Current Electricity Chapter-Wise Test 14

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A \( 36 \, \text{V} \) battery with negligible internal resistance is connected to a cubical network of 12 resistors, each \( 6 \, \Omega \). What is the total current?

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

Total current: \( I = \frac{V}{R_{\text{eq}}} = \frac{36}{5} = 7.2 \, \text{A} \).

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

A circuit has a \( 9 \, \text{V} \) battery with \( 1 \, \Omega \) internal resistance and two resistors \( 5 \, \Omega \) and \( 3 \, \Omega \) in parallel. What is the total current?

Parallel resistance: \( \frac{1}{R_p} = \frac{1}{5} + \frac{1}{3} = \frac{3 + 5}{15} = \frac{8}{15} \Rightarrow R_p = \frac{15}{8} = 1.875 \, \Omega \).

Total resistance: \( R_{\text{total}} = 1 + 1.875 = 2.875 \, \Omega \).

Current: \( I = \frac{\varepsilon}{R_{\text{total}}} = \frac{9}{2.875} \approx 3.13 \, \text{A} \).

\( 2.8 \, \text{A} \)
\( 3.0 \, \text{A} \)
\( 3.13 \, \text{A} \)
\( 3.5 \, \text{A} \)
3

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

Parallel resistance: \( \frac{1}{R_p} = \frac{1}{6} + \frac{1}{3} = \frac{1 + 2}{6} = \frac{3}{6} = 0.5 \Rightarrow R_p = 2 \, \Omega \).

Total resistance: \( R_{\text{total}} = 3 + 2 = 5 \, \Omega \).

Current: \( I = \frac{\varepsilon}{R_{\text{total}}} = \frac{12}{5} = 2.4 \, \text{A} \).

\( 2.0 \, \text{A} \)
\( 2.2 \, \text{A} \)
\( 2.4 \, \text{A} \)
\( 2.6 \, \text{A} \)
3

A \( 40 \, \text{V} \) battery with negligible internal resistance is connected to a cubical network of 12 resistors, each \( 4 \, \Omega \). What is the total current?

Equivalent resistance: \( R_{\text{eq}} = \frac{5}{6} R = \frac{5}{6} \times 4 = \frac{20}{6} \approx 3.33 \, \Omega \).

Total current: \( I = \frac{V}{R_{\text{eq}}} = \frac{40}{\frac{20}{6}} = 40 \times \frac{6}{20} = 12 \, \text{A} \).

\( 10.0 \, \text{A} \)
\( 11.0 \, \text{A} \)
\( 12.0 \, \text{A} \)
\( 13.0 \, \text{A} \)
3

In a circuit with a variable resistor, under what condition is the power delivered to the resistor maximized?

Power \( P = I^2 R = (\varepsilon / (R + r))^2 R \). Maximum power transfer occurs when \( R = r \) (internal resistance), balancing current and resistance for optimal energy delivery.

When resistance is zero
When resistance is infinite
When external resistance equals internal resistance
When voltage is halved
3

A \( 15 \, \text{V} \) battery with negligible internal resistance is connected to a \( 5 \, \Omega \) and \( 10 \, \Omega \) resistor in series. What is the power dissipated in the \( 5 \, \Omega \) resistor?

Total resistance: \( R = 5 + 10 = 15 \, \Omega \).

Current: \( I = \frac{V}{R} = \frac{15}{15} = 1 \, \text{A} \).

Power: \( P = I^2 R = 1^2 \times 5 = 5 \, \text{W} \).

\( 4 \, \text{W} \)
\( 5 \, \text{W} \)
\( 6 \, \text{W} \)
\( 7 \, \text{W} \)
2

In a conductor, if the number of free electrons per unit volume doubles while the electric field and relaxation time remain constant, what happens to the current density?

Current density \( j = n e v_d \), where \( v_d = e E \tau / m \). If \( n \) doubles and \( E \), \( \tau \), and \( m \) remain constant, \( v_d \) is unchanged, so \( j' = 2n e v_d = 2j \). Thus, current density doubles.

It doubles
It halves
It remains constant
It quadruples
1

A circuit with a \( 9 \, \text{V} \) battery and \( 1 \, \Omega \) internal resistance has three resistors: \( 2 \, \Omega \), \( 3 \, \Omega \), and \( 6 \, \Omega \) in parallel. What is the total current drawn from the battery?

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

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

Total current: \( I = \frac{\varepsilon}{R_{\text{total}}} = \frac{9}{2} = 4.5 \, \text{A} \).

\( 3.0 \, \text{A} \)
\( 4.0 \, \text{A} \)
\( 4.5 \, \text{A} \)
\( 5.0 \, \text{A} \)
3

A Wheatstone bridge has \( R_1 = 22 \, \Omega \), \( R_2 = 44 \, \Omega \), \( R_3 = 15 \, \Omega \). What is \( R_4 \) for balance?

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

Substitute: \( \frac{22}{44} = \frac{15}{R_4} \).

Solve: \( 0.5 = \frac{15}{R_4} \Rightarrow R_4 = \frac{15}{0.5} = 30 \, \Omega \).

\( 25 \, \Omega \)
\( 30 \, \Omega \)
\( 35 \, \Omega \)
\( 40 \, \Omega \)
2

Why does a semiconductor’s resistance decrease with increasing temperature, unlike a metal?

In semiconductors, as temperature rises, more electrons are thermally excited from the valence band to the conduction band, increasing the number of charge carriers (\( n \)). Since \( \rho = m / (n e^2 \tau) \), a significant increase in \( n \) outweighs the decrease in \( \tau \), reducing \( \rho \) and thus resistance.

Collision time increases
Number of charge carriers increases
Electron mass decreases
Conduction band shrinks
2

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