Current Electricity Chapter-Wise Test 3

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A circuit has a \( 28 \, \text{V} \) battery with \( 4 \, \Omega \) internal resistance and three resistors \( 4 \, \Omega \), \( 8 \, \Omega \), \( 16 \, \Omega \) in parallel. What is the current through the \( 8 \, \Omega \) resistor?

Parallel resistance: \( \frac{1}{R_p} = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} = \frac{4 + 2 + 1}{16} = \frac{7}{16} \Rightarrow R_p = \frac{16}{7} \approx 2.29 \, \Omega \).

Total resistance: \( R_{\text{total}} = 4 + 2.29 = 6.29 \, \Omega \).

Total current: \( I = \frac{\varepsilon}{R_{\text{total}}} = \frac{28}{6.29} \approx 4.45 \, \text{A} \).

Voltage across parallel: \( V = I R_p = 4.45 \times 2.29 \approx 10.19 \, \text{V} \).

Current through \( 8 \, \Omega \): \( I_8 = \frac{V}{8} = \frac{10.19}{8} \approx 1.27 \, \text{A} \).

\( 1.0 \, \text{A} \)
\( 1.2 \, \text{A} \)
\( 1.25 \, \text{A} \)
\( 1.27 \, \text{A} \)
4

A \( 6 \, \text{V} \) battery with \( 0.5 \, \Omega \) internal resistance is connected to a \( 5.5 \, \Omega \) resistor. What is the power dissipated in the external resistor?

Total resistance: \( R_{\text{total}} = 5.5 + 0.5 = 6 \, \Omega \).

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

Power: \( P = I^2 R = 1^2 \times 5.5 = 5.5 \, \text{W} \).

\( 4.5 \, \text{W} \)
\( 5.0 \, \text{W} \)
\( 5.25 \, \text{W} \)
\( 5.5 \, \text{W} \)
4

A \( 18 \, \text{V} \) battery with negligible internal resistance is connected to a \( 6 \, \Omega \) and \( 12 \, \Omega \) resistor in series. What is the power dissipated in the \( 6 \, \Omega \) resistor?

Total resistance: \( R = 6 + 12 = 18 \, \Omega \).

Current: \( I = \frac{V}{R} = \frac{18}{18} = 1 \, \text{A} \).

Power: \( P = I^2 R = 1^2 \times 6 = 6 \, \text{W} \).

\( 5 \, \text{W} \)
\( 6 \, \text{W} \)
\( 7 \, \text{W} \)
\( 8 \, \text{W} \)
2

A \( 10 \, \text{V} \) battery with \( 1 \, \Omega \) internal resistance is connected to a \( 9 \, \Omega \) resistor. What is the power dissipated in the internal resistance?

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

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

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

\( 0.5 \, \text{W} \)
\( 0.8 \, \text{W} \)
\( 0.9 \, \text{W} \)
\( 1.0 \, \text{W} \)
4

In a circuit with two resistors of different resistances in series, which resistor dissipates more power if the same current flows through both?

Power \( P = I^2 R \). With the same current (\( I \)) in series, power is proportional to resistance (\( R \)). The resistor with higher resistance dissipates more power.

The one with higher resistance
The one with lower resistance
Both dissipate equal power
Depends on voltage
1

A wire of length \( 3 \, \text{m} \) and cross-sectional area \( 3 \times 10^{-6} \, \text{m}^2 \) has a resistance of \( 5 \, \Omega \). What is the resistivity of the material?

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

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

Given: \( R = 5 \, \Omega \), \( A = 3 \times 10^{-6} \, \text{m}^2 \), \( l = 3 \, \text{m} \).

Substitute: \( \rho = \frac{5 \times 3 \times 10^{-6}}{3} = 5 \times 10^{-6} \, \Omega \text{m} \).

\( 5.0 \times 10^{-6} \, \Omega \text{m} \)
\( 6.0 \times 10^{-6} \, \Omega \text{m} \)
\( 7.0 \times 10^{-6} \, \Omega \text{m} \)
\( 8.0 \times 10^{-6} \, \Omega \text{m} \)
1

A cell of emf \( 7 \, \text{V} \) and internal resistance \( 1 \, \Omega \) is connected to a \( 6 \, \Omega \) resistor. What is the terminal voltage?

Total resistance: \( R_{\text{total}} = 6 + 1 = 7 \, \Omega \).

Current: \( I = \frac{\varepsilon}{R_{\text{total}}} = \frac{7}{7} = 1 \, \text{A} \).

Terminal voltage: \( V = \varepsilon - I r = 7 - 1 \times 1 = 6 \, \text{V} \).

\( 6.0 \, \text{V} \)
\( 6.5 \, \text{V} \)
\( 7.0 \, \text{V} \)
\( 7.5 \, \text{V} \)
1

A circuit has a \( 16 \, \text{V} \) battery with \( 2 \, \Omega \) internal resistance and three resistors \( 2 \, \Omega \), \( 4 \, \Omega \), \( 8 \, \Omega \) in parallel. What is the current through the \( 2 \, \Omega \) resistor?

Parallel resistance: \( \frac{1}{R_p} = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} = \frac{4 + 2 + 1}{8} = \frac{7}{8} \Rightarrow R_p = \frac{8}{7} \approx 1.14 \, \Omega \).

Total resistance: \( R_{\text{total}} = 2 + 1.14 = 3.14 \, \Omega \).

Total current: \( I = \frac{\varepsilon}{R_{\text{total}}} = \frac{16}{3.14} \approx 5.10 \, \text{A} \).

Voltage across parallel: \( V = I R_p = 5.10 \times 1.14 \approx 5.81 \, \text{V} \).

Current through \( 2 \, \Omega \): \( I_2 = \frac{V}{2} = \frac{5.81}{2} \approx 2.91 \, \text{A} \).

\( 2.7 \, \text{A} \)
\( 2.8 \, \text{A} \)
\( 2.9 \, \text{A} \)
\( 2.91 \, \text{A} \)
4

Two cells of emf \( 2.5 \, \text{V} \) and \( 3.5 \, \text{V} \) with internal resistances \( 0.5 \, \Omega \) and \( 1 \, \Omega \) are connected in series. What is the equivalent emf?

For series: \( \varepsilon_{\text{eq}} = \varepsilon_1 + \varepsilon_2 = 2.5 + 3.5 = 6.0 \, \text{V} \).

\( 6.0 \, \text{V} \)
\( 6.5 \, \text{V} \)
\( 7.0 \, \text{V} \)
\( 7.5 \, \text{V} \)
1

A \( 5 \, \Omega \) resistor carries a current of \( 2 \, \text{A} \) for \( 10 \, \text{s} \). How much energy is dissipated as heat?

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

Given: \( I = 2 \, \text{A} \), \( R = 5 \, \Omega \), \( t = 10 \, \text{s} \).

Substitute: \( W = (2)^2 \times 5 \times 10 = 4 \times 50 = 200 \, \text{J} \).

\( 150 \, \text{J} \)
\( 200 \, \text{J} \)
\( 250 \, \text{J} \)
\( 300 \, \text{J} \)
2

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