Explanation

(c) [image:2_image144.emf]
As the capacitors 4 $\mu F$ and 2 $\mu F$ are connected in parallel and are in series with 6$\ \mu F$ capacitor, their equivalent capacitance is
$\frac{(2 + 4) \times 6}{2 + 4 + 6} = 3\ \mu F$
Charge in the circuit
$\ \ \ \ \ \ \ \ \ \ Q = \ $ $3\ \mu F \times 12V = 36\mu C$
[image:2_image144.emf]
Since, the capacitors $4\mu F$ and 2$\mu F$ are connected in parallel, therefore potential difference across them is same.
$\Rightarrow$ $\frac{Q_{1}}{Q_{2}} = \frac{C_{1}}{C_{2}} = \frac{4}{2}or{\ Q}_{1} = 2Q_{2}\ $
Also $Q = Q_{1} + Q_{2}$
$\therefore\ \ \ 36\mu C = 2Q_{2} + Q_{2}$ or $Q_{2} = \frac{36\mu C}{3} = 12\mu C$
$\ \ \ \ \ \ \ \ \ \ \ \ Q_{1} = Q - Q_{2} = 36\mu C - 12\mu C$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 24\ \mu C = 24 \times 10^{- 6}C\ \ \ $