Question

In the circuit shown, the energy stored in the capacitor is \( n \, \mu \text{J} \). The value of \( n \) is: 

 

Answer 1

Correct Answer: 75

To find the energy stored in the capacitor, we first analyze the circuit to determine the potential difference across the capacitor. The total current in the circuit can be calculated as: \[ I_s = \frac{12}{3 + 9} = \frac{12}{12} = 1 \, \text{A}. \] The current through the upper branch (\( AB \)) is: \[ I_1 = \frac{12}{4 + 2} = \frac{12}{6} = 2 \, \text{A}. \] The voltage drop across the \( 3 \, \Omega \) resistor is: \[ V_A - V_C = 3I_1 = 3 \cdot 1 = 3 \, \text{V}. \] The voltage at point \( A \) with respect to point \( D \) is: \[ V_A - V_D = 3 \cdot 1 + 2 \cdot 2 = 3 + 4 = 8 \, \text{V}. \] The voltage across the capacitor is: \[ V_A - V_B = 5 \, \text{V}. \] The energy stored in the capacitor is given by the formula: \[ U = \frac{1}{2} C V^2, \] where \( C = 6 \, \mu \text{F} \) and \( V = 5 \, \text{V} \). Substitute the values: \[ U = \frac{1}{2} \cdot 6 \cdot 5^2 = \frac{1}{2} \cdot 6 \cdot 25 = 75 \, \mu \text{J}. \] Thus, the energy stored in the capacitor is \( \boxed{75 \, \mu \text{J}} \).