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