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Step-by-Step Solution
Step 1: Recall the formulas for combining capacitors
• When capacitors are connected in series, the reciprocal of the equivalent capacitance $C_{eq}$ is given by:
$$
\frac{1}{C_{eq}} \;=\; \frac{1}{C_1} \;+\; \frac{1}{C_2} \;+\; \dots
$$
• When capacitors are connected in parallel, the equivalent capacitance $C_{eq}$ is the sum of all individual capacitances:
$$
C_{eq} \;=\; C_1 + C_2 + \dots
$$
Step 2: Identify the required equivalent capacitance
We need three capacitors, each of 4 μF, to combine to get an equivalent capacitance of 6 μF.
Step 3: Examine possible combinations of three identical capacitors
1. All in series (three in series):
$$
\frac{1}{C_{eq}} \;=\; \frac{1}{4} + \frac{1}{4} + \frac{1}{4} \;=\; \frac{3}{4}
\quad \Rightarrow \quad
C_{eq} \;=\; \frac{4}{3} \,\mu F
$$
This gives $ \frac{4}{3} \,\mu F $, which is not 6 μF.
2. Two in series and one in parallel:
First, find the capacitance of two in series:
$$
\frac{1}{C_s} \;=\; \frac{1}{4} + \frac{1}{4} \;=\; \frac{2}{4} \;=\; \frac{1}{2}
\quad \Rightarrow \quad
C_s \;=\; 2 \,\mu F
$$
Then put this $2\,\mu F$ in parallel with the third capacitor (4 μF):
$$
C_{eq} \;=\; C_s + 4 \;=\; 2 + 4 \;=\; 6 \,\mu F
$$
This arrangement gives the desired 6 μF.
3. All in parallel (three in parallel):
$$
C_{eq} \;=\; 4 + 4 + 4 \;=\; 12 \,\mu F
$$
This is not 6 μF.
4. Two in parallel and one in series:
First, two in parallel:
$$
C_p \;=\; 4 + 4 \;=\; 8 \,\mu F
$$
Then in series with the third capacitor (4 μF):
$$
\frac{1}{C_{eq}} \;=\; \frac{1}{8} + \frac{1}{4} \;=\; \frac{1}{8} + \frac{2}{8} \;=\; \frac{3}{8}
\quad \Rightarrow \quad
C_{eq} \;=\; \frac{8}{3} \,\mu F
$$
This is not 6 μF either.
Step 4: Conclude the correct arrangement
From the above possibilities, connecting two capacitors in series and then this combination in parallel with the third capacitor yields an effective capacitance of 6 μF, which is the desired value.
