The equivalent capacitance for the circuit drawn above is
A.
18µf
B.
4µf
C.
2µf
D.
12µf
E.
13µf
Correct answer is B
capacitors 5microfarad and 7microfarad are parallel to each other, hence effective capacitance = 5 + 7 = 12microfarad
12microfarad and 6 microfarad are in series then effective capacitance =
\(\frac{1}{6} + \frac{1}{12}\)
= \(\frac{3}{12}\)
then effective capacitance = \(\frac{12}{3}\) = 4microfarad