Question

A capacitor of capacitance $2\mu \text{F}$ can withstand a maximum voltage of $10\text{kV}$. Another capacitor of capacitance $5\mu \text{F}$ can withstand a maximum voltage of $7\text{kV}$. If the capacitors are connected in series, the combination can withstand a maximum voltage of

• A. $10\text{kV}$
• B. $6\text{kV}$
• C. $14\text{kV}$
• D. $4\text{kV}$

Answer 1

The correct option is C $14\text{kV}$

Given:

$C_1=2\mu \text{F};V_1=10\text{kV}$

$C_2=5\mu \text{F};V_2=7\text{kV}$

The maximum charge the first capacitor can hold is,

$Q_1=C_1{V}_1=2\times {10}^{-6}\times 10000=2\times {10}^{-2}\text{C}$

The maximum charge the second capacitor can hold is,

$Q_2=C_2{V}_2=5\times {10}^{-6}\times 7000=3.5\times {10}^{-2}\text{C}$.

We know that in a series combination, the charge on each capacitor is the same.

Now, the first capacitor can only hold a maximum charge of $2\times {10}^{-2}\text{C}$.

Therefore, the charge on the second capacitor must be $2\times {10}^{-2}\text{C}$.

Hence, the voltage across the second capacitor is

$V_2^{\prime }=\frac{2\times {10}^{-2}\text{C}}{5\times {10}^{-6}\text{F}}=4\text{kV}$

Thus, the maximum voltage the system can withstand is

$(10\text{kV}+4\text{kV})=14\text{kV}$.

Hence, the option (b) is the correct answer.