Question

A capacitor of capacitance $1\mu \text{F}$ can withstand a maximum voltage of $6\text{kV}$. Another capacitor of capacitance $2\mu \text{F}$ can withstand a maximum voltage of $4\text{kV}$. If the capacitors are connected in series, the combination can withstand a maximum voltage of

• A. $10\text{kV}$
• B. $9\text{kV}$
• C. $8\text{kV}$
• D. $6\text{kV}$

Answer 1

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

Given:
$C_1=1\mu \text{F};V_1=6\text{kV}$

$C_2=2\mu \text{F};V_2=4\text{kV}$

The maximum charge the first capacitor can hold is
$Q_1=C_1{V}_1=1\times {10}^{-6}\times 6000=6\times {10}^{-3}\text{C}$

The maximum charge the second capacitor can hold is
$Q_2=C_2{V}_2=2\times {10}^{-6}\times 4000=8\times {10}^{-3}\text{C}$.

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 $6\times {10}^{-3}\text{C}$.

Therefore, the charge on the second capacitor must be $6\times {10}^{-3}\text{C}$.

Hence, the voltage across the second capacitor is

$V_2^{\prime }=\frac{6\times {10}^{-3}\text{C}}{2\times {10}^{-6}\text{F}}=3\text{kV}$

Thus, the maximum voltage the system can withstand is $9\text{kV}$$(=6\text{kV}+3\text{kV})$.

Therefore, the option $\left(b\right)$ is right.

Key concept-Breakdown voltage of capacitors with different capacity in their series connection.