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 \[\therefore \] \[\frac{1}{{{C}_{1}}}=\frac{1}{6}+\frac{1}{3}\] \[{{C}_{1}}=2\] \[{{C}_{1}}\]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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