Question

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

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

Answer 1

Answer: (D)

Initial energy $E_1 = \frac{1}{2}CV^2$
$= \frac{1}{2}C(50)^2$
When capacitor is discharged, then its potential drops by $10 \,V$. Therefore
Potential $V' = 50-10 = 40\, V$
and energy $E_2 = \frac{1}{2} CV'^2 $
$= \frac{1}{2}C(40)^2$
$\therefore $ The fraction of energy stored in the capacitor
$= \frac{E_2}{E_1}$
$ = \frac{\frac{1}{2}\times C \times (40)^2}{\frac{1}{2} \times C \times (50)^2}$
$ = 0.64$