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 — Write branch currents (node form)

Take the potential of Q as reference (0). Let Vp = 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 → Q is \[ I = \frac{V_p - E}{r}. \] (Derivation: the node between battery and resistor has potential V_p − E, and current through resistor = (V_p − E − V_Q)/r.)

Step 2 — Apply KCL at node P

Sum of currents (from P to Q) in all branches must be zero (no external injection at P): \[ I_{\text{top}} + I_{\text{mid}} + I_{\text{bottom}} = 0. \] In steady state I_mid = 0 (capacitor open), so \[ \frac{V_p - V}{R} \;+\; \frac{V_p - 2V}{2R} \;=\; 0. \]

Step 3 — Solve algebraically for V_p

Multiply by 2R to clear denominators: \[ 2(V_p - V) + (V_p - 2V) = 0. \] Combine terms: \[ 2V_p - 2V + V_p - 2V = 0 \quad\Longrightarrow\quad 3V_p - 4V = 0. \] Therefore \[ \boxed{\,V_p = \dfrac{4}{3}\,V\,} \quad\text{(potential difference between P and Q).} \]

Step 4 — Voltage across the capacitor

Middle branch (left-to-right): P → battery (emf V) → capacitor → Q. 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 capacitor (left plate minus right plate) is \[ V_C = V_A - V_Q = V_p - V. \] Substitute V_p = 4V/3: \[ V_C = \frac{4}{3}V - V = \frac{1}{3}V. \] Hence \[ \boxed{\,\text{Voltage across capacitor }C = \dfrac{V}{3}\,.} \]

Step 5 — Quick check via branch currents

Compute top-branch current: \[ I_{\text{top}}=\frac{V_p - V}{R}=\frac{\tfrac{4}{3}V - V}{R}=\frac{V}{3R}. \] Bottom-branch current: \[ I_{\text{bot}}=\frac{V_p - 2V}{2R}=\frac{\tfrac{4}{3}V - 2V}{2R}=-\frac{V}{3R}. \] So the two resistive branch currents are equal in magnitude and opposite in sign (their sum = 0), confirming the KCL balance. The middle branch carries no steady current.

Final answers (concise)

(i) Potential difference between P and Q: V_PQ = 4V/3.
(ii) Potential difference (magnitude) across capacitor C: V_C = V/3.

Answer 2

Step 1: Analyzing the Circuit. The circuit consists of three ideal cells and resistors in series with a capacitor. Since we are considering the steady state, the capacitor will act as an open circuit because in the steady state, the capacitor is fully charged.

Step 2: Simplifying the Circuit. In the steady state, the current will flow through the resistors, but no current will flow through the capacitor. The effective voltage of the battery is the sum of the voltages of the three cells. The total voltage is: \[ V_{\text{total}} = V + V + 2V = 4V \] The total resistance in the circuit is the sum of the resistances of the two resistors: \[ R_{\text{total}} = R + 2R = 3R \] 

Step 3: Current in the Circuit. The total current in the circuit is given by Ohm’s law: \[ I = \frac{V_{\text{total}}}{R_{\text{total}}} = \frac{4V}{3R} \] Step 4: Potential Difference Between P and Q. The potential difference between P and Q is across the capacitor and the second resistor. In the steady state, the capacitor has no current flowing through it, so the potential difference across the capacitor is equal to the potential difference across the second resistor (2R). The potential difference across the second resistor is: \[ V_{PQ} = I \times 2R = \frac{4V}{3R} \times 2R = \frac{8V}{3} \] Step 5: Potential Difference Across Capacitor C. Since the total voltage is \( 4V \) and the potential difference across the second resistor is \( \frac{8V}{3} \), the potential difference across the capacitor is the remaining voltage: \[ V_C = 4V - \frac{8V}{3} = \frac{12V}{3} - \frac{8V}{3} = \frac{4V}{3} \] Thus, the potential difference across the capacitor is \( \frac{4V}{3} \).