A) \[0.05\,J\]
B) 25
C) \[0.005\,J\]
D) \[0.001\,J\]
Correct Answer: A
Solution :
Energy stored in the capacitor is given by \[U=\frac{1}{2}C{{V}^{2}}\] .....(i) Here, \[C=10\mu F={{10}^{-5}}F.\] \[V=100\]volts. Putting the given value in equation (i) we get, \[\therefore \] \[U=\frac{1}{2}\times {{10}^{-5}}\times {{(100)}^{2}}=0.05\,J\]
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