Question

In given circuit, ratio of charge on $2\mu F$ capacitor to the charge on $3\mu F$ capacitor will be

• A. $2:3$
• B. $3:2$
• C. $1:9$
• D. $2:9$

Answer 1

The correct option is D $2:9$

$2:9$

$1\mu F$ and $2\mu F$ are in series, so the total capacitance reduces to

$\frac{2}{3}\mu F$

consider, $q_1$ to be the charge near $\frac{2}{3}\mu F$ and $q_2$ to be the charge near $3\mu F$

and $q$ be the overall charge.

$q_1=\frac{\frac{2}{3}}{\frac{2}{3}+3}\times q$

$q_2=\frac{3}{3+\frac{2}{3}}\times q$

$\frac{q_1}{q_2}=\frac{\frac{\frac{2}{3}}{\frac{2}{3}+3}\times q}{\frac{3}{3+\frac{2}{3}}\times q}$

$=\frac{\frac{2}{3}}{3}=\frac{2}{9}$

Therefore the ratio is given by,

$=2:9$