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Coefficient of coupling main simplified notes
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MCoefficient of coupling (k= L1 L2 )
• k, ratio of mutual inductance present in between the
two coils to the maximum possible value. (It shows
the extent at which two coils are coupled together)
• How to derive the value of k:
2
• Recall that self inductance, L =𝑙/𝜇𝑁𝜇
0 𝑟𝐴
• Therefore:
• two magnetically-coupled coils having N1 and N2 turns respectively
have the following individual coefficients of self-induction:
𝑁12 𝑁 22
L1 =
𝑙/𝜇0 𝜇𝑟𝐴
, L2= 𝑙/𝜇0 𝜇𝑟𝐴
…………………(i)
,• let ∅1 be flux produced by I1 amps, in coil 1:
𝐹
∅1 = R ,=𝑙/𝜇𝑁 𝜇𝐼 ֜
1 1
∅1
𝐼1
= 𝑙/𝜇𝑁𝜇
1 ……………………..(i)
0 𝑟𝐴 0 𝑟𝐴
• If a fraction (k1) of ∅1 links with coil 2, then:
∅ 𝑁1 𝑁1𝑁2 𝜇𝑟 𝜇0 𝐴𝑁2𝑁1
• M = 𝑘1 ∅𝐼1𝑁2 = 𝑘1 1 𝑁2 = 𝑘1 𝑁2 = 𝑘1 = 𝑘1 ( )
1 𝐼1 𝑙/𝜇0 𝜇𝑟𝐴 𝑙/𝜇0 𝜇𝑟𝐴 𝑙
……………………………..(ii)
Where 𝑘1 ≤ 1
• let ∅2 be flux produced by I2 amps, in coil 2:
∅2 =R𝐹 =𝑙/𝜇𝑁 𝜇𝐼
2 2 ֜ ∅𝐼 2 = 𝑙/𝜇𝑁𝜇 ……………………..(iii)
2
0 𝑟𝐴 2 0 𝑟𝐴
, • If a fraction (k2) of ∅2 links with coil 1, then:
𝑘 ∅ 𝑁
M= 2 2 1,
𝐼2
M =𝑘2 𝑙/𝜇𝑁1𝑁𝜇2 ………………(iv)
0 𝑟𝐴
Where𝑘2 ≤ 1
• Multiply (ii) and (iv) gives:
2
𝑁12 𝑁2
1 𝑘2 = 𝑘1 𝑘2L1L2
M 2=𝑘 𝑥
𝑙/𝜇0 𝜇𝑟𝐴 𝑙/𝜇0 𝜇𝑟𝐴
therefore:
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