# What is the total effective capacitance of two 0.25 μF capacitors connected in series

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What is the total effective capacitance of two 0.25 μF capacitors connected in series
1. 1.25 μF
2. 0.50 μF
3. 2.50 μF
4. 0.125 μF

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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 = C+ C2  + C+...  Cn

Calculation:

Here, Two 0.25 μF capacitors connected in series.

∴ C1 = C2 = 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

Ceq = 0.125 μF