Current Electricity Chapter-Wise Test 8

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A Wheatstone bridge with \( R_1 = 5 \, \Omega \), \( R_2 = 10 \, \Omega \), \( R_3 = 15 \, \Omega \), \( R_4 = 30 \, \Omega \) has a \( 18 \, \text{V} \) battery across AC and a galvanometer (\( 3 \, \Omega \)) across BD. What is the current through the galvanometer?

Check balance: \( \frac{R_1}{R_2} = \frac{5}{10} = 0.5 \), \( \frac{R_3}{R_4} = \frac{15}{30} = 0.5 \). 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

Two cells in parallel have emf \( 6 \, \text{V} \) and \( 3 \, \text{V} \) with internal resistances \( 3 \, \Omega \) and \( 1 \, \Omega \). What is the equivalent emf?

For parallel: \( \varepsilon_{\text{eq}} = \frac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1 + r_2} \).

Substitute: \( \varepsilon_{\text{eq}} = \frac{6 \times 1 + 3 \times 3}{3 + 1} = \frac{6 + 9}{4} = 3.75 \, \text{V} \).

\( 3.5 \, \text{V} \)
\( 3.75 \, \text{V} \)
\( 4.0 \, \text{V} \)
\( 4.5 \, \text{V} \)
2

A copper wire of cross-sectional area \( 1 \times 10^{-6} \, \text{m}^2 \) carries a current of \( 1.5 \, \text{A} \). 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: \( v_d = \frac{1.5}{8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times 1 \times 10^{-6}} \).

Calculate: \( v_d = \frac{1.5}{1.36 \times 10^4} \approx 1.1 \times 10^{-4} \, \text{m/s} \).

\( 0.9 \times 10^{-4} \, \text{m/s} \)
\( 1.1 \times 10^{-4} \, \text{m/s} \)
\( 1.3 \times 10^{-4} \, \text{m/s} \)
\( 1.5 \times 10^{-4} \, \text{m/s} \)
2

In a circuit with two resistors in series, why does the voltage divide proportionally to their resistances?

Voltage drop across each resistor is \( V = I R \). With the same current in series, \( V \propto R \), so the total voltage splits in proportion to the resistances (e.g., \( V_1 / V_2 = R_1 / R_2 \)).

Current doubles
Resistance cancels out
Voltage drop is proportional to resistance
Electric field varies
3

A wire of length \( 3 \, \text{m} \) and resistance \( 6 \, \Omega \) is stretched to \( 6 \, \text{m} \). What is the new resistance?

Volume constant: \( l A = l' A' \Rightarrow A' = \frac{A}{2} \).

New resistance: \( R' = \frac{\rho l'}{A'} = \frac{\rho (2l)}{\frac{A}{2}} = 4 \frac{\rho l}{A} = 4R = 4 \times 6 = 24 \, \Omega \).

\( 18 \, \Omega \)
\( 20 \, \Omega \)
\( 24 \, \Omega \)
\( 28 \, \Omega \)
3

A Wheatstone bridge has \( R_1 = 16 \, \Omega \), \( R_2 = 32 \, \Omega \), \( R_3 = 20 \, \Omega \). What is \( R_4 \) for balance?

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

Substitute: \( \frac{16}{32} = \frac{20}{R_4} \).

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

Why does the drift velocity of electrons in a conductor remain constant despite continuous acceleration by an electric field?

Electrons accelerate due to the electric field but collide with lattice ions, losing momentum. These collisions occur at random intervals, and the average time between collisions (\( \tau \)) stabilizes the drift velocity (\( v_d = e E \tau / m \)), balancing acceleration with energy loss.

Electric field cancels out
Collisions with ions limit acceleration
Electron mass increases
Field strength decreases over time
2

When two identical resistors are connected in series and then in parallel, how does the equivalent resistance change?

For resistors \( R \) in series: \( R_{\text{eq}} = R + R = 2R \). In parallel: \( 1/R_{\text{eq}} = 1/R + 1/R = 2/R \Rightarrow R_{\text{eq}} = R/2 \). The series resistance (2R) is four times the parallel resistance (R/2), as \( 2R / (R/2) = 4 \).

Series resistance is four times parallel resistance
Parallel resistance is four times series resistance
They become equal
Series resistance doubles
1

A wire of length \( 12 \, \text{m} \) and cross-sectional area \( 3 \times 10^{-6} \, \text{m}^2 \) has a resistance of \( 24 \, \Omega \). What is the resistivity of the material?

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

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

Substitute: \( \rho = \frac{24 \times 3 \times 10^{-6}}{12} = 6 \times 10^{-6} \, \Omega \text{m} \).

\( 6.0 \times 10^{-6} \, \Omega \text{m} \)
\( 6.5 \times 10^{-6} \, \Omega \text{m} \)
\( 7.0 \times 10^{-6} \, \Omega \text{m} \)
\( 7.5 \times 10^{-6} \, \Omega \text{m} \)
1

A circuit has a \( 15 \, \text{V} \) battery with \( 2 \, \Omega \) internal resistance and three resistors \( 3 \, \Omega \), \( 6 \, \Omega \), \( 12 \, \Omega \) in parallel. What is the current through the \( 6 \, \Omega \) resistor?

Parallel resistance: \( \frac{1}{R_p} = \frac{1}{3} + \frac{1}{6} + \frac{1}{12} = \frac{4 + 2 + 1}{12} = \frac{7}{12} \Rightarrow R_p = \frac{12}{7} \approx 1.71 \, \Omega \).

Total resistance: \( R_{\text{total}} = 2 + 1.71 = 3.71 \, \Omega \).

Total current: \( I = \frac{\varepsilon}{R_{\text{total}}} = \frac{15}{3.71} \approx 4.04 \, \text{A} \).

Voltage across parallel: \( V = I R_p = 4.04 \times 1.71 \approx 6.91 \, \text{V} \).

Current through \( 6 \, \Omega \): \( I_6 = \frac{V}{6} = \frac{6.91}{6} \approx 1.15 \, \text{A} \).

\( 1.0 \, \text{A} \)
\( 1.1 \, \text{A} \)
\( 1.12 \, \text{A} \)
\( 1.15 \, \text{A} \)
4

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