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One end of each of a resistance r capacitor C and resistance 2r 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 2E. The negative terminals of the batteries are then connected together. In this circuit, with steady current the potential drop across the capacitor is :

\[\begin {array} {1 1} (a)\;\frac{E}{3} & \quad (b)\;\frac{E}{2} \\ (c)\;\frac{2E}{3} & \quad (d)\;E \end {array}\]
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$ (a)\;\frac{E}{3}$
answered Nov 7, 2013 by pady_1

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