One end each of a resistance r,a capacitor C, | vedclass

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(A) In the steady state,no current flows through the capacitor branch.Let the common junction point of the resistors and capacitor be $A$ and the comm

Vedclass · Accepted Answer

$\frac{E}{3}$

Vedclass · Answer

(A) In the steady state,no current flows through the capacitor branch.Let the common junction point of the resistors and capacitor be $A$ and the common junction of the negative terminals of the batteries be $B$. Let the potential at $B$ be $0 	ext{ V}$.Let the potential at $A$ be $V_A$.The current $i$ flows through the top and bottom branches:$i = \frac{(2E - E)}{(r + 2r)} = \frac{E}{3r}$.The potential at $A$ can be calculated using the top branch:$V_A - 0 = E - i \cdot r = E - (\frac{E}{3r}) \cdot r = E - \frac{E}{3} = \frac{2E}{3}$.Since no current flows through the middle branch,the potential drop across the capacitor is equal to the potential difference between $A$ and the positive terminal of the battery $Q$ (which is at potential $E$ relative to $B$).Potential drop across capacitor $V_C = |E - V_A| = |E - \frac{2E}{3}| = \frac{E}{3}$.

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