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In this question, V1 is a dependent voltage source. enter image description here

The question asks us to calculate i(t) and ix(t). Assume that \$ i(0)= 10 A \$ for the inductor. I decided to make mesh analysis.

enter image description here

Mesh 1: $$0.5\frac{di_1}{dt} + 2(i_1-i_2)=0$$

$$\frac{di_1}{dt} = 4i_2 - 4i_1$$

Mesh 2:
$$2(i_2-i_1) - 3i + 4i_2 = 0 $$ …and at this point I am stuck.

What should I do for \$i\$ value? In the solution, \$i=-i_1\$ was assumed and \$i_2\$ became \$\frac{5}{6}i_1\$. But why do we get \$i_1\$ instead of \$i\$? Could you help me?

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    \$\begingroup\$ Suppose i is zero. Then the dependent voltage source's voltage difference is also zero. And everything is nice and stable and solved. Not so? \$\endgroup\$
    – jonk
    Apr 5 at 4:37
  • \$\begingroup\$ When I suppose \$ i \$ is zero, the answer becomes $$ i(t)= i(0).e^(-8t/3) $$ but the answer is $$ i(t)= i(0).e^(-2t/3) $$ in the solution. @jonk \$\endgroup\$
    – Mando
    Apr 5 at 11:55
  • \$\begingroup\$ It wasn't clear to me that you have to deal with non-zero initial conditions for the inductor. I was pressing you to improve the question and write something about that fact. \$\endgroup\$
    – jonk
    Apr 5 at 17:27

1 Answer 1

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Well, KCL we can see that:

$$\text{i}_\text{source}\left(t\right)=\text{i}_{\text{R}_1}\left(t\right)=\text{i}_{\text{R}_2}\left(t\right)+\text{i}_\text{L}\left(t\right)\tag1$$

Using Laplace transform we can see that:

  • $$\text{V}_\text{L}\left(\text{s}\right)=\text{sL}\cdot\text{I}_\text{L}\left(\text{s}\right)\tag2$$
  • $$\text{V}_\text{source}\left(\text{s}\right)-\text{V}_{\text{R}_1}\left(\text{s}\right)=\text{R}_1\cdot\text{I}_\text{source}\left(\text{s}\right)=\text{R}_1\cdot\text{I}_{\text{R}_1}\left(\text{s}\right)\tag3$$
  • $$\text{V}_{\text{R}_2}\left(\text{s}\right)=\text{R}_2\cdot\text{I}_{\text{R}_2}\left(\text{s}\right)\tag4$$

Using Laplace transform on \$(1)\$ and using the fact that \$\text{V}_x\left(\text{s}\right):=\text{V}_\text{L}\left(\text{s}\right)=\text{V}_{\text{R}_1}\left(\text{s}\right)=\text{V}_{\text{R}_2}\left(\text{s}\right)\$:

$$\frac{\text{V}_\text{source}\left(\text{s}\right)-\text{V}_x\left(\text{s}\right)}{\text{R}_1}=\frac{\text{V}_x\left(\text{s}\right)}{\text{R}_2}+\frac{\text{V}_x\left(\text{s}\right)}{\text{sL}}\tag5$$

Now, we know that \$\text{V}_\text{source}\left(\text{s}\right)\$ is given by:

$$\text{V}_\text{source}\left(\text{s}\right)=\text{n}\cdot\text{I}_\text{L}\left(\text{s}\right)=\text{n}\cdot\frac{\text{V}_\text{L}\left(\text{s}\right)}{\text{sL}}=\text{n}\cdot\frac{\text{V}_x\left(\text{s}\right)}{\text{sL}}\tag6$$

So, we end up with:

$$\frac{\text{n}\cdot\frac{\text{V}_x\left(\text{s}\right)}{\text{sL}}-\text{V}_x\left(\text{s}\right)}{\text{R}_1}=\frac{\text{V}_x\left(\text{s}\right)}{\text{R}_2}+\frac{\text{V}_x\left(\text{s}\right)}{\text{sL}}\tag7$$

Solving for \$\text{v}_x\left(t\right)\$, gives:

$$\text{v}_x\left(t\right)=0\tag8$$

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