Three capacitors connected in series have an effective capacitance $4\mu F$ . 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) $12\mu F$

Hint The equivalent capacitance of capacitors connected in parallel will be simply the individual sum of the capacitors. Whereas the equivalent capacitance of capacitors connected in series will be the inverse of the sum of the inverse of the individual capacitors. This calculation is the opposite of the calculation of the equivalent resistance for resistors connected in parallel and series.

Complete Step by step answer
Let the equivalent capacitance be denoted by $C$ . Let the three capacitors connected in series be denoted by ${{C}_{1}}$ , ${{C}_{2}}$ and ${{C}_{3}}$ . The equivalent capacitance for three capacitors connected in series is given by the inverse of the sum of the inverse of individual capacitors that are connected in series. Mathematically, we can write that as
$\dfrac{1}{C}=\dfrac{1}{{{C}_{1}}}+\dfrac{1}{{{C}_{2}}}+\dfrac{1}{{{C}_{3}}}$
It is given in the question that the equivalent capacitance for these three capacitors is $4\mu F$ .
Therefore,
$\dfrac{1}{4}=\dfrac{1}{{{C}_{1}}}+\dfrac{1}{{{C}_{2}}}+\dfrac{1}{{{C}_{3}}}$
Let the removed capacitor be ${{C}_{3}}$ . Therefore, by substituting the given value of the equivalent capacitance after removing one capacitor, we get
$\dfrac{1}{6}=\dfrac{1}{{{C}_{1}}}+\dfrac{1}{{{C}_{2}}}$
By substituting this equation that gives the sum of the inverse of the capacitors excluding the removed capacitor, into the original equation that showed the relation between the three capacitors and the original equivalent capacitance, we get
$\dfrac{1}{4}=\dfrac{1}{6}+\dfrac{1}{{{C}_{3}}}$
Simplifying the above equation gives us
$\dfrac{1}{{{C}_{3}}}=\dfrac{1}{12}$
Taking the inverse of the third capacitor,
$\Rightarrow {{C}_{3}}=12\mu F$
Hence, the option (D) is the correct answer.

Note
Here in this question, we have directly used and substituted the values given in the question to our equation. We can do this only if the given values are of the same units(microfarads in this case) and if we want the answer in the same units. If that is not the case, then the conversion of units has to be taken care of.