Confusing equivalent AC circuit (analysis)?

Question

Context

I'm doing some analysis on a dummy circuit. I've been explicitly told that the current generated by the source is a sinusoid: $$I_{0} = I_{p} cos(\omega t + \pi/2)$$. It also have been shown an equivalent form with some impedances. The problem is I cannot understand how they arrive at this equivalent. I explain why after the diagram

Circuit:

Analysis

The problem starts when I try to figure out the overall impedance of the circuit so that I can find the overall voltage across the circuit. I compute the impedance ($Z_{0}$) for a capacitor in parallel with a series of components. So in other words:

$$Z_{total} = Z_{C} || (Z_{R1} + Z_{L} + Z_{R2})$$

But before I solve this, I realize that the equivalent circuit I'm given has two impedance 'nodes' in series. How can this be? I don't understand how (supposing that the voltage $U$ marked in purple remains the same in both images) this is possible.

Ultimately I am confused about the circuit and cannot make sense of the equivalent circuit.

Answer 1

The current through the resistor \$\text{R}_2\$ is given by:

$$\underline{\text{I}}_{\space\text{R}_2}=\frac{\frac{1}{\text{j}\omega\text{C}}}{\frac{1}{\text{j}\omega\text{C}}+\text{R}_1+\text{j}\omega\text{L}+\text{R}_2}\cdot\underline{\text{I}}_{\space\text{in}}=\frac{1}{1-\omega^2\text{CL}+\left(\text{R}_1+\text{R}_2\right)\omega\text{C}\text{j}}\cdot\underline{\text{I}}_{\space\text{in}}\tag1$$

And we know that:

$$\underline{\text{I}}_{\space\text{in}}=\text{I}_\text{p}\cdot\exp\left(\frac{\pi}{2}\cdot\text{j}\right)\tag2$$

So, we get that:

$$\underline{\text{I}}_{\space\text{R}_2}=\frac{\text{I}_\text{p}\cdot\exp\left(\frac{\pi}{2}\cdot\text{j}\right)}{1-\omega^2\text{CL}+\left(\text{R}_1+\text{R}_2\right)\omega\text{C}\text{j}}\tag3$$

So, the voltage is given by:

$$\underline{\text{V}}_{\space\text{R}_2}=\underline{\text{I}}_{\space\text{R}_2}\cdot\underline{\text{Z}}_{\space\text{R}_2}=\frac{\text{I}_\text{p}\cdot\exp\left(\frac{\pi}{2}\cdot\text{j}\right)}{1-\omega^2\text{CL}+\left(\text{R}_1+\text{R}_2\right)\omega\text{C}\text{j}}\cdot\text{R}_2\tag4$$

The voltage is:

$$\text{V}_{\space\text{R}_2}\left(t\right)=\left|\underline{\text{V}}_{\space\text{R}_2}\right|\cos\left(\omega t+\arg\left(\underline{\text{V}}_{\space\text{R}_2}\right)\right)\tag5$$

And we can find:

$$\left|\underline{\text{V}}_{\space\text{R}_2}\right|=\frac{\text{I}_\text{p}\cdot\text{R}_2}{\sqrt{\left(1-\omega^2\text{CL}\right)^2+\left(\left(\text{R}_1+\text{R}_2\right)\omega\text{C}\right)^2}}\tag6$$

And:

$$\arg\left(\underline{\text{V}}_{\space\text{R}_2}\right)=\arg\left(\frac{\text{I}_\text{p}\cdot\exp\left(\frac{\pi}{2}\cdot\text{j}\right)}{1-\omega^2\text{CL}+\left(\text{R}_1+\text{R}_2\right)\omega\text{C}\text{j}}\cdot\text{R}_2\right)=$$ $$\arg\left(\text{I}_\text{p}\right)+\arg\left(\exp\left(\frac{\pi}{2}\cdot\text{j}\right)\right)-\arg\left(1-\omega^2\text{CL}+\left(\text{R}_1+\text{R}_2\right)\omega\text{C}\text{j}\right)+\arg\left(\text{R}_2\right)=$$ $$0+\frac{\pi}{2}-\arg\left(1-\omega^2\text{CL}+\left(\text{R}_1+\text{R}_2\right)\omega\text{C}\text{j}\right)-0=$$ $$\frac{\pi}{2}-\arg\left(1-\omega^2\text{CL}+\left(\text{R}_1+\text{R}_2\right)\omega\text{C}\text{j}\right)=$$ $$\frac{\pi}{2}- \begin{cases} \frac{\pi}{2},\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space1-\omega^2\text{CL}=0\\\ \\ \arctan\left(\frac{\left(\text{R}_1+\text{R}_2\right)\omega\text{C}}{1-\omega^2\text{CL}}\right),\space\space\space\space\space\space\space\space\space\space\space\space1-\omega^2\text{CL}>0\\ \\ \frac{\pi}{2}+\arctan\left(\frac{\left|1-\omega^2\text{CL}\right|}{\left(\text{R}_1+\text{R}_2\right)\omega\text{C}}\right),\space\space\space\space1-\omega^2\text{CL}<0 \end{cases} =$$ $$\begin{cases} 0,\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space1-\omega^2\text{CL}=0\\\ \\ \frac{\pi}{2}-\arctan\left(\frac{\left(\text{R}_1+\text{R}_2\right)\omega\text{C}}{1-\omega^2\text{CL}}\right),\space\space\space\space\space\space\space\space\space1-\omega^2\text{CL}>0\\ \\ -\arctan\left(\frac{\left|1-\omega^2\text{CL}\right|}{\left(\text{R}_1+\text{R}_2\right)\omega\text{C}}\right),\space\space\space\space\space\space\space\space\space\space\space\space\space1-\omega^2\text{CL}<0 \end{cases}\tag7$$

So, we also know:

$$\arg\left(\underline{\text{V}}_{\space\text{R}_2}\right)=\begin{cases} 0,\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space1-\omega^2\text{CL}=0\\\ \\ \frac{\pi}{2}-\arctan\left(\frac{\left(\text{R}_1+\text{R}_2\right)\omega\text{C}}{1-\omega^2\text{CL}}\right),\space\space\space\space\space\space\space\space\space1-\omega^2\text{CL}>0\\ \\ -\arctan\left(\frac{\left|1-\omega^2\text{CL}\right|}{\left(\text{R}_1+\text{R}_2\right)\omega\text{C}}\right),\space\space\space\space\space\space\space\space\space\space\space\space\space1-\omega^2\text{CL}<0 \end{cases}\tag8$$

Answer 2

Proceeding via the Thevenin equivalent, nominate \$ R_2\$ as the load, and hence remove it temporarily. Also, express the components in their Laplace forms: \$L\rightarrow sL\$, and \$C\rightarrow \large \frac{1}{sC}\$; this is easier than slogging through the analysis with \$j\omega L\$ and \$\large \frac{1}{j\omega C} \$

The Thevenin voltage is then $$ V_{th}=\frac{I}{sC}$$

The short circuit current is: $$I_{sc}=\frac{I}{1+sR_1 C+s^2LC} $$

Hence the Thevenin impedance is: $$Z_{th}=\frac{1+sR_1 C+s^2LC}{sC} $$

Applying the Thevenin source to \$R_2\$ gives: $$U=\frac{R_2\:I}{1+sC(R_1+R_2)+s^2LC} $$

Converting this expression to a frequency response, via \$s\rightarrow j\omega\$: $$U=\frac{R_2\:I}{\left( 1-\omega^2LC\right )+j\omega C(R_1+R_2)} $$