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Two coupled inductors of self-inductances L and 2L are connected in two different ways (Figure A and Figure B) with the indicated terminals, giving series-aiding and series-opposing combinations. The coupling coefficient is k = 1/sqrt(2). If the difference between the equivalent inductances of the two arrangements equals n*L, find n.
- 2
- 3
- 4
- 5
Correct answer: 4
Solution
For two coupled inductors in series, L_eq(aiding) = L1 + L2 + 2M and L_eq(opposing) = L1 + L2 - 2M. The difference is 4M. Here M = k*sqrt(L1*L2) = (1/sqrt(2))*sqrt(L*2L) = (1/sqrt(2))*sqrt(2)*L = L. So difference = 4M = 4L, giving n = 4.
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