A) 0.2 A
B) 40 mA
C) 2 A
D) 0.4 A
E) 0.8 A
Correct Answer: D
Solution :
Energy of charged capacitor = energy generated across inductor coil Given, \[C=16\mu F=16\times {{10}^{-6}}F,\]\[V=20\,\,\text{volt,}\] \[L=40\,mH=40\times {{10}^{-3}}H\] \[\therefore \]\[\frac{1}{2}\times 16\times {{10}^{-6}}\times {{(20)}^{2}}=\frac{1}{2}\times 40\times {{10}^{-3}}\times {{i}^{2}}\] \[\Rightarrow \] \[{{i}^{2}}=\frac{16}{100}\] \[\Rightarrow \] \[i=0.4\,A\]
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