Question

A circuit has a branch with 5 capacitors connected in parallel. The capacitance of those capacitors is 1 μF, 2 μF, 3 μF, 4 μF, 5 μF respectively. Calculate the equivalent capacitance of the entire branch.
1. 12 μF
2. 10 μF
3. 120 μF
4. 15 μF

Answer 1

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,

C₁ = 1 μF, C₂ = 2 μF, C₃ = 3 μF, C₄ = 4 μF, and C₅ = 5 μF are in parallel

C_eq =  C1 + C2 + C3 + C₄ + C₅

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