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,

Question

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,, $ \frac{E}{3} $, $ \frac{E}{2} $, $ \frac{2 E}{3} $, $ E $

MARKS App · Accepted Answer

In the steady state, no current flows through the capacitor branch. In the steady state, the middle branch having the capacitor acts as an open switch and hence, the effective emf of the circuit in the steady state is E + 2 E = 3 E and the current in the circuit is, i = n e t e m f n e t r e s i s t a n c e = 2 E - E r + 2 r = E 3 r So, potential drop across capacitor is, V = i r = E 3 r × r = E 3

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