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Question

A \( 4 \, \mu\text{F} \) capacitor charged to \( 150 \, \text{V} \) is connected to an uncharged \( 12
\, \mu\text{F} \) capacitor. What is the energy lost?

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Explanation

**Common potential** after connection is weighted average of initial potentials by capacitances. Energy loss ΔU = ½ C₁ C₂ (V₁-V₂)²/(C₁+C₂) dissipated as heat and spark, always positive unless V₁=V₂, explaining why energy reduces after sharing. Initial energy: U_i = (1/2) × 4 × 10⁻⁶ × (150)² = 0.045 J . Charge: Q = 4 × 10⁻⁶ × 150 = 6 × 10⁻⁴ C . Total C = 4 + 12 = 16 μF , V = (6 × 10⁻⁴/16 × 10⁻⁶) = 37.5 V . Final energy: U_f = (1/2) × 16 × 10⁻⁶ × (37.5)² = 0.01125 J . Loss: U_i - U_f =

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