Current Electricity Chapter-Wise Test 11

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A circuit has a \( 12 \, \text{V} \) battery with \( 2 \, \Omega \) internal resistance and three resistors \( 3 \, \Omega \), \( 6 \, \Omega \), \( 9 \, \Omega \) in parallel. What is the current through the \( 9 \, \Omega \) resistor?

Parallel resistance: \( \frac{1}{R_p} = \frac{1}{3} + \frac{1}{6} + \frac{1}{9} = \frac{6 + 3 + 2}{18} = \frac{11}{18} \Rightarrow R_p = \frac{18}{11} \approx 1.64 \, \Omega \).

Total resistance: \( R_{\text{total}} = 2 + 1.64 = 3.64 \, \Omega \).

Total current: \( I = \frac{\varepsilon}{R_{\text{total}}} = \frac{12}{3.64} \approx 3.3 \, \text{A} \).

Voltage across parallel: \( V = I R_p = 3.3 \times 1.64 \approx 5.41 \, \text{V} \).

Current through \( 9 \, \Omega \): \( I_9 = \frac{V}{9} = \frac{5.41}{9} \approx 0.6 \, \text{A} \).

\( 0.4 \, \text{A} \)
\( 0.5 \, \text{A} \)
\( 0.55 \, \text{A} \)
\( 0.6 \, \text{A} \)
4

A copper wire of cross-sectional area \( 4 \times 10^{-7} \, \text{m}^2 \) carries a current of \( 1.2 \, \text{A} \). 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: \( v_d = \frac{1.2}{8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times 4 \times 10^{-7}} \).

Calculate: \( v_d = \frac{1.2}{5.44 \times 10^3} \approx 2.21 \times 10^{-4} \, \text{m/s} \).

\( 2.0 \times 10^{-4} \, \text{m/s} \)
\( 2.21 \times 10^{-4} \, \text{m/s} \)
\( 2.4 \times 10^{-4} \, \text{m/s} \)
\( 2.6 \times 10^{-4} \, \text{m/s} \)
2

A cell of emf \( 4 \, \text{V} \) and internal resistance \( 0.5 \, \Omega \) is connected to a \( 3.5 \, \Omega \) resistor. What is the current in the circuit?

Total resistance: \( R_{\text{total}} = 3.5 + 0.5 = 4 \, \Omega \).

Current: \( I = \frac{\varepsilon}{R_{\text{total}}} = \frac{4}{4} = 1 \, \text{A} \).

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

A circuit has a \( 24 \, \text{V} \) battery with \( 3 \, \Omega \) internal resistance and three resistors \( 6 \, \Omega \), \( 12 \, \Omega \), \( 24 \, \Omega \) in parallel. What is the current through the \( 12 \, \Omega \) resistor?

Parallel resistance: \( \frac{1}{R_p} = \frac{1}{6} + \frac{1}{12} + \frac{1}{24} = \frac{4 + 2 + 1}{24} = \frac{7}{24} \Rightarrow R_p = \frac{24}{7} \approx 3.43 \, \Omega \).

Total resistance: \( R_{\text{total}} = 3 + 3.43 = 6.43 \, \Omega \).

Total current: \( I = \frac{\varepsilon}{R_{\text{total}}} = \frac{24}{6.43} \approx 3.73 \, \text{A} \).

Voltage across parallel: \( V = I R_p = 3.73 \times 3.43 \approx 12.79 \, \text{V} \).

Current through \( 12 \, \Omega \): \( I_{12} = \frac{V}{12} = \frac{12.79}{12} \approx 1.07 \, \text{A} \).

\( 0.9 \, \text{A} \)
\( 1.0 \, \text{A} \)
\( 1.05 \, \text{A} \)
\( 1.07 \, \text{A} \)
4

What happens to the current density in a conductor if the conductor is stretched to twice its original length without changing the applied voltage?

Stretching doubles length (\( l' = 2l \)) and halves area (\( A' = A/2 \)) due to volume conservation. Resistance becomes \( R' = \rho (2l) / (A/2) = 4R \). Current \( I' = V / (4R) = I/4 \). Current density \( j = I / A \), so \( j' = I' / A' = (I/4) / (A/2) = I / (2A) = j/2 \).

It doubles
It quadruples
It remains unchanged
It halves
4

A copper wire of cross-sectional area \( 4 \times 10^{-7} \, \text{m}^2 \) carries a current of \( 1.36 \, \text{A} \). 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: \( v_d = \frac{1.36}{8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times 4 \times 10^{-7}} \).

Calculate: \( v_d = \frac{1.36}{5.44 \times 10^3} = 2.5 \times 10^{-4} \, \text{m/s} \).

\( 2.2 \times 10^{-4} \, \text{m/s} \)
\( 2.5 \times 10^{-4} \, \text{m/s} \)
\( 2.7 \times 10^{-4} \, \text{m/s} \)
\( 3.0 \times 10^{-4} \, \text{m/s} \)
2

A \( 6 \, \Omega \) resistor carries a current of \( 4 \, \text{A} \) for \( 5 \, \text{s} \). What is the energy dissipated?

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

Substitute: \( W = 4^2 \times 6 \times 5 = 16 \times 30 = 480 \, \text{J} \).

\( 400 \, \text{J} \)
\( 450 \, \text{J} \)
\( 460 \, \text{J} \)
\( 480 \, \text{J} \)
4

A Wheatstone bridge with \( R_1 = 6 \, \Omega \), \( R_2 = 12 \, \Omega \), \( R_3 = 9 \, \Omega \), \( R_4 = 18 \, \Omega \) has a \( 15 \, \text{V} \) battery across AC and a galvanometer (\( 5 \, \Omega \)) across BD. What is the current through the galvanometer?

Check balance: \( \frac{R_1}{R_2} = \frac{6}{12} = 0.5 \), \( \frac{R_3}{R_4} = \frac{9}{18} = 0.5 \). Bridge is balanced.

Since balanced, \( I_g = 0 \, \text{A} \).

\( 0.1 \, \text{A} \)
\( 0.05 \, \text{A} \)
\( 0.02 \, \text{A} \)
\( 0 \, \text{A} \)
4

Why does the emf of a cell depend on its chemical composition rather than its size?

Emf is the potential difference generated by chemical reactions between electrodes and electrolyte, determined by the materials’ electrochemical properties, not physical dimensions like size.

Size increases resistance
Current depends on size
Voltage cancels out
Chemical reactions define potential
4

Why is high voltage used in power transmission lines to reduce energy loss?

Power loss in lines is \( P_{\text{loss}} = I^2 R \). For a given power (\( P = V I \)), increasing \( V \) reduces \( I \) (\( I = P / V \)), significantly lowering \( I^2 R \) losses.

Increases conductor resistance
Reduces line length
Increases current efficiency
Reduces current and thus resistive losses
4

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