Question

A capacitor of capacitance 1$\mu$F withstands a maximum voltage of 6 kV, while another capacitor of capacitance 2 $\mu$ F withstands a maximum voltage of 4 kV. If they are connected in series, the combination can withstand a maximum voltage of :

• A. $3kV$
• B. $6kV$
• C. $10kV$
• D. $9kV$

Answer 1

The correct option is D $9kV$

maximum charge on $1\mu F$
$Q_1=1\times {10}^{-6}\times 6\times {10}^3$
$Q_1=6\times {10}^{-3}C$
maximum charge on $2\mu F$
$Q_2=2\times {10}^{-6}\times 4\times {10}^3$
$Q_2=8\times {10}^{-3}$
Now when capacitor are connected in series charge present on them must be equal.
$\therefore$ charge present on both capacitor $=6\times {10}^{-3}$
$V=\frac{6\times {10}^{-3}}{1\times {10}^{-6}}+\frac{6\times {10}^{-3}}{2\times {10}^{-6}}$
$V=6\times {10}^3+3\times {10}^3$
$V=9\times {10}^{3}V$