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 + … + Cn
Calculation:
Given that,
C1 = 1 μF, C2 = 2 μF, C3 = 3 μF, C4 = 4 μF, and C5 = 5 μF are in parallel
Ceq = C1 + C2 + C3 + C4 + C5
= 1 μF + 2 μF + 3 μF + 4 μF + 5 μF = 15 μ