Calculate the equivalent capacitance of the circuit shown in Figure 3, given C = 2 ext extmu F and V = 20 ext V.

Capacitors $C_3$ and $C_4$ are in series.
Capacitors $C_5$ and $C_6$ are in series.
Capacitors $C_7$ and $C_8$ are in series.
For two capacitors in series, the equivalent capacitance $C_{eq}$ is given by $\frac{1}{C_{eq}} = \frac{1}{C_a} + \frac{1}{C_b}$ or $C_{eq} = \frac{C_a C_b}{C_a + C_b}$.

• Equivalent capacitance of $C_3$ and $C_4$:

  $$C_{34} = \frac{C_3 \times C_4}{C_3 + C_4} = \frac{2 \mu\text{F} \times 2 \mu\text{F}}{2 \mu\text{F} + 2 \mu\text{F}} = \frac{4 (\mu\text{F})^2}{4 \mu\text{F}} = 1 \mu\text{F}$$

• Equivalent capacitance of $C_5$ and $C_6$:

  $$C_{56} = \frac{C_5 \times C_6}{C_5 + C_6} = \frac{2 \mu\text{F} \times 2 \mu\text{F}}{2 \mu\text{F} + 2 \mu\text{F}} = 1 \mu\text{F}$$

• Equivalent capacitance of $C_7$ and $C_8$:

  $$C_{78} = \frac{C_7 \times C_8}{C_7 + C_8} = \frac{2 \mu\text{F} \times 2 \mu\text{F}}{2 \mu\text{F} + 2 \mu\text{F}} = 1 \mu\text{F}$$

The three equivalent capacitances $C_{34}$, $C_{56}$, and $C_{78}$ are in parallel.
For capacitors in parallel, the equivalent capacitance $C_{eq}$ is given by $C_{eq} = C_a + C_b + C_c$.

• Equivalent capacitance of the parallel section: $$C_{parallel} = C_{34} + C_{56} + C_{78} = 1 \mu\text{F} + 1 \mu\text{F} + 1 \mu\text{F} = 3 \mu\text{F}$$

The capacitor $C_1$, the parallel combination $C_{parallel}$, and the capacitor $C_2$ are all in series.
For capacitors in series, the equivalent capacitance $C_{total}$ is given by $\frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_{parallel}} + \frac{1}{C_2}$.

• Total equivalent capacitance: $$\frac{1}{C_{total}} = \frac{1}{2 \mu\text{F}} + \frac{1}{3 \mu\text{F}} + \frac{1}{2 \mu\text{F}}$$ $$\frac{1}{C_{total}} = \frac{3}{6 \mu\text{F}} + \frac{2}{6 \mu\text{F}} + \frac{3}{6 \mu\text{F}}$$ $$\frac{1}{C_{total}} = \frac{3 + 2 + 3}{6 \mu\text{F}} = \frac{8}{6 \mu\text{F}} = \frac{4}{3 \mu\text{F}}$$ $$C_{total} = \frac{3}{4} \mu\text{F} = 0.75 \mu\text{F}$$

The equivalent capacitance of the circuit is $\boxed{\text{0.75 \mu F}}$.

If the question also implies finding the total charge stored in the circuit:
The total charge $Q$ stored in the circuit is given by $Q = C_{total} \times V$.

$$Q = 0.75 \mu\text{F} \times 20 \text{ V}$$ $$Q = (0.75 \times 10^{-6} \text{ F}) \times 20 \text{ V}$$ $$Q = 15 \times 10^{-6} \text{ C} = 15 \mu\text{C}$$

The total charge stored in the circuit is $\boxed{\text{15 \mu C}}$.