Question

two capacitor of capacitance C and 3c are shown in diagram. when switch S is turned to position 2,the percentage of store energy dissipated is
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Answer 1

Answer:

ANSWER

Initial energy stored in capacitor 2 μF:U

i

=

2

1

2(V)

2

=V

2

Final voltage after switch 2 is on: V

f

=

C

1

+C

2

C

1

V

1

=

10

2V

=0.2 V

Final energy in both the capacitors, U

f

=

2

1

(C

1

+C

2

)V

f

2

=

2

1

×10×(

10

2V

)

2

=0.2 V

2

Therefore, energy dissipated =

V

2

V

2

−0.2V

2

×100=80%