Explanation

(d) [image:1_image242.emf]
Starting from the right end of the network, three $3\mu\ F$ capacitors are connected in series. The equivalent capacitance of these three capacitors is
$\frac{1}{C_{S}} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} \Rightarrow C_{S} = 1\mu\ F$
$C_{S}$ and $2\mu F$ are in parallel. The equivalent capacitance of these two capacitors is
$C_{P} = C_{S} + 2\mu\ F = 1\ \mu\ F + 2\mu\ F = 3\mu\ F$
Proceeding in this way, finally three $3\mu\ F$ are in series. Therefore, the equivalent capacitance between $A$ and $B$ is
$\frac{1}{C_{AB}} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} \Rightarrow C_{AB} = 1\mu\ F$