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

Options

A. 0.14
B. 0.25
C. `0.50`
D. 0.64

Answer

D

Explanation

Initial energy stored in the capacitor,
`U_(i) = (1)/(2)CV^(2) = (1)/(2) xx C xx (50)^(2) = (1)/(2)C(50)^(2)`
After 2s, when the potential drops by 10 V, the final potential is 40 V.
Final energy stored in the capacitor, `U_(f) = (1)/(2)C(40)^(2)`
Fraction of energy stored `= (U_(f))/(U_(i)) = ((1)/(2)C(40)^(2))/((1)/(2)C(50)^(2)) = 0.64`