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Update Ex07 Task2 according Ex07_Tasks_2nd_review_OW
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SevenOfNinePE committed Jan 29, 2025
1 parent 722242f commit 0889207
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3 changes: 2 additions & 1 deletion exercise/fig/ex07/Fig_ThreePhaseInverter_6StepMode.tex
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\begin{circuitikz}
% Add voltage U1p
\draw (0,0) coordinate (U1p) to [open, o-o, v = $\frac{U_1}{2}\hspace{0.5cm}$, voltage = straight] ++(0,-2.5) coordinate (Gnd)
(Gnd) node[rground, rotate = 270 ](){} ++(0.4,0)
(Gnd) ++ (-0.4,0) node[rground](){} coordinate (GndSymb)
(Gnd) to [short,-] (GndSymb)
(Gnd) to [open, -o, v = $\frac{U_1}{2}\hspace{0.5cm}$, voltage = straight] ++(0,-2.5) coordinate (U1m)
% Add current
(U1p) to [short, o-, i=$i_1(t)$] ++(2,0) coordinate (jT1c)
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2 changes: 1 addition & 1 deletion exercise/tex/exercise07.tex
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Expand Up @@ -153,7 +153,7 @@

\end{solutionblock}

\subtask{The internal voltages $u_\mathrm{2ai}(t)$, $u_\mathrm{2bi}(t)$ and $u_\mathrm{2ci}(t)$ are a symmetrical voltage system,
\subtask{The internal voltages $u_\mathrm{2ai}(t)$, $u_\mathrm{2bi}(t)$ and $u_\mathrm{2ci}(t)$ are from a symmetrical voltage system,
i.e., the following is always applicable: $u_\mathrm{2ai}(t)+u_\mathrm{2bi}(t)+u_\mathrm{2ci}(t)=\SI{0}{\volt}$.
Show that this equation is also applicable for the voltages $u_\mathrm{2a}(t)$, $u_\mathrm{2b}(t)$ and $u_\mathrm{2c}(t)$ under the same conditions.
}
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