11. Work done in charging a capacitor is ____________
A. QV
B. 1⁄2QV
C. 2QV
D. QV2
Answer: B
We know that work done = Q2/2C.
Substituting C as Q/V, we get work done = Q/2V.
12. Energy stored in a 2000mF capacitor charged to a potential difference of 10V is?
A. 100J
B. 200J
C. 300J
D. 400J
Answer: A
From the expression of Capacitor energy
WD = CV2/2 = 100J.
13. When do we get maximum energy from a set of capacitors?
A. When they are connected in parallel
B. When they are connected in series
C. Both in series and parallel
D. Insufficient information provided
Answer: A
We get maximum energy when capacitors are connected in parallel because the equivalent capacitance is larger than the largest individual capacitance when connected in parallel. The relation between capacitance and energy is:
Energy = CV2/2, hence as the capacitance increases, the energy stored in it also increases.
14. If the charge stored in a capacitor is 4C and the value of capacitance is 2F, calculate the energy stored in it.
A. 2J
B. 4J
C. 8J
D. 16J
Answer: B
The expression for finding the value of capacitor energy is:
U = Q2/2C = 4 × 4/(2 × 2) = 4J.
15. If the charge in a capacitor is 4C and the energy stored in it is 4J, find the value of capacitance.
A. 2F
B. 4F
C. 8F
D. 16F
Answer: A
The expression for finding the value of capacitance energy is:
U = Q2/2C.
Substituting the values of U and Q, we get C = 2F.
16. If the charge in a capacitor is 4C and the energy stored in it is 4J, calculate the voltage across its plates.
A. 2V
B. 4V
C. 8V
D. 16V
Answer: A
The expression for finding the value of capacitor energy is:
U = Q2/2C.
Substituting the values of U and Q, we get C = 2F.
V = Q/C, hence V = 4/2 = 2V.
17. Calculate the energy in the 2F capacitor in the given circuit below.
A. 8.6kJ
B. 64kJ
C. 64J
D. 6.4kJ
Answer: D
From the expression value of capacitor energy is
WD = CV2/2
= 2 × 802/2 = 6400J = 6.4kJ.
18. Calculate the energy in the 4F capacitor in the given circuit below.
A. 128kJ
B. 1.28kJ
C. 12.8kJ
D. 1280J
Answer: C
From the expression of capacitor energy
WD = CV2/2
= 4 × 802/2 = 12800J = 12.8kJ.
19. Calculate the energy stored in the combination of the capacitors.
A. 192kJ
B. 1.92kJ
C. 19.2kJ
D. 1920J
Answer: C
The equivalent capacitance is: Ceq = 4+2 = 6F.
From the expression of capacitor energy
WD = CV2/2
= 6 × 802/2 = 19200J = 19.2kJ.
20. Which of the following is the correct expression for the force between the plates of a parallel plate capacitor?
A. F = ε × A × (V/x)2/2
B. F = ε × A × (V/x)2/3
C. F = ε (V/x)2/2
D. F = ε (V/x)2/3
Answer: A
The force is proportional to the square of the potential gradient and the area.
