Current Electricity Study Materials for Competitive Exams

Current Electricity

Circuits & Kirchhoff Laws Ohms Law Potentiometer Resistivity Wheatstone Bridge

Q. A wire has a resistance of 10 ohms at 20°C. If the temperature coefficient of resistivity is 0.004/°C, what will be its resistance at 100°C?

A. 10.4 ohms
B. 12 ohms
C. 14 ohms
D. 16 ohms

Solution

R = R0(1 + α(T - T0)) = 10(1 + 0.004(100 - 20)) = 10(1 + 0.32) = 10.4 ohms.

Correct Answer: B — 12 ohms

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Q. A wire has a resistance of 10 Ω at 20°C. If the temperature coefficient of resistivity is 0.004/°C, what will be the resistance at 100°C?

A. 10 Ω
B. 12 Ω
C. 14 Ω
D. 16 Ω

Solution

R = R0(1 + α(T - T0)) = 10(1 + 0.004(100 - 20)) = 10(1 + 0.32) = 10(1.32) = 13.2 Ω.

Correct Answer: C — 14 Ω

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Q. A wire has a resistance of 10 Ω at 20°C. If the temperature coefficient of resistivity is 0.004/°C, what will be its resistance at 100°C?

A. 10 Ω
B. 12 Ω
C. 14 Ω
D. 16 Ω

Solution

R = R0(1 + α(T - T0)) = 10(1 + 0.004(100 - 20)) = 10(1 + 0.32) = 10(1.32) = 13.2 Ω.

Correct Answer: C — 14 Ω

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Q. A wire has a resistance of 12 Ω and is made of a material with a resistivity of 3 x 10^-6 Ω·m. If the length of the wire is 4 m, what is its cross-sectional area?

A. 0.5 mm²
B. 1 mm²
C. 2 mm²
D. 3 mm²

Solution

A = ρ * (L / R) = 3 x 10^-6 * (4 / 12) = 1 mm².

Correct Answer: B — 1 mm²

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Q. A wire has a resistance of 12 Ω. If it is stretched to double its length, what will be the new resistance assuming uniform cross-section?

A. 24 Ω
B. 48 Ω
C. 12 Ω
D. 6 Ω

Solution

Resistance R is proportional to length; if length doubles, resistance also doubles to 24 Ω.

Correct Answer: A — 24 Ω

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Q. A wire has a resistance of 5 Ω at 20°C. If the temperature coefficient of resistivity is 0.004/°C, what will be its resistance at 100°C?

A. 5.4 Ω
B. 6.4 Ω
C. 7.4 Ω
D. 8.4 Ω

Solution

R = R0(1 + α(T - T0)) = 5(1 + 0.004(100 - 20)) = 6.4 Ω.

Correct Answer: B — 6.4 Ω

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Q. A wire made of material A has a resistivity of 1.5 x 10^-8 Ω·m, while material B has a resistivity of 3.0 x 10^-8 Ω·m. If both wires have the same dimensions, which wire will have a higher resistance?

A. Wire A
B. Wire B
C. Both have the same resistance
D. Cannot be determined

Solution

Resistance is directly proportional to resistivity; hence, wire B with higher resistivity will have higher resistance.

Correct Answer: B — Wire B

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Q. A wire made of material A has twice the length and half the cross-sectional area of a wire made of material B. If the resistivity of A is ρ, what is the resistance of wire A in terms of the resistance of wire B?

A. 2R
B. 4R
C. R/2
D. R/4

Solution

Resistance R = ρ(L/A). For wire A, R_A = ρ(2L/(A/2)) = 4ρ(L/A) = 4R_B.

Correct Answer: B — 4R

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Q. A wire of length 10 m and cross-sectional area 2 mm² has a resistance of 3 Ω. What is the resistivity of the material?

A. 1.5 x 10^-6 Ω·m
B. 3 x 10^-6 Ω·m
C. 6 x 10^-6 Ω·m
D. 1.5 x 10^-5 Ω·m

Solution

Resistivity ρ = R * (A / L) = 3 * (2 x 10^-6 / 10) = 1.5 x 10^-6 Ω·m.

Correct Answer: A — 1.5 x 10^-6 Ω·m

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Q. According to Kirchhoff's Current Law, if three currents enter a junction as 2A, 3A, and 4A, what is the current leaving the junction?

A. 1A
B. 3A
C. 5A
D. 9A

Solution

According to Kirchhoff's Current Law, the sum of currents entering a junction equals the sum of currents leaving. Therefore, I_out = 2A + 3A + 4A = 9A.

Correct Answer: D — 9A

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Q. According to Kirchhoff's Current Law, if three currents enter a junction as 2A, 3A, and 1A, what is the total current leaving the junction?

A. 6A
B. 5A
C. 4A
D. 3A

Solution

According to Kirchhoff's Current Law, total current entering = total current leaving. 2A + 3A + 1A = 6A.

Correct Answer: B — 5A

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Q. According to Kirchhoff's Current Law, if three currents enter a junction as 2A, 3A, and 4A, what is the total current leaving the junction?

A. 1A
B. 3A
C. 5A
D. 9A

Solution

According to Kirchhoff's Current Law, the total current entering a junction equals the total current leaving it. Total entering = 2A + 3A + 4A = 9A.

Correct Answer: D — 9A

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Q. According to Kirchhoff's Current Law, if three currents entering a junction are 2A, 3A, and 5A, what is the total current leaving the junction?

A. 10A
B. 5A
C. 3A
D. 2A

Solution

According to Kirchhoff's Current Law, the total current entering a junction equals the total current leaving it. Therefore, total current leaving = 2A + 3A + 5A = 10A.

Correct Answer: A — 10A

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Q. According to Kirchhoff's Voltage Law, the sum of the potential differences around any closed loop in a circuit is equal to what?

A. Zero
B. The total current
C. The total resistance
D. The total power

Solution

Kirchhoff's Voltage Law states that the sum of the electromotive forces and potential differences in any closed loop is zero.

Correct Answer: A — Zero

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Q. According to Kirchhoff's Voltage Law, the sum of the potential differences around any closed loop in a circuit must equal what?

A. Zero
B. The total current
C. The total resistance
D. The total power

Solution

Kirchhoff's Voltage Law states that the sum of the electrical potential differences (voltage) around any closed circuit is zero.

Correct Answer: A — Zero

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Q. If a 10 ohm resistor is connected to a 20V battery, what is the energy consumed in 5 seconds?

A. 20 J
B. 40 J
C. 10 J
D. 50 J

Solution

Power P = V^2 / R = 20^2 / 10 = 40 W. Energy = Power * time = 40 W * 5 s = 200 J.

Correct Answer: B — 40 J

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Q. If a 10 ohm resistor is connected to a 30V battery, what is the energy consumed in 5 seconds?

A. 15 J
B. 30 J
C. 75 J
D. 150 J

Solution

Power P = V^2 / R = (30V)^2 / 10 ohms = 90 W. Energy = Power * Time = 90 W * 5 s = 450 J.

Correct Answer: C — 75 J

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Q. If a 10Ω resistor is connected to a 20V battery, what is the power dissipated by the resistor?

A. 20W
B. 40W
C. 100W
D. 200W

Solution

Power (P) can be calculated using P = V^2/R. Here, P = 20V^2 / 10Ω = 400 / 10 = 40W.

Correct Answer: B — 40W

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Q. If a 10Ω resistor is connected to a 20V source, what is the energy consumed in 5 seconds?

A. 20J
B. 40J
C. 50J
D. 100J

Solution

Power P = V^2 / R = 20V^2 / 10Ω = 40W. Energy = Power * time = 40W * 5s = 200J.

Correct Answer: B — 40J

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Q. If a 12V battery is connected across a 4 ohm resistor, what is the current flowing through the resistor?

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

Solution

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

Correct Answer: A — 2 A

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Q. If a 12V battery is connected across a 4 ohm resistor, what is the power dissipated in the resistor?

A. 12 W
B. 24 W
C. 36 W
D. 48 W

Solution

Power (P) can be calculated using P = V^2 / R = 12V^2 / 4Ω = 144 / 4 = 36 W.

Correct Answer: B — 24 W

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Q. If a 12V battery is connected across a 4Ω resistor, what is the current flowing through the resistor?

A. 3A
B. 2A
C. 4A
D. 1A

Solution

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

Correct Answer: B — 2A

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Q. If a 12V battery is connected to a 4 ohm resistor, what is the power dissipated by the resistor?

A. 12 W
B. 24 W
C. 36 W
D. 48 W

Solution

Power (P) can be calculated using P = V^2 / R = 12V^2 / 4Ω = 36 W.

Correct Answer: B — 24 W

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Q. If a 5 ohm resistor and a 10 ohm resistor are connected in series, what is the voltage across the 10 ohm resistor when a current of 2 A flows through the circuit?

A. 10 V
B. 20 V
C. 5 V
D. 15 V

Solution

Voltage across the 10 ohm resistor, V = I * R = 2 A * 10 Ω = 20 V.

Correct Answer: B — 20 V

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Q. If a 5 ohm resistor is connected to a 10V battery, what is the energy consumed in 2 seconds?

A. 10 J
B. 20 J
C. 5 J
D. 15 J

Solution

Power P = V^2 / R = 10^2 / 5 = 20 W. Energy = Power * time = 20 W * 2 s = 40 J.

Correct Answer: B — 20 J

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Q. If a 5 ohm resistor is connected to a 10V battery, what is the energy consumed in 2 minutes?

A. 60 J
B. 120 J
C. 240 J
D. 30 J

Solution

Power P = V^2/R = 10^2 / 5 = 20 W. Energy = Power * time = 20 W * 120 s = 2400 J.

Correct Answer: B — 120 J

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Q. If a 5Ω resistor is connected in parallel with a 10Ω resistor, what is the equivalent resistance?

A. 3.33Ω
B. 5Ω
C. 7.5Ω
D. 15Ω

Solution

R_eq = 1/(1/5 + 1/10) = 3.33Ω.

Correct Answer: A — 3.33Ω

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Q. If a 60W bulb operates at 120V, what is the resistance of the bulb?

A. 240 ohms
B. 120 ohms
C. 60 ohms
D. 30 ohms

Solution

Using P = V^2 / R, R = V^2 / P = (120V)^2 / 60W = 240 ohms.

Correct Answer: B — 120 ohms

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Q. If a 60W light bulb operates at 120V, what is the resistance of the bulb?

A. 240 ohms
B. 120 ohms
C. 60 ohms
D. 30 ohms

Solution

Power (P) = V^2 / R, so R = V^2 / P = (120V)^2 / 60W = 240 ohms.

Correct Answer: B — 120 ohms

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Q. If a 6V battery is connected to a circuit with a total resistance of 3 ohms, what is the current flowing through the circuit?

A. 1A
B. 2A
C. 3A
D. 4A

Solution

Using Ohm's law, I = V/R = 6V / 3 ohms = 2A.

Correct Answer: B — 2A

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Current Electricity

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