Improve ex04

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}