Question

A capacitor of capacitance $C_1=1\mu F$ can withstand a maximum voltage of $V_1=6kV$, and another capacitor of capacitance $C_2=2\mu F$ can withstand a maximum voltage of $V_2=4kV$. If they are connected in series, what maximum voltage in (kV) will the system withstand?

Answer 1

Maximum charge on $C_1$ is $Q_1=C_1{V}_1=1\times {10}^{-6}\times 6\times {10}^3=6mC$
Maximum charge on $C_2$ is $Q_2=C_2{V}_2=2\times {10}^{-6}\times 4\times {10}^3=8mC$
As they are connected in series so both will have same charge. The charge should be $Q_1$ for both because it is minimum.
For this charge voltage across $C_1$ is $V_1=6kV$ and voltage across $C_2$ becomes $V_2^{\prime }=Q_1/C_2=\frac{6\times {10}^{-3}}{2\times {10}^{-6}}=3kV$
Maximum voltage across the system $=6+3=9kV$