The charge on the capacitor of capacitance C | vedclass

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

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$\frac{C E R_2}{R_2+r}$

Vedclass · Answer

(C) In the steady state, the capacitor acts as an open circuit, meaning no current flows through the branch containing the capacitor.Therefore, the current $I$ flows only through the branch containing the resistor $R_2$ and the internal resistance $r$ of the battery.Using Ohm's law, the current in the circuit is $I = \frac{E}{R_2 + r}$.The potential difference $V$ across the capacitor is equal to the potential difference across the resistor $R_2$, because they are connected in parallel.Thus, $V = I R_2 = \frac{E R_2}{R_2 + r}$.The charge $Q$ on the capacitor is given by $Q = CV$.Substituting the value of $V$, we get $Q = C \left( \frac{E R_2}{R_2 + r} \right) = \frac{C E R_2}{R_2 + r}$.

The charge on the capacitor of capacitance $C$ shown in the figure below will be

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