Q. One end 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, $\mathcal{Q}$, 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
A) $\frac{E}{3}$
B) $\frac{E}{2}$
C) $\frac{2E}{3}$
D) E
Explanation
(a) In the steady state, no current flows through capacitor branch.
Current in the circuit
$i = \frac{net\ emf}{net\ resistance}$
$\ \ \ = \frac{2E - E}{r + 2r} = \frac{E}{3r}$
So, potential drop across capacitor
$V = ir = \frac{E}{3r} \times r = \frac{E}{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.