NEET | 2016Work, Power and EnergyHard

Question

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

Options

A.

0 %

B.

20 %

C.

75 %

D.

80 %

Solution


Q = 2V
Ui = 12×(2V)22=V2
 Vy = 1264V225×8
2V-q2=q8+124V225×2
8V - 4q = q
Uf = 5V225=V25
 q=8V5
Energy dissipated = 4V25

   % energy
Dissipated = 4V25V2×100 =80%

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