Question

A $100\mu F$ capacitor is charged to 100 V. After the charging. battery is disconnected. The capacitor is then connected in parallel to another capacitor. The final voltage is 20 V. If the capacity of the second capacitor is $X\mu F$ . Then $\frac{X}{100}$ will be

Answer 1

Charge, $q=CV={10}^4\mu C$

In parallel common potential is given by

$V=\frac{\text{Total charge}}{\text{Total capacity}}$

$20=\frac{\left({10}^4\mu C\right)}{(C+100)\mu C}$

Solving this equation we get

$C=400\mu F$