Two capacitors each of `1muF` capacitance are connected in parallel and are then charged by 200 V. The total energy of their charges is
A. 0.01 J
B. 0.02 J
C. 0.04 J
D. 0.06 J

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1 Answers

Correct Answer - C
`E=E_(1)+E_(2)=1/2(C_(1)+C_(2))V^(2)`
`=1/2xx2xx10^(-6)xx4xx10^(4)`
= 0.04J

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