Current Electricity
Q. A 10-ohm resistor has a voltage of 20 volts across it. What is the current flowing through the resistor?
A.
1 A
B.
2 A
C.
3 A
D.
4 A
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Solution
Using Ohm's Law, I = V / R = 20 V / 10 Ω = 2 A.
Correct Answer: B — 2 A
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Q. A 10-ohm resistor has a voltage of 50 volts across it. What is the current flowing through the resistor?
A.
5 A
B.
10 A
C.
15 A
D.
20 A
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Solution
Using Ohm's Law, I = V / R = 50 V / 10 Ω = 5 A.
Correct Answer: B — 10 A
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Q. A 10-ohm resistor is connected to a 20-volt battery. What is the power dissipated by the resistor?
A.
40 W
B.
20 W
C.
10 W
D.
2 W
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Solution
Power (P) can be calculated using P = V^2 / R = 20^2 / 10 = 400 / 10 = 40 W.
Correct Answer: A — 40 W
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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
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Solution
In series, the total resistance is R_total = R1 + R2 = 5 + 10 = 15 ohms.
Correct Answer: A — 15 ohms
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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
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Solution
Using P = V^2 / R, we rearrange to find R = V^2 / P = (120V)^2 / 60W = 240 ohms.
Correct Answer: B — 120 ohms
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Q. A circuit contains a 10Ω resistor and a 5Ω resistor in series with a 15V battery. What is the total current in the circuit?
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Solution
Total resistance R = 10Ω + 5Ω = 15Ω. Current I = V/R = 15V/15Ω = 1A.
Correct Answer: B — 2A
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Q. A circuit contains a 10Ω resistor and a 5Ω resistor in series. What is the total resistance?
A.
5Ω
B.
10Ω
C.
15Ω
D.
50Ω
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Solution
In series, total resistance R = R1 + R2 = 10Ω + 5Ω = 15Ω.
Correct Answer: C — 15Ω
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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
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Solution
Total resistance R = 4 + 8 = 12 ohms. Current I = V/R = 12 V / 12 ohms = 1 A.
Correct Answer: B — 1 A
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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
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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
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Q. A circuit contains a 9V battery and two resistors in series: 3Ω and 6Ω. What is the voltage across the 6Ω resistor?
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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
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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
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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
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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
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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
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Q. A circuit contains a 9V battery and two resistors of 3Ω and 6Ω in series. What is the voltage across the 6Ω resistor?
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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
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Q. A circuit has a total resistance of 10 ohms and a current of 5 A. What is the total voltage in the circuit?
A.
50 V
B.
10 V
C.
5 V
D.
2 V
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Solution
Using Ohm's law, V = I * R = 5 A * 10 ohms = 50 V.
Correct Answer: A — 50 V
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Q. A circuit has a voltage of 12 volts and a resistance of 4 ohms. What is the current flowing through the circuit?
A.
3 A
B.
4 A
C.
2 A
D.
1 A
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Solution
Using Ohm's Law, I = V / R = 12 V / 4 Ω = 3 A.
Correct Answer: A — 3 A
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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
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Solution
Using Ohm's Law, I = V/R = 12V / 4Ω = 3A.
Correct Answer: A — 3A
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Q. A circuit has a voltage of 24 volts and a current of 6 amperes. What is the resistance?
A.
4 Ω
B.
6 Ω
C.
8 Ω
D.
12 Ω
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Solution
Using Ohm's Law, R = V / I = 24 V / 6 A = 4 Ω.
Correct Answer: A — 4 Ω
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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
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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
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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
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Solution
Resistivity is an intrinsic property of the material and does not change with geometry.
Correct Answer: C — Remains the same
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Q. A cylindrical conductor has a length of 1 m and a radius of 0.01 m. If its resistivity is 2 x 10^-8 Ω·m, what is its resistance?
A.
0.01 Ω
B.
0.02 Ω
C.
0.03 Ω
D.
0.04 Ω
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Solution
Resistance R = ρ * (L / A) = 2 x 10^-8 * (1 / (π * (0.01)²)) = 0.02 Ω.
Correct Answer: B — 0.02 Ω
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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
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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
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Q. A potentiometer is used to compare two emf sources. If the first source gives a balance length of 60cm and the second gives 90cm, what is the ratio of their emfs?
A.
2:3
B.
3:2
C.
1:1
D.
4:5
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Solution
The ratio of emfs is equal to the ratio of the balance lengths, so it is 60cm:90cm = 2:3.
Correct Answer: B — 3:2
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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
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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
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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
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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
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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
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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
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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
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Solution
The potential gradient is calculated as V/L = 5 V / 10 m = 0.5 V/m.
Correct Answer: B — 1 V/m
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Q. A potentiometer wire has a uniform cross-section and a potential difference of 12V across it. If the length of the wire is 6m, what is the potential gradient?
A.
2 V/m
B.
4 V/m
C.
6 V/m
D.
8 V/m
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Solution
Potential gradient = Voltage / Length = 12V / 6m = 2 V/m.
Correct Answer: B — 4 V/m
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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
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Solution
The potential gradient is calculated as V/L = 5 V / 10 m = 0.5 V/m.
Correct Answer: B — 1 V/m
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Q. A potentiometer wire is made of constantan. What is the advantage of using this material?
A.
High conductivity
B.
Low temperature coefficient
C.
High resistance
D.
Low cost
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Solution
Constantan has a low temperature coefficient, making it stable for accurate measurements.
Correct Answer: B — Low temperature coefficient
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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
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Solution
Using Ohm's Law, I = V/R = 20V / 10Ω = 2 A.
Correct Answer: C — 2 A
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