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## Homework Statement

A capacitor of capacity C is charged with potential difference V. Another capacitor of capacity 2C is charged to 4V.they are connected with

**reverse polarity**after removing batteries. The heat produced during redistribution of charges is...

## Homework Equations

##Q = CV##

##U = 0.5CV^2##

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## The Attempt at a Solution

Let Q be charge in capacitor with capacitance C and other be q initially.

##Q=CV##

##q=8CV##

##U1 = Q^2/2C + q^2/(4C)##

##U2 = (Q+q)^2/2(C+4C)##

U2 is final energy... Net charge is Q+q and since potential difference becomes same when connected... C(net) = C1+C2

Heat = delta U

##delta U = C1C2(V1-V2)/(C1+C2)##

Here V gets added up...( I don't know why... But I got the answer)

Hence ##U = C1C2(V1+V2)/(C1+C2)##

You will get correct answer but why V1 + V2 is there some other method?