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A parallel plate capacitor is charged to...

A parallel plate capacitor is charged to a potential difference of 50 volts. It is then discharged through a resistance for 2 seconds and its potential drops by 10 volts. Calculate the fraction of energy stored in the capacitance.

A

`0.14 `

B

`0.25`

C

`0.50`

D

`0.64`

Text Solution

Verified by Experts

The correct Answer is:
D
V = 50 Volt, t = 2 sec, potential drop = 10 Volt
After 2 sec, V. = 40 Volt
`U_(i) =1/2 C(50)^2`
`U_f =1/2 C(40)^2`
`impliesU_f/U_i=((40)^2)/((50)^2)=16/25 impliesU_f=0.64U_i`
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