Q.
In the circuit arrangement shown in figure, the value of$\ \ C_{1} = C_{2} = C_{3} = 30$ pF and $C_{3} = 120$ pF. If the combination of capacitors is charged with 140V DC supply, the potential differences across the four capacitors will be respectively
A) 80, 40, 40 and 20 V
B) 20, 40, 40 and 80 V
C) 35, 35, 35 and 35 V
D) 80, 20, 20 and 20 V
Explanation
(a) Here, $C_{13} = 30 + 30 = 60\ pF.$ total equivalent capacitance is given by
$\frac{C}{1} = \frac{1}{C_{1}} + \frac{1}{C_{23}} + \frac{1}{C_{4}} = \frac{1}{30} + \frac{1}{60} + \frac{1}{120} = \frac{1}{170} \Rightarrow C = \frac{120}{7}pF$
$\therefore$ Total charge $Q = CV = \frac{120}{7} \times 140pC\ $= 2400pC
$\therefore\ \ V_{1} = \frac{Q}{C_{1}} = \frac{2400pC}{30pF} = 80V$
$V_{2} = V_{3} = V_{23} = \frac{Q}{C_{23}} = \frac{2400pC}{60pF} = 40V\ and\ $
$V_{4} = \frac{Q}{C_{4}} = \frac{2400pC}{120pF} = 20V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.