I have encountered this problem.

A coil A of 400 turns and another coil B of 500 turns lie near each other so that 20% of the flux produced in A links with B. It is found that a current of 5 A in coil A produces a flux of 0.33 Wb, while the same current in coil B produces a flux of 0.15 Wb. Determine the coefficient of coupling between the coils.

The solution given to me had the coefficient of coupling being different as it had to be computed. L1, L2 and M were all computed first and then solving for k.

I have not seen any source clear up this confusion. This is because an alternative formula for k is the ratio between the number of flux linkages between A and B over the flux linkage of A. So what is supposed to be the difference?

  • \$\begingroup\$ "being different" - being different to what? \$\endgroup\$
    – Andy aka
    Sep 6, 2020 at 10:54
  • 1
    \$\begingroup\$ The first thing to note is that more ampere-turns in B produces less flux : that tells you something about coil B compared to A. The definition of Weber may help that. \$\endgroup\$
    – user16324
    Sep 6, 2020 at 10:55
  • \$\begingroup\$ One of the definitions of the coefficient of coupling is that it is the flux linkages common to both two coils over the flux linkage in a single coil. I cannot clearly distinguish how that is different from the concept in a problem where they give a percentage of the flux linkages like 20% and then ask to compute for a coefficient of coupling like that 20% is not the answer already. \$\endgroup\$
    – AndroidV11
    Sep 6, 2020 at 11:44

1 Answer 1


The ratio of mutual flux to primary flux is the same as the coefficient of coupling. However, the flux linkage of the secondary is the mutual flux multiplied by the number of turns of the secondary. Therefore the flux linkage ratio is not the same as the coefficient of coupling and may have a value greater than 1 unlike the coefficient of coupling which is always less than 1 and greater than 0. Lets prove that out from some known laws.

So from Faraday's law we can derive the following relationship between the mutual inductance and the mutual flux:

\$ L_m = N_s \cdot \frac{\delta \phi_m}{\delta i_p} \$

the mutual flux is the portion of the flux from the primary that overlaps with the secondary. Therefore the following is always true:

\$ \phi_m \leq \phi_p\$

We also know flux as a function of current can be derived from the following equation:

\$ L = \frac{\phi \cdot N}{i} \$

Solving for the flux we get:

\$ \phi = \frac{L \cdot i}{N}\$

Lets also define the ratio of flux linkage with the secondary coil to primary coils flux:

\$ R_\lambda = \frac{\lambda_s}{\phi_p} \$

Solving for the flux linkage and substituting the primary flux with the earlier equation we get:

\$ \lambda_s = R_\lambda \cdot \phi_p \$

\$ \lambda_s = R_\lambda \cdot \frac{L_p \cdot i_p}{N_p} \$

\$ \lambda_s = \frac{R_\lambda \cdot L_p}{N_p} \cdot i_p \$

It is important to note here that the flux linkage, \$\lambda_s\$, is the mutual flux multiplied by the number of turns of the secondary winding:

\$\lambda_s = N_s \cdot \phi_m\$


\$\phi_m = \frac{\lambda_s}{N_s}\$

Since we know that flux linkage of the secondary is the portion of the primaries flux which is mutual multiplied by the number of turns in the secondary, then it stands to reason the ratio between the flux linkage and the primary flux can be any positive value, including values greater than 1.

Next we can take the derivative of our first equation by substitution:

\$ L_m = N_s \cdot \frac{\delta}{\delta i_p} (\frac{\lambda_s}{N_s}) \$

\$ L_m = N_s \cdot \frac{\delta}{\delta i_p} (\frac{\frac{R_\lambda \cdot L_p}{N_p} \cdot i_p}{N_s}) \$

\$ L_m = N_s \cdot \frac{\delta}{\delta i_p} (\frac{R_\lambda \cdot L_p}{N_p \cdot N_s} \cdot i_p) \$

\$ L_m = N_s \cdot \frac{R_\lambda \cdot L_p}{N_p \cdot N_s} \$

\$ L_m = \frac{R_\lambda \cdot L_p}{N_p} \$

We also know the relationship between mutual inductance and the coefficient of coupling with this equation:

\$ L_m = k \cdot \sqrt{L_p \cdot L_s} \$

We can now substitute this in to get an equation showing the relationship between the flux link ratio and the coefficient of coupling:

\$ k \cdot \sqrt{L_p \cdot L_s} = \frac{R_\lambda \cdot L_p}{N_p} \$

Now we can solve for the flux linking ratio and simplify:

\$ k \cdot \sqrt{L_p \cdot L_s} = R_\lambda \cdot \frac{L_p}{N_p} \$

\$ k \cdot \sqrt{L_p \cdot L_s} \cdot \frac{N_p}{L_p} = R_\lambda\$

\$ k \cdot \frac{N_p \cdot \sqrt{L_p \cdot L_s}}{L_p} = R_\lambda\$

Since all of the constants here are positive values we can further reduce this to:

\$ k \cdot N_p \cdot \sqrt{\frac{L_s}{L_p}} = R_\lambda\$

It should now be clear that the flux link percentage (\$R_\lambda\$) is not the same value as the Coefficient of Coupling (\$k\$).

While most of the derivations above I did myself you can find most of the equations I pulled in as laws (Like Faraday's Law in the first equation) in this wonderful online book that is worth taking a read for a deeper explanation of the subject.

  • \$\begingroup\$ Nice answer +1,name of that book ? \$\endgroup\$
    – user215805
    Sep 6, 2020 at 14:13
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    \$\begingroup\$ @user215805 Introductory Circuit Analysis by Robert L. Boyelstad. You can find the complete book as a collection of PDFs here: drb-m.org/eb1 \$\endgroup\$ Sep 6, 2020 at 14:15

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