Calculate the total effective capacitance of the circuit shown below. The values are 4uF, 4uF, 10uF, and 8uF.

For capacitors in series, the reciprocal of the equivalent capacitance is the sum of the reciprocals of individual capacitances.

$$\frac{1}{C_{s1}} = \frac{1}{4 \, \mu\text{F}} + \frac{1}{4 \, \mu\text{F}}$$ $$\frac{1}{C_{s1}} = \frac{2}{4 \, \mu\text{F}}$$ $$C_{s1} = \frac{4 \, \mu\text{F}}{2} = 2 \, \mu\text{F}$$

For capacitors in parallel, the equivalent capacitance is the sum of individual capacitances.

$$C_{p1} = C_{s1} + 10 \, \mu\text{F}$$ $$C_{p1} = 2 \, \mu\text{F} + 10 \, \mu\text{F} = 12 \, \mu\text{F}$$

$$\frac{1}{C_{eff}} = \frac{1}{C_{p1}} + \frac{1}{8 \, \mu\text{F}}$$ $$\frac{1}{C_{eff}} = \frac{1}{12 \, \mu\text{F}} + \frac{1}{8 \, \mu\text{F}}$$

Find a common denominator, which is 24.

$$\frac{1}{C_{eff}} = \frac{2}{24 \, \mu\text{F}} + \frac{3}{24 \, \mu\text{F}}$$ $$\frac{1}{C_{eff}} = \frac{5}{24 \, \mu\text{F}}$$ $$C_{eff} = \frac{24}{5} \, \mu\text{F}$$ $$C_{eff} = 4.8 \, \mu\text{F}$$

Comparing this result with the given options, option D is 4.8 $\mu$F.

The final answer is $\text{D. 4.8 } \mu\text{F}$.