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Mutual Inductance and coupled coils

  1. Oct 13, 2007 #1
    1. The problem statement, all variables and given/known data

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

    [tex]L_{eq} = \frac{L_1 L_2 - M^2}{L_1 + L_2 - 2M}[/tex]

    2. Relevant equations

    KVL (mesh analysis)

    3. 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:

    [tex]-V + (I_1 - I_2) j\omega L_1 + j\omega M I_2 = 0[/tex]

    [tex]j\omega L_1 I_1 + (j\omega M - j\omega L_1) I_2 = V[/tex]

    MESH 2:

    [tex]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[/tex]

    [tex](j\omega M + j\omega L_1) I_1 + (j\omega L_2 - j\omega L_1) I_2 = 0[/tex]

    So, there are 2 equations and 2 unknowns:

    [tex]\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}[/tex]

    I get

    [tex]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}[/tex]

    which means

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

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

    [tex]L_{eq} = \frac{L_1 L_2 - M^2}{L_2 - L_1}[/tex]

    Any help is highly appreciated

    Attached Files:

    Last edited: Oct 13, 2007
  2. jcsd
  3. Oct 13, 2007 #2
    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 [tex](I_2-I_1)j\omega L_1[/tex], 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 [tex]-2j\omega MI_2[/tex] 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.
  4. Oct 14, 2007 #3
    Yes, if

    [tex]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[/tex]


    [tex](j\omega M - j\omega L_1) I_1 + (j\omega L_1 + j\omega L_2 - 2j\omega M) I_2 = 0[/tex]

    which gives the right answer!

    [tex]L_{eq} = \frac{L_1 L_2 - M^2}{L_1 + L_2 - 2M}[/tex]

    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:

    [tex]j\omega L_2 I_2 - (I_2 - I_1) j\omega M[/tex]

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

    [tex]+ (I_2 - I_1)j\omega L_1 - I_2 j\omega M[/tex]

    Please clarify.
  5. Oct 14, 2007 #4
    Well, on the left we have the current [tex](I_2-I_1)[/tex] 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 [tex]I_2[/tex]entering 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 [tex]+(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[/tex]

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

    I hope this wasn't too convoluted an explanation.
  6. Oct 14, 2007 #5
    Thanks for the excellent explanation. I was confused with the dot notation, but now it's clear.
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