Question

Calculate the breakdown voltage of series combination of two capacitors having capacitance $2\mu \text{F}$ and $6\mu \text{F}$ as shown in the figure.

• A. $30\text{V}$
• B. $20\text{V}$
• C. $17\text{V}$
• D. $27\text{V}$

Answer 1

The correct option is D $27\text{V}$

Given: $C_1=2\mu \text{F},C_2=6\mu \text{F}$, $V_1=20\text{V},V_2=30\text{V}$

If the capacitors are connected in series then maximum charge that can be stored in all the capacitors in series is equal to the minimum charge of all the capacitors connected in series to avoid breakdown.

Here, $Q_1=C_1\times V_1=2\times 20=40\mu \text{C}$
$Q_2=C_2\times V_2=6\times 30=180\mu \text{C}$

So maximum permissible charge will be equal to $Q_1=40\mu \text{C}$

So, net breakdown voltage is, $V=\frac{Q_1}{C_1}+\frac{Q_1}{C_2}$

$\Rightarrow V=\frac{40}{2}+\frac{40}{6}$

$\Rightarrow V=20+6.67=26.67\approx 27\text{V}$