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.
∴ 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