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I used the device Variable Inductor Type 107N to built those circuits:

enter image description here

I have been told that:

$$M=\frac{|scale-308|}{2} mH$$ Where M is the Mutual Inductance, and scale is a scale on the devic.

Those are the result i got for a constant E with frequency of 1K Hz:

for the first circuit:

enter image description here

for the second one:

enter image description here

I have no idea how to explain the result of I, especially because M is symmetric around scale of 308.

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Well, let's do first some calculations:

$$\text{V}_{\space\text{in}}\left(t\right)=\text{I}_{\space\text{in}}'\left(t\right)\cdot\left(\text{L}_1+\text{L}_2\right)\tag1$$

Now, we know that:

  • $$\text{V}_{\space\text{in}}\left(t\right)=\text{V}\cdot\sin\left(2\pi\cdot1000\cdot t\right)\tag2$$
  • $$\text{L}_1+\text{L}_2=\frac{1}{1000}\cdot\frac{\left|\text{scale}-308\right|}{2}\tag3$$
  • Assuming that: $$\text{I}_{\space\text{in}}\left(0\right)=0\tag4$$

So, we can write:

$$ \begin{cases} \text{V}\cdot\sin\left(2\pi\cdot1000\cdot t\right)=\text{I}_{\space\text{in}}'\left(t\right)\cdot\frac{1}{1000}\cdot\frac{\left|\text{scale}-308\right|}{2}\\ \\ \text{I}_{\space\text{in}}\left(0\right)=0 \end{cases}\tag5 $$

Solving \$\left(5\right)\$ gives:

$$\text{I}_{\space\text{in}}\left(t\right)=\frac{2}{\pi}\cdot\text{V}\cdot\frac{\sin^2\left(1000\pi\cdot t\right)}{\left|\text{scale}-308\right|}\tag6$$

Now,in order to find your values of \$\text{I}\$ (that you put in the table), we need to find:

$$\overline{\text{I}}:=\lim_{\text{n}\to\infty}\sqrt{\frac{1}{\text{n}}\int_0^\text{n}\text{I}_{\space\text{in}}^2\left(t\right)\space\text{d}t}=\frac{\sqrt{\frac{3}{2}}}{\pi}\cdot\frac{\left|\text{V}\right|}{\left|\text{scale}-308\right|}\tag7$$

Let's try some values:

  1. \$\text{V}=0.817\$ and \$\text{scale}=100\$: $$\overline{\text{I}}=\frac{\sqrt{\frac{3}{2}}}{\pi}\cdot\frac{\left|0.817\right|}{\left|100-308\right|}\approx0.00153128\tag8$$
  2. \$\text{V}=0.85\$ and \$\text{scale}=200\$: $$\overline{\text{I}}=\frac{\sqrt{\frac{3}{2}}}{\pi}\cdot\frac{\left|0.85\right|}{\left|200-308\right|}\approx0.00306825\tag9$$
  3. \$\text{V}=1.76\$ and \$\text{scale}=100\$: $$\overline{\text{I}}=\frac{\sqrt{\frac{3}{2}}}{\pi}\cdot\frac{\left|1.76\right|}{\left|100-308\right|}\approx0.00329872\tag{10}$$
  4. \$\text{V}=1.76\$ and \$\text{scale}=500\$: $$\overline{\text{I}}=\frac{\sqrt{\frac{3}{2}}}{\pi}\cdot\frac{\left|1.76\right|}{\left|500-308\right|}\approx0.00357361\tag{11}$$
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