A) 0.02 J
B) 0.04 J
C) 0.01 J
D) 200 J
Correct Answer: A
Solution :
The energy stored is given by \[E=\frac{1}{2}C{{V}^{2}}\] When capacitors are connected in parallel, resultant capacitance is \[C'={{C}_{1}}+{{C}_{2}}=2\mu F+2\mu F=4\mu F,V=100\,Volt\] \[\therefore \] \[E=\frac{1}{2}\times 4\times {{10}^{-6}}\times {{(100)}^{2}}\] \[E=0.02\,J\]
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