Explanation

(c) [image:2_image94.emf]
Net emf in the circuit here
$\ \ \ \ \ \ \ \ \ E = E_{2} - E_{1} = 16 - 6 = 10$ volt
While the equivalent capacity
$\ \ \ \ \ \ \ \ C = \ \frac{C_{1}C_{2}}{C_{1} + C_{2}} = \ \frac{2 \times 3}{2 + 3} = \frac{6}{5}\mu F$
Charge on each capacitor
$\ \ \ \ \ q = CV = \frac{6}{5} \times 10 = 12\mu C$
$\therefore$ Potential difference across $2\mu F$ capacitor
$\ \ \ \ V_{1} = \frac{q}{C_{1}} = \frac{12}{2} = 6\ volt$