The net capacitance in the circuit above is
A.
80µF
B.
6.0µF
C.
4.0µF
D.
2.0µF
Correct answer is 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