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 the figure. In the steady state, find (i) the potential difference between P and Q, (ii) the potential difference across capacitor C.

Hint:

In the steady state, a fully charged capacitor behaves like an open circuit, and the current flows only through the resistors.

Answer 1

Step 1 — Branch Current Formulation (Node Analysis)

Designate node Q as the reference potential (0). Define the unknown potential difference between P and Q as Vp = V_P − V_Q. Assume all batteries have their positive terminals connected towards node P. For a branch containing an electromotive force E and series resistance r, with the battery positioned to the left and the resistor to the right, the steady-state current flowing from P to Q is given by: \[ I = \frac{V_p - E}{r}. \] (Justification: The potential at the node between the battery and the resistor is V_p − E. The current through the resistor is then calculated as (V_p − E − V_Q)/r.)

Step 2 — Kirchhoff's Current Law (KCL) at Node P

The sum of currents flowing from P to Q through all connected branches must be zero, as there is no external current injection at node P: \[ I_{\text{top}} + I_{\text{mid}} + I_{\text{bottom}} = 0. \] In the steady state, the current through the middle branch (capacitor) is zero, i.e., I_mid = 0. Therefore, the KCL equation simplifies to: \[ \frac{V_p - V}{R} \;+\; \frac{V_p - 2V}{2R} \;=\; 0. \]

Step 3 — Algebraic Solution for V_p

Multiply the equation by 2R to eliminate denominators: \[ 2(V_p - V) + (V_p - 2V) = 0. \] Combine like terms: \[ 2V_p - 2V + V_p - 2V = 0 \quad\Longrightarrow\quad 3V_p - 4V = 0. \] Solving for V_p yields: \[ \boxed{\,V_p = \dfrac{4}{3}\,V\,} \quad\text{(This is the potential difference between nodes P and Q).} \]

Step 4 — Voltage Across the Capacitor

Consider the middle branch: it consists of a battery with emf V and a capacitor, connected in series between nodes P and Q. Let node A be the junction between the battery and the capacitor. The potential at A is: \[ V_A = V_P - V = V_p - V. \] The capacitor is connected between node A and node Q. Therefore, the voltage across the capacitor (potential of the left plate minus the potential of the right plate) is: \[ V_C = V_A - V_Q = V_p - V. \] Substitute the value of V_p = 4V/3: \[ V_C = \frac{4}{3}V - V = \frac{1}{3}V. \] Consequently: \[ \boxed{\,\text{Voltage across capacitor }C = \dfrac{V}{3}\,.} \]

Step 5 — Verification Using Branch Currents

Calculate the current in the top resistive branch: \[ I_{\text{top}}=\frac{V_p - V}{R}=\frac{\tfrac{4}{3}V - V}{R}=\frac{V}{3R}. \] Calculate the current in the bottom resistive branch: \[ I_{\text{bot}}=\frac{V_p - 2V}{2R}=\frac{\tfrac{4}{3}V - 2V}{2R}=-\frac{V}{3R}. \] The currents in the two resistive branches are equal in magnitude and opposite in direction, summing to zero. This confirms the KCL balance at node P. The middle branch, containing the capacitor, carries no steady current.

Final Results (Summary)

(i) Potential difference between nodes P and Q: V_PQ = 4V/3.
(ii) Magnitude of the potential difference across capacitor C: V_C = V/3.