Topic: Network Theory
Under steady state condition, the energy stored in the circuit is:
- A. 8.22 × 10^-6 J
- B. 1.48 × 10^-6 J
- C. 2.88 × 10^-4 J
- D. 4.81 × 10^-4 J
Answer: C. 2.88 × 10^-4 J
Explanation: At steady state, capacitors act as open circuits. V_1 = V_2 = V_AD = 8 V. Energy stored = 1/2 C_1 V_1^2 + 1/2 C_2 V_2^2. With C_1 = C_2 = 2 μF, energy stored = 2 × (1/2 × 2 × 10^-6 × 8^2) = 2.88 × 10^-4 J.