Question

A parallel plate capacitor is charged to potential difference of $50V$ . It is then discharged through a resistance for $2s$ and its potential drops by $10V$ . 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 $E_1=\frac{1}{2}C{V}^2$
$=\frac{1}{2}C(50{)}^2$
When capacitor is discharged, then its potential drops by $10V$. Therefore
Potential $V^{\prime }=50-10=40V$
and energy $E_2=\frac{1}{2}C{V}^{\prime 2}$
$=\frac{1}{2}C(40{)}^2$
$\therefore$ The fraction of energy stored in the capacitor
$=\frac{E_2}{E_1}$
$=\frac{\frac{1}{2}\times C\times (40{)}^2}{\frac{1}{2}\times C\times (50{)}^2}$
$=0.64$