Question

In the circuit three ideal cells of e.m.f. V, V and 2V are connected to a resistor of resistance R, a capacitor of capacitance C and another resistor of resistance 2R as shown in figure. In the steady state find: 

1. the potential difference between P and Q and
2. potential difference across capacitor C.

Answer 1

Take the potential of Q as reference (0). Let V_p = V_p − V_Q be the unknown potential difference between P and Q that we must find. Assume each battery has its positive terminal connected to the left node P (as drawn). For a branch with emf E and series resistance r (battery on left, resistor on right), the steady current from P to Q is:

I = `(V_p - E)/r`

Sum of currents (from P to Q) in all branches must be zero (no external injection at P):

`I_"top" ​+ I_"mid"​ + I_"bottom"`​ = 0

i. In steady state I_mid = 0 (capacitor open), so,

`(V_p - V)/R + (V_p - 2V)/(2R)` = 0

Multiply by 2R to clear denominators:

2(V_p − V) + (V_p − 2V) = 0

Combine terms:

2V_p​ − 2V + V_p − 2V = 0

⇒ 3V_p​ − 4V = 0

⇒ V_p = `4/3 V`    ...(i)

ii. Let the node between the battery and capacitor be A. Then,

V_A​ = V_P​ − V

= V_p​ − V

The capacitor is between A and Q, so the voltage across the capacitor (left plate minus right plate) is:

V_C​ = V_A − V_Q​

V_C = V_p ​− V

V_C = `4/3 V - V`    ...[From equation (i)]

V_C = `1/3 V`

C = `V/3`