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Asked in GUJCET 2021 · Series and parallel combinations
Given: three capacitors, each C = 3 μF: two in parallel between A and the middle junction, and the third between that junction and B.
Idea: parallel capacitances add; series ones combine as 1/(Cₛ) = 1/(C₁) + 1/(C₂).
Parallel pair: 3 + 3 = 6 μF.
In series with the third: (6×3)/(6+3) = (18)/9 = 2 μF.
So C_AB = 2 μF.
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