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electrical_engineering_1:task_6tqttque1e2nf2c7_with_calculation [2023/03/31 08:01] – mexleadminelectrical_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}} 
-{{include_n>4000}} 
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-#@TaskTitle_HTML@#~~#@ee1_taskctr#~~  Equivalent linear Source       \\ <fs medium>(written test, approx. 14 % of a 60-minute written test, WS2022)</fs> 
-#@TaskText_HTML@#    
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-The circuit in the following has to be simplified.  
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-{{drawio>electrical_engineering_1:6tqttque1e2nf2c7Circuit.svg}} 
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-Calculated the internal resistance $R_\rm i$ and the source voltage $U_\rm s$ of an equivalent linear voltage source on the connectors $\rm A$ and $\rm B$. 
-\begin{align*} 
- R_1=5.0 ~\Omega, && U_2=6.0 ~\rm{V}, && R_3= 10 ~\Omega, \\ 
- I_4=4.2 ~\rm{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! 
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-#@PathBegin_HTML~1~@# 
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-The best thing is to re-think the wiring like rubber bands and adjust them: 
-{{drawio>electrical_engineering_1:6tqttque1e2nf2c7CircuitSolution1.svg}} \\ 
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-The linear voltage source of $U_2$ and $R_1$ can be transformed into a current source $I_2={{U_2}\over{R_1}}$ and $R_1$: 
-{{drawio>electrical_engineering_1:6tqttque1e2nf2c7CircuitSolution2.svg}} \\ 
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-Now a lot of them 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*} 
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-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_{\rm 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*} 
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-For the internal resistance $R_i$ the ideal voltage source is substituted by its resistance ($=0\Omega$, so a short-circuit): 
-\begin{align*} 
-R_{\rm AB} &= R_7 || ( R_6 + R_1||R_3||R_5) \\ 
-\end{align*} 
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-with $R_1||R_3||R_5 = 5 ~\Omega || 10 ~\Omega || 10 ~\Omega =  5 ~\Omega ||  5 ~\Omega = 2.5 ~\Omega$: 
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-\begin{align*} 
-U_{\rm AB} &= ({{6.0 ~\rm{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_{\rm AB} &= 15 ~\Omega|| ( 7.5 ~\Omega + 2.5 ~\Omega) \\ 
-\end{align*} 
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-#@PathEnd_HTML@# 
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-#@ResultBegin_HTML~1~@# 
-\begin{align*} 
-U_{\rm s} &= U_{\rm AB} &&= 4.5 ~\rm{V}\\ 
-R_{\rm i} &= R_{\rm AB} &&= 6   ~\Omega  
-\end{align*} 
-#@ResultEnd_HTML@# 
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-#@TaskEnd_HTML@#