Home
Class 12
PHYSICS
A 2muF capacitor is charged as shown in ...

A `2muF` capacitor is charged as shown in the figure. The percentage of its stored energy disispated after the switch S is turned to poistion 2 is

A

`0%`

B

`20%`

C

`75%`

D

`80%`

Text Solution

Verified by Experts

The correct Answer is:
D
`U_(i) = (1)/(2)CV^(2)`

`V_(C) = ("total charge")/("total capacity") = (CV)/(C+4C) = (V)/(5)`
`U_(f) = (1)/(2) (C +4C) xx ((V)/(5))^(2) = (1)/(5) xx (1)/(2)CV^(2) = (1)/(5)U_(i)`
`U_(f) = 20%` of `U_(i)`
so loss of enegry `=80%`.
Promotional Banner

Similar Questions

Explore conceptually related problems

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

For the arrangement shown in the figure, the charge on the 4muF capacitor is

If a 4muF capacitor is charged to 1 kV, then energy stored in the capacitor is

Initially the 900muF capacitor is charged to 100 V and the 100muF capacitor is uncharged in the figure shown. Then the switch S_(2) is closed for a time t_(1) , after which it is opened and at the same instant switch S_(1) is closed for a time t_(2) and then opened. It is now found that the 100muF capacitor is charged to 300 V. If t_(1) and t_(2) minimum possible values of the time intervals, then findout (t_(1))/(t_(2))

For the circuit shown in figure the charge on 4muF capacitor is

If a 4 muF capacitor is charged to 1KV, then energy stored is conductor is

The charge on 3 muF capacitor shown in the figure is