In the given figure,find the charge on the ca | vedclass

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(C) In the steady state,the capacitor acts as an open circuit,so no current flows through the branch containing the capacitor.Therefore,the entire cur

Vedclass · Accepted Answer

$\frac{CE R_2}{R_2 + r}$

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(C) In the steady state,the capacitor acts as an open circuit,so no current flows through the branch containing the capacitor.Therefore,the entire current $i$ from the battery flows through the resistor $R_2$.The total resistance of the circuit is $R_{total} = R_2 + r$.The current in the circuit is $i = \frac{E}{R_2 + r}$.The potential difference across the resistor $R_2$ is $V = i 	imes R_2 = \frac{E R_2}{R_2 + r}$.Since the capacitor is in parallel with the resistor $R_2$,the potential difference across the capacitor is the same as the potential difference across $R_2$.Thus,the charge on the capacitor is $Q = C 	imes V = \frac{C E R_2}{R_2 + r}$.

In the given figure,find the charge on the capacitor of capacitance $C$.

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