Dependent voltage source in RL circuit

Question

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?

Answer 1

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:

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