Solution:
Connecting two \(2\text{ }\mu\text{F}\) capacitors in series gives \(1\text{ }\mu\text{F}\). Adding the third \(2\text{ }\mu\text{F}\) capacitor in parallel to this combination yields \(1 + 2 = 3\text{ }\mu\text{F}\).
3 capacitors each of capacitance \(2\text{ }\mu\text{F}\) are to be connected to obtain an equivalent capacitance of \(3\text{ }\mu\text{F}\). Which of the following combination is possible?
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