Question

In the network shown below, the charge accumulated in the capacitor in steady state will be

• A. 10.3 μC
• B. 4.8 μC
• C. 7.2 μC
• D. 12 μC

Answer 1

The Correct Option is C

In steady state, no current passes through the capacitor. Hence, the capacitor acts as an open circuit. The circuit can be analyzed as follows: 1. Current \( i_2 = 0 \) through the capacitor. 2. The current \( i_1 \) in the resistors is determined by the total resistance of the branch excluding the capacitor. The total resistance of the \( 6 \, \Omega \) and \( 4 \, \Omega \) resistors in series is: \[ R_{\text{total}} = 6 + 4 = 10 \, \Omega. \] The current in the circuit is: \[ i_1 = \frac{3}{R_{\text{total}}} = \frac{3}{10} = 0.3 \, \text{A}. \] The potential difference across the \( 6 \, \Omega \) resistor is: \[ V = i_1 \cdot R = 0.3 \cdot 6 = 1.8 \, \text{V}. \] Since the capacitor is in parallel with the \( 6 \, \Omega \) resistor, the potential difference across the capacitor is also \( 1.8 \, \text{V} \). The charge accumulated in the capacitor is: \[ Q = C \cdot V, \] where \( C = 4 \, \mu\text{F} \) and \( V = 1.8 \, \text{V} \). Substitute the values: \[ Q = 4 \cdot 1.8 = 7.2 \, \mu\text{C}. \] Thus, the charge accumulated in the capacitor is \( \boxed{7.2 \, \mu\text{C}} \).