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Question
The primary and secondary coils of a transformer each have an inductance of 200 x 10-6 H. The mutual inductance between the windings is 4 x 10-6 H. What percentage of the flux from one coil reaches the other?
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Solution 1
Data: Lp = LS = 2 × 10-4 H, M= 4 × 10-6 H
M = K`sqrt("L"_"p""L"_"S")`
The coupling coefficient is
K = `"M"/sqrt("L"_"p""L"_"S") = (4 xx 10^-6)/sqrt((2xx10^-4)^2) = (4 xx 10^-6)/(2 xx 10^-4)` = 2 × 10-2
Therefore, the percentage of flux of the primary reaching the secondary is 0.02 × 100% = 2 %
Solution 2
Self-inductance, L = 200 × 10-6 H
Mutual inductance, M = 4 × 10-6 H
Percentage of flux transfer = `"M"/"L" xx 100`
= `(4 xx 10^-6)/(200 xx 10^-6) xx 100`
= 2%
2 % of the flux from one coil reaches the other.
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