Q.
In a circuit shown in figure, the potential difference across the capacitor of 2 F is
A) 8 V
B) 4 V
C) 12
D) 6 V
Explanation
(b) $\therefore\frac{1}{C_{s}} = \frac{1}{2} + \frac{1}{1} = \frac{3}{2}$
$C_{s} = \frac{2}{3}F$
$Q = C_{s}V = \frac{2}{3} \times 12 = 8C$
$V_{1} = \frac{Q}{C_{1}} = \frac{8}{2} = 4V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.