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--- electrical_engineering_1:task_jti0uzudcmg4u22t_with_calculation [2023/04/02 00:27] – mexleadmin
+++ electrical_engineering_1:task_jti0uzudcmg4u22t_with_calculation [Unbekanntes Datum] (aktuell) – gelöscht - Externe Bearbeitung (Unbekanntes Datum) 127.0.0.1
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-{{tag>complex_impedance exam_ee1_WS2022}}
-{{include_n>6000}}
-#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~  Analyzing complex Impedances  \\ <fs medium>(written test, approx. 14 % of a 60-minute written test, WS2022)</fs> #@TaskText_HTML@#
-A circuit with an ideal voltage source ($U=50 ~\rm V$, $f=330 ~\rm Hz$) and two components ($R$ and $\underline{X}_1$) shall be given. \\
-After analysis, the following formula for the impedance was extracted:
-\begin{align*}
-\underline{Z} = \left({{2}\over{3+4{\rm j}}}+5{\rm j} \right) \Omega
-\end{align*}
-. Calculate the physical values of the two components.
-#@PathBegin_HTML~1~@#
-\begin{align*}
-\underline{Z} &= \left({{2}\over{3+4{\rm j}}} + 5{\rm j} \right) ~\Omega \\
-              &= \left({{2}\over{3+4{\rm j}}} \cdot {{3-4{\rm j}}\over{3-4j}} + 5{\rm j} \right) ~\Omega \\
-              &= \left({{2}\over{9+16      }} \cdot  (3-4{\rm j}            ) + 5{\rm j} \right) ~\Omega \\
-              &= \left(0.24 - 0.32{\rm j}                                     + 5{\rm j} \right) ~\Omega \\
-              &= 0.24 ~\Omega + {\rm j} \cdot 4.68 ~\Omega \\
-              &= R + {\rm j} X_L \\
-\end{align*}
-With the complex part comes the physical value:
-\begin{align*}
-X_L &= \omega L \\
- L  &= {{X_L}\over{2\pi \cdot f}} \\
-    &= {{4.68 ~\Omega}\over{2\pi \cdot 300 ~\rm{Hz}}} \\
-\end{align*}
-#@PathEnd_HTML@#
-#@ResultBegin_HTML~1~@#
-\begin{align*}
- R  &= 0.24 ~\Omega \\
- L  &= 2.26 ~\rm{mH}
-\end{align*}
-#@ResultEnd_HTML@#
-. Calculate the phase and absolute value of complex current $\underline{I}$ through the circuit.
-#@PathBegin_HTML~2~@#
-\begin{align*}
-\underline{I} &= {{\underline{U}}\over{\underline{Z}}} \\
-              &= {{50 ~\rm{V}}\over{ 0.24 ~\Omega + {\rm j} \cdot 4.68 ~\Omega }} \\
-              &= {{50 ~\rm{V}}\over{ 0.24 ~\Omega + {\rm j} \cdot 4.68 ~\Omega }} \cdot {{ 0.24 ~\Omega - {\rm j} \cdot 4.68 ~\Omega }\over{ 0.24 ~\Omega - {\rm j} \cdot 4.68 ~\Omega }} \\
-              &= {{50 ~\rm{V}}\over{(0.24 ~\Omega)^2 + (4.68 ~\Omega)^2 }}        \cdot  ( 0.24 ~\Omega - {\rm j} \cdot 4.68 ~\Omega ) \\
-\end{align*}
-The absolute value $|\underline{I}|$ can be calculated as:
-\begin{align*}
-|\underline{I}| &= {|{\underline{U}|}\over{|\underline{Z}|}} \\
-                &= {{50 ~\rm{V}}\over{| 0.24 ~\Omega + {\rm j} \cdot 4.68 ~\Omega |}} \\
-                &= {{50 ~\rm{V}}\over{\sqrt{ (0.24 ~\Omega)^2 + (4.68 ~\Omega)^2 }}}
-\end{align*}
-The phase $\varphi_i$ can be calculated as
-\begin{align*}
-\varphi_i &= \arctan \left( {{\Im()}\over{\Re()}} \right) \\
-          &= \arctan \left( {{-4.68 ~\Omega}\over{0.24 ~\Omega}} \right) \\
-\end{align*}
-#@PathEnd_HTML@#
-#@ResultBegin_HTML~2~@#
-\begin{align*}
-|\underline{I}| &= 10.67 ~\rm{A} \\
-\varphi_i       &= -87.06°
-\end{align*}
-#@ResultEnd_HTML@#
-. Now an additional component $\underline{X}_2$ shall be added in series to the two components. \\
-This component shall be dimensioned in such a way that the current and voltage are in phase. Calculate these component value!
-#@PathBegin_HTML~3~@#
-The current and voltage are in phase once there is only a pure ohmic (= pure real) resulting impedance $\underline{Z} + \underline{X}_2$. \\
-Therefore, the component mus be a capacitor with the same absolute value of impedance: $|\underline{X}_C| = |\underline{X}_L| $
-\begin{align*}
-X_C &= {{1}\over{\omega \cdot C}} = X_L \\
- C  &= {{1}\over{\omega \cdot X_L}} \\
-    &= {{1}\over{2\pi \cdot 300 ~\rm{Hz} \cdot 4.68 ~\Omega}} \\
-\end{align*}
-#@PathEnd_HTML@#
-#@ResultBegin_HTML~3~@#
-\begin{align*}
- C  = 103 ~\rm{µF}
-\end{align*}
- \\
-#@ResultEnd_HTML@#
-#@TaskEnd_HTML@#

complex impedance exam ee1 ws2022