The numerical value of the charge on either p | vedclass

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) In a steady state,the capacitor acts as an open circuit,meaning no current flows through the branch containing the capacitor.The circuit effectively consists of the battery $E$ with internal resistance $r$ connected in series with the resistor $R_2$.The current $I$ flowing through the circuit is given by $I = \frac{E}{R_2 + r}$.The potential difference $V$ across the capacitor is equal to the potential difference across the resistor $R_2$,as 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 $q = C \left( \frac{E R_2}{R_2 + r} \right) = \frac{CE R_2}{R_2 + r}$.

The numerical value of the charge on either plate of the capacitor $C$ shown in the figure is

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