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--- electrical_engineering_1:introduction_in_alternating_current_technology [2023/07/17 02:41]
mexleadmin [6.5.2 Application on pure Loads]
+++ electrical_engineering_1:introduction_in_alternating_current_technology [2023/12/20 09:55] (aktuell)
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
@@ Zeile 476: / Zeile 476: @@
 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 $\varphi = \omega t + \varphi_u$.
@@ Zeile 504: / Zeile 504: @@
 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 $j\omega$ equals a phase shift of $+90°$
-  * A multiplication with $-j$ equals a phase shift of $-90°$
+  * A multiplication with ${{1}\over{j\omega}}$ equals a phase shift of $-90°$
 ===== 6.5 Complex Impedance =====
@@ Zeile 556: / Zeile 556: @@
 \\ \\
 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 $+90°$: \\ \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 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 neglectable since only AC values (without a DC value) are considered.)) <WRAP>
+  - 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}} \cdot {\rm e}^{{\rm j}(\omega t + \varphi_x)}
+  = {{1}\over{\rm j\omega}} \cdot {\rm e}^{{\rm j}(\omega t + \varphi_x)}
-  =         - {\rm j} \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>
@@ Zeile 817: / Zeile 817: @@
 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 $I$.
   - $I_R = 3~\rm A$, $I_L = 1  ~\rm A$, $I_C = 5  ~\rm A$, $I=?$
   - $I_R = ?$,       $I_L = 1.2~\rm A$, $I_C = 0.4~\rm A$, $I=1~\rm A$
@@ Zeile 824: / 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%>
-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!
   * $i_1(t) = \sqrt{2} \cdot 2 ~A \cdot \cos (\omega t + 20°)$
   * $i_2(t) = \sqrt{2} \cdot 5 ~A \cdot \cos (\omega t + 110°)$
@@ Zeile 832: / 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%>
 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 $\underline{Z}_2$?
+#@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>