A) Work done by the battery is half of the energy dissipated in the resistor
B) \[At\text{ }t=\tau ,\text{ }q=CV/2\]
C) \[~At\text{ }t=2\tau ,\text{ }q=CV(1-{{e}^{-2}})\]
D) \[At\text{ }t=\frac{\tau }{2},q=CV(1-{{e}^{-1}})\]
Correct Answer: C
Solution :
Case − 1 Its normal RC circuit \[{{W}_{bat}}=C{{V}^{2}}\] \[U=\frac{1}{2}C{{V}^{2}}\] \[H={{W}_{bat}}-U=\frac{1}{2}C{{V}^{2}}\] \[\Rightarrow \] \[({{W}_{ba}})=2(H)\] So (1) is wrong. \[q=CV(1-{{e}^{-t/\alpha }})\]\[\Rightarrow \]At \[t=\alpha ,\] at \[t=2\alpha ,\text{ }q=cv(1-{{e}^{2}})\] 3 is correct \[q=CV(1-{{e}^{-1}})\] 2 is wrong. At \[t=\frac{\alpha }{2}\] \[q=CV\left( 1-{{e}^{-1/2}} \right)\] \[\Rightarrow \](4) is wrong.You need to login to perform this action.
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