Question

A variable condenser is permanently connected to a $100V$ battery. If the capacity is changed from $2\mu F$ to $10\mu F$ then the ratio of charge stored in $10\mu F$ capacitor to that in $2\mu F$ capacitor is $\frac{x}{y}$. Find $(x+y)$.

Answer 1

Ratio of charges

Potential differences in both the cases will remain the same as the battery remains same

$Q_1=C_{1}V=(2\times {10}^{-6}C)\times \left(100V\right)$
$=2\times {10}^{-4}$

Also,

$Q_2=C_{2}V=(10\times {10}^{-6}C)\times \left(100V\right)$
$={10}^{-3}$

Thus,

$\frac{Q_2}{Q_1}=\frac{{10}^{-3}}{2\times {10}^{-4}}=\frac{5}{1}$

$\frac{x}{y}=\frac{5}{1}$

$x+y=6$