Q. A capacitor of capacitance 10 microfarads is charged to a voltage of 5 volts. What is the charge stored in the capacitor?
A.
50 microcoulombs
B.
5 microcoulombs
C.
0.5 microcoulombs
D.
100 microcoulombs
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Solution
Charge (Q) is given by Q = C * V. Thus, Q = 10 µF * 5 V = 50 µC.
Correct Answer:
A
— 50 microcoulombs
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Q. A capacitor of capacitance 5 µF is charged to a voltage of 10V. What is the charge stored in the capacitor? (2022)
A.
50 µC
B.
100 µC
C.
500 µC
D.
1000 µC
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Solution
Charge (Q) = Capacitance (C) × Voltage (V) = 5µF × 10V = 50µC.
Correct Answer:
B
— 100 µC
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Q. A capacitor stores 5 microfarads of charge at a voltage of 10 volts. What is the energy stored in the capacitor?
A.
0.25 mJ
B.
0.5 mJ
C.
0.75 mJ
D.
1 mJ
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Solution
The energy stored in a capacitor is given by the formula E = 0.5 * C * V^2. Here, E = 0.5 * 5 * 10^-6 F * (10 V)^2 = 0.5 * 5 * 10^-6 * 100 = 0.25 mJ.
Correct Answer:
B
— 0.5 mJ
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Q. A capacitor stores energy in the form of which type of field?
A.
Electric field
B.
Magnetic field
C.
Gravitational field
D.
Electromagnetic field
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Solution
A capacitor stores energy in the form of an electric field between its plates.
Correct Answer:
A
— Electric field
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Q. If a 10 ohm resistor is connected to a 5V battery, what is the current flowing through the resistor? (2020)
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 = 5V / 10Ω = 0.5 A.
Correct Answer:
B
— 1 A
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Q. If a 12 V battery is connected to a resistor of 4 ohms, what is the current flowing through the circuit?
A.
3 A
B.
4 A
C.
12 A
D.
0.33 A
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Solution
Using Ohm's Law (I = V/R), I = 12 V / 4 Ω = 3 A.
Correct Answer:
A
— 3 A
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Q. If a circuit has a resistance of 5 ohms and a current of 2 amperes, what is the voltage across the circuit?
A.
10 V
B.
5 V
C.
2 V
D.
1 V
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Solution
Using Ohm's Law (V = I * R), V = 2 A * 5 Ω = 10 V.
Correct Answer:
A
— 10 V
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Q. If a wire carries a current of 3 A and has a resistance of 2 ohms, what is the power dissipated in the wire?
A.
6 W
B.
9 W
C.
12 W
D.
15 W
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Solution
Power is given by P = I^2 * R. Here, P = (3 A)^2 * 2 Ω = 9 * 2 = 18 W.
Correct Answer:
C
— 12 W
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Q. If a wire carries a current of 4 A and has a resistance of 2 ohms, what is the power dissipated in the wire?
A.
8 W
B.
16 W
C.
4 W
D.
2 W
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Solution
Power (P) is given by P = I^2 * R. Thus, P = (4 A)^2 * 2 Ω = 16 W.
Correct Answer:
B
— 16 W
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Q. If the frequency of an AC circuit is doubled, what happens to the inductive reactance?
A.
It doubles
B.
It halves
C.
It remains the same
D.
It quadruples
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Solution
Inductive reactance (X_L) is given by X_L = 2πfL. If the frequency (f) is doubled, the inductive reactance also doubles.
Correct Answer:
A
— It doubles
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Q. If the frequency of an AC signal is doubled, what happens to its period?
A.
It doubles
B.
It halves
C.
It remains the same
D.
It quadruples
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Solution
The period T is the reciprocal of frequency f (T = 1/f). If frequency is doubled, the period halves.
Correct Answer:
B
— It halves
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Q. If the resistance of a conductor is doubled, what happens to the current if the voltage remains constant? (2020)
A.
Doubles
B.
Halves
C.
Remains the same
D.
Increases four times
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Solution
According to Ohm's Law, if the resistance is doubled while the voltage remains constant, the current will be halved.
Correct Answer:
B
— Halves
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Q. If the resistance of a wire is 10 ohms and the voltage across it is 20 volts, what is the current flowing through the wire?
A.
2 A
B.
5 A
C.
10 A
D.
20 A
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Solution
Using Ohm's Law (I = V/R), I = 20 V / 10 Ω = 2 A.
Correct Answer:
A
— 2 A
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Q. If the resistance of a wire is doubled, what happens to the current if the voltage remains constant? (2021)
A.
Doubles
B.
Halves
C.
Remains the same
D.
Increases
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Solution
According to Ohm's Law (I = V/R), if the resistance (R) is doubled while the voltage (V) remains constant, the current (I) will be halved.
Correct Answer:
B
— Halves
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Q. If the voltage across a resistor is doubled, what happens to the power consumed by the resistor? (2020)
A.
It halves
B.
It doubles
C.
It quadruples
D.
It remains the same
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Solution
Power (P) is given by P = V^2 / R. If V is doubled, P becomes (2V)^2 / R = 4V^2 / R, which is quadrupled.
Correct Answer:
C
— It quadruples
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Q. In a magnetic field, the direction of the magnetic force on a charged particle is given by which rule?
A.
Right-hand rule
B.
Left-hand rule
C.
Fleming's right-hand rule
D.
Fleming's left-hand rule
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Solution
The direction of the magnetic force on a charged particle moving in a magnetic field is determined by the right-hand rule.
Correct Answer:
A
— Right-hand rule
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Q. In a parallel circuit with two resistors of 6 ohms and 3 ohms, what is the total resistance?
A.
2 ohms
B.
4 ohms
C.
1.5 ohms
D.
9 ohms
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Solution
The total resistance in parallel is given by 1/R_total = 1/R1 + 1/R2. Thus, 1/R_total = 1/6 + 1/3 = 1/2, so R_total = 2 ohms.
Correct Answer:
A
— 2 ohms
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Q. In a parallel circuit, if one resistor fails, what happens to the rest of the circuit? (2022)
A.
All resistors fail
B.
Current stops flowing
C.
The circuit remains functional
D.
Voltage drops to zero
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Solution
In a parallel circuit, if one resistor fails, the other paths remain functional, allowing current to continue flowing.
Correct Answer:
C
— The circuit remains functional
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Q. In a parallel circuit, if one resistor fails, what happens to the total resistance? (2020)
A.
Increases
B.
Decreases
C.
Remains the same
D.
Becomes infinite
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Solution
In a parallel circuit, if one resistor fails, the total resistance decreases because there are fewer paths for current to flow.
Correct Answer:
B
— Decreases
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Q. In a series circuit with a 12 V battery and two resistors of 4 Ω and 8 Ω, 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_total = R1 + R2 = 4 Ω + 8 Ω = 12 Ω. Using Ohm's Law, I = V/R = 12 V / 12 Ω = 1 A.
Correct Answer:
B
— 1 A
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Q. What happens to the total current in a parallel circuit if more branches are added? (2023)
A.
Increases
B.
Decreases
C.
Remains the same
D.
Becomes zero
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Solution
In a parallel circuit, adding more branches increases the total current because each branch provides an additional path for current to flow.
Correct Answer:
A
— Increases
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Q. What is the direction of the magnetic field around a current-carrying conductor? (2019)
A.
From North to South
B.
From South to North
C.
Clockwise
D.
Counterclockwise
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Solution
The direction of the magnetic field around a current-carrying conductor is given by the right-hand rule, which indicates a counterclockwise direction when viewed from the positive current side.
Correct Answer:
D
— Counterclockwise
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Q. What is the equivalent resistance of two resistors of 4 ohms and 6 ohms connected in series?
A.
10 ohms
B.
24 ohms
C.
2.4 ohms
D.
0.67 ohms
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Solution
In series, the equivalent resistance is the sum: R_eq = R1 + R2 = 4 Ω + 6 Ω = 10 Ω.
Correct Answer:
A
— 10 ohms
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Q. What is the equivalent resistance of two resistors of 4 ohms and 6 ohms in series?
A.
10 ohms
B.
24 ohms
C.
2.4 ohms
D.
0.67 ohms
Show solution
Solution
In series, the equivalent resistance is the sum: R_eq = R1 + R2 = 4 Ω + 6 Ω = 10 Ω.
Correct Answer:
A
— 10 ohms
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Q. What is the equivalent resistance of two resistors, R1 = 4Ω and R2 = 6Ω, in parallel? (2022)
A.
2.4Ω
B.
10Ω
C.
24Ω
D.
1.5Ω
Show solution
Solution
The formula for equivalent resistance in parallel is 1/R_eq = 1/R1 + 1/R2. Thus, 1/R_eq = 1/4 + 1/6 = 5/12, so R_eq = 12/5 = 2.4Ω.
Correct Answer:
A
— 2.4Ω
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Q. What is the equivalent resistance of two resistors, R1 and R2, in parallel? (2022)
A.
R1 + R2
B.
R1 * R2 / (R1 + R2)
C.
1 / (1/R1 + 1/R2)
D.
R1 - R2
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Solution
The formula for equivalent resistance (R_eq) of two resistors in parallel is given by 1 / (1/R1 + 1/R2).
Correct Answer:
C
— 1 / (1/R1 + 1/R2)
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Q. What is the formula for calculating electric power? (2023)
A.
P = IV
B.
P = I^2R
C.
P = V^2/R
D.
All of the above
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Solution
The electric power can be calculated using multiple formulas: P = IV (power = current × voltage), P = I^2R (power = current squared × resistance), and P = V^2/R (power = voltage squared / resistance).
Correct Answer:
D
— All of the above
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Q. What is the formula for calculating the capacitance of a parallel plate capacitor?
A.
C = εA/d
B.
C = A/εd
C.
C = d/εA
D.
C = εd/A
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Solution
The capacitance (C) of a parallel plate capacitor is given by C = εA/d, where ε is the permittivity, A is the area of the plates, and d is the distance between them.
Correct Answer:
A
— C = εA/d
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Q. What is the formula for calculating the power consumed in an electrical circuit?
A.
P = IV
B.
P = I/R
C.
P = V/R
D.
P = IR
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Solution
The power consumed in an electrical circuit is given by the formula P = IV, where P is power, I is current, and V is voltage.
Correct Answer:
A
— P = IV
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Q. What is the formula for the capacitance of a parallel plate capacitor?
A.
C = εA/d
B.
C = A/εd
C.
C = d/εA
D.
C = εd/A
Show solution
Solution
The capacitance C of a parallel plate capacitor is given by the formula C = εA/d, where ε is the permittivity, A is the area of the plates, and d is the separation between them.
Correct Answer:
A
— C = εA/d
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