Current Electricity Chapter-Wise Test 12

Correct answer Carries: 4.

Wrong Answer Carries: -1.

What is the primary source of energy dissipation in a resistor carrying current?

Energy dissipation in a resistor occurs as electrons collide with lattice ions, transferring kinetic energy gained from the electric field into thermal energy (heat) via lattice vibrations.

Electron recombination
Collisions with lattice ions
Magnetic field generation
Radiation emission
2

In a Wheatstone bridge, what is the condition for no current to flow through the galvanometer?

No current flows when the bridge is balanced, i.e., \( R_1 / R_2 = R_3 / R_4 \), making the potentials at the galvanometer’s nodes equal, so the potential difference across it is zero.

Resistances are unequal
Voltage is zero
Ratio of resistances in opposite arms is equal
Current is maximum
3

In a circuit with a battery, why does the current establish almost instantaneously when the circuit is closed, despite the slow drift velocity of electrons?

The electric field propagates through the conductor at near-light speed, causing all free electrons to start moving simultaneously. Drift velocity is slow, but the field’s rapid establishment initiates current instantly.

Electric field propagates quickly
Electrons move at light speed
Resistance drops to zero
Voltage increases gradually
1

A \( 24 \, \text{V} \) 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: \( I = \frac{V}{R} = \frac{24}{12} = 2 \, \text{A} \).

Power: \( P = I^2 R = 2^2 \times 4 = 16 \, \text{W} \).

\( 12 \, \text{W} \)
\( 16 \, \text{W} \)
\( 20 \, \text{W} \)
\( 24 \, \text{W} \)
2

A \( 10 \, \text{V} \) battery with \( 2 \, \Omega \) internal resistance is connected to two resistors \( 4 \, \Omega \) and \( 6 \, \Omega \) in parallel. What is the current through the \( 6 \, \Omega \) resistor?

Parallel resistance: \( \frac{1}{R_p} = \frac{1}{4} + \frac{1}{6} = \frac{3 + 2}{12} = \frac{5}{12} \Rightarrow R_p = \frac{12}{5} = 2.4 \, \Omega \).

Total resistance: \( R_{\text{total}} = 2 + 2.4 = 4.4 \, \Omega \).

Total current: \( I = \frac{10}{4.4} \approx 2.27 \, \text{A} \).

Voltage across parallel: \( V = I R_p = 2.27 \times 2.4 \approx 5.45 \, \text{V} \).

Current through \( 6 \, \Omega \): \( I_6 = \frac{V}{6} = \frac{5.45}{6} \approx 0.91 \, \text{A} \).

\( 0.8 \, \text{A} \)
\( 0.91 \, \text{A} \)
\( 1.0 \, \text{A} \)
\( 1.2 \, \text{A} \)
2

In a circuit with resistors, why does the total power dissipated equal the power supplied by the source in steady state?

In steady state, energy is conserved. The power supplied by the source (\( P = V I \)) is fully dissipated as heat in the resistors (\( P = I^2 R_{\text{total}} \)), with no energy stored or lost elsewhere.

Current doubles
Voltage cancels out
Resistance decreases
Energy conservation
4

A \( 15 \, \Omega \) resistor dissipates \( 60 \, \text{W} \) of power. What is the voltage across it?

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

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

Substitute: \( V = \sqrt{60 \times 15} = \sqrt{900} = 30 \, \text{V} \).

\( 30 \, \text{V} \)
\( 32 \, \text{V} \)
\( 35 \, \text{V} \)
\( 40 \, \text{V} \)
1

A circuit has a \( 8 \, \text{V} \) battery with \( 1 \, \Omega \) internal resistance and two resistors \( 3 \, \Omega \) and \( 6 \, \Omega \) in parallel. What is the total current?

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

Total resistance: \( R_{\text{total}} = 1 + 2 = 3 \, \Omega \).

Current: \( I = \frac{\varepsilon}{R_{\text{total}}} = \frac{8}{3} \approx 2.67 \, \text{A} \).

\( 2.4 \, \text{A} \)
\( 2.5 \, \text{A} \)
\( 2.67 \, \text{A} \)
\( 2.8 \, \text{A} \)
3

A \( 7 \, \text{V} \) battery with \( 1 \, \Omega \) internal resistance is connected to a \( 6 \, \Omega \) resistor. What is the power dissipated in the internal resistance?

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

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

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

What is the significance of the relaxation time in the context of electron drift in a conductor?

Relaxation time (\( \tau \)) is the average time between electron collisions with lattice ions. It determines drift velocity (\( v_d = e E \tau / m \)), affecting how quickly electrons respond to the field and thus the conductor’s conductivity.

Time to reach maximum speed
Time for electrons to stop
Average time between collisions
Time for current to stabilize
3

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