Questions
A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. What is the common potential (in V) and energy lost (in J) after reconnection?
detailed solution
Correct option is A
VC=C1V1C1+C2=100 VEnergy lost =Ui−Uf=12C1V12−12C1+C2Vc2=6×10−6 JTalk to our academic expert!
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