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) 1⁄2QV
c) 2QV
d) QV2
View Answer
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
View Answer
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
View Answer
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.
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
View Answer
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
View Answer
Explanation: The expression for finding the value of energy is:
U=Q2/2C.
Substituting the values of U and Q, we get C=2F.
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
View Answer
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
View Answer
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
View Answer
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
View Answer
Explanation: The equivalent capacitance is: Ceq=4+2=6F.
From the expression:
WD = CV2/2 = 6*802/2 = 19200J = 19.2kJ.
Sanfoundry Global Education & Learning Series – Basic Electrical Engineering.
To practice all areas of Basic Electrical Engineering, here is complete set of 1000+ Multiple Choice Questions and Answers.
