Current Electricity Chapter-Wise Test 16

Correct answer Carries: 4.

Wrong Answer Carries: -1.

In a circuit obeying Kirchhoff’s loop rule, what does a zero sum of potential changes around a closed loop imply?

Kirchhoff’s loop rule states that the algebraic sum of potential differences around a closed loop is zero, reflecting conservation of energy, as the electric potential is a conservative field.

Current is zero
Resistance is infinite
Energy is conserved
Voltage sources cancel out
3

A cell of emf \( 6 \, \text{V} \) and internal resistance \( 2 \, \Omega \) is connected to an external resistor \( 4 \, \Omega \). What is the terminal voltage across the cell?

Current: \( I = \frac{\varepsilon}{R + r} = \frac{6}{4 + 2} = 1 \, \text{A} \).

Terminal voltage: \( V = \varepsilon - I r = 6 - 1 \times 2 = 4 \, \text{V} \).

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

A cell of emf \( 8 \, \text{V} \) and internal resistance \( 2 \, \Omega \) is connected to a \( 6 \, \Omega \) resistor. What is the terminal voltage?

Total resistance: \( R_{\text{total}} = 6 + 2 = 8 \, \Omega \).

Current: \( I = \frac{\varepsilon}{R_{\text{total}}} = \frac{8}{8} = 1 \, \text{A} \).

Terminal voltage: \( V = \varepsilon - I r = 8 - 1 \times 2 = 6 \, \text{V} \).

\( 6.0 \, \text{V} \)
\( 6.5 \, \text{V} \)
\( 7.0 \, \text{V} \)
\( 8.0 \, \text{V} \)
1

In a circuit with parallel branches, why does the branch with the lowest resistance carry the highest current?

In parallel, voltage across each branch is the same (\( V \)). Current \( I = V / R \), so the branch with the lowest \( R \) has the highest \( I \), as current is inversely proportional to resistance.

Voltage increases
Resistance doubles
Current cancels out
Current is inversely proportional to resistance
4

Why does the terminal voltage of a battery decrease when current increases?

Terminal voltage \( V = \varepsilon - I r \). As current (\( I \)) increases, the voltage drop across the internal resistance (\( I r \)) increases, reducing \( V \).

Emf decreases
External resistance increases
Internal voltage drop increases
Charge carriers slow down
3

A \( 8 \, \Omega \) resistor carries a current of \( 2 \, \text{A} \) for \( 15 \, \text{s} \). What is the energy dissipated?

Energy: \( W = I^2 R t \).

Substitute: \( W = 2^2 \times 8 \times 15 = 4 \times 120 = 480 \, \text{J} \).

\( 400 \, \text{J} \)
\( 450 \, \text{J} \)
\( 460 \, \text{J} \)
\( 480 \, \text{J} \)
4

A circuit has a \( 10 \, \text{V} \) battery with \( 1 \, \Omega \) internal resistance and three resistors \( 2 \, \Omega \), \( 4 \, \Omega \), \( 8 \, \Omega \) in parallel. What is the current through the \( 4 \, \Omega \) resistor?

Parallel resistance: \( \frac{1}{R_p} = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} = \frac{4 + 2 + 1}{8} = \frac{7}{8} \Rightarrow R_p = \frac{8}{7} \approx 1.14 \, \Omega \).

Total resistance: \( R_{\text{total}} = 1 + 1.14 = 2.14 \, \Omega \).

Total current: \( I = \frac{\varepsilon}{R_{\text{total}}} = \frac{10}{2.14} \approx 4.67 \, \text{A} \).

Voltage across parallel: \( V = I R_p = 4.67 \times 1.14 \approx 5.33 \, \text{V} \).

Current through \( 4 \, \Omega \): \( I_4 = \frac{V}{4} = \frac{5.33}{4} \approx 1.33 \, \text{A} \).

\( 1.0 \, \text{A} \)
\( 1.2 \, \text{A} \)
\( 1.3 \, \text{A} \)
\( 1.33 \, \text{A} \)
4

In a Wheatstone bridge, when the bridge is balanced, what is the potential difference across the galvanometer?

At balance, the potential at the two junctions connected to the galvanometer is equal (due to the ratio condition \( R_1 / R_2 = R_3 / R_4 \)), so the potential difference across the galvanometer is zero, and no current flows through it.

Zero
Equal to the battery voltage
Half the battery voltage
Dependent on galvanometer resistance
1

A Wheatstone bridge has \( R_1 = 12 \, \Omega \), \( R_2 = 24 \, \Omega \), \( R_3 = 18 \, \Omega \). What is \( R_4 \) for balance?

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

Substitute: \( \frac{12}{24} = \frac{18}{R_4} \).

Solve: \( 0.5 = \frac{18}{R_4} \Rightarrow R_4 = \frac{18}{0.5} = 36 \, \Omega \).

\( 30 \, \Omega \)
\( 36 \, \Omega \)
\( 40 \, \Omega \)
\( 45 \, \Omega \)
2

A \( 5 \, \text{V} \) battery with \( 0.5 \, \Omega \) internal resistance is connected to a \( 4.5 \, \Omega \) resistor. What is the power dissipated in the external resistor?

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

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

Power: \( P = I^2 R = 1^2 \times 4.5 = 4.5 \, \text{W} \).

\( 4.0 \, \text{W} \)
\( 4.2 \, \text{W} \)
\( 4.25 \, \text{W} \)
\( 4.5 \, \text{W} \)
4

Performance Summary

Score:

Category Details
Total Attempts:
Total Skipped:
Total Wrong Answers:
Total Correct Answers:
Time Taken:
Average Time Taken per Question:
Accuracy:
0