Mutual Inductance and coupled coils

[DivGradCurl]

Homework Statement

For coupled coils in this arrangement (illustration attached), show

L_{eq} = \frac{L_1 L_2 - M^2}{L_1 + L_2 - 2M}

Homework Equations

KVL (mesh analysis)

The Attempt at a Solution

As you can see, I get to a final answer - which is not exactly the same; there's a subtle difference! Please help me find the mistake. I guess my mesh equations are not correct, because the rest of the process is straight-forward. By the way, the meshes are in the same figure (attached).

MESH 1:

-V + (I_1 - I_2) j\omega L_1 + j\omega M I_2 = 0

j\omega L_1 I_1 + (j\omega M - j\omega L_1) I_2 = V

MESH 2:

j\omega L_2 I_2 - (I_2 - I_1) j\omega M - (I_2 - I_1)j\omega L_1 + I_2 j\omega M = 0

(j\omega M + j\omega L_1) I_1 + (j\omega L_2 - j\omega L_1) I_2 = 0

So, there are 2 equations and 2 unknowns:

\underbrace{\begin{bmatrix} \displaystyle j\omega L_1 & (j\omega M - j\omega L_1) \\ \displaystyle (j\omega M + j\omega L_1) & (j\omega L_2 - j\omega L_1)\end{bmatrix}}_{\displaystyle A} \underbrace{\begin{bmatrix} \displaystyle I_1 \\ I_2 \end{bmatrix}}_{\displaystyle X} = \underbrace{\begin{bmatrix} \displaystyle V \\ 0 \end{bmatrix}}_{\displaystyle B}

I get

X = A^{-1} B = \begin{bmatrix} \displaystyle \frac{(L_1 - L_2)V}{(L_1 \cdot L_2 - M^2)\omega} j\phantom{\Bigg(} \\ \displaystyle \frac{(L_1 +M)V}{(L_1 \cdot L_2 - M^2)\omega} j \end{bmatrix}

which means

I_1 = \frac{(L_1 - L_2)V}{(L_1 \cdot L_2 - M^2)\omega} j

But, Z = V/I and L_{eq} = Z_{eq}/(j\omega). Thus, it follows.

L_{eq} = \frac{L_1 L_2 - M^2}{L_2 - L_1}

Any help is highly appreciated

 

[huckmank]

I can't see your attachment as it isn't approved yet, but I think your second mesh is a bit off. I think your polarity is backwards on (I_2-I_1)j\omega L_1, which would also change the polarity of your voltage on the corresponding mutual inductance. Try switching those and I think you should get the correct answer. You should have -2j\omega MI_2 in your second mesh equation.

Keep in mind I can't see the circuit yet, but I'm fairly sure it's two inductors in parallel with dots at the top. Sorry if I assumed wrong.

 

[DivGradCurl]

Yes, if

j\omega L_2 I_2 - (I_2 - I_1) j\omega M + (I_2 - I_1)j\omega L_1 - I_2 j\omega M = 0

then

(j\omega M - j\omega L_1) I_1 + (j\omega L_1 + j\omega L_2 - 2j\omega M) I_2 = 0

which gives the right answer!

L_{eq} = \frac{L_1 L_2 - M^2}{L_1 + L_2 - 2M}

But there is something that does not make sense. As you said, I do have two inductors in parallel with dots at the top, but the sign convention does not seem to be consistent.

Inductor L_2 is on the right, and L_1 is on the left; this is the second mesh. I assigned a clockwise current I_2 there. For L_2, it appears that you're allowing a positive sign when current enters the dotted side:

j\omega L_2 I_2 - (I_2 - I_1) j\omega M

The converse is suggested for L_1, with (+) for non-dotted and (-) for dotted.

+ (I_2 - I_1)j\omega L_1 - I_2 j\omega M

Please clarify.

 

[huckmank]

Well, on the left we have the current (I_2-I_1) leaving the dot. The dot convention states that if this is the case, we place a - at the opposite dot for the mutual inductance. On the right, we have the current I_2entering the dot. The dot convention states that if this is the case we place a + at the opposite dot. So, using passive sign convention, for both inductors we have + facing into the current and for both mutual inductances we have - facing into the current.

So our equation should be +(I_2-I_1)j\omega L_1-I_2j\omega M+I_2j\omega L_2-(I_2-I_1)j\omega M=0

I guess to answer your question succinctly, the reason the left dot is positive, is because I_2 is entering the right dot. And the reason the right dot is negative is because the current (I_2-I_1) is leaving the left dot.

I hope this wasn't too convoluted an explanation.

 

[DivGradCurl]

Thanks for the excellent explanation. I was confused with the dot notation, but now it's clear.