Question

A capacitor is connected with a battery and stores energy U. After removing the battery, it is connected with another similar capacitor in parallel. The new stored energy in each capacitor will be

• A. $\frac{U}{2}$
• B. $U$
• C. $\frac{U}{4}$
• D. $\frac{3U}{2}$

Answer 1

The correct option is C $\frac{U}{4}$

When Q charge is present on capacitor energy stored in it is $U=\frac{1}{2}c{V}^2$

$V=\frac{Q}{c}$

Now when battery is disconnected total charge is constant

$q_1+q_2=Q$

$V_1=\frac{q_1}{c},V_2=\frac{q_2}{c}$

$V_1=V_2$

$\frac{q_1}{c}=\frac{q_2}{c}\therefore q_1=q_2=\frac{Q}{2}$

$V_1=V_2=\frac{Q}{2}=\frac{V}{2}$

$\therefore$ charge stored in each capacitor $=\frac{Q}{2}$

$\therefore totalenergy=\frac{(\frac{Q}{2}{)}^2}{2c}=\frac{1}{2}c{V}^2=\frac{U}{4}$