Unterschiede

Hier werden die Unterschiede zwischen zwei Versionen angezeigt.

Link zu dieser Vergleichsansicht

Beide Seiten der vorigen Revision Vorhergehende Überarbeitung
Nächste Überarbeitung		 Vorhergehende Überarbeitung
exam_infos:preparation [2023/02/02 10:38] – [Do physics right] mexleadminexam_infos:preparation [2023/02/02 13:28] (aktuell) – [Do math right] mexleadmin
Zeile 3: Zeile 3:
 ===== Do math right ===== ===== Do math right =====
  
-  * calculate fractions in fractions correct: \begin{align*} {{{A}\over{B}}\over{C}} \quad &\neq \quad {{A}\over{{B}\over{C}}} \\ {{{A}\over{B}}\over{C}} = {{A}\over{{B}\cdot{C}}} \quad &\neq \quad {{A}\over{{B}\over{C}}} = {{A\cdot C}\over{B}} \end{align*}+  * calculate fractions in fractions (complex fractions) correct 
 +    * Be aware which is the longest fraction line: \\  \begin{align*} {{{A}\over{B}}\over{C}} \quad &\neq \quad {{A}\over{{B}\over{C}}} \\ {{{A}\over{B}}\over{C}} = {{A}\over{{B}\cdot{C}}} \quad &\neq \quad {{A}\over{{B}\over{C}}} = {{A\cdot C}\over{B}} \end{align*} 
 +    * Be aware how to reduce complex fractions by multiplying numerator and denominator with a factor: \\ \begin{align*} { { {{5 \cdot {{1}\over{x}} } \over {5 + {{1}\over{x}} } }  } \over { 7 + {{2}\over{x}} } } &\xrightarrow{either} &&{ { {{5 \cdot {{1}\over{x}} } \over {5 + {{1}\over{x}} } } \cdot \color{blue}{{| \cdot x}\over{| \cdot x}} } \over { 7 + {{2}\over{x}} } } && = { { {{5 } \over {5x + 1 } } } \over { 7 + {{2}\over{x}} } } \\ &\xrightarrow{\quad or \quad} &&{ { {{5 \cdot {{1}\over{x}} } \over {5 + {{1}\over{x}} } } } \over { 7 + {{2}\over{x}} } } \cdot \color{blue}{{| \cdot x}\over{| \cdot x}} = { { {{5 \cdot {{1}\over{x}} } \over {5 + {{1}\over{x}} } } \cdot \color{blue}{{| \cdot x}\over{ }} } \over { 7 + {{2}\over{x}} } } \cdot \color{blue}{{ }\over{| \cdot x}} && = { { {{5 } \over {5 + {{1}\over{x}} } } } \over { 7x + 2 } } \end{align*} \\ So please never never multiply in such cases all numerators and denominator with the factor...
   * Rearrange fractions correct: based on $\beta = {{I_C}\over{I_B}}$ on **cannot** derive $I_B = {{\beta}\over{I_C}}$..   * Rearrange fractions correct: based on $\beta = {{I_C}\over{I_B}}$ on **cannot** derive $I_B = {{\beta}\over{I_C}}$..
 +
  
 ===== Do physics right ===== ===== Do physics right =====
Zeile 15: Zeile 18:
     * impedances are not (only) resistors, at least do not think of $R$ is given as $\underline{Z}$. \\ For purely ohmic components the other way around is correct: $\underline{Z}=R$.     * impedances are not (only) resistors, at least do not think of $R$ is given as $\underline{Z}$. \\ For purely ohmic components the other way around is correct: $\underline{Z}=R$.
     * However, impedances 'act' like ohmic resistors in formulars. \\ E.g. for the series circuit: $R_{eq}=R_1+R_2+R_3+... \rightarrow \underline{Z}_{eq}=\underline{Z}_1+\underline{Z}_2+\underline{Z}_3+... $     * However, impedances 'act' like ohmic resistors in formulars. \\ E.g. for the series circuit: $R_{eq}=R_1+R_2+R_3+... \rightarrow \underline{Z}_{eq}=\underline{Z}_1+\underline{Z}_2+\underline{Z}_3+... $
 +  * A resistor does not have a continuous line through: \\ {{drawio>exam_infos:wrongresistor.svg}} 
 +  * circuits have to be closed, at least with giving voltages with arrows. \\ So, the in the following imgage the left side circuit is not correct: \\ {{drawio>exam_infos:wrongcircuit.svg}} 
  
 ===== Do exam right ===== ===== Do exam right =====
-  * Do not miss out questions. Sometimes there is more than one answer required+  * Do not miss out questions. Sometimes there is more than one answer required. 
 +  * Please add your solution path. Once only a wrong final result is given the grading will be 0. \\ When you at least give the correct basic formula there is the chance that you can get at least some reward..