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ee2:task_jfzlmsucghsqvop5_with_calculation [2024/07/03 10:51] (aktuell)
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 +{{tag>magnetic_voltage exam_ee2_SS2021}}{{include_n>1210}}
 + 
 +#@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: 
 +  * $I_1 = 5 {~\rm A}$
 +  * $I_2 = 2 {~\rm A}$
 +  * $I_3 = 1 {~\rm A}$
 +  * $I_4 = 4 {~\rm A}$
 +
 +{{drawio>ee2:jfZlmSUcgHSQVop5_question1.svg}}
 +
 +Specify which magnetic voltages $\theta_{(1)}$, $\theta_{(2)}$, and $\theta_{(3)}$ result. \\
 +Note the direction of the path in each case! 
 +
 +#@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~@#
 +
 +#@HiddenBegin_HTML~jfZlmSUcgHSQVop5_12,Result~@#
 +\begin{align*}
 +\theta_{(1)} &= -4 {~\rm A} \\
 +\theta_{(2)} &= 0 {~\rm A} \\
 +\theta_{(3)} &= -5 {~\rm A} \\
 +\end{align*}
 +
 +#@HiddenEnd_HTML~jfZlmSUcgHSQVop5_12,Result~@#
 +
 +#@TaskEnd_HTML@#