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  4. Current Electricity Chapter-Wise Test 2

Current Electricity Chapter-Wise Test 2

Correct answer Carries: 4.

Wrong Answer Carries: -1.

What is the primary reason a battery’s terminal voltage is less than its emf when supplying current?

Terminal voltage \( V = \varepsilon - I r \). When current flows, the voltage drop across the internal resistance (\( I r \)) reduces the voltage available at the terminals below the emf (\( \varepsilon \)).

Voltage drop across internal resistance
Decrease in emf
Increase in external resistance
Loss of charge carriers
1

A Wheatstone bridge has \( R_1 = 15 \, \Omega \), \( R_2 = 45 \, \Omega \), \( R_3 = 10 \, \Omega \). What is \( R_4 \) for balance?

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

Substitute: \( \frac{15}{45} = \frac{10}{R_4} \).

Solve: \( \frac{1}{3} = \frac{10}{R_4} \Rightarrow R_4 = 10 \times 3 = 30 \, \Omega \).

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

A \( 12 \, \text{V} \) battery with negligible internal resistance is connected to a \( 4 \, \Omega \) and \( 8 \, \Omega \) resistor in series. What is the power dissipated in the \( 4 \, \Omega \) resistor?

Total resistance: \( R = 4 + 8 = 12 \, \Omega \).

Current: \( I = \frac{V}{R} = \frac{12}{12} = 1 \, \text{A} \).

Power: \( P = I^2 R = 1^2 \times 4 = 4 \, \text{W} \).

\( 3 \, \text{W} \)
\( 4 \, \text{W} \)
\( 5 \, \text{W} \)
\( 6 \, \text{W} \)
2

In a circuit with two cells of different emfs connected in parallel, why is the equivalent emf not simply the sum of the individual emfs?

In parallel, the equivalent emf depends on both emfs and internal resistances (\( \varepsilon_{\text{eq}} = (\varepsilon_1 r_2 + \varepsilon_2 r_1) / (r_1 + r_2) \)). The lower-emf cell can draw current from the higher-emf cell, resulting in a weighted average rather than a sum.

Cells cancel each other
Internal resistances affect the net emf
Current flows in reverse
Voltage doubles
2

Why does the current density in a conductor remain uniform across its cross-section under steady-state conditions?

Current density (\( j = I / A \)) is uniform if the current distributes evenly. In steady state, charge conservation (Kirchhoff’s junction rule) ensures a constant current through a uniform conductor, making \( j \) consistent across the area.

Voltage increases
Resistance varies
Charge flows uniformly
Electric field cancels out
3

A conductor has a resistivity of \( 5 \times 10^{-8} \, \Omega \text{m} \) and \( \alpha = 4 \times 10^{-3} \, ^\circ\text{C}^{-1} \) at \( 30^\circ \text{C} \). What is its resistivity at \( 100^\circ \text{C} \)?

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

Substitute: \( \rho_t = 5 \times 10^{-8} [1 + 4 \times 10^{-3} (100 - 30)] \).

Calculate: \( \rho_t = 5 \times 10^{-8} [1 + 0.28] = 5 \times 10^{-8} \times 1.28 = 6.4 \times 10^{-8} \, \Omega \text{m} \).

\( 6.0 \times 10^{-8} \, \Omega \text{m} \)
\( 6.2 \times 10^{-8} \, \Omega \text{m} \)
\( 6.4 \times 10^{-8} \, \Omega \text{m} \)
\( 6.6 \times 10^{-8} \, \Omega \text{m} \)
3

A \( 12 \, \Omega \) resistor dissipates \( 48 \, \text{W} \) of power. What is the current through it?

Power: \( P = I^2 R \).

Rearrange: \( I = \sqrt{\frac{P}{R}} \).

Substitute: \( I = \sqrt{\frac{48}{12}} = \sqrt{4} = 2 \, \text{A} \).

\( 2.0 \, \text{A} \)
\( 2.5 \, \text{A} \)
\( 3.0 \, \text{A} \)
\( 4.0 \, \text{A} \)
1

What happens to the current in a conductor if its cross-sectional area is doubled while keeping the potential difference and length constant?

Resistance \( R = \rho l / A \). If \( A \) doubles, \( R \) becomes \( R/2 \). Current \( I = V / R \), so if \( R \) halves and \( V \) is constant, \( I \) doubles.

It doubles
It halves
It remains unchanged
It quadruples
1

A Wheatstone bridge has \( R_1 = 26 \, \Omega \), \( R_2 = 52 \, \Omega \), \( R_3 = 20 \, \Omega \). What is \( R_4 \) for balance?

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

Substitute: \( \frac{26}{52} = \frac{20}{R_4} \).

Solve: \( 0.5 = \frac{20}{R_4} \Rightarrow R_4 = \frac{20}{0.5} = 40 \, \Omega \).

\( 35 \, \Omega \)
\( 40 \, \Omega \)
\( 45 \, \Omega \)
\( 50 \, \Omega \)
2

A copper wire carries \( 4.5 \, \text{A} \) with a drift speed of \( 1.8 \times 10^{-4} \, \text{m/s} \). 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: \( A = \frac{4.5}{8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times 1.8 \times 10^{-4}} \).

Calculate: \( A = \frac{4.5}{2.448 \times 10^5} \approx 1.84 \times 10^{-5} \, \text{m}^2 \).

\( 1.6 \times 10^{-5} \, \text{m}^2 \)
\( 1.7 \times 10^{-5} \, \text{m}^2 \)
\( 1.8 \times 10^{-5} \, \text{m}^2 \)
\( 1.84 \times 10^{-5} \, \text{m}^2 \)
4

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