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.
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Approach Solution -2
The circuit shows four capacitors arranged as two branches in series (3 μF with 6 μF, and 2 μF with 2 μF) with the two branches joined in parallel between A and B. Let's test each of the given options against this configuration.
13 μF: This value would arise only if all four capacitors were mistakenly treated as being in parallel with each other, since \( 3 + 6 + 2 + 2 = 13 \). But the circuit clearly groups the capacitors into two series pairs first, so simply summing all four is not correct.
3 μF: Reducing each series pair first gives the upper branch as \( \frac{3 \times 6}{3+6} = 2\ \mu\text{F} \) and the lower branch as \( \frac{2 \times 2}{2+2} = 1\ \mu\text{F} \). These two branch-equivalents are then in parallel across A and B, so they simply add: \( 2 + 1 = 3\ \mu\text{F} \). This is consistent with correctly treating the series and parallel groupings in the right order.
9 μF: This number would appear if the capacitor values were combined without first reducing each branch through the series formula, for example by adding raw values from only part of the circuit.
12 μF: This would result from treating both the 3-6 and 2-2 pairs as if they were in parallel rather than in series, which contradicts how the branches are actually wired.
Reducing each series branch correctly before combining the branches in parallel is the only approach consistent with the circuit's actual wiring, and it gives an effective capacitance of 3 μF.
