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I know that inductors oppose the change of the current. So for two inductors with different direction in series (one in clockwise and the other in counterclockwise), the equivalent inductor is $L_{eq} = L_1 + L_2 -2M$.

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(the inductors here have the same direction, don't mind it)

But the definition of an inductor is $L=N\Phi_{B}/i$. And the magnetic fields two inductors produce are in opposite direction. So $\Phi_B = |\Phi_1 - \Phi_2|$, which means the equivalent inductor should be $L_\text{eq}=|(L_1 -M)-(L_2-M)|=|L_1 - L_2|$. It seems to contradict to the conclusion before. I am quite confused about that.

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  • $\begingroup$ Your answer seems rather strange in that If the coils are very close to one another and they both have the same dimensions, ie equivalent to one coil, the equivalent inductance would be zero! $\endgroup$
    – Farcher
    Commented Jun 5 at 9:36
  • $\begingroup$ Yeah,i think the reason is that the definition of aninductor is not perfect.A restriction like''only one direction of the coil is allowed'' should be added.If we use$\oint E\cdot ds=-\frac{d\Phi_B}{dt}$,there is only one direction of integration for the left of the equation using the right-hand rule.So when you integrate along the coils,there is always a coil with plus sign and the other with minus sign for the electric field,whose result corresponds to$\Phi_B=|\Phi_1 -\Phi_2|$ .But for two individual coils,their emf are in the same direction.So i think the definition should be more complete. $\endgroup$
    – Jarry
    Commented Jun 5 at 15:01
  • $\begingroup$ @Farcher Can you elaborate on why this is strange? If the coils are wound in opposite directions, this is what I would expect, as their magnetic fields would cancel out. $\endgroup$
    – Puk
    Commented Jun 5 at 15:28
  • $\begingroup$ @Jarry Regardless of which way the coils are wound, you need to add the fluxes ($\Phi_B=\Phi_1 + \Phi_2$), and let the sign of $M$ take care of any cancellations. The problem in your case is that the flux cancellation is already taken into account by the negative sign in front of $M$, but you also attempt to do so in defining $\Phi_B = |\Phi_1 - \Phi_2|$, leading to an incorrect result. $\endgroup$
    – Puk
    Commented Jun 5 at 15:35
  • $\begingroup$ @Jarry In the example I have suggested that $L_{\rm eq}= L_1-L_1=0$. $\endgroup$
    – Farcher
    Commented Jun 5 at 19:44

1 Answer 1

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Let $N_1\Phi_{11}$ be the flux linkage with coil 1 due to current in coil 1, and $N_1\Phi_{12}$ be the flux linkage with coil 1 due to current in coil 2 and so on. Therefore, $$N_1\Phi_{11}=L_1I_1\ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ N_1\Phi_{12}=MI_2$$ Likewise for the second coil $$N_2\Phi_{21}=MI_1\ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ N_2\Phi_{22}=L_2I_2$$ Adding together the effective inductances of the two coils, $$L=\frac{N_1(\Phi_{11}+\Phi_{12})}{I_1}+\frac{N_2(\Phi_{21}+\Phi_{22})}{I_2}=\frac{L_1I_1+MI_2}{I_1}+\frac{MI_1+L_2I_2}{I_2}.$$

Putting $I_2=-I_1$, to take care of the currents being in opposite senses, we get $$L=L_1+L_2-2M.$$ Special case: perfect coupling (all flux linked with coil 1 is linked with coil 2 and vice versa. In that case we can show that $M=\sqrt{L_1L_2}$, so our formula boils down to $L=\left(\sqrt{L_1}-\sqrt{L_2}\right)^2$. When $L_1=L_2$, then $L=0$, which is what we'd expect.

Additional check: If the coils are wound in the same sense, $I_1=I_2$ and we get $L=L_1+L_2+2M.$ With identical coils and perfect coupling this gives $L=4L_1$, just as if we'd doubled the number of turns on an inductor with no 'flux leakage'.

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  • $\begingroup$ Why$I_1=I_2$?The two inductors are in series,so i think they are the same current.I suppose 'Equivalent' means that if you put all of them in a balck box ,you can see no difference,so there is only one current. $\endgroup$
    – Jarry
    Commented Jun 5 at 14:40
  • $\begingroup$ I wrote $I_2=-I_1$ because the currents in the two coils are in opposite senses. This makes the fluxes they produce in opposite directions, according to $N_1\Phi_{11}=L_1I_1$ and so on. $\endgroup$
    – Philip Wood
    Commented Jun 5 at 15:05
  • $\begingroup$ It does make sense in that way. But what if you regard them as a whole?Current enters it and leaves it.You cannot know the details and you only know the total flux$\Phi_B$.That's what 'equivalent' means,needless to go about the details. $\endgroup$
    – Jarry
    Commented Jun 5 at 15:11
  • $\begingroup$ I'd guess that there's more than one valid way of handling signs in this case. The way I chose seems to give a sensible result! $\endgroup$
    – Philip Wood
    Commented Jun 5 at 15:20
  • $\begingroup$ I agree with you : ),so maybe the defintion of inductors is not perfect. $\endgroup$
    – Jarry
    Commented Jun 5 at 15:25

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