Explanation

(a) In the given circuit capacitor’s (1) (2) and (3) are connected in series, hence equivalent capacitance is
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$\ \ \ \ \ \frac{1}{C^{''}} = \frac{1}{C} + \frac{1}{C} + \frac{1}{C} = \frac{3}{C}$
$\Rightarrow C^{''} = \frac{C}{3}$
This is connected in parallel with (4).
$\therefore\ \ C^{''''} = C^{''} + C = \frac{C}{3} + C = \frac{4C}{3}$
The three capacitor’s (5),$\frac{4C}{3},$ (6) are now connected in series.
$\therefore$ Equivalent capacitance is
$\ \ \ \ \ \frac{1}{C\ ''''''} = \frac{1}{C} + \frac{3}{4C} + \frac{1}{C}$
$\ \ \ \ \ \frac{1}{C''''''} = \frac{11}{4C}$
$\Rightarrow \ \ C^{''''''} = \frac{4C}{11}$