Question

Three capacitors connected in series have an
effective capacitance of 4 uF. If one of the
capacitor is removed, the net capacitance of the
capacitor increases to 6 uF. The removed capacitor
has a capacitance of
(1) 12 uF
(2) 10 uF
(3) 4 MF
(4) 2uF​

Answer 1

Let the capacitors be $\sf C_1, C_2 \ and \ C_3$.

The capacitors are connected in series, the equivalent capacitance is 4uF.

$\sf \dfrac{1}{C_1} + \dfrac{1}{C_2} + \dfrac{1}{C_3} = \dfrac{1}{4}--------(1)$

Now, one of the capacitor is removed, thus it's capacitance is considered zero. Also, equivalent capacitance becomes 6uF.

$\sf \dfrac{1}{C_1} + \dfrac{1}{C_2} = \dfrac{1}{6}-------(2)$

Subtracting (1) and (2),

$\sf \dfrac{1}{C_3} = \dfrac{1}{4} - \dfrac{1}{6} \\ \\ \longrightarrow \sf \dfrac{1}{C_3} = \dfrac{6 - 4}{24} \\ \\ \longrightarrow \sf \dfrac{1}{C_3} = \dfrac{2}{24} \\ \\ \longrightarrow \sf \dfrac{1}{C_3} = \dfrac{1}{12} \\ \\ \longrightarrow \boxed{\boxed{\sf C_3 = 12uF}}$

Option (1) is correct.

Answer 2

Answer:-

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

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

In the second case, $\frac{1}{C_{eq1}} = \frac{1}{C_2} + \frac{1}{C_3} ...(2)$

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

$= > \frac{1}{4} = \frac{1}{C_1} + \frac{1}{6}$

$= > C_1 = 12μF$

Option (1) 12μF is the correct answer

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