A) \[2\mu J\]
B) \[4\mu J\]
C) \[8\mu J\]
D) \[16\mu J\]
Correct Answer: C
Solution :
\[6\mu F\] and \[3\,\mu F\] capacitors are in series \[\therefore \] \[\frac{1}{{{C}_{1}}}=\frac{1}{6}+\frac{1}{3}\] \[{{C}_{1}}\] is parallel to \[2\mu F\] capacitor \[\therefore \] \[{{C}_{eq}}=2+2=4\mu F\] Total energy, \[U=\frac{1}{2}C{{V}^{2}}=\frac{1}{2}\times 4\times {{(2)}^{2}}=8\mu J\]
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