Question

A capacitor 1mF withstands a maximum voltage of 6KV, while another capacitor 2mF withstands a maximum voltage of 4KV. If the capacitors are connected in series, the system will withstand a maximum voltage of?

• A. 2 KV
• B. 4 KV
• C. 6 KV
• D. 9 KV

Answer 1

The correct option is (D) 9 KV

Maximum charge capacitance $C_1$ can withstand is $Q_1=C_1{V}_1$ while capacitance $C_2$ can withstand maximum charge$Q_2=C_2{V}_2$

When $C_1$ and $C_2$ are connected in series the charge on each capacitor is equal and is equal to the maximum charge the combination can withstand,

$Q_{max}=C_1{V}_1$

Since $C_1{V}_1<C_2{V}_2$ For given values

The resultant capacitance of the combination is $C_R=\frac{C_2{C}_1}{C_1+C_2}$

$Q_{max}=6\times {10}^{-3}Cand{C}_R=\frac{2}{3}\times {10}^{-6}F$

The maximum voltage that the combination of capacitance can withstand is

$V_{max}=\frac{Q_{max}}{C_R}$

After putting values

$V_{max}=\frac{6\times {10}^{-3}}{\frac{2}{3}\times {10}^{-6}}$

After solving

$V_{max}=9kV$

Hence the maximum voltage is 9 kV.