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Two capacitors of capacitance `C_(1)` and `C_(2)` are connected in parallel and a charge Q is delivered to the combination. The two are then disconnected and reconnected in series. What are the new potential differences and charge on the capacitors?

Answer» Correct Answer - `V_(1)=Q//C_(1),V_(2)=Q//C_(2),` charges are `q_(1),q_(2)-q_(2),+q_(2)-q_(2),q_(1),` where `q_(2)=Q//2` and `q_(1)=Q(C_(1)-C_(2))//(C_(1)+C_(2))`


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