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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

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Best answer
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

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