NEET-XII-Physics
exam-1 year:2016
- #5
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:
(1) 0%
(2) 20%
(3) 75%
(4) 80%digAnsr: 4Ans : 4
Sol. Initial energy stored in capacitor 2 µF
Ui = =
2 212(V) V
2
Final voltage after switch 2 is ON
Vf = +
1 1
1 2
C V
C C
=
2V
10
= 0.2 V
Final energy in both the capacitors
Uf = +
2
1 2 f
1
(C C )V
2
=
2
1 2V
10
2 10
= 0.2 V2
So energy dissipated =
2 2
2
V 0.2V
100
V
= 80%