Current Electricity Study Materials for Competitive Exams

Current Electricity

Circuits & Kirchhoff Laws Ohms Law Potentiometer Resistivity Wheatstone Bridge

Q. If a resistor has a resistance of 5 ohms and a current of 2 amperes flows through it, what is the voltage across the resistor?

A. 10 V
B. 5 V
C. 2.5 V
D. 1 V

Solution

Using Ohm's Law, V = I * R = 2 A * 5 Ω = 10 V.

Correct Answer: A — 10 V

Learn More →

Q. If a resistor is connected to a 9V battery and the current flowing through it is 3A, what is the resistance of the resistor?

A. 1 Ω
B. 3 Ω
C. 9 Ω
D. 27 Ω

Solution

Using Ohm's Law, R = V / I = 9 V / 3 A = 3 Ω.

Correct Answer: C — 9 Ω

Learn More →

Q. If a wire's length is doubled while keeping its cross-sectional area constant, how does its resistance change?

A. Remains the same
B. Doubles
C. Halves
D. Quadruples

Solution

Resistance is directly proportional to length; doubling the length doubles the resistance.

Correct Answer: B — Doubles

Learn More →

Q. If one of the resistors in a Wheatstone bridge is replaced with a variable resistor, what is the effect on the balance condition?

A. It cannot be balanced
B. It can be balanced by adjusting the variable resistor
C. It will always be unbalanced
D. It will short-circuit the bridge

Solution

The variable resistor allows for adjustment to achieve balance in the bridge.

Correct Answer: B — It can be balanced by adjusting the variable resistor

Learn More →

Q. If R1 = 10Ω, R2 = 15Ω, R3 = 5Ω, what should R4 be for the Wheatstone bridge to be balanced?

A. 7.5Ω
B. 10Ω
C. 12.5Ω
D. 15Ω

Solution

Using the balance condition R1/R2 = R3/R4, we find R4 = (R2 * R3) / R1 = (15 * 5) / 10 = 7.5Ω.

Correct Answer: C — 12.5Ω

Learn More →

Q. If R1 = 10Ω, R2 = 20Ω, and R3 = 30Ω in a Wheatstone bridge, what should R4 be for the bridge to be balanced?

A. 15Ω
B. 20Ω
C. 25Ω
D. 30Ω

Solution

For balance, R1/R2 = R3/R4, thus R4 = (R2 * R3) / R1 = (20 * 30) / 10 = 60Ω.

Correct Answer: B — 20Ω

Learn More →

Q. If R1 = 10Ω, R2 = 20Ω, R3 = 30Ω, and R4 = 60Ω in a Wheatstone bridge, is the bridge balanced?

A. Yes
B. No
C. Depends on the voltage
D. Depends on the current

Solution

The bridge is not balanced because 10/20 ≠ 30/60.

Correct Answer: B — No

Learn More →

Q. If R1 = 2Ω, R2 = 3Ω, and R3 = 6Ω in a Wheatstone bridge, what is the value of R4 for balance?

A. 4Ω
B. 5Ω
C. 6Ω
D. 9Ω

Solution

Using the balance condition R1/R2 = R3/R4, we have 2/3 = 6/R4. Solving gives R4 = 4Ω.

Correct Answer: A — 4Ω

Learn More →

Q. If R1 = 2Ω, R2 = 3Ω, and R3 = 6Ω, what is the value of R4 for the Wheatstone bridge to be balanced?

A. 4Ω
B. 5Ω
C. 6Ω
D. 8Ω

Solution

Using the balance condition R1/R2 = R3/R4, we have 2/3 = 6/R4. Solving gives R4 = 4Ω.

Correct Answer: A — 4Ω

Learn More →

Q. If R1 = 5Ω, R2 = 10Ω, and R3 = 15Ω in a Wheatstone bridge, what is the value of R4 for balance?

A. 7.5Ω
B. 10Ω
C. 15Ω
D. 20Ω

Solution

Using the balance condition R1/R2 = R3/R4, we find R4 = (R2 * R3) / R1 = (10 * 15) / 5 = 30Ω.

Correct Answer: B — 10Ω

Learn More →

Q. If R1 = 5Ω, R2 = 15Ω, R3 = 10Ω, what should R4 be for the Wheatstone bridge to be balanced?

A. 30Ω
B. 10Ω
C. 15Ω
D. 5Ω

Solution

For balance, R4 must be 30Ω since 5/15 = 10/30.

Correct Answer: B — 10Ω

Learn More →

Q. If the balance length of a potentiometer is 50cm for a cell of unknown emf, and the potential gradient is 4 V/m, what is the emf of the cell?

A. 2V
B. 4V
C. 6V
D. 8V

Solution

The emf is calculated as V = potential gradient × length = 4 V/m × 0.5 m = 2 V.

Correct Answer: B — 4V

Learn More →

Q. If the balancing length of a potentiometer is found to be 40 cm for a cell of emf 2V, what is the potential gradient if the total length of the wire is 100 cm?

A. 5 V/m
B. 2 V/m
C. 4 V/m
D. 3 V/m

Solution

The potential gradient is V/L = 2V/0.4m = 5 V/m.

Correct Answer: A — 5 V/m

Learn More →

Q. If the cross-sectional area of a wire is doubled, how does its resistance change?

A. Doubles
B. Halves
C. Remains the same
D. Increases four times

Solution

Resistance is inversely proportional to cross-sectional area; doubling the area halves the resistance.

Correct Answer: B — Halves

Learn More →

Q. If the cross-sectional area of a wire is doubled, what happens to its resistance?

A. Doubles
B. Halves
C. Remains the same
D. Increases four times

Solution

Resistance (R) is inversely proportional to cross-sectional area (A). Doubling A halves the resistance.

Correct Answer: B — Halves

Learn More →

Q. If the current in a circuit is 5 A and the resistance is 2 ohms, what is the power consumed in the circuit?

A. 10 W
B. 25 W
C. 5 W
D. 2.5 W

Solution

Power (P) can be calculated using P = I^2R = (5 A)^2 * 2 Ω = 25 W.

Correct Answer: B — 25 W

Learn More →

Q. If the current in a circuit is doubled while the resistance remains constant, what happens to the voltage?

A. It doubles
B. It halves
C. It remains the same
D. It quadruples

Solution

According to Ohm's Law, if the current (I) is doubled and resistance (R) remains constant, the voltage (V) must also double.

Correct Answer: A — It doubles

Learn More →

Q. If the current in a circuit is halved, what happens to the power consumed by the circuit?

A. It doubles
B. It halves
C. It remains the same
D. It quadruples

Solution

Power P = I^2 * R. If I is halved, P becomes (1/2)^2 * R = 1/4 * P, so the power is reduced to a quarter, which means it halves.

Correct Answer: B — It halves

Learn More →

Q. If the current in a circuit is split into two branches with resistances R1 = 2Ω and R2 = 4Ω, what is the current through R1 if the total current is 6A?

A. 2A
B. 3A
C. 4A
D. 6A

Solution

Using the current division rule, I1 = I_total * (R2 / (R1 + R2)) = 6A * (4Ω / (2Ω + 4Ω)) = 6A * (4/6) = 4A.

Correct Answer: B — 3A

Learn More →

Q. If the current through a 10Ω resistor is 2A, what is the voltage across the resistor?

A. 5V
B. 10V
C. 20V
D. 30V

Solution

Using Ohm's law, V = I * R = 2A * 10Ω = 20V.

Correct Answer: C — 20V

Learn More →

Q. If the current through a conductor is doubled while the resistance remains constant, what happens to the power consumed?

A. It doubles
B. It triples
C. It remains the same
D. It halves

Solution

Power P = I^2 * R. If I is doubled, P becomes (2I)^2 * R = 4I^2 * R, which is four times the original power.

Correct Answer: A — It doubles

Learn More →

Q. If the current through a conductor is doubled, what happens to the power dissipated in the conductor?

A. It doubles
B. It quadruples
C. It remains the same
D. It halves

Solution

Power P = I^2 * R. If I is doubled, P becomes (2I)^2 * R = 4I^2 * R, which is quadrupled.

Correct Answer: B — It quadruples

Learn More →

Q. If the current through a conductor is doubled, what happens to the resistance if the voltage remains constant?

A. It doubles
B. It halves
C. It remains the same
D. It quadruples

Solution

According to Ohm's law, V = I * R. If V is constant and I is doubled, R must halve to maintain the equation.

Correct Answer: B — It halves

Learn More →

Q. If the current through a resistor is 3A and the resistance is 4Ω, what is the voltage across the resistor?

A. 6V
B. 9V
C. 12V
D. 15V

Solution

Using Ohm's law, V = I * R = 3A * 4Ω = 12V.

Correct Answer: C — 12V

Learn More →

Q. If the current through a resistor is doubled, what happens to the power consumed by the resistor?

A. Increases by 2 times
B. Increases by 4 times
C. Remains the same
D. Decreases by 2 times

Solution

Power P = I^2 * R. If I is doubled, P becomes (2I)^2 * R = 4I^2 * R, which is 4 times the original power.

Correct Answer: B — Increases by 4 times

Learn More →

Q. If the current through a resistor is doubled, what happens to the power dissipated in the resistor?

A. It halves
B. It doubles
C. It quadruples
D. It remains the same

Solution

Power P = I^2 * R. If I is doubled, P becomes (2I)^2 * R = 4I^2 * R, which is quadrupled.

Correct Answer: C — It quadruples

Learn More →

Q. If the emf of a cell is 12 V and the potentiometer wire is 20 m long, what is the potential gradient if the wire is calibrated to give a reading of 0.6 V/m?

A. 12 V
B. 0.6 V/m
C. 0.3 V/m
D. 0.5 V/m

Solution

The potential gradient is given as 0.6 V/m, which is the calibration value for the potentiometer.

Correct Answer: B — 0.6 V/m

Learn More →

Q. If the emf of a cell is 2V and the potential gradient of the potentiometer wire is 4 V/m, what will be the balancing length?

A. 0.5 m
B. 1 m
C. 0.25 m
D. 0.75 m

Solution

The balancing length is calculated as L = V / (potential gradient) = 2V / 4 V/m = 0.5 m.

Correct Answer: B — 1 m

Learn More →

Q. If the galvanometer in a Wheatstone bridge has a high resistance, what is the effect on the sensitivity of the bridge?

A. Increases sensitivity
B. Decreases sensitivity
C. No effect
D. Depends on the resistances

Solution

A high resistance in the galvanometer increases the sensitivity of the bridge.

Correct Answer: A — Increases sensitivity

Learn More →

Q. If the galvanometer in a Wheatstone bridge has a resistance of 5Ω and the bridge is balanced, what is the current through the galvanometer?

A. Zero.
B. 5A.
C. 1A.
D. 10A.

Solution

In a balanced Wheatstone bridge, the current through the galvanometer is zero.

Correct Answer: A — Zero.

Learn More →

Showing 121 to 150 of 607 (21 Pages)

Current Electricity

On a scale of 0–10, how likely are you to recommend The Soulshift Academy?