A capacitor of capacitance `C_(1)=1 muF` can withstand maximum voltage `V_(1)=6 kV` and another capacitor of capacitance `C_(2)=3 muF` 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

Options

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

Answer

C

Explanation

As `Q=CV,(Q_(1))_("max")=10^(-6)xx6xx10^(3)=6mC`
While `(Q_(2))_("max")=3xx10^(-6)xx4xx10^(3)=12mC`
However in series charge in same so maximum charge on `C_(2)` will also be 6mC (and not 12 mC) and potential difference across it `V_(2)=6mC//3 muF=2kV` and as in series `V=V_(1)+V_(2)` so `V_("max")=6kV+2kV=8kV`