Question

How do you arrange four equal capacitors of 4 µF to get effective capacity of 3 µF?

• A. three in series, one in parallel
• B. two in parallel, two in series
• C. three in parallel, one in series
• D. all four in series

Answer 1

Answer: (C)

three in parallel, one in series

Answer 2

Solution:

1. Step 1: Form the Parallel Combination

   Connect three 4 µF capacitors in parallel. Their capacitances add:

   $$C_{\text{parallel}} = 4 + 4 + 4 = 12\,\mu\text{F}$$

2. Step 2: Form the Series Combination

   Now connect this 12 µF combination in series with the remaining 4 µF capacitor. For two capacitors in series:

   $$\frac{1}{C_{\text{eq}}} = \frac{1}{12} + \frac{1}{4} = \frac{1}{12} + \frac{3}{12} = \frac{4}{12} = \frac{1}{3}$$

   Thus,

   $$C_{\text{eq}} = 3\,\mu\text{F}$$

Quick Explanation:

Parallel combination of three 4 µF gives 12 µF; series with another 4 µF results in an equivalent capacitance of 3 µF.