The given circuit contains three capacitors $C_1 = 3.00 \, \mu F$, $C_2 = 4.00 \, \mu F$, and $C_3 = 2.00 \, \mu F$, and a voltage source of $V = 21.0 \, V$. The capacitors $C_1$ and $C_2$ are connected in series, and this series combination is connected in parallel with $C_3$.

Steps to find the net capacitance:

1. Capacitors in Series: For capacitors in series, the reciprocal of the equivalent capacitance $C_{12}$ is given by: $\frac{1}{C_{12}} = \frac{1}{C_1} + \frac{1}{C_2}$ Substituting the values: $\frac{1}{C_{12}} = \frac{1}{3.00} + \frac{1}{4.00} \, (\mu F^{-1})$ Simplifying: $\frac{1}{C_{12}} = \frac{4}{12} + \frac{3}{12} = \frac{7}{12}$ Therefore: $C_{12} = \frac{12}{7} \approx 1.71 \, \mu F$

2. Capacitors in Parallel: The equivalent capacitance of capacitors in parallel is simply the sum of their capacitances. So, the total capacitance $C_{eq}$ of $C_{12}$ in parallel with $C_3$ is: $C_{eq} = C_{12} + C_3$ Substituting the values: $C_{eq} = 1.71 + 2.00 = 3.71 \, \mu F$

Thus, the net capacitance of the circuit is $C_{eq} = 3.71 \, \mu F$.

Would you like more details on any part of this solution or have further questions?

Follow-up Questions:

1. What would happen if all three capacitors were connected in series?
2. How does the voltage distribute across $C_1$ and $C_2$ in this configuration?
3. How do you calculate the total energy stored in the circuit?
4. What is the charge on each capacitor if the total voltage is 21.0 V?
5. How would changing $C_3$ affect the overall capacitance?

Tip: For capacitors in series, the equivalent capacitance is always smaller than the smallest individual capacitor.