Current Electricity Chapter-Wise Test 4

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A Wheatstone bridge with \( R_1 = 4 \, \Omega \), \( R_2 = 8 \, \Omega \), \( R_3 = 6 \, \Omega \), \( R_4 = 12 \, \Omega \) has a \( 12 \, \text{V} \) battery across AC and a galvanometer (\( 2 \, \Omega \)) across BD. What is the current through the galvanometer?

Check balance: \( \frac{R_1}{R_2} = \frac{4}{8} = 0.5 \), \( \frac{R_3}{R_4} = \frac{6}{12} = 0.5 \). 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

A circuit has a \( 12 \, \text{V} \) battery with \( 1 \, \Omega \) internal resistance and two resistors \( 2 \, \Omega \) and \( 6 \, \Omega \) in parallel. What is the total current?

Parallel resistance: \( \frac{1}{R_p} = \frac{1}{2} + \frac{1}{6} = \frac{3 + 1}{6} = \frac{4}{6} = \frac{2}{3} \Rightarrow R_p = \frac{3}{2} = 1.5 \, \Omega \).

Total resistance: \( R_{\text{total}} = 1 + 1.5 = 2.5 \, \Omega \).

Current: \( I = \frac{\varepsilon}{R_{\text{total}}} = \frac{12}{2.5} = 4.8 \, \text{A} \).

\( 4.0 \, \text{A} \)
\( 4.5 \, \text{A} \)
\( 4.8 \, \text{A} \)
\( 5.0 \, \text{A} \)
3

A wire has a resistance of \( 12 \, \Omega \) at \( 25^\circ \text{C} \) and \( 13.2 \, \Omega \) at \( 85^\circ \text{C} \). What is the temperature coefficient of resistivity?

Use: \( R_t = R_0 [1 + \alpha (T - T_0)] \).

Substitute: \( 13.2 = 12 [1 + \alpha (85 - 25)] \).

Solve: \( 13.2 = 12 + 720\alpha \Rightarrow 720\alpha = 1.2 \Rightarrow \alpha = \frac{1.2}{720} \approx 1.67 \times 10^{-3} \, ^\circ\text{C}^{-1} \).

\( 1.67 \times 10^{-3} \, ^\circ\text{C}^{-1} \)
\( 1.8 \times 10^{-3} \, ^\circ\text{C}^{-1} \)
\( 2.0 \times 10^{-3} \, ^\circ\text{C}^{-1} \)
\( 2.2 \times 10^{-3} \, ^\circ\text{C}^{-1} \)
1

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

Power: \( P = \frac{V^2}{R} \).

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

Substitute: \( V = \sqrt{48 \times 12} = \sqrt{576} = 24 \, \text{V} \).

\( 24 \, \text{V} \)
\( 26 \, \text{V} \)
\( 28 \, \text{V} \)
\( 30 \, \text{V} \)
1

A Wheatstone bridge with \( R_1 = 16 \, \Omega \), \( R_2 = 32 \, \Omega \), \( R_3 = 24 \, \Omega \), \( R_4 = 48 \, \Omega \) has a \( 20 \, \text{V} \) battery across AC and a galvanometer (\( 5 \, \Omega \)) across BD. What is the current through the galvanometer?

Check balance: \( \frac{R_1}{R_2} = \frac{16}{32} = 0.5 \), \( \frac{R_3}{R_4} = \frac{24}{48} = 0.5 \). 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

A \( 18 \, \text{V} \) battery with negligible internal resistance is connected to a cubical network of 12 resistors, each \( 3 \, \Omega \). What is the total current?

Equivalent resistance: \( R_{\text{eq}} = \frac{5}{6} R = \frac{5}{6} \times 3 = 2.5 \, \Omega \).

Total current: \( I = \frac{V}{R_{\text{eq}}} = \frac{18}{2.5} = 7.2 \, \text{A} \).

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

A Wheatstone bridge with \( R_1 = 10 \, \Omega \), \( R_2 = 20 \, \Omega \), \( R_3 = 15 \, \Omega \), \( R_4 = 30 \, \Omega \) has a \( 15 \, \text{V} \) battery across AC and a galvanometer (\( 10 \, \Omega \)) across BD. What is the current through the galvanometer?

Check balance: \( \frac{R_1}{R_2} = \frac{10}{20} = 0.5 \), \( \frac{R_3}{R_4} = \frac{15}{30} = 0.5 \). Bridge is balanced.

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

\( 0.05 \, \text{A} \)
\( 0.02 \, \text{A} \)
\( 0.01 \, \text{A} \)
\( 0 \, \text{A} \)
4

Why does the drift velocity of electrons in a conductor remain much smaller than their thermal velocity?

Drift velocity (\( v_d = e E \tau / m \)) is small because \( E \) is typically weak and \( \tau \) is short due to frequent collisions, whereas thermal velocity arises from random motion at high speeds (proportional to \( \sqrt{k T / m} \)), unaffected by the field.

Electric field cancels thermal motion
Electrons lose charge
Resistance stops motion
Frequent collisions limit net motion
4

A cell of emf \( 5 \, \text{V} \) and internal resistance \( 0.5 \, \Omega \) is connected to a \( 4.5 \, \Omega \) resistor. What is the current in the circuit?

Total resistance: \( R_{\text{total}} = 4.5 + 0.5 = 5 \, \Omega \).

Current: \( I = \frac{\varepsilon}{R_{\text{total}}} = \frac{5}{5} = 1 \, \text{A} \).

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

Why does a non-ohmic device, like a diode, exhibit a non-linear current-voltage relationship?

In a diode, resistance depends on the direction and magnitude of the applied voltage due to its semiconductor junction properties. Forward bias reduces resistance, while reverse bias increases it, leading to a non-linear \( I \)-versus-\( V \) curve.

Temperature fluctuates
Resistance varies with voltage
Current is constant
Conductor length changes
2

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