Explanation

(b) The points $C\ and\ D$ will be at same potentials since,$\frac{3}{6} = \frac{4}{8}.$
Therefore, capacitance of 2$\mu F$ will be unaffected.
So, the equivalent circuit can be shown as.
[image:2_image145.emf]
The effective capacitance in upper arm in series, is given by
$\ \ \ \ \ \ \ C_{1} = \frac{3 \times 6}{3 + 6} = \frac{18}{9} = 2\ \mu F$
The effective capacitance in lower arm in series is given by
$\ \ \ \ \ \ \ C_{2} = \ \frac{4 \times 8}{4 + 8} = \frac{32}{12} = \frac{8}{3}\mu F$
Hence, the resultant capacitance in parallel is given by
$\ \ \ \ C = C_{1} + C_{2} = 2 + \frac{8}{3} = \frac{14}{3}\mu F$