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In the circuit shown, find C if the effective capacitance of the whole circuit is to be 0.5 $$\mu F$$. All values in the circuit are in $$\mu F$$.
The given circuit simplifies to:

$$C_{\text{total-inner}} = 1 + \frac{4}{3} = \frac{7}{3}\ \mu\text{F}$$
$$\frac{C \times \frac{7}{3}}{C + \frac{7}{3}} = 0.5$$
$$\frac{7C}{3C + 7} = \frac{1}{2}$$
$$14C = 3C + 7$$
$$C = \frac{7}{11}\ \mu\text{F}$$
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