Question

In a lab experiment, three capacitors each of capacitance $C$ and of breakdown voltage $V$ are joined in series. Then capacitance and breakdown voltage of the combination will be?

• A. $\frac{C}{3},3V$
• B. $\frac{C}{3},1.5V$
• C. $\frac{C}{3},\frac{V}{3}$
• D. $3C,3V$

Answer 1

The correct option is A $\frac{C}{3},3V$

Given: $C_1=C_2=C_3=C$

These three capacitors are connected in series for the input voltage of $V\text{volt}$.
The equivalent capacitance of the series combination is,

$\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$

i.e. $\frac{1}{C_{eq}}=\frac{1}{C}+\frac{1}{C}+\frac{1}{C}$

Therefore, $\frac{1}{C_{eq}}=\frac{3}{C}$

So, $C_{eq}=\frac{C}{3}$

Now, the required breakdown voltage,

$V^{\prime }=\frac{Q_{max}}{C_{eq}}=\frac{CV}{C/3}=3V$

Option (a) is the correct answer.