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electrical_engineering_1:introduction_in_alternating_current_technology [2023/03/27 09:21]
mexleadmin 		electrical_engineering_1:introduction_in_alternating_current_technology [2023/12/20 09:53]
mexleadmin
Zeile 1: Zeile 1:
-====== 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
Zeile 152: Zeile 152:
 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>
Zeile 175: Zeile 176:
    
 \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*}
  
Zeile 204: Zeile 205:
 \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)} \\
Zeile 295: Zeile 296:
 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*}
  
Zeile 309: Zeile 310:
 \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*}
  
Zeile 353: Zeile 354:
 \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*}
  
Zeile 366: Zeile 367:
 \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*}
  
Zeile 495: Zeile 496:
 \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*}
Zeile 541: Zeile 542:
   * $X = Z \sin \varphi$   * $X = Z \sin \varphi$
  
-==== 6.5.2 Application on pure Loads ====+value - and therefore a phasor - can simply ==== 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: 
Zeile 555: Zeile 556:
 \\ \\ \\ \\
 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 "$\cdot {\rm j}$", \\ 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 "$\cdot (-{\rm j})$", \\ 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}} \cdot {\rm e}^{{\rm j}(\omega t + \varphi_x)}  
 +  =         - {\rm j} \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.
Zeile 810: Zeile 817:
  
 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 $I$.
   - $I_R = 3~\rm A$, $I_L = 1  ~\rm A$, $I_C = 5  ~\rm A$, $I=?$   - $I_R = 3~\rm A$, $I_L = 1  ~\rm A$, $I_C = 5  ~\rm A$, $I=?$
   - $I_R = ?$,       $I_L = 1.2~\rm A$, $I_C = 0.4~\rm A$, $I=1~\rm A$   - $I_R = ?$,       $I_L = 1.2~\rm A$, $I_C = 0.4~\rm A$, $I=1~\rm A$
Zeile 817: Zeile 824:
  
 <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!
   * $i_1(t) = \sqrt{2} \cdot 2 ~A \cdot \cos (\omega t + 20°)$   * $i_1(t) = \sqrt{2} \cdot 2 ~A \cdot \cos (\omega t + 20°)$
   * $i_2(t) = \sqrt{2} \cdot 5 ~A \cdot \cos (\omega t + 110°)$   * $i_2(t) = \sqrt{2} \cdot 5 ~A \cdot \cos (\omega t + 110°)$
Zeile 825: Zeile 832:
 <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~@#
 +
 </WRAP></WRAP></panel> </WRAP></WRAP></panel>