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A capacitor of capacitance C is charged to voltage V then connected through a resistor R to an uncharged capacitor of same C. Final voltage across each:
AV/2
BV
C0
DV/4
Answer & Solution
Correct answer: A. V/2
Charge conservation: CV initial. After sharing equally: each has CV/2 → V/2 voltage. Energy lost as heat in R = (1/2)CV² - 2 × (1/2)C(V/2)² = CV²/4 (half lost regardless of R).
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