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ee2:task_jfzlmsucghsqvop5_with_calculation [2024/07/03 10:51] – angelegt mexleadminee2:task_jfzlmsucghsqvop5_with_calculation [2025/11/22 19:26] (aktuell) – gelöscht mexleadmin
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-{{tag>magnetic_voltage exam_ee2_SS2021}}{{include_n>1210}} 
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-#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~ Magnetic Voltage \\ 
-<fs medium>(written test, approx. 6 % of a 120-minute written test, SS2021)</fs> #@TaskText_HTML@# 
  
-The following images show cross-sections of electrical cables. \\ 
-A closed path is shown as a dashed line. The magnetic voltage $\theta$ on these paths shall be analyzed. \\ 
-The following values are given for the currents:  
-  * $I_1 = 5 {~\rm A}$ 
-  * $I_2 = 2 {~\rm A}$ 
-  * $I_3 = 1 {~\rm A}$ 
-  * $I_4 = 4 {~\rm A}$ 
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-{{drawio>ee2:jfZlmSUcgHSQVop5_question1.svg}} 
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-Specify which magnetic voltages $\theta_{(1)}$, $\theta_{(2)}$, and $\theta_{(3)}$ result. \\ 
-Note the direction of the path in each case!  
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-#@HiddenBegin_HTML~jfZlmSUcgHSQVop5_11,Path~@# 
-For the resulting current the direction of the path has to be considered with the right-hand rule: 
-  * $I_{(1)} = +I_2 - I_1 - I_3 \quad \rightarrow \quad \theta_{(1)} = 2 {~\rm A} - 5 {~\rm A} - 1 {~\rm A} $  
-  * $I_{(2)} = +I_3 + I_4 - I_1 \quad \rightarrow \quad \theta_{(2)} = 1 {~\rm A} + 4 {~\rm A} - 5 {~\rm A} $   
-  * $I_{(3)} = +I_3 - I_4 - I_2 \quad \rightarrow \quad \theta_{(3)} = 1 {~\rm A} - 4 {~\rm A} - 2 {~\rm A} $  
-#@HiddenEnd_HTML~jfZlmSUcgHSQVop5_11,Path~@# 
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-#@HiddenBegin_HTML~jfZlmSUcgHSQVop5_12,Result~@# 
-\begin{align*} 
-\theta_{(1)} &= -4 {~\rm A} \\ 
-\theta_{(2)} &= 0 {~\rm A} \\ 
-\theta_{(3)} &= -5 {~\rm A} \\ 
-\end{align*} 
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-#@HiddenEnd_HTML~jfZlmSUcgHSQVop5_12,Result~@# 
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-#@TaskEnd_HTML@#