# Find $v_0(t), t>0$

In the circuit given below: $$v_s(t)=4e^{-2t}u(t), i(0)=1, v_0(0)=2$$ Find $$v_0(t), t>0$$

I assumed the current through capacitor is $$\i_1\$$. By KVL in first loop we have: $$v_s(t)-2(i(t)+i_1(t))-\frac{di}{dt}=0$$ and taking laplace transforms we get $$V_s(s)-2(I(s)+I_1(s))-sI(s)+i(0)=0$$

and for second loop we have: $$2I(s)-\frac{1}{s}I_1(s)+sI(s)=0$$

Also $$\V_s(s)=\frac{4}{s+2}\$$

Eliminating $$\I(s)\$$ we get: $$I_{1}(s)=\frac{s(s+6)}{(s+2)(2 s+1)}$$

$$\\implies\$$

$$V_{0}(s)=\frac{1}{s} I_{1}(s)=\frac{s+6}{(s+2)(2s+1)}$$

Taking inverse laplace transform we get: $$v_0(t)=\left(\frac{-4}{3} e^{-t}+\frac{11}{3} e^{-t / 2}\right) u(t).$$

But the answer given is different. Where i went wrong?

Well, we have the following circuit:

simulate this circuit – Schematic created using CircuitLab

Using KCL, we can see that:

$$\text{i}_1\left(t\right)=\text{i}_2\left(t\right)+\text{i}_3\left(t\right)\tag1$$

Using the voltage-current relations in the elements we can write:

• $$\text{v}_\text{i}\left(t\right)-\text{v}_1\left(t\right)=\text{i}_1\left(t\right)\cdot\text{R}\tag2$$
• $$\text{v}_1\left(t\right)=\text{i}_2'\left(t\right)\cdot\text{L}\tag3$$
• $$\text{i}_3\left(t\right)=\text{v}_2'\left(t\right)\cdot\text{C}\tag4$$

We can also see that:

$$\text{v}_2\left(t\right)-\text{v}_1\left(t\right)=\text{n}\cdot\text{i}_2\left(t\right)\tag5$$

So, taking the derivative of equation $$\(1)\$$ we get:

$$\text{i}_1'\left(t\right)=\text{i}_2'\left(t\right)+\text{i}_3'\left(t\right)\tag6$$

Using $$\(2),(3)\$$ and $$\(4)\$$, we can rewrite $$\(6)\$$ as follows:

$$\frac{\text{v}_\text{i}'\left(t\right)-\text{v}_1'\left(t\right)}{\text{R}}=\frac{\text{v}_1\left(t\right)}{\text{L}}+\text{v}_2'\left(t\right)\cdot\text{C}\tag7$$

Taking the derivative of equation $$\(5)\$$ we get:

$$\text{v}_2'\left(t\right)-\text{v}_1'\left(t\right)=\text{n}\cdot\text{i}_2'\left(t\right)\tag8$$

Using $$\(3)\$$, we can rewrite $$\(8)\$$ as follows:

$$\text{v}_2'\left(t\right)-\text{v}_1'\left(t\right)=\text{n}\cdot\frac{\text{v}_1\left(t\right)}{\text{L}}\tag9$$

So, we need to solve the following set of DE:

$$\begin{cases} \frac{\text{v}_\text{i}'\left(t\right)-\text{v}_1'\left(t\right)}{\text{R}}=\frac{\text{v}_1\left(t\right)}{\text{L}}+\text{v}_2'\left(t\right)\cdot\text{C}\\ \\ \text{v}_2'\left(t\right)-\text{v}_1'\left(t\right)=\text{n}\cdot\frac{\text{v}_1\left(t\right)}{\text{L}} \end{cases}\tag{10}$$

Using that, you'll find:

$$\text{v}_2\left(t\right)=\frac{2}{3} e^{-2 t} \left(5 e^{\frac{3 t}{2}}-2\right)\tag{11}$$