A capacitor of 2 μF is charged as shown in figure, and when switch S is shifted to position 2 the percentage energy dissipated is: [Diagram: Image]
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A capacitor of 2 μF is charged as shown in figure, and when switch S is shifted to position 2 the percentage energy dissipated is: [Diagram: Image]

G1 Physics · Accepted Answer

Step 1: Initially capacitor 2 μF is charged by 1 V source; U(initial) = ½×2 μF×(1 V)² = 1 μJ. Step 2: After changing switch, the equivalent network shares charge and source solution gives final voltage 0.2 V. Step 3: Compute final energy with connected capacitances using ½ΣCᵢV(final)², then dissipated fraction = (U(initial)−U(final))/U(initial). Step 4: Source printed choice (C) = 80%.