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One end each of a resistance $r$ capacitor $C$ and resistance $2 r$ are connected together. The other ends are respectively connected to the positive terminals of batteries, $P, Q, R$ having respectively emf's $E, E$ and $2 E$. The negative terminals of the batteries are then connected together. In this circuit, with steady current the potential drop across the capacitor is :
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The correct answer is:
$\frac{E}{3}$
In the steady state, no current flows through capacitor branch.
Current in the circuit
$i=\frac{\text { net emf }}{\text { net resistance }}=\frac{2 E-E}{r+2 r}$
$=\frac{E}{3 r}$
So, potential drop across capacitor
$V=i r=\frac{E}{3 r} \times r=\frac{E}{3}$
Current in the circuit
$i=\frac{\text { net emf }}{\text { net resistance }}=\frac{2 E-E}{r+2 r}$
$=\frac{E}{3 r}$
So, potential drop across capacitor
$V=i r=\frac{E}{3 r} \times r=\frac{E}{3}$
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