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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:

enter image description here

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.

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  • \$\begingroup\$ Can you get it down to a single element? Consider that you know that \$V_{R2} = V_{Z2}\$ and see where you can go from there. \$\endgroup\$ – Hearth Sep 26 at 22:24
  • \$\begingroup\$ @Hearth I cannot get it down no. I must compute overall impedance over two parallel components, and one of those parallel components has R2 in the path. So I have to account for it. I can't get the overall impedance without it. Which is why it can't be a node on its own. If you wanted to do that, you would need to connect C above R2 and below L \$\endgroup\$ – Micrified Sep 26 at 22:34
  • \$\begingroup\$ Do a Thevenin equivalent of the circuit with \$R_2\$ removed as the eventual load. Then let \$Z_1 = Z_{th}\$ and \$Z_2=R_2\$. \$\endgroup\$ – Chu Sep 26 at 23:14
  • \$\begingroup\$ @Chu Is that correct to do in this circuit? The load R2 is then treated like it was in series with the Thevenin of everything else, when it formed part of the parallel circuit. In other words, can I actually create a thevenin equivalent here by "moving" where R2 appears (that seems to no longer be an equivalent circuit no?) \$\endgroup\$ – Micrified Sep 26 at 23:18
  • \$\begingroup\$ I can't see anything wrong with it. \$\endgroup\$ – Chu Sep 26 at 23:41
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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:

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

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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\$: $$U=\frac{R_2\:I}{\left( 1-\omega^2LC\right )+j\omega C(R_1+R_2)} $$

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