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electrical_engineering_1:task_x357drkaqv84jnsc_with_calculation [2023/02/12 06:41] – mexleadminelectrical_engineering_1:task_x357drkaqv84jnsc_with_calculation [Unbekanntes Datum] (aktuell) – gelöscht - Externe Bearbeitung (Unbekanntes Datum) 127.0.0.1
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-{{tag>  exam_ee1_WS2022}} 
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-<panel type="info" > <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> 
-<fs x-large>**Exercise ~~#~~ : Pure Resistor Network Simplification ** \\ (written test, approx. 13% of a 60-minute written test, WS2022) \\ \\ </fs> 
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-The following circuit with $R_1=200 \Omega$, $R_2=R_3=100 \Omega$ and the switch $S$ is given. \\ 
-{{drawio>electrical_engineering_1:70jjg4yzwctqarsqCircuit.svg}} 
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-1. The switch shall now be open. Calculate the equivalent resistance $R_{eq}$ between $A$ and $B$. 
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-<button size="xs" type="link" collapse="X357DRKAQV84JNSC_1_path">{{icon>eye}} Solution</button><collapse id="X357DRKAQV84JNSC_1_path" collapsed="true"> 
-<callout type="tip" icon="true">With the switch open the resistor $R_3$ dies not take part into the resulting resistor. 
-{{drawio>electrical_engineering_1:70jjg4yzwctqarsqCircuitSolution1.svg}} 
-The equivalent resistor is given by a parallel configuation of resistors in series: 
-\begin{align*} 
-R_{eq} &= (R_2 + R_1 + R_1)||(R_2 + R_2)\\ 
-R_{eq} &= (100 \Omega + 200 \Omega + 200 \Omega )&&||(100 \Omega + 100 \Omega ) &&\\ 
-R_{eq} &= (500 \Omega )&&||(200 \Omega )&&\\ 
-R_{eq} &= {{500 \Omega \cdot 200 \Omega }\over{500 \Omega + 200 \Omega}}&&\\ 
-\end{align*} 
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-</callout></collapse> 
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-<button size="xs" type="link" collapse="X357DRKAQV84JNSC_1_solution">{{icon>eye}} Final result</button><collapse id="X357DRKAQV84JNSC_1_solution" collapsed="true"> 
-<callout type="tip" icon="true"> 
-\begin{align*} 
-R_{eq} &= 142.8 \Omega \\ 
-\end{align*} 
- \\ 
-</callout></collapse> 
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-2. The switch shall now be closed. Calculate the equivalent resistance $R_{eq}$ between $A$ and $B$. 
- 
-<button size="xs" type="link" collapse="X357DRKAQV84JNSC_2_path">{{icon>eye}} Solution</button><collapse id="X357DRKAQV84JNSC_2_path" collapsed="true"> 
-<callout type="tip" icon="true"> 
-Now a wye-delta transformation is necessary. 
-{{drawio>electrical_engineering_1:70jjg4yzwctqarsqCircuitSolution2.svg}} 
-Since $R_2=R_3$ and based on the equations for the transformation, the transformed $R_Y$ is given as: 
-\begin{align*} 
-R_{Y} &= {{R_2 \cdot R_2}\over{R_2 + R_2 + R_2}} \\ 
-      &= {{(100 \Omega)^2}\over{3 \cdot 100 \Omega}} \\ 
-      &= {{1}\over{3}} \cdot 100 \Omega = 33.33 \Omega 
-\end{align*} 
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-The equivalent resistor is given by a parallel configuation of resistors in series: 
-\begin{align*} 
-R_{eq} &= R_Y + (R_Y + R_1 + R_1)||(R_Y + R_2)\\ 
-R_{eq} &= 33.33 \Omega + (33.33 \Omega + 400 \Omega)||(33.33 \Omega + 100 \Omega)\\ 
-\end{align*} 
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-</callout></collapse> 
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-<button size="xs" type="link" collapse="X357DRKAQV84JNSC_2_solution">{{icon>eye}} Final result</button><collapse id="X357DRKAQV84JNSC_2_solution" collapsed="true"> 
-<callout type="tip" icon="true"> 
-\begin{align*} 
-R_{eq} &= 135.3 \Omega \\ 
-\end{align*} 
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-</callout></collapse> 
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