# Dependent voltage source in RL circuit

In this question, V1 is a dependent voltage source.

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.

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?

• 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?
– jonk
Commented Apr 5, 2022 at 4:37
• 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 Commented Apr 5, 2022 at 11:55
• 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.
– jonk
Commented Apr 5, 2022 at 17:27

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:

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