1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook