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 \[=\frac{1}{2}C{{V}^{2}}\] \[=\frac{1}{2}400\times {{10}^{-12}}\times {{({{10}^{2}})}^{2}}\] \[=\frac{1}{2}\times 4\times {{10}^{-10}}\times {{10}^{4}}\] \[=2\times {{10}^{-6}}J\] \[\Rightarrow \] \[Q=CV=4\times {{10}^{-10}}\times 100\] \[=4\times {{10}^{-8}}C\] \[\Rightarrow \] 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, Q' will be same for both, \[\Rightarrow \] So \[Q'=\frac{Q}{2}=2\times {{10}^{-8}}C\] \[\Rightarrow \]So, energy stored in both the capacitors \[=\left( \frac{{{(Q')}^{2}}}{2C} \right)=\frac{{{(Q')}^{2}}}{2C}\] \[=\frac{(2\times {{10}^{-8}})}{8\times {{10}^{-10}}}={{10}^{-6}}J\] 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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