typo corr

exam/summer2024/tex/task01.tex

@@ -32,7 +32,7 @@
 \begin{solutionblock}
     Faraday's law of induction reads 
     $$u_\mathrm{i} =\oint_{\partial\mathcal{S}} \bm{E} \cdot \mathrm{d}\bm{s}$$ 
-    where $\bm{E}$ is the electric field and $\partial\mathcal{S}$ is the boundary of the surface $\mathcal{S}$, which is the copper disk in this case. The electric field is a resulting of the rotating magnetic field and points from the center to the edge of the disk along the radial direction:
+    where $\bm{E}$ is the electric field and $\partial\mathcal{S}$ is the boundary of the surface $\mathcal{S}$, which is the copper disk in this case. The electric field is a result of the rotating copper disk within the static magnetic field pointing from the center to the edge of the disk along the radial direction:
     $$\bm{E} = \bm{v} \times \bm{B} = v(r) B \bm{e}_\mathrm{r} .$$
     Here, $\bm{v}(r) = 2 \pi r n_\mathrm{d}$ is the velocity field of the disk for a given radius element $r$ and $\bm{e}_\mathrm{r}$ is the unit vector in the radial direction. The induced voltage is then
     \begin{align*}

exam/summer2024/tex/task03.tex

@@ -49,7 +49,7 @@
     $$n_0 = \frac{\omega_{\mathrm{r,el}}}{2 \pi p } \cdot \SI{60}{\second\per\minute} = \SI{1800}{\per\minute}.$$ 
 \end{solutionblock}
 
-\subtask{Calculate the nominal torque $T_\mathrm{n}$ and nominal slips $s_\mathrm{n}$.}{2}
+\subtask{Calculate the nominal torque $T_\mathrm{n}$ and nominal slip $s_\mathrm{n}$.}{2}
 \subtaskGerman{Berechnen Sie das Nennmoment $T_\mathrm{n}$ und den Nennschlupf $s_\mathrm{n}$.}
 
 \begin{solutionblock}

exam/summer2024/tex/task04.tex

@@ -98,16 +98,16 @@
     $$T = \frac{P}{2\pi f/p} = \frac{3\cdot\SI{-4}{\mega\watt}}{2\pi \cdot \SI{50}{\hertz}} = \SI{-38.2}{\kilo\newton\meter}.$$
     Furthermore, the torque is also defined via 
     $$
-    T = 3 p \frac{U_\mathrm{s}U_\mathrm{i}}{\omega_\mathrm{s} X_\mathrm{s}}\sin(\delta) = T_\mathrm{max} \sin(\delta)
+    T = 3 p \frac{U_\mathrm{s}U_\mathrm{i}}{\omega_\mathrm{s} X_\mathrm{s}}\sin(\theta) = T_\mathrm{max} \sin(\theta)
     $$ 
-    where $\delta$ is the load angle. The maximum torque is then 
+    where $\theta$ is the load angle. The maximum torque is then 
     $$
-    T_\mathrm{max} = \frac{T}{\sin(\delta)} = \frac{\SI{-38.2}{\kilo\newton\meter}}{\sin(\SI{-26.5}{\degree})} = \SI{85.6}{\kilo\newton\meter}.
+    T_\mathrm{max} = \frac{T}{\sin(\theta)} = \frac{\SI{-38.2}{\kilo\newton\meter}}{\sin(\SI{-26.5}{\degree})} = \SI{85.6}{\kilo\newton\meter}.
     $$
 \end{solutionblock}
 
-\subtask{Determine a modified field excitation current $I_\mathrm{f}$ which delivers the same active power but reduces the reactive power to zero. Which load angle $\delta$ results?}{2}
-\subtaskGerman{Bestimmen Sie den Felderregungsstrom $I_\mathrm{f}$, der die gleiche Wirkleistung liefert, aber die Blindleistung auf null reduziert. Welcher Lastwinkel $\delta$ ergibt sich?}
+\subtask{Determine a modified field excitation current $I_\mathrm{f}$ which delivers the same active power but reduces the reactive power to zero. Which load angle $\theta$ results?}{2}
+\subtaskGerman{Bestimmen Sie den Felderregungsstrom $I_\mathrm{f}$, der die gleiche Wirkleistung liefert, aber die Blindleistung auf null reduziert. Welcher Lastwinkel $\theta$ ergibt sich?}
 
 \begin{solutionblock}
     The described scenario renders $\varphi=0$. The resulting stator current for this new scenario is
@@ -125,9 +125,9 @@
     $$
     I_\mathrm{f} = \frac{\SI{2.62}{\kilo\volt}}{\SI{3}{\kilo\volt}} \cdot\SI{100}{\ampere} = \SI{87.28}{\ampere}.
     $$
-    The new load angle $\delta$ is then
+    The new load angle $\theta$ is then
     $$
-    \delta = -\arccos\left(\frac{U_\mathrm{s}}{U_\mathrm{i}}\right) = -\arccos\left(\frac{\SI{2.24}{\kilo\volt}}{\SI{2.62}{\kilo\volt}}\right) = \SI{-31.24}{\degree}.
+    \theta = -\arccos\left(\frac{U_\mathrm{s}}{U_\mathrm{i}}\right) = -\arccos\left(\frac{\SI{2.24}{\kilo\volt}}{\SI{2.62}{\kilo\volt}}\right) = \SI{-31.24}{\degree}.
     $$
 \end{solutionblock}
 

exam/summer2024/tex/titlepage.tex

@@ -53,10 +53,10 @@ \section*{\hspace*{2mm}Instructions:}
 \begin{germanblock}
     Antworten müssen vollständig und nachvollziehbar formuliert werden. Antworten ohne jeglichen Kontext werden nicht berücksichtigt. Antworten werden in Deutsch und Englisch akzeptiert.
 \end{germanblock}
-\item Permitted tools are (exclusively): black/blue pens (indelible ink), triangle, a non-programmable calculator without graphic display and a single DINA4 cheat sheet.
+\item Permitted tools are (exclusively): black/blue pens (indelible ink), triangle, a non-programmable calculator without graphic display and a two DIN A4 cheat sheet.
 
 \begin{germanblock}
-    Erlaubte Hilfsmittel sind (exklusiv): schwarze / blaue Stifte (dokumentenecht), Geodreieck, ein nicht programmierbarer Taschenrechner ohne Grafikdisplay und ein DINA4-Notizzettel.
+    Erlaubte Hilfsmittel sind (exklusiv): schwarze / blaue Stifte (dokumentenecht), Geodreieck, ein nicht programmierbarer Taschenrechner ohne Grafikdisplay und zwei DIN A4 Notizblätter.
 \end{germanblock}
 \item The exam time is 90 minutes.
 

lecture/main.tex

@@ -220,7 +220,7 @@
 
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 %%%Lecture Include Onlys%%%
-\includeonly{tex/Lecture07}
+%\includeonly{tex/Lecture07}
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 
 \makeglossaries