# 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 :

$\begin {array} {1 1} (a)\;\frac{E}{3} & \quad (b)\;\frac{E}{2} \\ (c)\;\frac{2E}{3} & \quad (d)\;E \end {array}$

$(a)\;\frac{E}{3}$
answered Nov 7, 2013 by