The net capacitance in the circuit above is - JAMB Physics 2017 Question
A
80µF
B
6.0µF
C
4.0µF
D
2.0µF
correct option: d
For capacitance i n parallel, 2µF and 2µF are in parallel,
their equivalence is 2µF and 2µF = 4µF
The 4µF generated is now in series with the remaining 4µF.
The net capacitance for series connection is
\(\frac{1}{C}\) = \(\frac{1}{4}\) + \(\frac{1}{4}\) = 1 + \(\frac{1}{4}\) = \(\frac{2}{4}\)
C = \(\frac{4}{2}\)
= 2µF
their equivalence is 2µF and 2µF = 4µF
The 4µF generated is now in series with the remaining 4µF.
The net capacitance for series connection is
\(\frac{1}{C}\) = \(\frac{1}{4}\) + \(\frac{1}{4}\) = 1 + \(\frac{1}{4}\) = \(\frac{2}{4}\)
C = \(\frac{4}{2}\)
= 2µF
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