Question

A capacitor of capacitance $C_1=1.0\mu \text{F}$ withstand a maximum voltage $E_1=6.0\text{kV}$ while a capacitor of capacitance $C_2=2.0\mu \text{F}$ can withstand the maximum voltage $E_2=4.0\text{kV}$. For what voltage will the system of these two capacitors with stand , if they both are connected in series ?

• A. $9\text{kV}$
• B. $10\text{kV}$
• C. $11\text{kV}$
• D. $12\text{kV}$

Answer 1

The correct option is A $9\text{kV}$

$q_1=C_1{E}_1=1\times {10}^{-6}\times 6\times {10}^3=6000\mu \text{C}$

$q_2=C_2{E}_2=2\times {10}^{-6}\times 4\times {10}^3=8000\mu \text{C}$

The maximum charge should be equal to the the smaller one in series connection.


In series combination , $C_s=\frac{C_1{C}_2}{C_1+C_2}$
$C_s=\frac{1\times 2}{1+2}=\frac{2}{3}\mu \text{F}$ ; $q_{max}=6000\mu \text{C}$

The maximum voltage , $E_{max}=\frac{q_{max}}{C_s}=\frac{6000\times {10}^{-6}}{\left(\frac{2}{3}\right)\times {10}^{-6}}$

$E_{max}=9000\text{V}\text{or}9\text{kV}$

Hence, option (a) is the correct answer.