Q. A 5 ohm resistor and a 10 ohm resistor are connected in series. What is the total resistance?
A.15 ohms
B.5 ohms
C.10 ohms
D.2 ohms
Solution
In series, the total resistance is R_total = R1 + R2 = 5 + 10 = 15 ohms.
Correct Answer: A — 15 ohms
Q. A 60W bulb is connected to a 120V supply. 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, we rearrange to find R = V^2 / P = (120V)^2 / 60W = 240 ohms.
Correct Answer: B — 120 ohms
Q. A circuit contains a 12 V battery and two resistors of 4 ohms and 8 ohms in series. What is the current flowing through the circuit?
A.0.5 A
B.1 A
C.1.5 A
D.2 A
Solution
Total resistance R = 4 + 8 = 12 ohms. Current I = V/R = 12 V / 12 ohms = 1 A.
Correct Answer: B — 1 A
Q. A circuit contains a 12V battery and two resistors of 4 ohms and 8 ohms in series. What is the total current in the circuit?
A.1 A
B.0.5 A
C.2 A
D.3 A
Solution
Total resistance R = R1 + R2 = 4 + 8 = 12 ohms. Current I = V/R = 12V / 12 ohms = 1 A.
Correct Answer: B — 0.5 A
Q. A circuit contains a 9V battery and two resistors in series: 3Ω and 6Ω. What is the voltage across the 6Ω resistor?
A.6V
B.3V
C.9V
D.0V
Solution
The total resistance R_total = 3Ω + 6Ω = 9Ω. The current I = V/R_total = 9V / 9Ω = 1A. Voltage across 6Ω, V = I * R = 1A * 6Ω = 6V.
Correct Answer: A — 6V
Q. A circuit contains a 9V battery and two resistors of 3 ohms and 6 ohms in series. What is the voltage drop across the 6 ohm resistor?
A.3 V
B.6 V
C.9 V
D.4.5 V
Solution
Total resistance R_total = 3 + 6 = 9 ohms. Current I = V/R_total = 9V / 9Ω = 1 A. Voltage drop across 6 ohm resistor = I * R = 1 A * 6Ω = 6 V.
Correct Answer: B — 6 V
Q. A circuit contains a 9V battery and two resistors of 3 ohms and 6 ohms in series. What is the voltage across the 6 ohm resistor?
A.6V
B.3V
C.9V
D.4.5V
Solution
Total resistance R_total = 3 + 6 = 9 ohms. Current I = V/R_total = 9V / 9Ω = 1 A. Voltage across 6 ohm resistor = I * R = 1 A * 6Ω = 6V.
Correct Answer: A — 6V
Q. A circuit contains a 9V battery and two resistors of 3Ω and 6Ω in series. What is the voltage across the 6Ω resistor?
A.6V
B.3V
C.9V
D.0V
Solution
The total resistance R_total = 3Ω + 6Ω = 9Ω. The current I = V/R_total = 9V / 9Ω = 1A. Voltage across 6Ω, V = I * R = 1A * 6Ω = 6V.
Correct Answer: A — 6V
Q. A circuit has a voltage of 12V and a resistance of 4Ω. What is the current flowing through the circuit?
A.3A
B.4A
C.12A
D.48A
Solution
Using Ohm's Law, I = V/R = 12V / 4Ω = 3A.
Correct Answer: A — 3A
Q. A copper wire has a resistivity of 1.68 x 10^-8 Ω·m. What is the resistance of a 100 m long wire with a cross-sectional area of 1 mm²?
A.1.68 Ω
B.0.168 Ω
C.0.0168 Ω
D.16.8 Ω
Solution
Resistance (R) = ρ * (L / A) = 1.68 x 10^-8 * (100 / 1 x 10^-6) = 1.68 Ω.
Correct Answer: A — 1.68 Ω
Q. A cylindrical conductor has a length L and radius r. If the radius is doubled while keeping the length constant, how does the resistivity change?
A.Doubles
B.Halves
C.Remains the same
D.Increases four times
Solution
Resistivity is an intrinsic property of the material and does not change with geometry.
Correct Answer: C — Remains the same
Q. A cylindrical wire has a length of 1 m and a radius of 0.5 mm. If its resistivity is 1.68 x 10^-8 Ω·m, what is its resistance?
A.0.0212 Ω
B.0.0424 Ω
C.0.0848 Ω
D.0.168 Ω
Solution
Resistance R = ρ(L/A) = 1.68 x 10^-8 * (1 / (π(0.5 x 10^-3)²)) = 0.0424 Ω.
Correct Answer: B — 0.0424 Ω
Q. A potentiometer is used to compare two EMFs. If the known EMF is 6V and the length of the wire is 120 cm, what is the potential gradient if the length of the wire is used to balance an unknown EMF of 4V?
A.0.05 V/cm
B.0.03 V/cm
C.0.04 V/cm
D.0.02 V/cm
Solution
The potential gradient is calculated as (6V / 120 cm) = 0.05 V/cm. For the unknown EMF of 4V, the length used would be (4V / 0.05 V/cm) = 80 cm.
Correct Answer: C — 0.04 V/cm
Q. A potentiometer wire has a length of 10 m and a potential difference of 5 V across it. What is the potential gradient?
A.0.5 V/m
B.1 V/m
C.2 V/m
D.5 V/m
Solution
The potential gradient is calculated as the potential difference divided by the length of the wire: 5 V / 10 m = 0.5 V/m.
Correct Answer: A — 0.5 V/m
Q. A potentiometer wire has a resistance of 10 ohms and is connected to a 5 V battery. What is the current flowing through the wire?
A.0.5 A
B.1 A
C.2 A
D.5 A
Solution
Using Ohm's law (V = IR), the current can be calculated as I = V/R = 5 V / 10 ohms = 0.5 A.
Correct Answer: B — 1 A
Q. A potentiometer wire has a uniform cross-section and a length of 10 m. If a potential difference of 5 V is applied, what is the potential gradient?
A.0.5 V/m
B.1 V/m
C.2 V/m
D.5 V/m
Solution
The potential gradient is calculated as V/L = 5 V / 10 m = 0.5 V/m.
Correct Answer: B — 1 V/m
Q. A potentiometer wire has a uniform cross-section and a total length of 10 m. If a potential difference of 5 V is applied across it, what is the potential gradient?
A.0.5 V/m
B.1 V/m
C.2 V/m
D.5 V/m
Solution
The potential gradient is calculated as V/L = 5 V / 10 m = 0.5 V/m.
Correct Answer: B — 1 V/m
Q. A resistor of 10 ohms is connected to a 20V battery. What is the current flowing through the resistor?
A.0.5 A
B.1 A
C.2 A
D.5 A
Solution
Using Ohm's Law, I = V/R = 20V / 10Ω = 2 A.
Correct Answer: C — 2 A
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?
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
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
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
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
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
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
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
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
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
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.
