A) \[{{10}^{-5}}J\]
B) \[{{10}^{-6}}J\]
C) \[{{10}^{-7}}J\]
D) \[{{10}^{-4}}J\]
Correct Answer: B
Solution :
\[C=400\times {{10}^{-12}}F\] \[V=100\,V\] Initially energy stored\[=\frac{1}{2}C{{V}^{2}}\] \[=\frac{1}{2}400\times {{10}^{-12}}\times {{({{10}^{2}})}^{2}}\] \[=\frac{1}{2}\times 4\times {{10}^{-10}}\times {{10}^{4}}\] \[\Rightarrow \] \[Q=CV=4\times {{10}^{-10}}\times 100\] \[=4\times {{10}^{-8}}C\] \[\Rightarrow \]By again connecting the charged capacitor with a uncharged ideal capacitor. \[\Rightarrow \] \[\Rightarrow \] Now, the charge Q will divide into two capacitors. \[\Rightarrow \]As V and C are same for both, 0 will be same for both, \[\Rightarrow \]So \[Q=\frac{Q}{2}=2\times {{10}^{-8}}C\] \[\Rightarrow \]So, energy stored in both the capacitors \[=\left( \frac{{{(Q)}^{2}}}{2C} \right)=\frac{{{(Q)}^{2}}}{2C}\] \[=\frac{{{(2\times {{10}^{-8}})}^{2}}}{8\times {{10}^{-10}}}\] \[={{10}^{-6}}J\] So energy lost \[=2\times {{10}^{-6}}-0.25\times {{10}^{-6}}\] \[=0.75\times {{10}^{-6}}J\approx {{10}^{-6}}J\]You need to login to perform this action.
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