The given circuit requires calculating the equivalent capacitance $C_{ab}$ between points $a$ and $b$. Let's solve it step by step:

Step 1: Analyze the circuit structure

1. Capacitors $6\mu\text{F}$ and $1\mu\text{F}$ are in series:

   • For capacitors in series: $\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2}$
   • Substituting: $\frac{1}{C_{6,1}} = \frac{1}{6} + \frac{1}{1} = \frac{1}{6} + 1 = \frac{7}{6}$ $C_{6,1} = \frac{6}{7} \, \mu\text{F}$

2. Capacitors $C_{6,1}$ and $9\mu\text{F}$ are in parallel:

   • For capacitors in parallel: $C_p = C_1 + C_2$
   • Substituting: $C_{6,1 \parallel 9} = \frac{6}{7} + 9 = \frac{6}{7} + \frac{63}{7} = \frac{69}{7} \, \mu\text{F}$ $C_{6,1 \parallel 9} = 9.857 \, \mu\text{F}$

3. Capacitors $C_{6,1 \parallel 9}$ and $4\mu\text{F}$ are in series:

   • Again for series: $\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2}$
   • Substituting: $\frac{1}{C_{(6,1 \parallel 9) \text{,4}}} = \frac{1}{9.857} + \frac{1}{4}$ $\frac{1}{C_{(6,1 \parallel 9) \text{,4}}} = 0.1015 + 0.25 = 0.3515$ $C_{(6,1 \parallel 9) \text{,4}} = \frac{1}{0.3515} = 2.845 \, \mu\text{F}$

4. Final Equivalent Capacitance $C_{ab}$:

   • $C_{ab}$ is the result from Step 3: $C_{ab} = 2.845 \, \mu\text{F}$

Final Answer:

The equivalent capacitance $C_{ab} = 2.845 \, \mu\text{F}$.

Would you like a detailed breakdown of any step, or do you have further questions?

Related Questions:

1. What happens to the equivalent capacitance if the $6\mu\text{F}$ capacitor is replaced by a $12\mu\text{F}$ capacitor?
2. How does capacitance change if all capacitors are doubled in value?
3. What is the physical meaning of capacitors in series and parallel combinations?
4. How would the equivalent capacitance change if the $9\mu\text{F}$ capacitor is removed?
5. Can you explain the role of the voltage source $\mathcal{E}_B$ in this circuit?

Tip:

When combining capacitors, always start by identifying series and parallel groups systematically to simplify complex circuits step by step.