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Electricity

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19 Feb 2007 11:49:38 IST

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two identical capacitors,have the same capacitance C.one of them is charged to potential V1 and the other to potential V2.the negative ends of the capacitors are connected together .when the positive ends are also connected the decrease in the energy of the system is?

1/4C(v1-v2)²this is the ans but how?

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avdesh100

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Joined: 9 Feb 2007

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19 Feb 2007 12:10:05 IST

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yes the answer is correct..........
when capacitor are connected, the plates which are isolated form another capacitor and the charge Q1=CV1 and Q2=CV2 will get induced and thus net charge on each plate will become ,
(CV1-CV2)/2 and thus you can calculate the energy........

Bipin Dubey

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Joined: 23 Jan 2007

Posts: 7942

21 Feb 2007 07:58:17 IST

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Initially charges on capacitors are CV₁ and CV₂
Let the charges be q₁and q₂ after connection.
When they are connected their potential drops would become same.
=> q₁/C = q₂/C
=> q₁ = q₂
conserving charges gives   q₁ + q₂ = CV₁ + CV₂
Hence q₁ = q₂ = C(V₁+V₂)/2
Energy loss = initial energy - final energy
                  = CV₁² /2 + CV₂² /2 - [C(V₁+V₂)/2]²/2C - [C(V₁+V₂)/2]²/2C
                  = (1/4C)(V₁-V₂)²

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