Correct Answer - Option 4 : 0.125 μF

__Concept__:

Equivalent capacitance of capacitors:

**Connected in series:** When n capacitors C1, C2, C3, ... Cn are connected in series, the net capacitance (Cs) is given by:

\(⇒ \dfrac{1}{C_s} = \dfrac{1}{C_1} + \dfrac{1}{C_2}+ \dfrac{1}{C_3} + ... \dfrac{1}{C_n}\) ---(1)

**Connected in parallel:** When n capacitors C1, C2, C3, ... Cn are connected in parallel, the net capacitance (Cp) is given by:

⇒ Cp = C1 + C2 + C3 +... Cn

__Calculation:__

Here, Two 0.25 μF capacitors connected in series.

∴ C_{1} = C_{2} = 0.25 μF

From equation (1);

\( \dfrac{1}{C_s} = \dfrac{1}{C_1} + \dfrac{1}{C_2}\)

\( \dfrac{1}{C_s} = \dfrac{1}{0.25μ F} + \dfrac{1}{0.25μ F}\)

On calculating we'll get

C_{eq} = 0.125 μF