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--- electrical_engineering_and_electronics_2:block03 [2026/03/05 03:10] – mexleadmin
+++ electrical_engineering_and_electronics_2:block03 [2026/03/05 03:12] (aktuell) – mexleadmin
@@ Zeile 230: / Zeile 230: @@
 ==== 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>
+<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 =====