A) \[12\times {{10}^{-6}}J\]
B) \[16\times {{10}^{-4}}J\]
C) \[8\times {{10}^{-6}}J\]
D) \[4\times {{10}^{-4}}J\]
Correct Answer: B
Solution :
The total amount of work done in charging a capacitor, is stored up in the capacitor in the form of electric potential energy U. \[U=\frac{1}{2}C\,({{V}_{2}}^{2}-{{V}_{1}}^{2})\] Given, \[C=8\mu F=8\times {{10}^{-6}}F,\,{{V}_{2}}=25\] volt, \[{{V}_{1}}=15\] volt \[\therefore \] \[U=\frac{1}{2}\times 8\times {{10}^{-6}}\times [{{(25)}^{2}}-{{(15)}^{2}}]\] \[=4\times {{10}^{6}}\times 400=16\times {{10}^{-4}}J\].
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