In the given circuit,with steady current,the | vedclass

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Question

(D) In a steady state,the capacitor acts as an open circuit,so no current flows through the middle branch.The current $I$ flows through the outer loop

Vedclass · Accepted Answer

$2V/3$

Vedclass · Answer

(D) In a steady state,the capacitor acts as an open circuit,so no current flows through the middle branch.The current $I$ flows through the outer loop containing the $V$ and $2V$ batteries and the $R$ and $2R$ resistors.Applying Kirchhoff's voltage law to the outer loop:$2V - V = I(R + 2R)$$V = I(3R)$$I = \frac{V}{3R}$The potential difference across the capacitor $C$ is equal to the potential difference between the two nodes to which it is connected.Let the left node be $A$ and the right node be $B$. The potential difference $V_{AB}$ across the capacitor branch is:$V_{AB} = V - I 	imes R$ (or $V_{AB} = 2V - I 	imes 2R$)$V_{AB} = V - (\frac{V}{3R}) 	imes R = V - \frac{V}{3} = \frac{2V}{3}$Thus,the potential drop across the capacitor is $\frac{2V}{3}$.

In the given circuit,with steady current,the potential drop across the capacitor must be

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