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?
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a
100,6×10−6
b
200,6×10−5
c
200,5×10−6
d
100,6×10−5
answer is A.
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Detailed Solution
VC=C1V1C1+C2=100V Energy lost =Ui−Uf =12C1V12−12C1+C2Vc2 =6×10−6J