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--- electrical_engineering_1:task_6tqttque1e2nf2c7_with_calculation [2023/02/12 06:24] – mexleadmin
+++ electrical_engineering_1:task_6tqttque1e2nf2c7_with_calculation [Unbekanntes Datum] (aktuell) – gelöscht - Externe Bearbeitung (Unbekanntes Datum) 127.0.0.1
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-{{tag>dc_network_analysis pure_resistor_network_simplification delta_wye_transformation exam_ee1_WS2022}}
-<panel type="info" > <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
-<fs x-large>**Exercise ~~#@ee1_taskctr.#~~ : Equivalent Linear Source ** \\ (written test, approx. 14% of a 60-minute written test, WS2022) \\ \\ </fs>
-The circuit in the following has to be simplified.
-{{drawio>electrical_engineering_1:6tqttque1e2nf2c7Circuit.svg}}
-Calculated the internal resistance $R_i$ and the source voltage $U_s$ of an equivalent linear voltage source on the connectors $A$ and $B$.
-\begin{align*}
- R_1=5.0 \Omega, && U_2=6.0 V,      && R_3=10 \Omega, \\
- I_4=4.2 A,      && R_5=10 \Omega , && R_6=7.5 \Omega, \\
- R_7=15 \Omega   &&                 &&
-\end{align*}
-Use equivalent sources in order to simplify the circuit!
-<button size="xs" type="link" collapse="6tqttque1e2nf2c7_1_path">{{icon>eye}} Solution</button><collapse id="6tqttque1e2nf2c7_1_path" collapsed="true">
-Best thing is to re-think the wiring like rubber bands and adjust them:
-{{drawio>electrical_engineering_1:6tqttque1e2nf2c7CircuitSolution1.svg}} \\
-The linear voltage source of $U_2$ and $R_1$ can be transormed into a current source $I_2={{U_2}\over{R_1}}$ and $R_1$:
-{{drawio>electrical_engineering_1:6tqttque1e2nf2c7CircuitSolution2.svg}} \\
-Now a lot of can be combined. The resistors $R_1$, $R_3$, $R_5$ are in parallel, like also $I_2$ and $I_4$:
-\begin{align*}
-R_{135} &= R_1||R_3||R_5\\
-I_{24} &= I_2 - I_4 = {{U_2}\over{R_1}} - I_4
-\end{align*}
-The resulting circuit can again be transformed:
-{{drawio>electrical_engineering_1:6tqttque1e2nf2c7CircuitSolution3.svg}} \\
-Here, the $U_{24}$ is calculated by $I_{24}$ as the following:
-\begin{align*}
-U_{24} &= R_{135} \cdot I_{24} \\
-       &= ({{U_2}\over{R_1}} - I_4) \cdot R_1||R_3||R_5
-\end{align*}
-On the right side of the last circuit there is a voltage divider given by $R_{135}$, $R_6$ and $R_7$. \\
-Therefore the voltage between $A$ and $B$ is given as:
-\begin{align*}
-U_{AB} &= U_{24} \cdot {{R_7}\over{R_6 + R_7 + R_1||R_3||R_5}} \\
-       &= ({{U_2}\over{R_1}} - I_4) \cdot {{R_7 \cdot R_1||R_3||R_5}\over{R_6 + R_7 + R_1||R_3||R_5}} \\
-\end{align*}
-For the internal resistance $R_i$ the ideal voltage source is substituted by its resistance ($=0\Omega$, so a short-circuit):
-\begin{align*}
-R_{AB} &= R_7 || ( R_6 + R_1||R_3||R_5) \\
-\end{align*}
-with $R_1||R_3||R_5 = 5 \Omega || 10 \Omega || 10 \Omega =  5 \Omega ||  5 \Omega = 2.5 \Omega$:
-\begin{align*}
-U_{AB} &= ({{6.0 V}\over{5.0 \Omega}} - 4.2 \Omega) \cdot {{15 \Omega \cdot 2.5 \Omega}\over{7.5 \Omega + 15 \Omega + 2.5 \Omega}} \\
-R_{AB} &= 15 \Omega|| ( 7.5 \Omega + 2.5 \Omega) \\
-\end{align*}
-</collapse>
-<button size="xs" type="link" collapse="6tqttque1e2nf2c7_1_solution">{{icon>eye}} Final result</button><collapse id="6tqttque1e2nf2c7_1_solution" collapsed="true">
-\begin{align*}
-U_{AB} &= 4.5 V\\
-R_{AB} &= 6 \Omega \\
-\end{align*}
- \\
-</collapse>
-</WRAP></WRAP></panel>

dc network analysis pure resistor network simplification delta wye transformation exam ee1 ws2022