Practice question

Given: A 6 μF capacitor charged to 150 V is connected to an uncharged 9 μF capacitor. What is the energy lost? These values define the system as per NCERT data. Formula: Initial energy: U_i = 1/2 × 6 × 10⁻⁶ × (150)² = 0.0675 J. This is standard NCERT relation. Substitution & Calculation: Charge: Q = 6 × 10⁻⁶ × 150 = 9 × 10⁻⁴C . Total C = 6 + 9 = 15 μF, V = frac9 × 10⁻⁴¹⁵ × 10⁻⁶= 60 V . Final energy: U_f = 1/2 × 15 × 10⁻⁶ × (60)² = 0.027 J . Loss: U_i - U_f = 0.0675 - 0.027 = 0.0405 J . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.