A capacitor of 2  μF is charged as shown in the diagram. When the switch S is turned to position 2 , the percentage of its stored energy dissipated is

Step-by-step solution

First position, C 1 = 2     μF C 2 = 8     μF Q = C 1 V U i = Q 2 2 C 1 Second position, Q − q C 1 = q C 2 C 2 Q − C 2 q = C 1 q q = C 2 Q C 1 + C 2 & Q − q = C 1 C 1 + C 2 Q Energy on 2   μF = 1 2 ( Q − q ) 2 C 1 ( U f ) 1 = 1 2 C 1 ( C 1 + C 2 ) 2 Q 2 Energy on 8   μF = 1 2 q 2 C ( U f ) 2 = 1 2 C 2 ( C 1 + C 2 ) 2 Q 2 Total energy = 1 2 C 1 ( C 1 + C 2 ) 2 Q 2 + 1 2 C 2 ( C 1 + C 2 ) 2 Q 2 = 1 2 Q 2 ( C 1 + C 2 ) Energy dissipated