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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)\$.

enter image description here

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.

ocrdu
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ricardovaras_99
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    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 Sep 23 '22 at 07:22
  • Your formula for \$k\$ is incorrect. You forgot the square root. – Andy aka Sep 23 '22 at 08:05
  • @Andyaka just a misstype, the value was calculated with the sqrt. – ricardovaras_99 Sep 23 '22 at 21:23
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    @jonk thank you. I think that was the problem. – ricardovaras_99 Sep 23 '22 at 21:28
  • @ricardovaras_99 I think so, too. I'm glad I may have helped somewhat. And thanks for letting me know your thoughts about this. Appreciated. – jonk Sep 24 '22 at 00:52

2 Answers2

1

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

enter image description here

is given.

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

Tesla23
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-1

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{equation} \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 \end{equation}

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

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

Jan Eerland
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