Practice question

Given: A 4 μF capacitor charged to 300 V is connected to an uncharged 8 μF capacitor. What is the energy lost? Formula: Initial energy: U_i = 1/2 × 4 × 10⁻⁶ × (300)² = 0.18 J. Substitution & Calculation: Charge: Q = 4 × 10⁻⁶ × 300 = 1.2 × 10⁻³C . Total C = 4 + 8 = 12 μF, V = frac1.2 × 10⁻³¹² × 10⁻⁶= 100 V . Final energy: U_f = 1/2 × 12 × 10⁻⁶ × (100)² = 0.06 J . Loss: U_i - U_f = 0.18 - 0.06 = 0.12 J . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.