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Homework Statement
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]
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:
[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
Attachments
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