Is a unit coupling coefficient for mutual inductance (so $M=\sqrt{L_1L_2}$) compatible with different self-inductances $L_1\neq L_2$?

Question

We considered mutual inductance in my lectures and the definition $M=k\sqrt{L_1L_2}$ where the coupling constant $k$ can vary between 0 and 1. When covering transformers, my lecture notes "assume that all of the flux couples to both coils, such that the coupling coefficient is equal to unity". However I am not sure how this is the case, unless also the self inductance of each coil is the same (which certainly is not the case in the rest of my notes!).

In particular, the flux through each coil is

$$\phi_1 = L_1I_1 + MI_2 = L_1I_1 + k\sqrt{L_1L_2}I_2$$

and

$$\phi_2 = L_2I_2 + k\sqrt{L_1L_2}I_1$$

Equating these and rearranging, I find that

$$\sqrt{\frac{L_1}{L_2}}I_1 + \sqrt{\frac{L_2}{L_1}}I_2 = (I_1-I_2)k$$

so $$k=1 \iff L_1=L_2$$

Have I misunderstood what it means for 'all of the flux to couple to both coils'?

Answer 1

You are making two errors. First, you cannot equate $\phi_1$ and $\phi_2$ because those are the flux linkages not the flux in the core (the flux linkage is the flux in the core multiplied by the turns number of a coil); second, even assuming that $\phi_1$ and $\phi_2$ can be equated, your conclusion would be incorrect.

So, let's start with the right equations and see what we can obtain.

Let $\lambda_1=N_1\phi$ and $\lambda_2=N_2\phi$ be the flux linkages, $\phi$ being the flux in the core, and $N_1$ and $N_2$ the turns numbers of the two windings.

Then,

$$\lambda_1=N_1\phi = L_1i_1 + k\sqrt{L_1L_2}i_2$$

and

$$\lambda_2=N_2\phi = L_2i_2 + k\sqrt{L_1L_2}i_1.$$

By equating $\phi$ from the equations above and setting $k=1$ we get

$$\frac{L_1i_1 + \sqrt{L_1L_2}i_2}{N_1} = \frac{L_2i_2 + \sqrt{L_1L_2}i_1}{N_2}$$

from which

$$\frac{L_1}{N_1}\left(i_1+\sqrt{\frac{L_2}{L_1}}i_2\right) = \frac{\sqrt{L_1L_2}}{N_2}\left(i_1+\sqrt{\frac{L_2}{L_1}}i_2\right).$$

Cancellation of the two equal terms on both sides yields

$$\sqrt{\frac{L_2}{L_1}}=\frac{N_2}{N_1}.$$

That is, when $k=1$, the ratio of the two inductances equals the turns ratio squared.

Furthermore, recalling that the voltages across the windings are the derivatives of the flux linkages, with $k=1$ we get

$$v_1 = \frac{\mathrm{d}\lambda_1}{\mathrm{d}t} = L_1 \frac{\mathrm{d}}{\mathrm{d}t}\left(i_1+\sqrt{\frac{L_2}{L_1}}i_2\right)$$

and

$$v_2 = \frac{\mathrm{d}\lambda_2}{\mathrm{d}t} = \sqrt{L_1L_2} \frac{\mathrm{d}}{\mathrm{d}t}\left(i_1+\sqrt{\frac{L_2}{L_1}}i_2\right).$$

Taking the ratio of the last two equations yields

$$\frac{v_2}{v_1} = \sqrt{\frac{L_2}{L_1}}=\frac{N_2}{N_1},$$

which is the well-known formula relating the voltage ratio in an ideal transformer.

Finally, take the equation

$$v_1 = L_1 \frac{\mathrm{d}}{\mathrm{d}t}\left(i_1+\sqrt{\frac{L_2}{L_1}}i_2\right)$$

and divide both sides by $L_1$ getting

$$\frac{v_1}{L_1} = \frac{\mathrm{d}}{\mathrm{d}t}\left(i_1+\sqrt{\frac{L_2}{L_1}}i_2\right).$$

Now let $L_1$ tend to infinity; the above equation yields

$$\frac{\mathrm{d}}{\mathrm{d}t}\left(i_1+\sqrt{\frac{L_2}{L_1}}i_2\right) = 0$$

or

$$i_1+\sqrt{\frac{L_2}{L_1}}i_2=\text{const.}$$

If this constant is initially zero (zero initial flux linkage), the above equation reduces to

$$i_1 = -\sqrt{\frac{L_2}{L_1}}i_2 = -\frac{N_2}{N_1}i_2,$$

which is the second constitutive equation of the ideal transformer.

The answer to your titular question is thus: yes, a unit coupling coefficient is compatible with different self-inductances; as you can see from the derivations above, the consequences of $k=1$ are that i) the ratio of the two inductances equals the turns ratio squared and that ii) the voltage ratio equals the turns ratio.