A) \[0.02\text{ }J\]
B) \[0.04\text{ }J\]
C) \[0.01\text{ }J\]
D) \[200\text{ }J\]
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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