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Energy of a charged capacitor is U. Now it is removed from the battery and then connected in parallel to another uncharged capacitor having capacitance twice that of the first one. The energies of the first and second capacitors respectively are ______.

Asked in GUJCET 2016 · Redistribution of charge

Answer: (1) 1/9U, 2/9U

Step-by-step solution

Given: capacitor C with energy U = (Q²)/(2C), isolated, then joined in parallel with an uncharged capacitor 2C.

Idea: the charge Q is conserved, and both capacitors end at the same voltage.

Common voltage: V' = Q/(C + 2C) = Q/(3C).

First: U₁ = 1/2C(Q/(3C))² = 1/9·(Q²)/(2C) = 1/9U.

Second: U₂ = 1/2(2C)(Q/(3C))² = 2/9U.

(Together 1/3U; the other 2/3U is lost as heat and radiation while the charge flows.)

So the energies are 1/9U and 2/9U.

Why the other options are wrong

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