To solve this, we need to find the equivalent capacitance between points A and B for the given group of capacitors. Here's the strategy:

1. Capacitor configuration: The circuit shows $C_1$ in series with the combination of $C_2$ and $C_3$, which are in parallel.

2. Capacitance in parallel: Capacitors $C_2$ and $C_3$ are in parallel, so their combined capacitance is the sum: $C_{\text{parallel}} = C_2 + C_3 = 9.0 \, \mu F + 20 \, \mu F = 29.0 \, \mu F$

3. Capacitance in series: Now, $C_1$ is in series with this parallel combination. For series capacitors, the formula is: $\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_{\text{parallel}}}$ Substituting the values: $\frac{1}{C_{\text{eq}}} = \frac{1}{12 \, \mu F} + \frac{1}{29 \, \mu F}$

4. Calculating $C_{\text{eq}}$: $\frac{1}{C_{\text{eq}}} = 0.0833 + 0.03448 = 0.11778$ $C_{\text{eq}} = \frac{1}{0.11778} = 8.49 \, \mu F$

Thus, the equivalent capacitance is approximately 8.5 µF to two significant figures.

Would you like more details or have any questions?

Here are 5 related questions to expand your understanding:

1. How do you calculate the total capacitance in a series circuit?
2. What happens to the overall capacitance when capacitors are connected in parallel?
3. How does increasing the capacitance of one component affect the total capacitance in a series circuit?
4. What are practical examples of series and parallel capacitors in electronics?
5. How would the equivalent capacitance change if $C_1$ were much larger?

Tip: In a series circuit, the total capacitance is always less than the smallest individual capacitor's value.