, \Omega \) and \( 4 \, \Omega \) in parallel. What is the total current?

Parallel resistance: \( \frac{1}{R_p} = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = 0.5 \Rightarrow R_p = 2 \, \Omega \).

Total resistance: \( R_{\text{total}} = 2 + 2 = 4 \, \Omega \).

Current: \( I = \frac{\varepsilon}{R_{\text{total}}} = \frac{8}{4} = 2 \, \text{A} \).

\( 1.5 \, \text{A} \)
\( 1.8 \, \text{A} \)
\( 2.0 \, \text{A} \)
\( 2.5 \, \text{A} \)
3

A \( 30 \, \text{V} \) battery with negligible internal resistance is connected to a cubical network of 12 resistors, each \( 5 \, \Omega \). What is the total current?

Equivalent resistance: \( R_{\text{eq}} = \frac{5}{6} R = \frac{5}{6} \times 5 = \frac{25}{6} \approx 4.17 \, \Omega \).

Total current: \( I = \frac{V}{R_{\text{eq}}} = \frac{30}{\frac{25}{6}} = 30 \times \frac{6}{25} = 7.2 \, \text{A} \).

\( 6.0 \, \text{A} \)
\( 6.5 \, \text{A} \)
\( 7.2 \, \text{A} \)
\( 8.0 \, \text{A} \)
3

A \( 9 \, \text{V} \) battery with \( 0.5 \, \Omega \) internal resistance is connected to a \( 8.5 \, \Omega \) resistor. What is the power dissipated in the external resistor?

Total resistance: \( R_{\text{total}} = 8.5 + 0.5 = 9 \, \Omega \).

Current: \( I = \frac{\varepsilon}{R_{\text{total}}} = \frac{9}{9} = 1 \, \text{A} \).

Power: \( P = I^2 R = 1^2 \times 8.5 = 8.5 \, \text{W} \).

\( 8.0 \, \text{W} \)
\( 8.2 \, \text{W} \)
\( 8.25 \, \text{W} \)
\( 8.5 \, \text{W} \)
4

A \( 12 \, \text{V} \) battery with \( 2 \, \Omega \) internal resistance is connected to a \( 10 \, \Omega \) resistor. What is the power dissipated in the internal resistance?

Total resistance: \( R_{\text{total}} = 10 + 2 = 12 \, \Omega \).

Current: \( I = \frac{\varepsilon}{R_{\text{total}}} = \frac{12}{12} = 1 \, \text{A} \).

Power: \( P = I^2 r = 1^2 \times 2 = 2 \, \text{W} \).

\( 1.5 \, \text{W} \)
\( 1.8 \, \text{W} \)
\( 1.9 \, \text{W} \)
\( 2.0 \, \text{W} \)
4

A wire of length \( 10 \, \text{m} \) and cross-sectional area \( 1 \times 10^{-6} \, \text{m}^2 \) has a resistance of \( 20 \, \Omega \). What is the resistivity of the material?

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

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

Substitute: \( \rho = \frac{20 \times 1 \times 10^{-6}}{10} = 2 \times 10^{-6} \, \Omega \text{m} \).

\( 2.0 \times 10^{-6} \, \Omega \text{m} \)
\( 2.5 \times 10^{-6} \, \Omega \text{m} \)
\( 3.0 \times 10^{-6} \, \Omega \text{m} \)
\( 3.5 \times 10^{-6} \, \Omega \text{m} \)
1

A wire has a resistance of \( 35 \, \Omega \) at \( 20^\circ \text{C} \) and \( 36.4 \, \Omega \) at \( 60^\circ \text{C} \). What is the temperature coefficient of resistivity?

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

Substitute: \( 36.4 = 35 [1 + \alpha (60 - 20)] \).

Solve: \( 36.4 = 35 + 1400\alpha \Rightarrow 1400\alpha = 1.4 \Rightarrow \alpha = \frac{1.4}{1400} = 1.0 \times 10^{-3} \, ^\circ\text{C}^{-1} \).

\( 1.0 \times 10^{-3} \, ^\circ\text{C}^{-1} \)
\( 1.2 \times 10^{-3} \, ^\circ\text{C}^{-1} \)
\( 1.5 \times 10^{-3} \, ^\circ\text{C}^{-1} \)
\( 1.8 \times 10^{-3} \, ^\circ\text{C}^{-1} \)
1

In a circuit, a \( 20 \, \text{V} \) battery with negligible internal resistance is connected across a cubical network of 12 resistors, each \( 2 \, \Omega \). What is the total current?

Equivalent resistance of cube network: \( R_{\text{eq}} = \frac{5}{6} R = \frac{5}{6} \times 2 = \frac{10}{6} = \frac{5}{3} \, \Omega \).

Total current: \( I = \frac{V}{R_{\text{eq}}} = \frac{20}{\frac{5}{3}} = 20 \times \frac{3}{5} = 12 \, \text{A} \).

\( 10 \, \text{A} \)
\( 12 \, \text{A} \)
\( 14 \, \text{A} \)
\( 16 \, \text{A} \)
2

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