Q. A capacitor of capacitance $C_{1} = 1\ \mu F$ can with stand maximum voltage $V_{1} = 6kV\ (kilo - volt)$ and another capacitor of capacitance $C_{2} = 3\mu F$ can withstand maximum voltage $V_{2} = 4\ kV.$ When the two capacitors are connected in series, the combined system can withstand a maximum voltage of
A) $4kV$
B) $6kV$
C) $8kV$
D) $10kV$
Explanation
(c) As $Q = CV,\ \left( Q_{1} \right)_{\max} = 10^{- 6} \times 6 \times 10^{3} = 6mC$
While $\left( Q_{2} \right)_{\max} = 3 \times 10^{- 6} \times 4 \times 10^{3} = 12mC$
However in series charge is same so maximum charge on $C_{2}$ will also be $6\ mC$ (and not $12\ mC$) and potential difference across it $V_{2} = 6mC/3\ \mu F = 2KV$ and as in series $V = V_{1} + V_{2}$ so $V_{\max} = 6KV + 2KV = 8KV$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.