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To determine the current I have to calculate the total impedance, here named \$Z_{Ges}\$, in the circuit which can be seen in the first picture. Picture 1

First I summed \$R_1 \$, \$C \$ and \$R_2 \$ as a parallel impedance \$Z_{1C2} = (Z_{R_1} || Z_C) || Z_{R_2} = (\frac{R_1 * \frac{1}{j\omega C}}{R_1 + \frac{1}{j \omega C}}) || R_2 = (\frac{R_1}{1+j \omega R_1 C})||R_2 = \frac{\frac{R_1}{1+j \omega R_1 C}R_2}{\frac{R_1}{1+j \omega R_1 C}+R_2} = ... = \frac{R_1^2 R_2 + R_1 R_2^2}{(R_1 + R_2)^2 + \omega^2 R_1^2 R_2^2 C^2} - j\frac{\omega R_1^2 R_2^2 C}{(R_1 + R_2)^2 + \omega^2 R_1^2 R_2^2 C^2}\$.

This leads to the simplified circuit with only \$Z_{1C2}\$, \$Z_{3}\$ and \$Z_{L}\$. Then I add this in a series and get

\$Z_{Ges} = Z_{1C2} + Z_{3} + Z_{L} = \frac{R_1^2 R_2 + R_1 R_2^2}{(R_1 + R_2)^2 + \omega^2 R_1^2 R_2^2 C^2} - j\frac{\omega R_1^2 R_2^2 C}{(R_1 + R_2)^2 + \omega^2 R_1^2 R_2^2 C^2} + R_3 + j\omega L = (\frac{R_1^2 R_2 + R_1 R_2^2}{(R_1 + R_2)^2 + \omega^2 R_1^2 R_2^2 C^2} + R_3)+ j(\omega L - \frac{\omega R_1^2 R_2^2 C}{(R_1 + R_2)^2 + \omega^2 R_1^2 R_2^2 C^2})\$.

After inserting the respective values I get \$ Z_{ges} = 0,6 * (3+j) [\Omega]\$

But when I want to write the calculated impedance in polar form (see picture 2) I get a somehow "strange" result in the exponent, namely \$ e^{j\frac{\pi}{540}} \$. I get this as follows:

\$ 3 + j = \sqrt{10}(\cos{x} + j\sin{x}) \$ and \$ x \$ evaluates to \$ \frac{1}{3} \$. Converting this to radians I get \$ \frac{\pi}{540} \$

Picture 2

Are the steps, particularly where I calculated the parallel impedance \$Z_{12C}\$, correct?

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    \$\begingroup\$ Your handwriting is as bad as mine and that's why I try always to use latex for formulas. \$\endgroup\$ Commented May 9, 2023 at 10:35
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    \$\begingroup\$ Please crop away any whitespace or preferably, use latex. \$\endgroup\$ Commented May 9, 2023 at 10:39
  • \$\begingroup\$ @Andyaka I added the LaTeX equations in the editg \$\endgroup\$ Commented May 9, 2023 at 11:07
  • \$\begingroup\$ @winny I did it \$\endgroup\$ Commented May 9, 2023 at 11:09

1 Answer 1

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Well, notice that the input impedance of your circuit is given by:

\begin{equation} \begin{split} \underline{\text{Z}}_{\space\text{i}}\left(\omega\right)&=\underline{\text{Z}}_{\space\text{R}_1}+\left[\left(\underline{\text{Z}}_{\space\text{C}}\space\text{||}\space\underline{\text{Z}}_{\space\text{R}_2}\right)\space\text{||}\space\left(\underline{\text{Z}}_{\space\text{R}_3}+\underline{\text{Z}}_{\space\text{L}}\right)\right]\\ \\ &=\text{R}_1+\frac{\displaystyle1}{\displaystyle\frac{\displaystyle1}{\displaystyle\frac{\displaystyle1}{\displaystyle\text{j}\omega\text{C}}}+\frac{\displaystyle1}{\displaystyle\text{R}_2}+\frac{\displaystyle1}{\displaystyle\text{R}_3+\text{j}\omega\text{L}}}\\ \\ &=\text{R}_1+\frac{\displaystyle1}{\displaystyle\text{j}\omega\text{C}+\frac{\displaystyle1}{\displaystyle\text{R}_2}+\frac{\displaystyle1}{\displaystyle\text{R}_3+\text{j}\omega\text{L}}}\\ \\ &=\text{R}_1+\frac{\displaystyle1}{\displaystyle\text{j}\omega\text{C}+\frac{\displaystyle1}{\displaystyle\text{R}_2}+\frac{\displaystyle1}{\displaystyle\text{R}_3+\text{j}\omega\text{L}}}\cdot\frac{\displaystyle\text{R}_3+\text{j}\omega\text{L}}{\displaystyle\text{R}_3+\text{j}\omega\text{L}}\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_3+\text{j}\omega\text{L}}{\displaystyle\text{j}\omega\text{C}\left(\text{R}_3+\text{j}\omega\text{L}\right)+\frac{\displaystyle\text{R}_3+\text{j}\omega\text{L}}{\displaystyle\text{R}_2}+\frac{\displaystyle\text{R}_3+\text{j}\omega\text{L}}{\displaystyle\text{R}_3+\text{j}\omega\text{L}}}\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_3+\text{j}\omega\text{L}}{\displaystyle\text{j}\omega\text{C}\left(\text{R}_3+\text{j}\omega\text{L}\right)+\frac{\displaystyle\text{R}_3+\text{j}\omega\text{L}}{\displaystyle\text{R}_2}+1}\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_3+\text{j}\omega\text{L}}{\displaystyle\text{j}\omega\text{C}\left(\text{R}_3+\text{j}\omega\text{L}\right)+\frac{\displaystyle\text{R}_3+\text{j}\omega\text{L}}{\displaystyle\text{R}_2}+1}\cdot\frac{\displaystyle\text{R}_2}{\displaystyle\text{R}_2}\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_2\left(\text{R}_3+\text{j}\omega\text{L}\right)}{\displaystyle\text{j}\omega\text{CR}_2\left(\text{R}_3+\text{j}\omega\text{L}\right)+\frac{\displaystyle\text{R}_2\left(\text{R}_3+\text{j}\omega\text{L}\right)}{\displaystyle\text{R}_2}+\text{R}_2}\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_2\left(\text{R}_3+\text{j}\omega\text{L}\right)}{\displaystyle\text{j}\omega\text{CR}_2\left(\text{R}_3+\text{j}\omega\text{L}\right)+\text{R}_3+\text{j}\omega\text{L}+\text{R}_2}\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_2\left(\text{R}_3+\text{j}\omega\text{L}\right)}{\displaystyle\text{R}_2+\text{R}_3+\text{j}\omega\text{CR}_2\text{R}_3+\text{j}\omega\text{CR}_2\text{j}\omega\text{L}+\text{j}\omega\text{L}}\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_2\left(\text{R}_3+\text{j}\omega\text{L}\right)}{\displaystyle\text{R}_2+\text{R}_3-\text{CLR}_2\omega^2+\omega\left(\text{CR}_2\text{R}_3+\text{L}\right)\text{j}}\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_2\left(\text{R}_3+\text{j}\omega\text{L}\right)}{\displaystyle\text{R}_2+\text{R}_3-\text{CLR}_2\omega^2+\omega\left(\text{CR}_2\text{R}_3+\text{L}\right)\text{j}}\cdot\frac{\displaystyle\text{R}_2+\text{R}_3-\text{CLR}_2\omega^2-\omega\left(\text{CR}_2\text{R}_3+\text{L}\right)\text{j}}{\displaystyle\text{R}_2+\text{R}_3-\text{CLR}_2\omega^2-\omega\left(\text{CR}_2\text{R}_3+\text{L}\right)\text{j}}\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_2\left(\text{R}_3+\text{j}\omega\text{L}\right)\left(\text{R}_2+\text{R}_3-\text{CLR}_2\omega^2-\omega\left(\text{CR}_2\text{R}_3+\text{L}\right)\text{j}\right)}{\displaystyle\left(\text{R}_2+\text{R}_3-\text{CLR}_2\omega^2\right)^2+\left(\omega\left(\text{CR}_2\text{R}_3+\text{L}\right)\right)^2} \end{split}\tag1 \end{equation}

Where \$\alpha\space\text{||}\space\beta:=\frac{\displaystyle\alpha\beta}{\displaystyle\alpha+\beta}\$.

Using your values, I found:

$$\underline{\text{Z}}_{\space\text{i}}\left(1\right)=3\space\Omega\tag2$$

So, we see that: \$\displaystyle\Im\left(\underline{\text{Z}}_{\space\text{i}}\left(1\right)\right)=0\space\Omega\$.

EDIT, the voltage across the inductor is given by:

$$\text{V}_\text{L}\left(t\right)=\hat{\text{u}}\cos\left(t+\varphi\right)\tag3$$

Where:

  • $$\hat{\text{u}}=\left|\text{j}\omega\text{L}\cdot\underbrace{\frac{\displaystyle\hat{\text{u}}_\text{i}\exp\left(\varphi_\text{i}\right)}{\displaystyle\underline{\text{Z}}_{\space\text{i}}\left(\omega\right)}\cdot\underbrace{\frac{\displaystyle\text{R}_2\space\text{||}\space\frac{1}{\text{j}\omega\text{C}}}{\displaystyle\text{R}_3+\text{j}\omega\text{L}+\left(\text{R}_2\space\text{||}\space\frac{1}{\text{j}\omega\text{C}}\right)}}_{=\space\text{current divider}}}_{=\space\text{current through inductor}}\right|\tag4$$
  • $$\varphi=\arg\left(\text{j}\omega\text{L}\cdot\frac{\displaystyle\hat{\text{u}}_\text{i}\exp\left(\varphi_\text{i}\right)}{\displaystyle\underline{\text{Z}}_{\space\text{i}}\left(\omega\right)}\cdot\frac{\displaystyle\text{R}_2\space\text{||}\space\frac{1}{\text{j}\omega\text{C}}}{\displaystyle\text{R}_3+\text{j}\omega\text{L}+\left(\text{R}_2\space\text{||}\space\frac{1}{\text{j}\omega\text{C}}\right)}\right)\tag5$$

Using your values, we find: trough

  • $$\hat{\text{u}}=\frac{4 \sqrt{2}}{3}\approx1.88562\space\text{V}\tag6$$
  • $$\varphi=\frac{\pi}{2}\tag7$$
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  • \$\begingroup\$ Thank You!! But why is my attempt wrong? Why can't I add \$ Z_{R_1} \$ and \$ C \$ parallel, then add \$ Z_{R_2} \$ to that parallel, and in the end the \$ Z_{1C2} \$ in series to \$ Z_{R_3} \$ and \$ Z_L \$ ? \$\endgroup\$ Commented May 9, 2023 at 11:30
  • \$\begingroup\$ and can you tell me how to determine the voltage drop on the inductor? \$\endgroup\$ Commented May 9, 2023 at 12:36
  • \$\begingroup\$ @syphracos You're welcome! Take a look at my edit. \$\endgroup\$ Commented May 9, 2023 at 12:59
  • \$\begingroup\$ again, thank You very much! Is it possible to get the inductor current by using the fact that \$ i_L = i_{tot} - i_{C} - i_{R_2} \$ ? \$\endgroup\$ Commented May 9, 2023 at 13:17
  • \$\begingroup\$ where \$ i_{tot} = \frac{u_q}{Z_{tot}} = \frac{8 \cdot e^j{\frac{\pi}{4}}}{3} \$ , \$ i_C = \frac{u_{tot} - u_{R_1}}{Z_C} \$ and \$ i_{R_2} = \frac{u_{tot} - u_{R_1}}{Z_{R_2}}\$ Because my understand is, first we have a voltage drop over \$ R_1 \$ and, because the rest of the circuit is parallel, the votage over the capacitor, the other two resistors and the inductor is \$ u_{tot} - u_{R_1} \$. So then just inserting these values we calculate \$ i_L \$. And then \$ u_L = Z_L \cdot i_L \$ \$\endgroup\$ Commented May 9, 2023 at 13:21

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