Question

Capacity of a capacitor is $48\mu F$ . When it is charged from $0.1C$ to $0.5C$ , change in the energy stored is

• A. $2500J$
• B. $2.5\times 10^{-3}J$
• C. $2.5\times 10^{6}J$
• D. $2.42\times 10^{-2}J$

Answer 1

Answer: (A)

Change in energy $\Delta U=\frac{1}{2}\left[\frac{q_1^2-q_2^2}{C}\right]$
$=\frac{1}{2}\left[\frac{(0.5{)}^2-(0.1{)}^2}{48\times 10^{-6}}\right]$
$=\frac{1}{2}\left[\frac{0.25-0.01}{48\times 10^{-6}}\right]$
$=\frac{1}{2}\left[\frac{0.24}{48\times 10^{-6}}\right]$
$=\frac{1}{2}\left[\frac{10^4}{2}\right]$
$=2500J$