# Deriving currents of parallel mutual inductors

This is possibly just a math question, not sure.

In Irwin's textbook "Basic Engineering Circuit Analysis," 7e, there's a quick derivation for current in parallel inductors I don't follow:

The KVL equations are fine and make sense but the solutions for I1 and I2 don't. I generate a half page of algebra (when trying with substitution and subtraction of the linear systems of equations for V) and don't get there and Wolfram Alpha also doesn't come up with that simple of an answer. Is there a simple algebraic answer here?

• I believe that's figure (e) on the bottom right? Look at that circuit instead of (d). It's the same circuit just drawn differently.
– user103380
Commented Oct 9, 2018 at 4:06

There are a few ways to solve this, but I guess to me the most natural one for me would be:

\begin{align} j\omega L_1I_1 + j\omega MI_2 &= j\omega MI_1 + j\omega L_2I_2 \\ &\Downarrow \\ L_1I_1 + MI_2 &= MI_1 + L_2I_2 \\ &\Downarrow \\ (L_1 - M)I_1 &= (L_2 - M)I_2 \end{align}

So in order to find the relationship between $$\V\$$ and $$\I_1\$$, you use the substitution

\begin{align} V &= j\omega L_1I_1 + j\omega MI_2\\ &\Downarrow\\ V &= j\omega L_1I_1 + j\omega M\cdot \frac{L_1 - M}{L_2 - M}I_1 \\ &= j\omega I_1 \left(L_1 + M\frac{L_1 - M}{L_2 - M} \right) \\ &= j\omega I_1 \left( \frac{L_1(L_2 - M) + M(L_1 - M)}{L_2 - M} \right) \\ &= j\omega I_1 \left(\frac{L_1L_2 - M^2}{L_2 - M}\right) \end{align}

You can do similar steps to find the relationship $$\V\$$ and $$\I_2\$$.

• I saw the step, I was trying to find a common denominator at the wrong stage and going in loops. Thanks! Marked as answer. Commented Oct 9, 2018 at 20:46

What if the coupling is perfect? Then you have $$M=\sqrt{L_1 L_2}$$ and $$j\omega I_1 \left( \frac{L_1 L_2 - L_1 L_2}{L_2-\sqrt{L_1 L_2}} \right) = 0$$ for all $$L_1 \ne L_2$$