Question

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

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

Answer 1

The correct option is D $0.64$

Initial energy stored in capacitor,

$U_i=\frac{1}{2}C{V}_i^2=\frac{1}{2}C(50{)}^2$

When the capacitor is discharged for $2\text{seconds}$, the final potential difference across it will be,

$V_f=50-10=40\text{V}$

So, final energy stored,

$U_f=\frac{1}{2}C{V}_f^2=\frac{1}{2}C(40{)}^2$

Thus, fractional energy stored in the capacitor will be

$\frac{U_f}{U_i}=\frac{{40}^2}{{50}^2}=0.64$

Thus, option $\left(D\right)$ is correct.