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Two capacitors of 3muF and 6 muF are connected in series and a potential difference of 900 V is applied across the combination. They are then disconnected and reconnected inparallel. The potential difference across the combination is |
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Answer» Zero `thereforeCalpha1/VthereforeC_(1)/C_(2)=V_(2)/V_(1)rArr3/6=V_(2)/V_(1)rArrV_(1)=2V_(2)` ALSO `V_(1)+V_(2)=900V` `V_(2)=300VandV_(2)=600V` COMMON potential V = `(C_(1)V_(1)+C_(2)V_(2))/(C_(1)+C_(2))` = `(3xx10^(-6)xx600+6xx10^(-6)xx300)/(3xx10^(-6)+6xx10^(-6))` = `(1800+1800xx10^(-6))/(9xx10^(-6))=3600/9` = 400 V 
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