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electrical_engineering_1:task_jti0uzudcmg4u22t_with_calculation [2023/02/12 06:34] – mexleadminelectrical_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}} 
- 
-<panel type="info" > <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> 
-<fs x-large>**Exercise ~~#@ee1_taskctr.#~~ : Analyzing complex Impedances ** \\ (written test, approx. 14% of a 60-minute written test, WS2022) \\ \\ </fs> 
- 
-A circuit with an ideal voltage source ($U=50 V$, $f=330 Hz$) 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+4j}}+5j \right) \Omega 
-\end{align*} 
- 
-1. Calculate the physical values of the two components. 
- 
-<button size="xs" type="link" collapse="jti0uzudcmg4u22t_1_path">{{icon>eye}} Solution</button><collapse id="jti0uzudcmg4u22t_1_path" collapsed="true"> 
-<callout type="tip" icon="true"> 
- 
-\begin{align*} 
-\underline{Z} &= \left({{2}\over{3+4j}} + 5j \right) \Omega \\ 
-              &= \left({{2}\over{3+4j}} \cdot {{3-4j}\over{3-4j}} + 5j \right) \Omega \\ 
-              &= \left({{2}\over{9+16}} \cdot (3-4j) + 5j \right) \Omega \\ 
-              &= \left(0.24 - 0.32j + 5j \right) \Omega \\ 
-              &= 0.24 \Omega + j \cdot 4.68 \Omega \\ 
-              &= R + 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 Hz}} \\ 
-\end{align*} 
-</callout></collapse> 
- 
-<button size="xs" type="link" collapse="jti0uzudcmg4u22t_1_solution">{{icon>eye}} Final result</button><collapse id="jti0uzudcmg4u22t_1_solution" collapsed="true"><callout type="tip" icon="true"> 
- 
-\begin{align*} 
- R  &= 0.24 \Omega \\ 
- L  &= 2.26 mH  
-\end{align*}</callout> 
- \\ 
-</collapse> 
- 
-2. Calculate the phase and absolute value of complex current $\underline{I}$ through the circuit. 
- 
-<button size="xs" type="link" collapse="jti0uzudcmg4u22t_2_path">{{icon>eye}} Solution</button><collapse id="jti0uzudcmg4u22t_2_path" collapsed="true"> 
-<callout type="tip" icon="true"> 
-\begin{align*} 
-\underline{I} &= {{\underline{U}}\over{\underline{Z}}} \\ 
-              &= {{50 V}\over{ 0.24 \Omega + j \cdot 4.68 \Omega }} \\ 
-              &= {{50 V}\over{ 0.24 \Omega + j \cdot 4.68 \Omega }} \cdot {{ 0.24 \Omega - j \cdot 4.68 \Omega }\over{ 0.24 \Omega - j \cdot 4.68 \Omega }} \\ 
-              &= {{50 V}\over{ (0.24 \Omega)^2 + (4.68 \Omega)^2 }} \cdot ( 0.24 \Omega - 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 V}\over{| 0.24 \Omega + j \cdot 4.68 \Omega |}} \\ 
-                &= {{50 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*} 
- 
-</callout></collapse> 
- 
-<button size="xs" type="link" collapse="jti0uzudcmg4u22t_2_solution">{{icon>eye}} Final result</button><collapse id="jti0uzudcmg4u22t_2_solution" collapsed="true"><callout type="tip" icon="true"> 
- 
-\begin{align*} 
-|\underline{I}| &= 10.67 A \\ 
-\varphi_i       &= -87.06°  
-\end{align*}</callout> 
- \\ 
-</collapse> 
- 
-3. 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! 
- 
-<button size="xs" type="link" collapse="jti0uzudcmg4u22t_3_path">{{icon>eye}} Solution</button><collapse id="jti0uzudcmg4u22t_3_path" collapsed="true"> 
-<callout type="tip" icon="true"> 
-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 Hz \cdot 4.68 \Omega}} \\ 
-\end{align*} 
- 
-</callout></collapse> 
- 
-<button size="xs" type="link" collapse="jti0uzudcmg4u22t_3_solution">{{icon>eye}} Final result</button><collapse id="jti0uzudcmg4u22t_3_solution" collapsed="true"> 
-<callout type="tip" icon="true"> 
-\begin{align*} 
- C  = 103 \mu F 
-\end{align*}</callout> 
- \\ 
-</collapse> 
- 
- 
-</WRAP></WRAP></panel>