For two identical capacitors in series, the equivalent capacitance is simply half the value of one ($2/2 = 1$). For non-identical ones, use "Product over Sum" ($18/9 = 2$).
Step 1: Understanding the Concept:
Capacitors in series combine like resistors in parallel, and capacitors in parallel combine like resistors in series. In this circuit, we have two parallel branches, each containing two capacitors in series. Step 2: Key Formula or Approach:
For Series: \[ \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} \]
For Parallel: \[ C_p = C_1 + C_2 \] Step 3: Detailed Explanation:
1. Upper Branch ($C_{up}$): $3\mu\text{F}$ and $6\mu\text{F}$ in series.
\[ C_{up} = \frac{3 \times 6}{3 + 6} = \frac{18}{9} = 2\mu\text{F} \]
2. Lower Branch ($C_{low}$): $2\mu\text{F}$ and $2\mu\text{F}$ in series.
\[ C_{low} = \frac{2 \times 2}{2 + 2} = \frac{4}{4} = 1\mu\text{F} \]
3. Total Capacitance ($C_{AB}$): The two branches are in parallel.
\[ C_{AB} = C_{up} + C_{low} = 2\mu\text{F} + 1\mu\text{F} = 3\mu\text{F} \] Step 4: Final Answer:
The effective capacitance between A and B is 3μF.
