Question

Three capacitors connected in series have an effective capacitance of $4\mu F$. If one of the capacitance is removed, the net capacitance of the capacitor increases to $6\mu F$. The removed capacitor has a capacitance of

• A. $2\mu F$
• B. $4\mu F$
• C. $10\mu F$
• D. $12\mu F$
• E. $24\mu F$

Answer 1

The correct option is D $12\mu F$

Let there are three capacitors with capacitances $C_1,C_2,C_3$ respectively and $C_1$ is removed.

In first case, $\frac{1}{C_{eq1}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$ ...(1)

In second case, $\frac{1}{C_{eq2}}=\frac{1}{C_2}+\frac{1}{C_3}$ ...(2)

From (1) and (2), $\frac{1}{C_{eq1}}=\frac{1}{C_1}+\frac{1}{C_{eq2}}$

or $1/4=1/C_1+1/6$

or $C_1=12\mu F$