Correct Answer - Option 4 : 15 μF

__Concept__:

When the capacitors are connected in series, then the equivalent capacitance is given by

\(\frac{1}{{Ceq}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} + \frac{1}{{{C_3}}} + \ldots + \frac{1}{{{C_n}}}\)

When the capacitors are connected in parallel, then the equivalent capacitance is given by

Ceq = C1 + C2 + C3 + … + C_{n}

**Calculation:**

Given that,

C_{1} = 1 μF, C_{2} = 2 μF, C_{3} = 3 μF, C_{4} = 4 μF, and C_{5} = 5 μF are in parallel

C_{eq} = C1 + C2 + C3 + C_{4}_{ }+ C_{5}

= 1 μF + 2 μF + 3 μF + 4 μF + 5 μF = 15 μ