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--- electrical_engineering_and_electronics_2:block04 [2026/04/11 07:30] – removed - external edit (Unknown date) 127.0.0.1
+++ electrical_engineering_and_electronics_2:block04 [2026/04/21 02:12] (current) – mexleadmin
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+====== Block 04 — Complex Calculus in EE ======
+===== Learning objectives =====
+<callout>
+After this 90-minute block, you
+  * know how sine variables can be symbolized by a vector.
+  * know which parameters can determine a sinusoidal quantity.
+  * graphically derive a pointer diagram for several existing sine variables.
+  * can plot the phase shift on the vector and time plots.
+  * can add sinusoidal quantities in vector and time representation.
+  * know and apply the impedance of components.
+  * know the frequency dependence of the impedance of the components. In particular, you should know the effect of the ideal components at very high and very low frequencies and be able to apply it for plausibility checks.
+  * are able to draw and read pointer diagrams.
+  * know and apply the complex value formulas of impedance, reactance, and resistance.
+</callout>
+===== Preparation at Home =====
+Well, again
+  * read through the present chapter and write down anything you did not understand.
+  * Also here, there are some clips for more clarification under 'Embedded resources' (check the text above/below, sometimes only part of the clip is interesting).
+For checking your understanding please do the following exercises:
+  * ...
+===== 90-minute plan =====
+  - Warm-up (x min):
+    - ....
+  - Core concepts & derivations (x min):
+    - ...
+  - Practice (x min): ...
+  - Wrap-up (x min): Summary box; common pitfalls checklist.
+===== Conceptual overview =====
+<callout icon="fa fa-lightbulb-o" color="blue">
+  - ...
+</callout>
+===== Core content =====
+==== Representation and Interpretation  ====
+Up to now, we used for the AC signals the formula $x(t)= \sqrt{2} X \cdot \sin (\omega t + \varphi_x)$ - which was quite obvious. \\
+However, there is an alternative way to look at the alternating sinusoidal signals.
+For this, we look first at a different, but already a familiar problem (see <imgref pic06>).
+  - A mechanical, linear spring with the characteristic constant $D$ is displaced due to a mass $m$ in the Earth's gravitational field. The deflection only based on the gravitational field is $X_0$.
+  - At the time $t_0=0$ , we deflect this spring a bit more to $X_0 + x(t_0)=X_0 + \hat{X}$ and therefore induce energy into the system.
+  - When the mass is released, the mass will spring up and down for $t>0$. The signal can be shown as a shadow when the mass is illuminated sideways. \\ For $t>0$, the energy is continuously shifted between potential energy (deflection $x(t)$ around $X_0$) and kinetic energy (${{\rm d}\over{{\rm d}t}}x(t)$)
+  - When looking onto the course of time of $x(t)$, the signal will behave as: $x(t)= \hat{X} \cdot \sin (\omega t + \varphi_x)$
+  - The movement of the shadow can also be created by the sideways shadow of a stick on a rotating disc. \\ This means, that a two-dimensional rotation is reduced down to a single dimension.
+<WRAP>
+<imgcaption pic06 | interpretation of sinusoidal deflection of a spring>
+</imgcaption>
+\\ {{drawio>deflectionspring.svg}} \\
+</WRAP>
+The transformation of the two-dimensional rotation to a one-dimensional sinusoidal signal is also shown in <imgref BildNr00>.
+<panel>
+<imgcaption BildNr00 | Creation of the sinusoidal signal from a rotational movement>
+</imgcaption> \\
+<collapse id="geogebra" collapsed="true"><well>
+{{url>https://www.geogebra.org/material/iframe/id/wSHXZMyy/width/1600/height/800/border/888888/rc/false/ai/false/sdz/false/smb/false/stb/false/stbh/true/ld/false/sri/true/at/preferhtml5 500,250 noborder}}
+Click on the box "animate?"
+</well></collapse>
+<collapse id="geogebra" collapsed="false">
+<button type="warning" collapse="geogebra">press here for the animation</button>
+</collapse>
+</panel>
+The two-dimensional rotation can be represented with a complex number in Euler's formula.
+It combines the exponential representation with real part $\Re$ and imaginary part $\Im$ of a complex value:
+$$ \underline{x}(t)=\hat{X}\cdot {\rm e}^{{\rm j}(\omega t + \varphi_x)} = \Re(\underline{x}) + {\rm j}\cdot \Im(\underline{x})$$
+For the imaginary unit ${\rm i}$ the letter ${\rm j}$ is used in electrical engineering since the letter ${\rm i}$ is already taken for currents.
+<WRAP>
+<imgcaption pic07 | representation of a phasor on the complex plane>
+</imgcaption>
+\\ {{drawio>phasorcomplexplane.svg}} \\
+</WRAP>
+==== Complex Current and Voltage ====
+The concepts of complex numbers shall now be applied to voltages and currents.
+Up to now, we used the following formula to represent alternating voltages:
+$$u(t)= \sqrt{2} U \cdot \sin (\varphi)$$
+This is now interpreted as the instantaneous value of a complex vector $\underline{u}(t)$, which rotates given by the time-dependent angle $\varphi = \omega t + \varphi_u$.
+<WRAP>
+<imgcaption pic08 | representation of a voltage phasor on the complex plane>
+</imgcaption>
+\\ {{drawio>voltagephasorcomplexplane.svg}} \\
+</WRAP>
+The parts on the complex plane are then given by:
+  - The real part      $\Re{(\underline{u}(t))} = \sqrt{2}U \cdot \cos (\omega t + \varphi_u)$
+  - The imaginary part $\Im{(\underline{u}(t))} = \sqrt{2}U \cdot \sin (\omega t + \varphi_u)$
+This is equivalent to the complex phasor $\underline{u}(t)=\sqrt{2}U \cdot {\rm e} ^{{\rm j}  (\omega t + \varphi_u)}$
+The complex phasor can be separated:
+\begin{align*}
+\underline{u}(t) &=\sqrt{2}                          U \cdot {\rm e}^{{\rm j} (\omega t + \varphi_u)} \\
+                 &=\sqrt{2}\color{blue}             {U \cdot {\rm e}^{{\rm j} \varphi_u}}
+                                                       \cdot {\rm e}^{{\rm j} \omega t} \\
+                 &=\sqrt{2}\color{blue}{\underline{U}} \cdot {\rm e}^{{\rm j} \omega t} \\
+\end{align*}
+The **fixed phasor** (in German: //komplexer Festzeiger//) of the voltage is given by $\color{blue}{\underline{U}}= \color{blue}{U  \cdot e ^{j  \varphi_u}} $
+Generally, from now on not only the voltage will be considered as a phasor, but also the current $\underline{I}$ and derived quantities like the impedance $\underline{X}$. \\
+Therefore, the known properties of complex numbers from Mathematics 101 can be applied:
+  * A multiplication with $j$ equals a phase shift of $+90°$
+  * A multiplication with ${{1}\over{j}}$ equals a phase shift of $-90°$
+==== Complex Impedance ====
+=== Introduction to Complex Impedance ===
+The complex impedance is "nearly" similar calculated like the resistance. In the subchapters before, that impedance $Z$ was calculated by $Z=\frac{U}{I}$. \\
+Now the complex impedance is:
+\begin{align*}
+\underline{Z}&=\frac{\underline{U}}{\underline{I}} \\
+             &= \Re{(\underline{Z})} + {\rm j} \cdot \Im{(\underline{Z})} \\
+             &= R                    + {\rm j} \cdot X \\
+             &= Z \cdot {\rm e}^{{\rm j} \varphi} \\
+             &= Z \cdot (\cos \varphi + j \cdot \sin \varphi )
+\end{align*}
+With
+  * the resistance $R$ (in German: //Widerstand//)       as the pure real part
+  * the reactance  $X$ (in German: //Blindwiderstand//)  as the pure imaginary part
+  * the impedance  $Z$ (in German: //Scheinwiderstand//) as the complex number given by the __complex__ addition of resistance and the reactance as a complex number
+The impedance can be transformed from Cartesian to polar coordinates by:
+  * $Z=\sqrt{R^2 + X^2}$
+  * $\varphi = \arctan  \frac{X}{R} $
+The other way around it is possible to transform by:
+  * $R = Z \cos \varphi$
+  * $X = Z \sin \varphi$
+=== Application on pure Loads ===
+With the complex impedance in mind, the <tabref tab01> can be expanded to:
+<tabcaption tab02 | Formulas for the different pure loads>
+^ Load      $\phantom{U\over I}$  ^                        ^ integral representation $\phantom{U\over I}$  ^ complex impedance $\underline{Z}={{\underline{U}}\over{\underline{I}}}$  ^ impedance $Z \phantom{U\over I}$    ^ phase $\varphi \phantom{U\over I}$            ^
+| Resistance                      | $R\phantom{U\over I}$  | ${u} = R \cdot {i}$                           | $Z_R = R $                                                                     | $Z_R = R $                          | $\varphi_R = 0$                               |
+| Capacitance                     | $C$                    | ${u} ={{1}\over{C}}\cdot \int {\rm i} dt$     | $Z_C = {{1}\over{{\rm j}\omega \cdot C}} = {{-{\rm j}}\over{\omega \cdot C}}$  | $Z_C = {{1}\over{\omega \cdot C}}$  | $\varphi_C = -{{1}\over{2}}\pi \hat{=} -90°$  |
+| Inductance                      | $L$                    | ${u} = L \cdot {{\rm d}\over{{\rm d}t}} {i}$  | $Z_L = {\rm j} \omega \cdot L                                        $         | $Z_L = \omega \cdot L            $  | $\varphi_L = +{{1}\over{2}}\pi \hat{=} +90°$  |
+</tabcaption>
+\\ \\
+The relationship between ${\rm j}$ and integral calculus should be clear:
+  - The derivative of a sinusoidal value - and therefore a phasor - can simply be written as "$\cdot {\rm j}\omega$", \\ which also means a phase shift of $+90°$: \\ \begin{align*}{{\rm d}\over{{\rm d}t}} {\rm e}^{{\rm j}(\omega t + \varphi_x)} = {\rm j} \cdot {\rm e}^{{\rm j}(\omega t + \varphi_x)}\end{align*}
+  - The integral of a sinusoidal value - and therefore a phasor - can simply be written as "$\cdot (-{{1}\over{ {\rm j}\omega}})$", \\ which also means a phase shift of $-90°$.((in general, here the integration constant must be considered. This is however often neglectable since only AC values (without a DC value) are considered.)) <WRAP>
+\begin{align*}
+                     \int {\rm e}^{{\rm j}(\omega t + \varphi_x)}
+  = {{1}\over{\rm j\omega}} \cdot {\rm e}^{{\rm j}(\omega t + \varphi_x)}
+  = -{{\rm j}\over{\omega}} \cdot {\rm e}^{{\rm j}(\omega t + \varphi_x)}
+\end{align*}
+</WRAP>
+Once a fixed input voltage is given, the voltage phasor $\underline{U}$, the current phasor $\underline{I}$, and the impedance phasor $\underline{Z}$. In <imgref pic10> these phasors are shown.
+<WRAP>
+<imgcaption pic10 | phasors of the pure loads>
+</imgcaption>
+\\ {{drawio>phasorspureloads.svg}} \\
+</WRAP>
+=== Application on Impedance Networks ===
+\\
+== Simple Networks ==
+In the chapter [[:electrical_engineering_1:simple_circuits#Kirchhoff's Circuit Laws]] we already had a look at simple networks like a series or parallel circuit of resistors. \\
+These formulas not only apply to ohmic resistors but also to impedances:
+<WRAP>
+<imgcaption pic11 | Simple Networks>
+</imgcaption>
+\\ {{drawio>SimpleNetworks.svg}} \\
+</WRAP>
+Similarly, the voltage divider, the current divider, the star-delta transformation, and the Thevenin and Northon Theorem can be used, by substituting resistances with impedances.
+This means for example, every linear source can be represented by an output impedance $\underline{Z}_o$ and an ideal voltage source $\underline{U}$.
+== More "complex" Networks ==
+For more complex problems having AC values in circuitries, the following approach is beneficial. \\
+This concept will be used in the next chapter and in circuit design.
+<WRAP>
+<imgcaption pic12 | Approach for AC cicruitries>
+</imgcaption>
+\\ {{drawio>ACapproach.svg}} \\
+</WRAP>
+\\
+<callout icon="fa fa-exclamation" color="red" title="Notice:">
+<WRAP>
+For a complex number are always two values are needed. These are either
+  - the real part (e.g. the resistance) and the imaginary part (e.g. the reactance), or
+  - the absolute value (e.g. the absolute value of the impedance) and the phase
+Therefore, instead of the form $\underline{Z}=Z\cdot {\rm e}^{{\rm j}\varphi}$ for the phasors often the form $Z\angle{\varphi}$ is used.
+</WRAP>
+</callout>
+===== Common pitfalls =====
+  * ...
+===== Exercises =====
+==== Worked examples ====
+<panel type="info" title="Exercise 6.5.1 Two voltage sources"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
+Two ideal AC voltage sources $1$ and $2$ shall generate the RMS voltage drops $U_1 = 100~\rm V$ and $U_2 = 120~\rm V$. \\
+The phase shift between the two sources shall be $+60°$. The phase of source $1$ shall be $\varphi_1=0°$. \\
+The two sources shall be located in series.
+<WRAP indent> 1. Draw the phasor diagram for the two voltage phasors and the resulting phasor.
+<WRAP indent><button size="xs" type="link" collapse="Loesung_6_5_1_Endergebnis">{{icon>eye}} Solution 1</button><collapse id="Loesung_6_5_1_Endergebnis" collapsed="true">
+The phasor diagram looks roughly like this:
+{{drawio>phasordiagram6511.svg}}
+</collapse></WRAP></WRAP><WRAP indent>2. Calculate the resulting voltage and phase.
+<WRAP indent><button size="xs" type="link" collapse="Loesung_6_5_1_2_Solution">{{icon>eye}} Solution 2</button><collapse id="Loesung_6_5_1_2_Solution" collapsed="true">
+By the law of cosine, we get:
+\begin{align*}
+U&= \sqrt{{{U_1  }^2}+{{U_1  }^2}-2{U_1       } \cdot{U_2  }\cdot \cos(180°- {\varphi_2})} \\
+ &= \sqrt{{{100~V}^2}+{{120~V}^2}-2\cdot{100~V} \cdot{120~V}\cdot \cos(120°)}
+\end{align*}
+The angle is by the tangent of the relation of the imaginary part to the real part of the resulting voltage.
+\begin{align*}
+\varphi&= \arctan 2 \left(\frac{Im\{\underline{U}\}}                        {Re\{\underline{U}\}} \right)\\
+       &= \arctan 2 \left(\frac{Im\{\underline{U}_1\}+Im\{\underline{U}_2\}}{Re\{\underline{U}_1\}+\{\underline{U}_2\}} \right)\\
+       &= \arctan 2 \left(\frac{{U}_2      \cdot\sin(\varphi_2)}{{U}_1      +{U}_2\cos(\varphi_2)} \right)\\
+       &= \arctan 2 \left(\frac{120~{\rm V}\cdot\sin(60°)}      {100~{\rm V}+120~V\cos(60°)} \right)
+\end{align*}
+</collapse> <button size="xs" type="link" collapse="Loesung_6_5_1_2_Endergebnis">{{icon>eye}} Final value</button><collapse id="Loesung_6_5_1_2_Endergebnis" collapsed="true">
+\begin{align*}
+U      &=190.79~{\rm V} \\
+\varphi&=33°
+\end{align*}
+</collapse></WRAP></WRAP><WRAP indent>3. Is the resulting voltage the RMS value or the amplitude?
+<WRAP indent><button size="xs" type="link" collapse="Loesung_6_5_1_3_Solution">{{icon>eye}} Solution 3</button><collapse id="Loesung_6_5_1_3_Solution" collapsed="true">
+The resulting voltage is the RMS value. \\ \\
+</collapse></WRAP></WRAP> \\ The source $2$ shall now be turned around (the previous plus pole is now the minus pole and vice versa).
+<WRAP indent> 4. Draw the phasor diagram for the two voltage phasors and the resulting phasor for the new circuit.
+<WRAP indent><button size="xs" type="link" collapse="Loesung_6_5_1_4_Solution">{{icon>eye}} Solution 4</button><collapse id="Loesung_6_5_1_4_Solution" collapsed="true">
+The phasor diagram looks roughly like this. \\
+But have a look at the solution for question 5!
+{{drawio>phasordiagram6514.svg}}
+</collapse></WRAP></WRAP><WRAP indent>5. Calculate the resulting voltage and phase.
+<WRAP indent><button size="xs" type="link" collapse="Loesung_6_5_1_5_Solution">{{icon>eye}} Solution 5</button><collapse id="Loesung_6_5_1_5_Solution" collapsed="true">
+By the law of cosine, we get:
+\begin{align*}
+U&= \sqrt{{{U_1  }^2}+{{U_1  }^2}-2{U_1       } \cdot{U_2  }\cdot \cos(180°- {\varphi_1})}
+ &= \sqrt{{{100~V}^2}+{{120~V}^2}-2\cdot{100~V} \cdot{120~V}\cdot \cos(60°)}
+\end{align*}
+The angle is by the tangent of the relation of the imaginary part to the real part of the resulting voltage.
+\begin{align*}
+\varphi&= \arctan 2 \left(\frac{Im\{\underline{U}\}}                        {Re\{\underline{U}\}} \right)\\
+       &= \arctan 2 \left(\frac{Im\{\underline{U}_1\}+Im\{\underline{U}_2\}}{Re\{\underline{U}_1\}+\{\underline{U}_2\}} \right)\\
+       &= \arctan 2 \left(\frac{-{U}_2      \cdot\sin(\varphi_2)}{{U}_1      -{U}_2\cos(\varphi_2)} \right)\\
+       &= \arctan 2 \left(\frac{-120~{\rm V}\cdot\sin(60°)      }{100~{\rm V}-120~V\cos(60°)} \right) \\
+       &= \arctan 2 \left(\frac{-103.92...~{\rm V}              }{+40~{\rm V}} \right)
+\end{align*}
+The calculated (positive) horizontal and (negative) vertical dimension for the voltage indicates a phasor in the fourth quadrant. Does it seem right? \\
+The phasor diagram which was shown in answer 4. cannot be correct. \\
+With the correct lengths and angles, the real phasor diagram looks like this:
+{{drawio>phasordiagram6515.svg}}
+Here the phasor is in the fourth quadrant with a negative angle. \\
+</collapse> <button size="xs" type="link" collapse="Loesung_6_5_1_5_Endergebnis">{{icon>eye}} Final value</button><collapse id="Loesung_6_5_1_5_Endergebnis" collapsed="true">
+\begin{align*}
+U      &=111.355~{\rm V}\\
+\varphi&=-68.948°
+\end{align*}
+</collapse></WRAP>
+</WRAP>
+<callout icon="fa fa-exclamation" color="red" title="Notice:">
+Be aware that some of the calculators only provide $\tan^{-1}$ or $\arctan$ and not $\arctan2$! \\
+Therefore, you have always to check whether the solution lies in the correct quadrant.
+</callout>
+</WRAP></WRAP></panel>
+<panel type="info" title="Exercise 6.5.2 oscilloscope plot"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
+The following plot is visible on an oscilloscope (= plot tool for voltages and current).
+{{drawio>ExampleScope.svg}}
+  - What is the RMS value of the current and the voltage? What is the frequency $f$ and the phase $\varphi$? Does the component under test behave ohmic, capacitive, or inductive?
+  - How would the equivalent circuit look like, when it is built by two series components?
+  - Calculate the equivalent component values ($R$, $C$ or $L$) of the series circuit.
+  - How would the equivalent circuit look like, when it is built by two parallel components?
+  - Calculate the equivalent component values ($R$, $C$ or $L$)  of the parallel circuit.
+</WRAP></WRAP></panel>
+<panel type="info" title="Exercise 6.5.3 Series Circuit"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
+The following circuit shall be given. \\
+{{drawio>ExampleSeriesCircuit.svg}}
+This circuit is used with different component values, which are given in the following. \\
+Calculate the RMS value of the missing voltage and the phase shift $\varphi$ between $U$ and $I$.
+<WRAP indent>1. $U_R = 10~\rm V$, $U_L = 10~\rm V$, $U_C = 20~\rm V$, $U=\rm ?$
+<WRAP indent>
+<button size="xs" type="link" collapse="Loesung_6_5_3_1_Endergebnis">{{icon>eye}} Solution</button><collapse id="Loesung_6_5_3_1_Endergebnis" collapsed="true">
+The drawing of the voltage pointers is as follows:{{drawio>SeriesPhasor.svg}}
+The voltage U is determined by the law of Pythagoras
+\begin{align*}
+U &= \sqrt{{{U_R     } ^2}+{({U_L       }-{U_C       }})^2} \\
+  &= \sqrt{(10~{\rm V})^2+ {({10~{\rm V}}-{20~{\rm V}}})^2}
+\end{align*}
+The phase shift angle is calculated by simple geometry.
+\begin{align*}
+\tan(\varphi)&=\frac{{U_L       }-{U_C       }}{U_R}\\
+             &=\frac{{10~{\rm V}}-{20~{\rm V}}}{10~{\rm V}}
+\end{align*}
+Considering that the angle is in the fourth quadrant we get:
+</collapse><button size="xs" type="link" collapse="Loesung_6_5_3_2_Endergebnis">{{icon>eye}} Final value</button><collapse id="Loesung_6_5_3_2_Endergebnis" collapsed="true">
+\begin{equation*}
+U=\sqrt{2}\cdot 10~{\rm V} = 14.1~{\rm V} \qquad \varphi=-45°
+\end{equation*}
+</collapse>
+</WRAP>
+</WRAP><WRAP indent>2. $U_R = ?$, $U_L = 150~\rm V$, $U_C = 110~\rm V$, $U=50~\rm V$
+<WRAP indent>
+<button size="xs" type="link" collapse="Loesung_6_5_3_3_Solution">{{icon>eye}} Solution 2</button><collapse id="Loesung_6_5_3_3_Solution" collapsed="true">
+The drawing of the voltage pointers is as follows: {{drawio>SeriesPhasor2.svg}}
+The voltage $U_R$ is determined by the law of Pythagoras
+\begin{align*}
+U_R&=\sqrt{{U        ^2}+{({U_L}     -{U_C}}    )^2}\\
+   &=\sqrt{(50~\rm V)^2 +{(150~\rm V -110~\rm V})^2}
+\end{align*}
+The phase shift angle is calculated by simple geometry.
+\begin{align*}
+\tan(\varphi)&=\frac{{U_L}      -{U_C}      }{U_R}\\
+             &=\frac{{150~\rm V}-{110~\rm V}}{30~\rm V}
+\end{align*}
+Considering that the angle is in the fourth quadrant we get:
+</collapse>
+<button size="xs" type="link" collapse="Loesung_6_5_3_4_Endergebnis">{{icon>eye}} Final value</button><collapse id="Loesung_6_5_3_4_Endergebnis" collapsed="true">
+\begin{equation*}
+U_R= 30~{\rm V}\qquad \varphi=53.13°
+\end{equation*}
+</collapse>
+</WRAP>
+</WRAP>
+</WRAP></WRAP></panel>
+<panel type="info" title="Exercise 6.5.4 Parallel Circuit"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
+The following circuit shall be given.
+{{drawio>ExampleParallelCircuit.svg}}
+in the following, some of the numbers are given.
+Calculate the RMS value of the missing currents and the phase shift $\varphi$ between $U$ and $I$.
+  - $I_R = 3~\rm A$, $I_L = 1  ~\rm A$, $I_C = 5  ~\rm A$, $I=?$
+  - $I_R = ?$,       $I_L = 1.2~\rm A$, $I_C = 0.4~\rm A$, $I=1~\rm A$
+</WRAP></WRAP></panel>
+<panel type="info" title="Exercise 6.5.5 Complex Calculation I"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
+The following two currents with similar frequencies, but different phases have to be added. Use complex calculation!
+  * $i_1(t) = \sqrt{2} \cdot 2 ~A \cdot \cos (\omega t + 20°)$
+  * $i_2(t) = \sqrt{2} \cdot 5 ~A \cdot \cos (\omega t + 110°)$
+</WRAP></WRAP></panel>
+<panel type="info" title="Exercise 6.5.6 Complex Calculation II"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
+Two complex impedances $\underline{Z}_1$ and $\underline{Z}_2$ are investigated.
+The resulting impedance for a series circuit is   $60~\Omega + \rm j \cdot 0 ~\Omega $.
+The resulting impedance for a parallel circuit is $25~\Omega + \rm j \cdot 0 ~\Omega $.
+What are the values for $\underline{Z}_1$ and $\underline{Z}_2$?
+#@HiddenBegin_HTML~656Sol,Solution~@#
+It's a good start to write down all definitions of the given values:
+  * the given values for the series circuit ($\square_\rm s$) and the parallel circuit ($\square_\rm p$) are: \begin{align*} R_\rm s = 60 ~\Omega , \quad X_\rm s = 0 ~\Omega \\ R_\rm p = 25 ~\Omega , \quad X_\rm p = 0 ~\Omega \\ \end{align*}
+  * the series circuit and the parallel circuit results into: \begin{align*}  R_{\rm s} = \underline{Z}_1 + \underline{Z}_2 \tag{1} \\ R_{\rm p} = \underline{Z}_1 || \underline{Z}_2  \tag{2} \\ \end{align*}
+  * the unknown values of the two impedances are: \begin{align*} \underline{Z}_1 = R_1 + {\rm j}\cdot X_1  \tag{3} \\ \underline{Z}_2 = R_2 + {\rm j}\cdot X_2 \tag{4} \\ \end{align*}
+Based on $(1)$,$(3)$ and $(4)$:
+\begin{align*}
+R_\rm s         &= \underline{Z}_1     &&+ \underline{Z}_2  \\
+                &= R_1 + {\rm j}\cdot X_1    &&+ R_2 + {\rm j}\cdot X_2  \\
+\rightarrow 0   &= R_1 + R_2 - R_\rm s &&+ {\rm j}\cdot (X_1 + X_2)  \\
+\end{align*}
+Real value and imaginary value must be zero:
+\begin{align*}
+R_1 &= R_{\rm s} - R_2  \tag{5} \\
+X_1 &= - X_2  \tag{6}
+\end{align*}
+Based on $(2)$ with $R_\rm s = \underline{Z}_1 + \underline{Z}_2$  $(1)$:
+\begin{align*}
+R_{\rm p}                  &= {{\underline{Z}_1 \cdot \underline{Z}_2}\over{\underline{Z}_1 + \underline{Z}_2}} \\
+                           &= {{\underline{Z}_1 \cdot \underline{Z}_2}\over{R_\rm s}} \\ \\
+R_{\rm p} \cdot R_{\rm s}  &=   \underline{Z}_1 \cdot \underline{Z}_2 \\
+                           &= (R_1 + {\rm j}\cdot X_1)\cdot (R_2 + {\rm j}\cdot X_2)     \\
+                           &= R_1 R_2 + {\rm j}\cdot (R_1 X_2 + R_2 X_1) - X_1 X_2     \\
+\end{align*}
+Substituting $R_1$ and $X_1$ based on $(5)$ and $(6)$:
+\begin{align*}
+R_{\rm p} \cdot R_{\rm s}  &=  (R_{\rm s} - R_2 )  R_2 + {\rm j}\cdot ((R_{\rm s} - R_2 )  X_2 - R_2 X_2) + X_2 X_2     \\
+\rightarrow 0 &=  R_{\rm s} R_2 - R_2^2  + X_2^2 - R_{\rm p} \cdot R_{\rm s}  + {\rm j}\cdot ((R_{\rm s} - R_2 )  X_2 - R_2 X_2)      \\
+\end{align*}
+Again real value and imaginary value must be zero:
+\begin{align*}
+&=  j\cdot ((R_{\rm s} - R_2 )  X_2 - R_2 X_2)     \\
+  &=           R_{\rm s}X_2 - 2 \cdot R_2 X_2        \\
+\rightarrow    R_2 = {{1}\over{2}} R_{\rm s} \tag{7} \\ \\
+&= R_{\rm s} R_2 - R_2^2  + X_2^2 - R_{\rm p} \cdot R_{\rm s}  \\
+  &= R_{\rm s} ({{1}\over{2}} R_{\rm s}) - ({{1}\over{2}} R_{\rm s})^2  - X_2^2 - R_{\rm p} \cdot R_{\rm s}  \\
+  &= {{1}\over{4}} R_{\rm s}^2 + X_2^2 - R_{\rm p} \cdot R_{\rm s}  \\
+\rightarrow    X_2 = \pm \sqrt{R_{\rm p} \cdot R_{\rm s}  - {{1}\over{4}} R_{\rm s}^2 } \tag{8} \\ \\
+\end{align*}
+The concluding result is:
+\begin{align*}
+(5)+(7): \quad R_1 &= {{1}\over{2}} R_{\rm s} \\
+(7): \quad R_2 &= {{1}\over{2}} R_{\rm s} \\
+(6)+(8)  \quad X_1 &= \mp \sqrt{R_{\rm p} \cdot R_{\rm s}  - {{1}\over{4}} R_{\rm s}^2 } \\
+(8): \quad X_2 &= \pm \sqrt{R_{\rm p} \cdot R_{\rm s}  - {{1}\over{4}} R_{\rm s}^2 }
+\end{align*}
+#@HiddenEnd_HTML~656Sol,Solution ~@#
+#@HiddenBegin_HTML~656Res,Result~@#
+\begin{align*}
+R_1 &= 30~\Omega \\
+R_2 &= 30~\Omega \\
+X_1 &= \mp \sqrt{600}~\Omega \approx \mp 24.5~\Omega \\
+X_2 &= \pm \sqrt{600}~\Omega \approx \pm 24.5~\Omega \\
+\end{align*}
+#@HiddenEnd_HTML~656Res,Result~@#
+</WRAP></WRAP></panel>
+<panel type="info" title="Exercise 6.5.7 real Coils I"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
+A real coil has both ohmic and inductance behavior.
+At DC voltage the resistance is measured as $9 ~\Omega$.
+With an AC voltage of $5~\rm V$ at $50~\rm Hz$ a current of $0.5~\rm A$ is measured.
+What is the value of the inductance $L$?
+</WRAP></WRAP></panel>
+<panel type="info" title="Exercise 6.5.8 real Coils II"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
+A real coil has both ohmic and inductance behavior.
+This coil has at $100~\rm Hz$ an impedance of $1.5~\rm k\Omega$ and a resistance $1~\rm k\Omega$.
+What is the value of the reactance and inductance?
+</WRAP></WRAP></panel>
+<panel type="info" title="Exercise 6.5.9  Capacitors and Resistance I"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
+An ideal capacitor is in series with a resistor $R=1~\rm k\Omega$.
+The capacitor shows a similar voltage drop to the resistor for $100~\rm Hz$.
+What is the value of the capacitance?
+</WRAP></WRAP></panel>
+#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~  EMI Input Check of an AC Power Module: Parallel Resistor-Capacitor Branch                    #@TaskText_HTML@#
+<WRAP right>{{drawio>electrical_engineering_and_electronics_2:EMIcheckExerciseV01.svg}}</WRAP>
+An AC power module contains a damping resistor and an EMI capacitor connected in parallel at its input.
+For commissioning, the complex input behavior at mains frequency shall be evaluated.
+Data:
+\begin{align*}
+u_1 &= \hat U_1 \cos(\omega t) \\
+\hat U_1 &= 325~{\rm V} \\
+f &= 50~{\rm Hz} \\
+R &= 220~{\rm \Omega} \\
+C &= 4.7~{\rm \mu F}
+\end{align*}
+. How large is $\omega C$?
+<WRAP group>
+<WRAP half column rightalign>
+#@PathBegin_HTML~401~@#
+<WRAP leftalign>
+\begin{align*}
+\omega &= 2\pi f = 2\pi \cdot 50~{\rm s^{-1}} = 314.16~{\rm s^{-1}} \\
+\omega C &= 314.16~{\rm s^{-1}} \cdot 4.7 \cdot 10^{-6}~{\rm F} \\
+         &= 1.48 \cdot 10^{-3}~{\rm {1}\over{\Omega}} \\
+         &= 1.48 \cdot 10^{-3}~{\rm S}
+\end{align*}
+</WRAP>
+#@PathEnd_HTML@#
+</WRAP>
+<WRAP half column>
+#@ResultBegin_HTML~401~@#
+\begin{align*}
+\omega C = 1.48~{\rm mS}
+\end{align*}
+#@ResultEnd_HTML@#
+</WRAP>
+</WRAP>
+. Calculate the complex input impedance.
+<WRAP group>
+<WRAP half column rightalign>
+#@PathBegin_HTML~402~@#
+<WRAP leftalign>
+For the parallel circuit, it is easiest to start with the reciprocal of the impedance (=admittance):
+\begin{align*}
+{1}\over{\underline{Z}_{\rm in}} &= \frac{1}{R} + j\omega C \\
+                       &= \frac{1}{220~\Omega} + j\,1.48~{\rm mS} \\
+                       &= 4.55~{\rm mS} + j\,1.48~{\rm mS}
+\end{align*}
+The reciprocal magnitude and phase are
+\begin{align*}
+\left|{1}\over{\underline{Z}_{\rm in}} \right| &= 4.78~{\rm mS} \\
+\varphi_{{1}\over{Z}} &= \arctan\left(\frac{1.48}{4.55}\right) \approx +18.0^\circ
+\end{align*}
+Now invert:
+\begin{align*}
+\underline{Z}_{\rm in} &= \frac{1}{4.78~{\rm mS}} \\
+                       &= 209.2~{\rm \Omega} \\
+\varphi_Z &= \approx -18.0^\circ
+\end{align*}
+In polar form:
+\begin{align*}
+\underline{Z}_{\rm in} &= 209.2~{\rm \Omega}\angle -18.0^\circ
+\end{align*}
+</WRAP>
+#@PathEnd_HTML@#
+</WRAP>
+<WRAP half column>
+#@ResultBegin_HTML~402~@#
+\begin{align*}
+\underline{Y}_{\rm in} &= 4.55~{\rm mS} + j\,1.48~{\rm mS}
+= 4.78~{\rm mS}\angle 18.0^\circ \\
+\underline{Z}_{\rm in} &= 199.0 - j\,64.6~{\rm \Omega}
+= 209.2~{\rm \Omega}\angle -18.0^\circ
+\end{align*}
+#@ResultEnd_HTML@#
+</WRAP>
+</WRAP>
+. Calculate the input current by magnitude and phase.
+<WRAP group>
+<WRAP half column rightalign>
+#@PathBegin_HTML~403~@#
+<WRAP leftalign>
+\begin{align*}
+\underline{\hat I}_{\rm in} &= \underline{\hat U}_1 \cdot {{1}\over{\underline{Z}_{\rm in}}}
+\end{align*}
+With $\underline{\hat U}_1 = 325~{\rm V}\angle 0^\circ$:
+\begin{align*}
+\left|\underline{\hat I}_{\rm in}\right|
+&= 325~{\rm V} \cdot 4.78~{\rm mS}
+= 1.55~{\rm A}
+\end{align*}
+The current phase is equal to the admittance phase (negative phase of impedance):
+\begin{align*}
+\varphi_I &= 18.0^\circ
+\end{align*}
+So the current leads the voltage by about $18^\circ$.
+</WRAP>
+#@PathEnd_HTML@#
+</WRAP>
+<WRAP half column>
+#@ResultBegin_HTML~403~@#
+\begin{align*}
+\underline{\hat I}_{\rm in} = 1.55~{\rm A}\angle 18.0^\circ
+\end{align*}
+#@ResultEnd_HTML@#
+</WRAP>
+</WRAP>
+. Draw the phasor diagram for all currents and voltages.
+<WRAP group>
+<WRAP half column rightalign>
+#@PathBegin_HTML~404~@#
+<WRAP leftalign>
+Choose the input voltage as reference:
+\begin{align*}
+\underline{\hat U}_1 &= 325~{\rm V}\angle 0^\circ
+\end{align*}
+Branch currents:
+\begin{align*}
+\underline{\hat I}_R &= \frac{325}{220}~{\rm A}\angle 0^\circ
+= 1.48~{\rm A}\angle 0^\circ \\
+\underline{\hat I}_C &= 325 \cdot \omega C~{\rm A}\angle 90^\circ
+= 0.48~{\rm A}\angle 90^\circ
+\end{align*}
+Total current:
+\begin{align*}
+\underline{\hat I}_{\rm in}
+= \underline{\hat I}_R + \underline{\hat I}_C
+= 1.55~{\rm A}\angle 18.0^\circ
+\end{align*}
+All branch voltages are equal in a parallel circuit:
+\begin{align*}
+\underline{\hat U}_R = \underline{\hat U}_C = \underline{\hat U}_1
+\end{align*}
+</WRAP>
+#@PathEnd_HTML@#
+</WRAP>
+<WRAP half column>
+#@ResultBegin_HTML~404~@#
+\begin{align*}
+\underline{\hat U}_1 &= \underline{\hat U}_R = \underline{\hat U}_C
+= 325~{\rm V}\angle 0^\circ \\
+\underline{\hat I}_R &= 1.48~{\rm A}\angle 0^\circ \\
+\underline{\hat I}_C &= 0.48~{\rm A}\angle 90^\circ \\
+\underline{\hat I}_{\rm in} &= 1.55~{\rm A}\angle 18.0^\circ
+\end{align*}
+{{drawio>electrical_engineering_and_electronics_2:diagramsolution404.svg}}
+#@ResultEnd_HTML@#
+</WRAP>
+</WRAP>
+#@TaskEnd_HTML@#
+#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~  AC Magnet Valve Branch: Parallel Resistor-Inductor Circuit                    #@TaskText_HTML@#
+<WRAP right>{{drawio>electrical_engineering_and_electronics_2:ACcontrolunitV01.svg}}</WRAP>
+An AC control unit contains a magnetizing branch modeled as a resistor in parallel with an inductor.
+The complex input behavior at operating frequency shall be determined.
+Data:
+\begin{align*}
+u_1 &= \hat U_1 \cos(\omega t) \\
+\hat U_1 &= 170~{\rm V} \\
+f &= 60~{\rm Hz} \\
+R &= 220~{\rm \Omega} \\
+L &= 325~{\rm mH}
+\end{align*}
+. How large is $\omega L$?
+. Calculate the complex input impedance.
+. Calculate the input current by magnitude and phase.
+. Draw the phasor diagram for all currents and voltages.
+#@TaskEnd_HTML@#
+#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~  AC Sensor Front-End: Composite Reactive Network                    #@TaskText_HTML@#
+<WRAP right>{{drawio>electrical_engineering_and_electronics_2:ACsensorFrontendV01.svg}}</WRAP>
+A sensor front-end in an industrial AC cabinet is modeled by a series capacitor, a series inductance, and a parallel output branch consisting of a resistor and another inductance.
+The complex input behavior shall be analyzed.
+Data:
+\begin{align*}
+\omega C &= 0.01~{\rm S} \\
+\omega L_1 &= 50~{\rm \Omega} \\
+\omega L_2 &= 200~{\rm \Omega} \\
+R &= 100~{\rm \Omega} \\
+\hat U_1 &= 325~{\rm V}
+\end{align*}
+. Determine the complex input impedance of the circuit.
+<WRAP group>
+<WRAP half column rightalign>
+#@PathBegin_HTML~421~@#
+<WRAP leftalign>
+First determine the individual impedances:
+\begin{align*}
+\underline{Z}_C &= \frac{1}{j\,\omega C}
+= -j\,\frac{1}{\omega C}
+= -j\,100~{\rm \Omega} \\
+\underline{Z}_{L_1} &= j\,\omega L_1
+= j\,50~{\rm \Omega} \\
+\underline{Z}_{L_2} &= j\,\omega L_2
+= j\,200~{\rm \Omega}
+\end{align*}
+Now the parallel branch:
+\begin{align*}
+\underline{Z}_{R\parallel L_2}
+&= \frac{R \cdot \underline{Z}_{L_2}}{R+\underline{Z}_{L_2}} \\
+&= \frac{100 \cdot j\,200}{100+j\,200}
+= 80~{\rm \Omega} + j\,40~{\rm \Omega}
+\end{align*}
+Total input impedance:
+\begin{align*}
+\underline{Z}_{\rm in}
+&= \underline{Z}_C + \underline{Z}_{L_1} + \underline{Z}_{R\parallel L_2} \\
+&= (-j\,100~{\rm \Omega}) + (j\,50~{\rm \Omega}) + (80~{\rm \Omega}+j\,40~{\rm \Omega}) \\
+&= 80~{\rm \Omega} - j10~{\rm \Omega}
+\end{align*}
+In polar form:
+\begin{align*}
+\underline{Z}_{\rm in}
+= 80.6~{\rm \Omega}\angle -7.13^\circ
+\end{align*}
+</WRAP>
+#@PathEnd_HTML@#
+</WRAP>
+<WRAP half column>
+#@ResultBegin_HTML~421~@#
+\begin{align*}
+\underline{Z}_{\rm in} &= 80 - j10~{\rm \Omega} \\
+                       &= 80.6~{\rm \Omega}\angle -7.13^\circ
+\end{align*}
+#@ResultEnd_HTML@#
+</WRAP>
+</WRAP>
+. Calculate the input current by magnitude and phase.
+<WRAP group>
+<WRAP half column rightalign>
+#@PathBegin_HTML~422~@#
+<WRAP leftalign>
+\begin{align*}
+\underline{\hat I}_{\rm in}
+&= \frac{\underline{\hat U}_1}{\underline{Z}_{\rm in}}
+\end{align*}
+With $\underline{\hat U}_1 = 325~{\rm V}\angle 0^\circ$:
+\begin{align*}
+\left|\underline{\hat I}_{\rm in}\right|
+&= \frac{325}{80.6}
+= 4.03~{\rm A}
+\end{align*}
+Because the impedance angle is $-7.13^\circ$, the current angle is
+\begin{align*}
+\varphi_I = +7.13^\circ
+\end{align*}
+So the current leads the source voltage slightly.
+</WRAP>
+#@PathEnd_HTML@#
+</WRAP>
+<WRAP half column>
+#@ResultBegin_HTML~422~@#
+\begin{align*}
+\underline{\hat I}_{\rm in} = 4.03~{\rm A}\angle 7.13^\circ
+\end{align*}
+#@ResultEnd_HTML@#
+</WRAP>
+</WRAP>
+. Draw the phasor diagram for all currents and voltages.
+<WRAP group>
+<WRAP half column rightalign>
+#@PathBegin_HTML~423~@#
+<WRAP leftalign>
+Use the source voltage as reference:
+\begin{align*}
+\underline{\hat U}_1 = 325~{\rm V}\angle 0^\circ
+\end{align*}
+Input current:
+\begin{align*}
+\underline{\hat I}_{\rm in} = 4.03~{\rm A}\angle 7.13^\circ
+\end{align*}
+Voltage drops in the series elements:
+\begin{align*}
+\underline{\hat U}_C
+&= \underline{\hat I}_{\rm in}\,\underline{Z}_C
+= 403.1~{\rm V}\angle -82.9^\circ \\
+\underline{\hat U}_{L_1}
+&= \underline{\hat I}_{\rm in}\,\underline{Z}_{L_1}
+= 201.6~{\rm V}\angle 97.1^\circ
+\end{align*}
+Voltage across the parallel branch:
+\begin{align*}
+\underline{\hat U}_{R\parallel L_2}
+&= \underline{\hat I}_{\rm in}\,(80+j40) \\
+&= 360.6~{\rm V}\angle 33.7^\circ
+\end{align*}
+Branch currents:
+\begin{align*}
+\underline{\hat I}_R
+&= \frac{\underline{\hat U}_{R\parallel L_2}}{R}
+= 3.61~{\rm A}\angle 33.7^\circ \\
+\underline{\hat I}_{L_2}
+&= \frac{\underline{\hat U}_{R\parallel L_2}}{j200}
+= 1.80~{\rm A}\angle -56.3^\circ
+\end{align*}
+Check:
+\begin{align*}
+\underline{\hat I}_{\rm in}
+= \underline{\hat I}_R + \underline{\hat I}_{L_2}
+\end{align*}
+</WRAP>
+#@PathEnd_HTML@#
+</WRAP>
+<WRAP half column>
+#@ResultBegin_HTML~423~@#
+\begin{align*}
+\underline{\hat I}_{\rm in} &= 4.03~{\rm A}\angle 7.13^\circ \\
+\underline{\hat U}_C &= 403.1~{\rm V}\angle -82.9^\circ \\
+\underline{\hat U}_{L_1} &= 201.6~{\rm V}\angle 97.1^\circ \\
+\underline{\hat U}_{R\parallel L_2} &= 360.6~{\rm V}\angle 33.7^\circ \\
+\underline{\hat I}_R &= 3.61~{\rm A}\angle 33.7^\circ \\
+\underline{\hat I}_{L_2} &= 1.80~{\rm A}\angle -56.3^\circ
+\end{align*}
+{{drawio>electrical_engineering_and_electronics_2:diagramsolution423.svg}}
+#@ResultEnd_HTML@#
+</WRAP>
+</WRAP>
+#@TaskEnd_HTML@#
+#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~  Compensated Oscilloscope Probe on a Long Cable                    #@TaskText_HTML@#
+<WRAP right>{{drawio>electrical_engineering_and_electronics_2:scopeExerciseV01.svg}}</WRAP>
+A sinusoidal measurement signal shall be observed through a long cable and an oscilloscope probe.
+The oscilloscope input is characterized by the input resistance $R_E$ and input capacitance $C_E$.
+The cable is represented by its capacitance $C_K$.
+The probe itself consists of a resistor $R_T$ in parallel with an adjustable capacitor $C_T$.
+. Give the complex impedance of the probe.
+. Draw the circuit diagram for the given arrangement.
+. What is the ratio between measured voltage and oscilloscope input voltage,
+$v_U=\hat U_M/\hat U_E$, as a function of the given resistances and capacitances?
+. How must the probe capacitor $C_T$ be adjusted so that $v_U$ becomes frequency-independent?
+What is $v_U$ under this condition?
+#@TaskEnd_HTML@#
+#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~  Phase-Shift Bridge for Trigger and Synchronization Signals                    #@TaskText_HTML@#
+<WRAP right>{{drawio>electrical_engineering_and_electronics_2:phaseShiftBridgeExerciseV01.svg}}</WRAP>
+A phase-shift bridge is used in a control cabinet to generate a shifted AC reference signal for triggering and synchronization.
+One branch contains a potentiometer $R_P$ and a capacitor, while the other branch is a symmetric resistor divider.
+The output voltage $u_2$ is taken between the two center nodes.
+Data:
+\begin{align*}
+R &= 1~{\rm k\Omega} \\
+\omega C &= 1~{\rm mS} \\
+\hat U_1 &= 325~{\rm V} \\
+f &= 50~{\rm Hz}
+\end{align*}
+. The potentiometer is set to $R_P=R$. Draw the phasor diagram for all currents and voltages.
+Give amplitude and phase of $u_2$.
+. How must $R_P$ be adjusted so that $u_2$ lags by $45^\circ$?
+. What is the complex phasor $\underline{\hat U}_2$ as a function of the resistance $R_P$?
+#@TaskEnd_HTML@#
+<panel type="info" title="Exercise 6.6.1 Impedance in Series Circuit of multiple Components I"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
+{{youtube>ZSDpIpnlTbY}}
+</WRAP></WRAP></panel>
+<panel type="info" title="Exercise 6.6.2 Impedance in Series Circuit of multiple Components II"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
+{{youtube>LM2G3cunKp4?start=196}}
+</WRAP></WRAP></panel>
+<panel type="info" title="Exercise 6.6.3 Impedance in Parallel Circuit of multiple Components I"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
+{{youtube>8MMzeeHNjIw}}
+</WRAP></WRAP></panel>
+<panel type="info" title="Exercise 6.6.4 Impedance in Mixed Parallel and Series Circuit of multiple Components I"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
+{{youtube>u6lE4gIIfBw}}
+</WRAP></WRAP></panel>
+<panel type="info" title="Exercise 6.6.5 Impedance in Mixed Parallel and Series Circuit of multiple Components II"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
+{{youtube>RPK0wkyLyMY?stop=705}}
+</WRAP></WRAP></panel>
+<panel type="info" title="Exercise 6.6.6 Impedance in Mixed Parallel and Series Circuit of multiple Components III"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
+{{youtube>oueHAbLOSPA}}
+</WRAP></WRAP></panel>
+===== Embedded resources =====
+<WRAP column half>
+The following two videos explain the basic terms of the complex AC calculus: Impedance, Reactance, Resistance
+{{youtube>fSPcuOu_bf8}}
+</WRAP>
+<WRAP column half>
+This does the same
+{{youtube>WmTlioVfS78}}
+</WRAP>
+~~PAGEBREAK~~ ~~CLEARFIX~~