A) \[8\,\mu J\]
B) \[16\,\mu J\]
C) \[2\,\mu J\]
D) \[4\,\mu J\]
Correct Answer: A
Solution :
(a) \[8\,\mu J\] \[3\,\mu F\] and \[6\,\mu F\]are in series. \[\therefore \,\,\,\,\,\,\,\,\,\,\frac{3\times 6}{\left( 3+6 \right)}=2\mu F\] This is in parallel with \[2\,\mu F\]. \[\therefore \]Total capacitance in the circuit is\[4\mu F\], \[Q=CV\]. Energy \[=\frac{1}{2}QV=\left( 1/2 \right)\,{{V}^{2}}C\] \[=\left( 1/2 \right)\times {{2}^{2}}\times 4\times {{10}^{-6}}\,J=8\,\mu J\]
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