A) 5 J
B) 0.5 J
C) 50 J
D) 500 J
Correct Answer: B
Solution :
(b) 0.5 J Here, \[C=25\,\mu F=25\times {{10}^{-6}}F,\] \[L=20\,mH=20\times {{10}^{-3}}H,{{q}_{0}}=5\,mC=5\times {{10}^{-3}}C\] Total energy stored in the circuit initially is \[U=\frac{q_{0}^{2}}{2C}=\frac{{{\left( 5\times {{10}^{-3}} \right)}^{2}}}{2\times 25\times {{10}^{-6}}}\] \[=\frac{25\times {{10}^{-6}}}{2\times 25\times {{10}^{-6}}}=\frac{1}{2}=0.5\,J\]
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