A) Fraction of stored energy after 1 second is 16/25
B) Potential difference between the plates after 2 seconds will be 32 V
C) Potential difference between the plates after 2 seconds will be 20 V
D) Fraction of stored energy after 1 second is 4/5
Correct Answer: A , B
Solution :
By using \[V={{V}_{0}}{{e}^{-t/CR}}\Rightarrow 40=50\,{{e}^{-1/CR}}\Rightarrow {{e}^{-1/CR}}=4/5\] Potential difference after 2 sec \[{V}'={{V}_{0}}{{e}^{-2/CR}}=50{{({{e}^{-1/CR}})}^{2}}=50{{\left( \frac{4}{5} \right)}^{2}}=32\,V\] Fraction of energy after 1 sec \[=\,\]\[\frac{\frac{1}{2}C{{({{V}_{f}})}^{2}}}{\frac{1}{2}C{{({{V}_{i}})}^{2}}}={{\left( \frac{40}{50} \right)}^{2}}=\frac{16}{25}\]Solution :
Fraction of stored energy after 1 second is 16/25 (a, b) By using \[V={{V}_{0}}{{e}^{-t/CR}}\Rightarrow 40=50\,{{e}^{-1/CR}}\Rightarrow {{e}^{-1/CR}}=4/5\] Potential difference after 2 sec \[{V}'={{V}_{0}}{{e}^{-2/CR}}=50{{({{e}^{-1/CR}})}^{2}}=50{{\left( \frac{4}{5} \right)}^{2}}=32\,V\] Fraction of energy after 1 sec \[=\,\]\[\frac{\frac{1}{2}C{{({{V}_{f}})}^{2}}}{\frac{1}{2}C{{({{V}_{i}})}^{2}}}={{\left( \frac{40}{50} \right)}^{2}}=\frac{16}{25}\]
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