Question

Three capacitors connected in series have an effective capacitance of $2\mu \text{F}$. If one of the capacitors is removed, the effective capacitance becomes $3\mu \text{F}$. The capacitance of the capacitor that is removed is

• A. $13\mu \text{F}$
• B. $\frac{3}{2}\mu \text{F}$
• C. $\frac{2}{3}\mu \text{F}$
• D. $6\mu \text{F}$

Answer 1

The correct option is D $6\mu \text{F}$

Let the capacitance of capacitors be $C_1,C_2\&C_3$.

Effective capacitance of series combination is given by

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

Here, $C_{eq}=2\mu \text{F}$

$\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}=\frac{1}{2}......\left(1\right)$

Let's say $C_3$ is removed,

$\frac{1}{C_1}+\frac{1}{C_2}=\frac{1}{3}.........\left(2\right)$

Subtracting eq.$\left(1\right)$ to $\left(2\right)$, we get

$\frac{1}{C_3}=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}$

$\therefore C_3=6\mu \text{F}$

Hence, option $\left(d\right)$ is correct.