A capacitor of capacitance `C_(1)=1 muF` withstand a maximum voltage of `V_(1)=6 KV`, and another capacitor of capacitance `C_(2)=2 muF`, can with stand a maximum voltage of `V_(2)=4KV`. If they are connected in series, what maximum voltage will the system withstand?

Answer

`9 KV`

Explanation

Maximum charge `C_(1)` can hold
`Q_(1)=C_(1)V_(1)=1xx10^(-6)xx6xx10^(3)=6xx10^(-3)C`
and maximum charge `C_(2)` can hold
`Q_(2)=C_(2)V_(2)=2xx10^(-6)xx4xx10^(3)=8xx10^(3)C`
When connected in series, both will have equal charges and so each can have charge `Q_(1)`, which is smaller of the two In this case, voltage across `C_(1)` is `6kV`. And voltage aross `C_(2)` is
`Q_(1)/C_(2)=(6xx10^(-3))/(2xx10^(-6))=3 kV`.
Therefore, the maximum voltage across the system is `6+3=9kV`.