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ee2:task_jfzlmsucghsqvop5_with_calculation [2024/07/03 10:51] (aktuell) mexleadmin angelegt |
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+ | {{tag> | ||
+ | |||
+ | # | ||
+ | <fs medium> | ||
+ | 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> | ||
+ | |||
+ | Specify which magnetic voltages $\theta_{(1)}$, | ||
+ | Note the direction of the path in each case! | ||
+ | |||
+ | # | ||
+ | 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} $ | ||
+ | # | ||
+ | |||
+ | # | ||
+ | \begin{align*} | ||
+ | \theta_{(1)} &= -4 {~\rm A} \\ | ||
+ | \theta_{(2)} &= 0 {~\rm A} \\ | ||
+ | \theta_{(3)} &= -5 {~\rm A} \\ | ||
+ | \end{align*} | ||
+ | |||
+ | # | ||
+ | |||
+ | # |