# Basic Electrical Engineering Questions and Answers – Energy Stored in a Charged Capacitor

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This set of Basic Electrical Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Energy Stored in a Charged Capacitor”.

1. Work done in charging a capacitor is ____________
a) QV
b) 12QV
c) 2QV
d) QV2

Explanation: We know that work done= Q2/2C.
Substituting C as Q/V, we get work done = Q/2V.

2. Energy stored in 2000mF capacitor charged to a potential difference of 10V is?
a) 100J
b) 200J
c) 300J
d) 400J

Explanation: From the expression:
WD = CV2/2 = 100J.

3. 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

Explanation: 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.
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4. 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

Explanation: The expression for finding the value of energy is:
U=Q2/2C = 4*4/(2*2) = 4J.

5. 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

Explanation: The expression for finding the value of energy is:
U=Q2/2C.
Substituting the values of U and Q, we get C=2F.
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6. 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

Explanation: The expression for finding the value of energy is:
U=Q2/2C.
Substituting the values of U and Q, we get C=2F.
V=Q/C, hence V=4/2=2V.

7. Calculate the energy in the 2F capacitor.

a) 8.6kJ
b) 64kJ
c) 64J
d) 6.4kJ

Explanation: From the expression:
WD= CV2/2 = 2*802/2=6400J=6.4kJ.

8. Calculate the energy in the 4F capacitor.

a) 128kJ
b) 1.28kJ
c) 12.8kJ
d) 1280J

Explanation: From the expression:
WD = CV2/2 = 4*802/2 = 12800J = 12.8kJ.

9. Calculate the energy stored in the combination of the capacitors.

a) 192kJ
b) 1.92kJ
c) 19.2kJ
d) 1920J

Explanation: The equivalent capacitance is: Ceq=4+2=6F.
From the expression:
WD = CV2/2 = 6*802/2 = 19200J = 19.2kJ.