Q. Six identical capacitors are joined in parallel, charged to a potential difference of 10 V, separated and then connected in series, $ie,$ the positive plate of one is connected to negative plate of other. Then potential difference between free plates is
A) 10 V
B) 30 V
C) 60 V
D) $\frac{10}{6}V$
Explanation
(c) If $C$ is capacity of each condenser, then charge on each capacitor$\ = 10\ C$
$(\because V = 10V)$
When connected in series, potential difference between free plates =$\ \frac{total\ charge}{\begin{array}{r}
total\ capacity \\
\
\end{array}}$
$= \frac{10\ C}{C/6} = 60\ V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.