Electronic – RLC Circuits at t=0 and infinity

Since the R and L are in parallel to begin with the voltage across them is going to be the same

1. start by taking the voltage across the parallel branch as your reference.
2. Ir is going to be in phase with your reference and Il is going to lag it by 90 degrees.
3. Then the capacitor is in series with the parallel branch to vectorially add Ir and Il. That will give you Ic.
4. Vc will simply lag Ic by 90 degrees.

That is the general procedure, you can find the Magnitudes by simply applying ohm's law.

Well, according to 'Faraday's law' in a series RLC-circuit:

$$\text{V}_{\space\text{C}}\left(t\right)+0+\text{V}_{\space\text{R}}\left(t\right)-\text{V}_{\space\text{in}}\left(t\right)=-\text{V}_{\space\text{L}}\left(t\right)\tag1$$

Now, we know a few things:

• $$\text{I}_{\space\text{C}}\left(t\right)=\text{I}_{\space\text{in}}\left(t\right)=\text{V}_{\space\text{C}}'\left(t\right)\cdot\text{C}\tag2$$
• $$\text{V}_{\space\text{R}}\left(t\right)=\text{I}_{\space\text{R}}\left(t\right)\cdot\text{R}=\text{I}_{\space\text{in}}\left(t\right)\cdot\text{R}\tag3$$
• $$\text{V}_{\space\text{L}}\left(t\right)=\text{I}_{\space\text{L}}'\left(t\right)\cdot\text{L}=\text{I}_{\space\text{in}}'\left(t\right)\cdot\text{L}\tag4$$

So, we get:

$$\text{V}_{\space\text{C}}'\left(t\right)+0+\text{V}_{\space\text{R}}'\left(t\right)-\text{V}_{\space\text{in}}'\left(t\right)=-\text{V}_{\space\text{L}}'\left(t\right)\space\Longleftrightarrow\space$$ $$\text{I}_{\space\text{in}}\left(t\right)\cdot\frac{1}{\text{C}}+0+\text{I}_{\space\text{in}}'\left(t\right)\cdot\text{R}-\text{V}_{\space\text{in}}'\left(t\right)=-\text{I}_{\space\text{in}}''\left(t\right)\cdot\text{L}\space\Longleftrightarrow\space$$ $$\text{V}_{\space\text{in}}'\left(t\right)=\text{I}_{\space\text{in}}\left(t\right)\cdot\frac{1}{\text{C}}+\text{I}_{\space\text{in}}'\left(t\right)\cdot\text{R}+\text{I}_{\space\text{in}}''\left(t\right)\cdot\text{L}\space\Longleftrightarrow\space$$ $$\text{V}_{\space\text{in}}'\left(t\right)=\text{I}_{\space\text{in}}''\left(t\right)\cdot\text{L}+\text{I}_{\space\text{in}}'\left(t\right)\cdot\text{R}+\text{I}_{\space\text{in}}\left(t\right)\cdot\frac{1}{\text{C}}\tag5$$

Assuming that the initial conditions are equal to \$0\$, and using Laplace transform:

$$\text{s}\cdot\text{v}_{\space\text{in}}\left(\text{s}\right)=\text{s}^2\cdot\text{i}_{\space\text{in}}\left(\text{s}\right)\cdot\text{L}+\text{s}\cdot\text{i}_{\space\text{in}}\left(\text{s}\right)\cdot\text{R}+\text{i}_{\space\text{in}}\left(t\right)\cdot\frac{1}{\text{C}}\space\Longleftrightarrow\space$$ $$\text{i}_{\space\text{in}}\left(\text{s}\right)=\frac{\text{s}\cdot\text{v}_{\space\text{in}}\left(\text{s}\right)}{\text{L}\cdot\text{s}^2+\text{s}\cdot\text{R}+\frac{1}{\text{C}}}\tag6$$

Using the 'Convolution Theorem' of the Laplace transform:

$$\text{I}_{\space\text{in}}\left(t\right)=\int_0^t\mathscr{L}_\text{s}^{-1}\left[\text{v}_{\space\text{in}}\left(\text{s}\right)\right]_{\left(\text{y}\right)}\cdot\mathscr{L}_\text{s}^{-1}\left[\frac{\text{s}}{\text{L}\cdot\text{s}^2+\text{s}\cdot\text{R}+\frac{1}{\text{C}}}\right]_{\left(t-\text{y}\right)}\space\text{d}\text{y}=$$ $$\int_0^t\text{V}_{\space\text{in}}\left(\text{y}\right)\cdot\mathscr{L}_\text{s}^{-1}\left[\frac{\text{s}}{\text{L}\cdot\text{s}^2+\text{s}\cdot\text{R}+\frac{1}{\text{C}}}\right]_{\left(t-\text{y}\right)}\space\text{d}\text{y}\tag7$$

And \$\mathscr{L}_\text{s}^{-1}\left[\frac{\text{s}}{\text{L}\cdot\text{s}^2+\text{s}\cdot\text{R}+\frac{1}{\text{C}}}\right]_{\left(t-\text{y}\right)}\$, equals: