Question

A capacitor has capacitance of $6nF$, maximum voltage it can sustain is $40$ volt. It is in series with other capacitor $3nF$ and this capacitor can sustain $30$ volt, what is maximum voltage that can be applied in series.

• A. $70$
• B. $50$
• C. $60$
• D. $10$

Answer 1

The correct option is C $60$


$q_{supplied}=C_{eff}ϵ=\left(\frac{C_1{C}_2}{C_1+C_2}\right)ϵ$

$\Delta V_1=\frac{q}{C_1}=\frac{C_2ϵ}{C_1+C_2}=\frac{6ϵ}{9}=\frac{2ϵ}{3}$

$\left(\begin{array}{c}{C}_1=6\\ C_2=3\end{array}\right)$

$\Delta V_2=\frac{q}{C_2}=\frac{C_1ϵ}{C_1+C_2}=\frac{3ϵ}{9}=\frac{ϵ}{3}$

$\begin{array}{cc}\Delta V_1<40& \Delta V_2<30\\ 2\frac{ϵ}{3}<40& \frac{ϵ}{3}<30\\ ϵ<60& ϵ<90\\ & ϵ<90\end{array}$

${ϵ}_{max}=60$