Three 2-µF capacitors are arranged as seen in the circuit above . The effective capacitance between E and F is
A.
0.75µF
B.
1.33µF
C.
3.00µF
D.
6.00µF
Correct answer is B
\(C_T = \frac{2 \times (2 + 2)}{2 + (2 + 2)}\)
= \(\frac{8}{6}\)
= 1.33 \(\mu\)F