From 14c8cf8800adbad648013f252590f29be94f20f1 Mon Sep 17 00:00:00 2001
From: "[SilasElter]" <[SilasElter]>
Date: Thu, 28 Nov 2024 15:40:44 +0100
Subject: [PATCH] Improve ex04

---
 exercise/tex/exercise04.tex | 4 ++--
 1 file changed, 2 insertions(+), 2 deletions(-)

diff --git a/exercise/tex/exercise04.tex b/exercise/tex/exercise04.tex
index 331b677..68faea5 100644
--- a/exercise/tex/exercise04.tex
+++ b/exercise/tex/exercise04.tex
@@ -126,11 +126,11 @@
     \end{equation}
     The mean value for the current $i_\mathrm{D}$ and the RMS value for  $i_\mathrm{D}$ are determined as already explained:
     \begin{equation}
-        \overline{i}_\mathrm{D} = \frac{1}{T_\mathrm{S}}\frac{1}{2}\hat i_\mathrm{2}T_\mathrm{off}=\frac{1}{\SI{20}{\micro\s}}\cdot\frac{1}{2}\cdot\SI{6.28}{\ampere}\cdot\SI{12.7}{\micro\s}=\SI{2}{\ampere},
+        \overline{i}_\mathrm{D} = \frac{1}{T_\mathrm{S}}\frac{1}{2}\hat i_\mathrm{2}T_\mathrm{off}'=\frac{1}{\SI{20}{\micro\s}}\cdot\frac{1}{2}\cdot\SI{6.28}{\ampere}\cdot\SI{12.7}{\micro\s}=\SI{2}{\ampere},
     \end{equation}
 
     \begin{equation}
-        I_\mathrm{D} = \hat i_\mathrm{2} \sqrt{\frac{T_\mathrm{off}}{3T_\mathrm{S}}}= \SI{6.28}{\ampere}\cdot\sqrt{\frac{\SI{12.7}{\micro\s}}{3\cdot\SI{20}{\micro\s}}}= \SI{2.89}{\ampere}.
+        I_\mathrm{D} = \hat i_\mathrm{2} \sqrt{\frac{T_\mathrm{off}'}{3T_\mathrm{S}}}= \SI{6.28}{\ampere}\cdot\sqrt{\frac{\SI{12.7}{\micro\s}}{3\cdot\SI{20}{\micro\s}}}= \SI{2.89}{\ampere}.
     \end{equation}
     The voltage  $u_\mathrm{T,max}$ is calculated as in \eqref{eq:voltageTransistorTask1 ex04} as:
     \begin{equation}