Problem finding voltage in a circuit with mutual inductance

I'm trying to solve an example from Hayt's Engineering Circuit Analysis:

let $$\i_1=4i_2= 20\cos(500t-20°) \mathrm{\ mA}\$$. Determine $$\v_1(0)\$$.

Using $$\k=\frac{M}{\sqrt{L_1L_2}}\$$, I got $$\M=0.6\ \mathrm{H}\$$, which is the same value obtained by the authors. Now, if the equation for $$\v_1\$$ is $$v_1(t)=L_1\frac{di_1}{dt}+M\frac{di_2}{dt}$$

For the values of $$\di_1/dt\$$ and $$\di_2/dt\$$ I got, $$\frac{di_1}{dt}= -10\sin -20° A$$ $$\frac{di_2}{dt}= -40\sin -20° A$$ There's no problem in $$\di_1/dt\$$, got the same expression from the book, just a derivative. But for $$\di_2/dt\$$ I get a different value.

The expression shown in the book for $$\v_1(0)\$$ is, $$v_1(0)=0.4[-10\sin -20°]+0.6[-2.5\sin -20°]=1.881 \ \mathrm{V}$$

So, the only problem here is that $$\2.5\$$. And the only place I see that $$\2.5\$$ could come from is $$\L_2=2.5 \ \mathrm{H}\$$, but I can't see how or why.

• The current in L1 is 4 times the current in L2, so the magnitude of the current in L2 must be 4 times smaller. 10/4 = 2.5. Not so? So $\frac{\text{d}}{\text{d} t}i_2=\frac14 \frac{\text{d}}{\text{d} t}i_1$?
– jonk
Commented Sep 23, 2022 at 7:22
• Your formula for $k$ is incorrect. You forgot the square root. Commented Sep 23, 2022 at 8:05
• @Andyaka just a misstype, the value was calculated with the sqrt. Commented Sep 23, 2022 at 21:23
• @jonk thank you. I think that was the problem. Commented Sep 23, 2022 at 21:28
– jonk
Commented Sep 24, 2022 at 0:52

There's no problem in $$\di_1/dt\$$, got the same expression from the book, just a derivative. But for $$\di_2/dt\$$ I get a different value.

There does seem to be an error in the solution you posted as

is given.

so $$\di_1/dt = 4 di_2/dt\$$ , they have it the other way around.

Well, when we know that:

1. $$\text{I}_1\left(t\right)=\hat{\text{u}}\cos\left(\omega t+\varphi\right)\tag1$$
2. $$\text{I}_2\left(t\right)=\text{n}\hat{\text{u}}\cos\left(\omega t+\varphi\right)\tag2$$

We also know that:

1. $$\text{I}_1'\left(t\right)=-\omega\hat{\text{u}}\sin\left(\omega t+\varphi\right)\tag3$$
2. $$\text{I}_2'\left(t\right)=-\omega\text{n}\hat{\text{u}}\sin\left(\omega t+\varphi\right)\tag4$$

And we have:

$$\text{k}=\frac{\text{m}}{\sqrt{\text{L}_1\text{L}_2}}\space\Longleftrightarrow\space\text{m}=\text{k}\sqrt{\text{L}_1\text{L}_2}\tag5$$

So, we get:

$$$$\begin{split} \text{V}_1\left(t\right)&=\text{L}_1\cdot\text{I}_1'\left(t\right)+\text{m}\cdot\text{I}_2'\left(t\right)\\ \\ &=\text{L}_1\left(-\omega\hat{\text{u}}\sin\left(\omega t+\varphi\right)\right)+\text{k}\sqrt{\text{L}_1\text{L}_2}\cdot\left(-\omega\text{n}\hat{\text{u}}\sin\left(\omega t+\varphi\right)\right)\\ \\ &=-\omega\hat{\text{u}}\sin\left(\omega t+\varphi\right)\left(\text{L}_1+\text{n}\text{k}\sqrt{\text{L}_1\text{L}_2}\right) \end{split}\tag6$$$$

So, when $$\t=0\$$ we get:

$$\text{V}_1\left(0\right)=-\omega\hat{\text{u}}\sin\left(\varphi\right)\left(\text{L}_1+\text{n}\text{k}\sqrt{\text{L}_1\text{L}_2}\right)\tag7$$

• The question is 'where did the factor of 2.5 come from`? The OP already has the solution in the desired form. Commented Sep 26, 2022 at 17:55