In the circuit shown,the steady state voltage | vedclass

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

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$R_1$ and $R_2$ only

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(B) In the steady state,the capacitor $C$ acts as an open circuit,meaning no current flows through the branch containing the capacitor. Therefore,the circuit simplifies to a series combination of resistors $R_1$ and $R_2$ connected across the battery $V$. The current $I$ in the circuit is given by $I = \frac{V}{R_1 + R_2}$. The voltage across the resistor $R_2$ is $V_{R_2} = I \cdot R_2 = \left( \frac{R_2}{R_1 + R_2} \right) V$. Since the capacitor is in parallel with $R_2$,the steady state voltage across the capacitor $C$ is equal to the voltage across $R_2$. Thus,the voltage across the capacitor is $V_C = \left( \frac{R_2}{R_1 + R_2} \right) V$. This fraction depends only on $R_1$ and $R_2$. Therefore,the correct option is $(b)$.

In the circuit shown,the steady state voltage across capacitor $C$ is a fraction of the battery e.m.f. $V$. The fraction is decided by:

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