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electrical_engineering_1:introduction_in_alternating_current_technology [2023/03/27 09:15] – mexleadminelectrical_engineering_1:introduction_in_alternating_current_technology [2024/12/04 14:43] (current) – [Bearbeiten - Panel] mexleadmin
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-====== 6. Introduction to Alternating Current Technology ======+====== 6 Introduction to Alternating Current Technology ======
  
 Up to now, we had analyzed DC signals (chapters 1. -  4.) and abrupt voltage changes for (dis)charging capacitors (chapter 5.). In households, we use alternating voltage (AC) instead of a constant voltage (DC). This is due to at least three main facts Up to now, we had analyzed DC signals (chapters 1. -  4.) and abrupt voltage changes for (dis)charging capacitors (chapter 5.). In households, we use alternating voltage (AC) instead of a constant voltage (DC). This is due to at least three main facts
-  - Often the voltage given by the **power plant is AC**. This is true for example in all power plants which use electric generators. In these, the mechanical energy of a rotating system is transformed into electric energy by means of moving magnets, which induce an alternating electric voltage. Some modern plants, like photovoltaic plants, do not primarily generate AC voltages.+  - Often the voltage given by the **power plant is AC**. This is true for example in all power plants which use electric generators. In these, the mechanical energy of a rotating system is transformed into electric energy using moving magnets, which induce an alternating electric voltage. Some modern plants, like photovoltaic plants, do not primarily generate AC voltages.
   - For long-range power transfer the power losses $P_{\rm loss}$ can be reduced by reducing the currents $I$ since $P_{\rm loss}=R\cdot I^2$. Therefore, for constant power transfer, the voltage has to be increased. This is much easier done with AC voltages: **AC enables the transformation of a lower voltage to a higher** by the use of alternating magnetic fields in a transformer.    - For long-range power transfer the power losses $P_{\rm loss}$ can be reduced by reducing the currents $I$ since $P_{\rm loss}=R\cdot I^2$. Therefore, for constant power transfer, the voltage has to be increased. This is much easier done with AC voltages: **AC enables the transformation of a lower voltage to a higher** by the use of alternating magnetic fields in a transformer. 
   - AC signals have **at least one more value** which can be used for understanding the situation of the source or load. This simplifies the power and load management in a complex power network.   - AC signals have **at least one more value** which can be used for understanding the situation of the source or load. This simplifies the power and load management in a complex power network.
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 Besides the applications in power systems AC values are also important in communication engineering. Acoustic and visual signals like sound and images can often be considered as wavelike AC signals. Additionally, also for signal transfer like Bluetooth, RFID, and antenna design AC signals are important. Besides the applications in power systems AC values are also important in communication engineering. Acoustic and visual signals like sound and images can often be considered as wavelike AC signals. Additionally, also for signal transfer like Bluetooth, RFID, and antenna design AC signals are important.
  
-In order to understand these systems a bit more, we will start this chapter with a first introduction to AC systems. +To understand these systems a bit more, we will start this chapter with a first introduction to AC systems. 
  
 <callout> <callout>
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 </callout> </callout>
  
-In order to analyze AC signals more, often different types of averages are taken into account. The most important values are:+To analyze AC signals more, often different types of averages are taken into account. The most important values are:
   - the arithmetic mean $\overline{X}$   - the arithmetic mean $\overline{X}$
   - the rectified value $\overline{|X|}$   - the rectified value $\overline{|X|}$
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 Since $sin(\omega t)\geq0$ for $t\in [0,\pi]$, the integral can be changed and the absolute value bars can be excluded like the following  \\ Since $sin(\omega t)\geq0$ for $t\in [0,\pi]$, the integral can be changed and the absolute value bars can be excluded like the following  \\
 \begin{align*} \begin{align*}
-\overline{|X|}    &= {{1}\over{T}}\cdot 2 \cdot \int_{t=0}^{T/2} \hat{X}\cdot \sin( {{2\pi}\over{T}} t ) {\rm d}t \\ +\overline{|X|}    &= {{1}\over{T}}\cdot 2 \cdot \int_{t=0}^{T/2}        \hat{X}\cdot   \sin( {{2\pi}\over{T}} t ) {\rm d}t \\ 
-               &= 2 \cdot {{1}\over{T}}\cdot [-\hat{X}\cdot {{T}\over{2\pi}}\cdot \cos( {{2\pi}\over{T}} t )]_{t=0}^{T/2} \\ +                  &= 2 \cdot {{1}\over{T}}\cdot [-\hat{X}\cdot {{T}\over{2\pi}}\cdot   \cos( {{2\pi}\over{T}} t )]_{t=0}^{T/2} \\ 
-               &= 2 \cdot {{1}\over{T}}\cdot {{T}\over{2\pi}}\cdot \hat{X}\cdot [-\cos( {{2\pi}\over{T}} t )]_{t=0}^{T/2} \\ +                  &= 2 \cdot {{1}\over{T}}\cdot {{T}\over{2\pi}}\cdot   \hat{X}\cdot [-\cos( {{2\pi}\over{T}} t )]_{t=0}^{T/2} \\ 
-               &= {{1}\over{\pi}}\cdot \hat{X} \cdot [1+1] \\ +                  &= {{1}\over{\pi}}\cdot \hat{X} \cdot [1+1] \\ 
-\boxed{\overline{|X|} = {{2}\over{\pi}}\cdot \hat{X} \approx 0.6366 \cdot \hat{X}}\\+\boxed{\overline{|X|}  
 +                   = {{2}\over{\pi}}\cdot \hat{X} \approx 0.6366 \cdot \hat{X}}\\
 \end{align*} \end{align*}
 </callout> </callout>
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 \begin{align*} \begin{align*}
-             P_{\rm DC} &= P_{\rm AC} \\ +             P_{\rm DC}   &= P_{\rm AC} \\ 
-U_{DC} \cdot I_{\rm DC} &= {{1}\over{T}} \int_{0}^{T} u(t) \cdot i(t) {\rm d}t \\ +U_{DC} \cdot I_{\rm DC}   &= {{1}\over{T}} \int_{0}^{T} u(t) \cdot i(t)   {\rm d}t \\ 
-   R \cdot I_{\rm DC}^2 &= {{1}\over{T}} \int_{0}^{T} R \cdot i^2(t) {\rm d}t \\ +     R \cdot I_{\rm DC}^2 &= {{1}\over{T}} \int_{0}^{T} R    \cdot i^2(t) {\rm d}t \\ 
-           I_{\rm DC}^2 &= {{1}\over{T}} \int_{0}^{T} i^2(t) {\rm d}t \\ +             I_{\rm DC}^2 &= {{1}\over{T}} \int_{0}^{T}            i^2(t) {\rm d}t \\ 
-\rightarrow I_{\rm DC} &= \sqrt{{{1}\over{T}} \int_{0}^{T} i^2(t) {\rm d}t}  +\rightarrow  I_{\rm DC}   &= \sqrt{{{1}\over{T}} \int_{0}^{T}      i^2(t) {\rm d}t}  
 \end{align*} \end{align*}
  
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 \begin{align*} \begin{align*}
 X &=  \sqrt{{{1}\over{T}}\cdot \int_{t=t_0}^{t_0 + T} x^2(t)  {\rm d}t} \\ X &=  \sqrt{{{1}\over{T}}\cdot \int_{t=t_0}^{t_0 + T} x^2(t)  {\rm d}t} \\
-  &=  \sqrt{{{1}\over{T}}\cdot \int_{t=0}  ^{T}      \hat{X}^2\cdot \sin^2(\omega t)  {\rm d}t} \\ +  &=  \sqrt{{{1}\over{T}}\cdot \int_{t=0}  ^{T}      \hat{X}^2\cdot                        \sin^2(     \omega t)  {\rm d}t} \\ 
-  &=  \sqrt{{{1}\over{T}}\cdot \int_{t=0}  ^{T}      \hat{X}^2\cdot {{1}\over{2}}\cdot (1- \cos(2\cdot \omega t))  {\rm d}t} \\+  &=  \sqrt{{{1}\over{T}}\cdot \int_{t=0}  ^{T}      \hat{X}^2\cdot {{1}\over{2}}\cdot (1- \cos(2\cdot \omega t)) {\rm d}t} \\
   &=  \sqrt{{{1}\over{T}}\cdot \hat{X}^2\cdot {{1}\over{2}}\cdot [t + {{1}\over{2\omega }}\cdot \sin(2\cdot \omega t)]_{0}^{T}} \\   &=  \sqrt{{{1}\over{T}}\cdot \hat{X}^2\cdot {{1}\over{2}}\cdot [t + {{1}\over{2\omega }}\cdot \sin(2\cdot \omega t)]_{0}^{T}} \\
   &=  \sqrt{{{1}\over{T}}\cdot \hat{X}^2\cdot {{1}\over{2}}\cdot (T - 0  + 0 - 0)} \\   &=  \sqrt{{{1}\over{T}}\cdot \hat{X}^2\cdot {{1}\over{2}}\cdot (T - 0  + 0 - 0)} \\
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 The following simulation shows the different values for averaging a rectangular, a sinusoidal, and a triangular waveform. \\ The following simulation shows the different values for averaging a rectangular, a sinusoidal, and a triangular waveform. \\
-Be aware that one has to wait for a full period in order to see the resulting values on the right outputs of the average generating blocks. +Be aware that one has to wait for a full period to see the resulting values on the right outputs of the average generating blocks. 
  
 <WRAP>  <WRAP> 
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 Now, we insert the functions representing the instantaneous signals and calculate the derivative: Now, we insert the functions representing the instantaneous signals and calculate the derivative:
 \begin{align*} \begin{align*}
- \sqrt{2}{I}\cdot \sin(\omega t + \varphi_i) &= {{\rm d}\over{{\rm d}t}}\left( C \cdot \sqrt{2}{U}\cdot \sin(\omega t + \varphi_u)  \right) \\ + \sqrt{2}{I}\cdot \sin(\omega t + \varphi_i) &= {{\rm d}\over{{\rm d}t}}\left( C \cdot \sqrt{2}{U}\cdot              \sin(\omega t + \varphi_u)  \right) \\ 
-                                             &=  C \cdot \sqrt{2}{U}\cdot \omega \cdot \cos(\omega t + \varphi_u) \\ \\ +                                             &=                                C \cdot \sqrt{2}{U}\cdot \omega \cdot \cos(\omega t + \varphi_u) \\ \\ 
-        {I}\cdot \sin(\omega t + \varphi_i)  &=  C \cdot {U}\cdot \omega \cdot \sin(\omega t + \varphi_u + {{1}\over{2}}\pi)  \tag{6.3.1}+        {I}\cdot \sin(\omega t + \varphi_i)  &=  C \cdot {U}\cdot \omega \cdot                                       \sin(\omega t + \varphi_u + {{1}\over{2}}\pi)  \tag{6.3.1}
 \end{align*} \end{align*}
  
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 \omega t + \varphi_i &= \omega t + \varphi_u + {{1}\over{2}}\pi \\ \omega t + \varphi_i &= \omega t + \varphi_u + {{1}\over{2}}\pi \\
            \varphi_i &=            \varphi_u + {{1}\over{2}}\pi \\            \varphi_i &=            \varphi_u + {{1}\over{2}}\pi \\
-\varphi_u -\varphi_i &=            - {{1}\over{2}}\pi +\varphi_u -\varphi_i &=                      - {{1}\over{2}}\pi 
 \end{align*} \end{align*}
  
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 \begin{align*} \begin{align*}
  \sqrt{2}{U}\cdot \sin(\omega t + \varphi_u) &= L \cdot {{\rm d}\over{{\rm d}t}}\left( \sqrt{2}{I}\cdot \sin(\omega t + \varphi_i)  \right) \\  \sqrt{2}{U}\cdot \sin(\omega t + \varphi_u) &= L \cdot {{\rm d}\over{{\rm d}t}}\left( \sqrt{2}{I}\cdot \sin(\omega t + \varphi_i)  \right) \\
-                                             &= L \cdot \sqrt{2}{I}\cdot \omega \cdot \cos(\omega t + \varphi_i) \\ \\ +                                             &= L \cdot                   \sqrt{2}{I}\cdot \omega \cdot \cos(\omega t + \varphi_i) \\ \\ 
-         {U}\cdot \sin(\omega t + \varphi_u) &= L \cdot {I}\cdot \omega \cdot \sin(\omega t + \varphi_i + {{1}\over{2}}\pi)  \tag{6.3.2}+         {U}\cdot \sin(\omega t + \varphi_u) &= L \cdot                           {I}\cdot \omega \cdot \sin(\omega t + \varphi_i + {{1}\over{2}}\pi)  \tag{6.3.2}
 \end{align*} \end{align*}
  
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 \omega t + \varphi_u &= \omega t + \varphi_i + {{1}\over{2}}\pi \\ \omega t + \varphi_u &= \omega t + \varphi_i + {{1}\over{2}}\pi \\
            \varphi_u &=            \varphi_i + {{1}\over{2}}\pi \\            \varphi_u &=            \varphi_i + {{1}\over{2}}\pi \\
-\boxed{\varphi = \varphi_u -\varphi_i =            + {{1}\over{2}}\pi }+\boxed{\varphi = \varphi_u -\varphi_i =      + {{1}\over{2}}\pi }
 \end{align*} \end{align*}
  
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 </callout> </callout>
  
-For the concept of AC two-terminal networks, we are also able to use the DC methods of network analysis in order to solve AC networks.+For the concept of AC two-terminal networks, we are also able to use the DC methods of network analysis to solve AC networks.
  
 ===== 6.4 Complex Values in Electrical Engineering ===== ===== 6.4 Complex Values in Electrical Engineering =====
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 Up to now, we used the following formula to represent alternating voltages: Up to now, we used the following formula to represent alternating voltages:
  
-$$u(t)= \sqrt{2} \hat{U} \cdot \sin (\varphi)$$+$$u(t)= \sqrt{2} U \cdot \sin (\varphi)$$
  
 This is now interpreted as the instantaneous value of a complex vector $\underline{u}(t)$, which rotates given by the time-dependent angle $\varphi = \omega t + \varphi_u$. This is now interpreted as the instantaneous value of a complex vector $\underline{u}(t)$, which rotates given by the time-dependent angle $\varphi = \omega t + \varphi_u$.
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 <imgcaption pic08 | representation of a voltage phasor on the complex plane>  <imgcaption pic08 | representation of a voltage phasor on the complex plane> 
 </imgcaption>  </imgcaption> 
-\\ {{drawio>voltagephasorcomplexplane,svg}} \\ +\\ {{drawio>voltagephasorcomplexplane.svg}} \\ 
 </WRAP> </WRAP>
  
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 \underline{u}(t) &=\sqrt{2}                          U \cdot {\rm e}^{{\rm j} (\omega t + \varphi_u)} \\ \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}}                   &=\sqrt{2}\color{blue}             {U \cdot {\rm e}^{{\rm j} \varphi_u}} 
-                                                       \cdot {\rm e}^{{\rm j}  \omega t } \\+                                                       \cdot {\rm e}^{{\rm j} \omega t} \\
                  &=\sqrt{2}\color{blue}{\underline{U}} \cdot {\rm e}^{{\rm j} \omega t} \\                  &=\sqrt{2}\color{blue}{\underline{U}} \cdot {\rm e}^{{\rm j} \omega t} \\
 \end{align*} \end{align*}
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 Therefore, the known properties of complex numbers from Mathematics 101 can be applied: 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 $j$ equals a phase shift of $+90°$
-  * A multiplication with $-j$ equals a phase shift of $-90°$+  * A multiplication with ${{1}\over{j}}$ equals a phase shift of $-90°$
  
 ===== 6.5 Complex Impedance ===== ===== 6.5 Complex Impedance =====
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   * $X = Z \sin \varphi$   * $X = Z \sin \varphi$
  
-==== 6.5.2 Application on pure Loads ====+y ==== 6.5.2 Application on pure Loads ====
  
 With the complex impedance in mind, the <tabref tab01> can be expanded to:  With the complex impedance in mind, the <tabref tab01> can be expanded to: 
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 \\ \\ \\ \\
 The relationship between ${\rm j}$ and integral calculus should be clear:  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 "$\cdot {\rm j}$",  which also means a phase shift of $+90°$: \\ ${{\rm d}\over{{\rm d}t}} {\rm e}^{{\rm j}(\omega t + \varphi_x)} = {\rm j} \cdot {\rm e}^{{\rm j}(\omega t + \varphi_x)}$ +  - 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 "$\cdot (-{\rm j})$", which also means a phase shift of $-90°$.((in general, here the integration constant must be considered. This is however often be neglectable since only AC values (without a DC value) are considered.)) \\ $\int {\rm e}^{{\rm j}(\omega t + \varphi_x)} = {{1}\over{\rm j}} \cdot {\rm e}^{{\rm j}(\omega t + \varphi_x)} = - {\rm j} \cdot {\rm e}^{{\rm j}(\omega t + \varphi_x)}$+  - 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. 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.
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 <panel type="info" title="Exercise 6.3.1 Impedance of single Components I"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> <panel type="info" title="Exercise 6.3.1 Impedance of single Components I"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
-A coil has a reactance of $80\Omega$ at a frequency of $500 ~\rm Hz$. At which frequencies the impedance will have the following values?+A coil has a impedance of $80~\Omega$ at a frequency of $500 ~\rm Hz$. At which frequencies the impedance will have the following values?
   - $85  ~\Omega$   - $85  ~\Omega$
   - $120 ~\Omega$   - $120 ~\Omega$
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 <panel type="info" title="Exercise 6.3.2 Impedance of single Components II"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> <panel type="info" title="Exercise 6.3.2 Impedance of single Components II"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
-A capacitor with $5 ~{\rm µF}$ is connected to a voltage source which generates $U_\sim = 200 ~{\rm V}$. At which frequencies the following currencies can be measured?+A capacitor with $5 ~{\rm µF}$ is connected to a voltage source which generates $U_\sim = 200 ~{\rm V}$. At which frequencies the following currents can be measured?
   - $0.5 ~\rm A$   - $0.5 ~\rm A$
   - $0.8 ~\rm A$   - $0.8 ~\rm A$
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 <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">  <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:  The phasor diagram looks roughly like this: 
-{{drawio>phasordiagram6511}}  
 {{drawio>phasordiagram6511.svg}}  {{drawio>phasordiagram6511.svg}} 
  
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 The phasor diagram looks roughly like this. \\  The phasor diagram looks roughly like this. \\ 
 But have a look at the solution for question 5! But have a look at the solution for question 5!
-{{drawio>phasordiagram6514}}  
 {{drawio>phasordiagram6514.svg}}  {{drawio>phasordiagram6514.svg}} 
  
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 The calculated (positive) horizontal and (negative) vertical dimension for the voltage indicates a phasor in the fourth quadrant. Does it seem right? \\ 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. \\ The phasor diagram which was shown in answer 4. cannot be correct. \\
-With the correct lengths and angles, the real phasor diagram actually looks like this: +With the correct lengths and angles, the real phasor diagram looks like this:
-{{drawio>phasordiagram6515}}+
 {{drawio>phasordiagram6515.svg}} {{drawio>phasordiagram6515.svg}}
 Here the phasor is in the fourth quadrant with a negative angle. \\ Here the phasor is in the fourth quadrant with a negative angle. \\
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 <WRAP indent> <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"> <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}}{{drawio>SeriesPhasor.svg}}+The drawing of the voltage pointers is as follows:{{drawio>SeriesPhasor.svg}}
 The voltage U is determined by the law of Pythagoras  The voltage U is determined by the law of Pythagoras 
 \begin{align*}  \begin{align*} 
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 <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"> <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}}{{drawio>SeriesPhasor2.svg}}+The drawing of the voltage pointers is as follows: {{drawio>SeriesPhasor2.svg}}
 The voltage $U_R$ is determined by the law of Pythagoras  The voltage $U_R$ is determined by the law of Pythagoras 
 \begin{align*}  \begin{align*} 
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 in the following, some of the numbers are given.  in the following, some of the numbers are given. 
-Calculate the RMS value of the missing voltage and the phase shift $\varphi$ between $U$ and $I$.+Calculate the RMS value of the missing currents and the phase shift $\varphi$ between $U$ and $I$.
   - $I_R = 3~\rm A$, $I_L = 1  ~\rm A$, $I_C = 5  ~\rm A$, $I=?$   - $I_R = 3~\rm A$, $I_L = 1  ~\rm A$, $I_C = 5  ~\rm A$, $I=?$
   - $I_R = ?$,       $I_L = 1.2~\rm A$, $I_C = 0.4~\rm A$, $I=1~\rm A$   - $I_R = ?$,       $I_L = 1.2~\rm A$, $I_C = 0.4~\rm A$, $I=1~\rm A$
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 <panel type="info" title="Exercise 6.5.5 Complex Calculation I"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> <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 calulation!+The following two currents with similar frequencies, but different phases have to be added. Use complex calculation!
   * $i_1(t) = \sqrt{2} \cdot 2 ~A \cdot \cos (\omega t + 20°)$   * $i_1(t) = \sqrt{2} \cdot 2 ~A \cdot \cos (\omega t + 20°)$
   * $i_2(t) = \sqrt{2} \cdot 5 ~A \cdot \cos (\omega t + 110°)$   * $i_2(t) = \sqrt{2} \cdot 5 ~A \cdot \cos (\omega t + 110°)$
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 <panel type="info" title="Exercise 6.5.6 Complex Calculation II"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> <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.  Two complex impedances $\underline{Z}_1$ and $\underline{Z}_2$ are investigated. 
-The resulting impedance for a series circuit is $60~\Omega$.  +The resulting impedance for a series circuit is   $60~\Omega + \rm j \cdot 0 ~\Omega $.  
-The resulting impedance for a parallel circuit is $25~\Omega$.+The resulting impedance for a parallel circuit is $25~\Omega + \rm j \cdot 0 ~\Omega $.
  
 What are the values for $\underline{Z}_1$ and $\underline{Z}_2$? 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*}
 +0 &=  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} \\ \\
 +
 +0 &= 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~@#
 +
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