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: 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 \( \frac{4V}{3} \).