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 potential of the junction where the resistor $r$,capacitor $C$,an

Vedclass · Accepted Answer

$\frac{E}{3}$

Vedclass · Answer

(A) In the steady state,no current flows through the capacitor branch.Let the common potential of the junction where the resistor $r$,capacitor $C$,and resistor $2r$ meet be $V_x$,and the potential of the junction where the negative terminals of the batteries meet be $0 	ext{ V}$.The potential at the positive terminals of the batteries $P$,$Q$,and $R$ are $E$,$E$,and $2E$ respectively.Since no current flows through the capacitor branch,the potential at the capacitor plate connected to the junction is $V_x$,and the potential at the other plate is $E$.Applying Kirchhoff's Current Law at the junction $V_x$:$\frac{V_x - E}{r} + \frac{V_x - 2E}{2r} + 0 = 0$Multiplying by $2r$:$2(V_x - E) + (V_x - 2E) = 0$$2V_x - 2E + V_x - 2E = 0$$3V_x = 4E$$V_x = \frac{4E}{3}$The potential drop across the capacitor is $|V_x - E| = |\frac{4E}{3} - E| = \frac{E}{3}$.

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