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--- electrical_engineering_1:task_pdkggtyexxy1ktu3_with_calculation [2023/03/10 17:30] – 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}}
-<panel type="info" > <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
-<fs x-large>**Exercise ~~#@ee1_taskctr.#~~ : Impedances at Frequencies ** \\ (written test, approx. 18 % of a 60-minute written test, WS2022) \\ \\ </fs>
-Calculate the **resistor values** which have to be used the following circuits.
-. A resistor $R_1$ shall have the same absolute value of the impedance like a capacitor $C_1=40 ~\rm{nF}$ at $f_1=4 ~\rm{MHz}$.
-<button size="xs" type="link" collapse="pdkggtyexxy1ktu3_1_path">{{icon>eye}} Solution</button><collapse id="pdkggtyexxy1ktu3_1_path" collapsed="true">
-<callout type="tip" icon="true">
-\begin{align*}
-R_1 &= |\underline{X}_{C1}| \\
-    &= {{1}\over{2\pi \cdot f \cdot C_1}} \\
-    &= {{1}\over{2\pi \cdot 4 ~MHz \cdot 40 ~nF}} \\
-\end{align*}
-</callout></collapse>
-<button size="xs" type="link" collapse="pdkggtyexxy1ktu3_1_solution">{{icon>eye}} Final result</button><collapse id="pdkggtyexxy1ktu3_1_solution" collapsed="true"><callout type="tip" icon="true">
-\begin{align*}
- R_1  &= 1.00 ~\Omega \\
-\end{align*}</callout>
- \\
-</collapse>
-. A $RL$ series circuit with $L_2=4.7 ~\mu H$, where an AC voltage source of $U_2=1.0 ~V$ with $f_2=450 kHz$ generates a current $I_2=60 ~mA$.
-<button size="xs" type="link" collapse="pdkggtyexxy1ktu3_2_path">{{icon>eye}} Solution</button><collapse id="pdkggtyexxy1ktu3_2_path" collapsed="true">
-<callout type="tip" icon="true">
-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 + j \cdot \omega L
-\end{align*}
-Since $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*}
-</callout></collapse>
-<button size="xs" type="link" collapse="pdkggtyexxy1ktu3_2_solution">{{icon>eye}} Final result</button><collapse id="pdkggtyexxy1ktu3_2_solution" collapsed="true"><callout type="tip" icon="true">
-\begin{align*}
- R_2  &= 10.0 ~\Omega \\
-\end{align*}</callout>
- \\
-</collapse>
-. A $RC$ parallel circuit with $C_3=4.7 ~nF$ on an AC current source ($I_{3S}=1.3 ~A$,$f_3=200 ~kHz$), which generates a current of $I_{3R}=1.0 ~A$ through $R_3$.
-<button size="xs" type="link" collapse="pdkggtyexxy1ktu3_3_path">{{icon>eye}} Solution</button><collapse id="pdkggtyexxy1ktu3_3_path" collapsed="true">
-<callout type="tip" icon="true">
-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} = -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  $-j\cdot {X}_{3C}$, 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 on 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*}
-</callout>
-</collapse>
-<button size="xs" type="link" collapse="pdkggtyexxy1ktu3_3_solution">{{icon>eye}} Final result</button><collapse id="pdkggtyexxy1ktu3_3_solution" collapsed="true"><callout type="tip" icon="true">
-\begin{align*}
- R_3  &= 70.0 ~\Omega \\
-\end{align*}</callout>
- \\
-</collapse>
-</WRAP></WRAP></panel>

complex impedance exam ee1 ws2022