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--- electrical_engineering_1:task_pdkggtyexxy1ktu3_with_calculation [2023/04/02 00:27] – mexleadmin
+++ electrical_engineering_1:task_pdkggtyexxy1ktu3_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>7000}}
-#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~  Impedances at different Frequencies  \\ <fs medium>(written test, approx. 18 % of a 60-minute written test, WS2022)</fs> #@TaskText_HTML@#
-Calculate the **resistor values** which have to be used in the following circuits.
-. A resistor $R_1$ shall have the same absolute value of the impedance as a capacitor $C_1=40 ~{\rm nF}$ at $f_1=4 ~{\rm MHz}$.
-#@PathBegin_HTML~1~@#
-\begin{align*}
-R_1 &= |\underline{X}_{C1}| \\
-    &= {{1}\over{2\pi \cdot f \cdot C_1}} \\
-    &= {{1}\over{2\pi \cdot 4 ~{\rm MHz} \cdot 40 ~\rm nF}} \\
-\end{align*}
-#@PathEnd_HTML@#
-#@ResultBegin_HTML~1~@#
-\begin{align*}
- R_1  &= 1.00 ~\Omega \\
-\end{align*}
-#@ResultEnd_HTML@#
-. A $RL$ series circuit with $L_2=4.7 ~\rm µH$, where an AC voltage source of $U_2=1.0 ~\rm V$ with $f_2=450 ~\rm kHz$ generates a current $I_2=60 ~\rm mA$.
-#@PathBegin_HTML~2~@#
-A series circuit means that the current is constant on every component. \\
-The equivalent impedance for $R$ and $L$ combined is given by
-\begin{align*}
-{{\underline{U}}\over{\underline{I}}} &= R_2 + \underline{X}_{L2} \\
-                                      &= R_2 + {\rm j} \cdot \omega L
-\end{align*}
-Since ${\rm j} \cdot \omega L $ is perpendicular to $R_2$ this can be simplified to:
-\begin{align*}
-\left| {{\underline{U}}\over{\underline{I}}} \right|^2 &= |R_2|^2 + |\underline{X}_{L2}|^2 \\
-\left( {{U}\over{I}} \right)^2 &= {R_2}^2 + {X_{L2}}^2 \\
-\end{align*}
-This can be rearranged to get $R_2$:
-\begin{align*}
-R_2 &= \sqrt{ \left( {{U }\over{I   }} \right)^2 - X_{L2}^2 } \\
-    &= \sqrt{ \left( {{1~{\rm V}}\over{60~\rm mA}} \right)^2 - (2\pi \cdot 450~{\rm kHz} \cdot 4.7 ~ {\rm µH})^2 } \\
-\end{align*}
-#@PathEnd_HTML@#
-#@ResultBegin_HTML~2~@#
-\begin{align*}
- R_2  &= 10.0 ~\Omega \\
-\end{align*}
-#@ResultEnd_HTML~2~@#
-. A $RC$ parallel circuit with $C_3=4.7 ~\rm nF$ on an AC current source ($I_{3S}=1.3 ~\rm A$,$f_3=200 ~\rm kHz$), which generates a current of $I_{3R}=1.0 ~\rm A$ through $R_3$.
-#@PathBegin_HTML~3~@#
-Parallel circuit means that the voltage is the same on $R_3$ and $C_3$: \\
-\begin{align*}
- \underline{U}_3 = R_3 \cdot \underline{I}_{3R} = -{\rm j}\cdot {X}_{3C} \cdot \underline{I}_{3C}
-\end{align*}
-So it gets clear, that $\underline{I}_{3R}$ is perpendicular to $\underline{I}_{3C}$ (It has to, since $R_3$ is perpendicular to  $-{\rm j}\cdot {X}_{3C}$, too). \\
-Therefore, the resulting current of the parallel circuit is given as:
-\begin{align*}
- \underline{I}_{3}     &=  \underline{I}_{3R}    +  \underline{I}_{3C} \\
- |\underline{I}_{3}|^2 &= |\underline{I}_{3R}|^2 + |\underline{I}_{3C}|^2 \\
-  {I}_{3C}    &= \sqrt{|{I}_{3}|^2 - |{I}_{3R}|^2}
-\end{align*}
-Back to the first formula:
-\begin{align*}
-R_3 \cdot {I}_{3R} &= {X}_{3C} \cdot {I}_{3C} \\
-R_3  &=  {X}_{3C} \cdot {{{I}_{3C}}\over{{I}_{3R}}} \\
-     &=  {{1}\over{2\pi \cdot f \cdot C_3}} \cdot {{\sqrt{|{I}_{3}|^2 - |{I}_{3R}|^2}}\over{{I}_{3R}}} \\
-\end{align*}
-#@PathEnd_HTML~3~@#
-#@ResultBegin_HTML~3~@#
-\begin{align*}
- R_3  &= 70.0 ~\Omega \\
-\end{align*}
-#@ResultEnd_HTML~3~@#
-#@TaskEnd_HTML@#

complex impedance exam ee1 ws2022