Question

A capacitor of capacitance $C_1=1\mu F$ can withstand maximum voltage $V_1=6kV$ (kilo-volt) and another capacitor of capacitance $C_2=3\mu F$ can withstand maximum voltage $V_2=4kV$. When the two capacitors are connected in series, the combined system can withstand a maximum voltage of

• A. $4kV$
• B. $6kV$
• C. $8kV$
• D. $10kV$

Answer 1

Answer: (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 $6mC$ (and not $12mC$ ) and potential difference across it $V_2=6mC/3\mu F=2kV$ and as in series $V$
$=V_1+V_2$ so $V_{\max }=6kV+2kV=8kV$