A network of three capacitors, each of capacitance `2muF` is formed as shown in the figure. Calculate the resultant capacitance between A and B.

Explanation

Here, first two `2muF` and `2muF` capacitors are in series and they are in parallel with third `2muF` capacitor.
So, in series, `C_("eq")^(.)=(C_(1)C_(2))/(C_(1)+C_(2))=(2xx2)/(2+2)=1muF`
`C_("eq")^(.)` is in parallel to third `2muF` capacitor,
`:." "C_("eq")=C_("eq")^(.)+2muF=1+2=3muF`
Hence, the capacitance of the ntework is `3muF`.