Differences

This shows you the differences between two versions of the page.

Link to this comparison view

Both sides previous revision Previous revision
Next revision
Previous revision
electrical_engineering_and_electronics_2:block09 [2026/05/18 03:02] mexleadminelectrical_engineering_and_electronics_2:block09 [2026/05/19 02:55] (current) mexleadmin
Line 324: Line 324:
 \] \]
  
-Often, the self-inductances are shortened:+Often, the self-inductances are abbreviated:
 \[ \[
 \begin{align*} \begin{align*}
Line 753: Line 753:
 </WRAP> </WRAP>
  
-The technical voltage ratio is often defined from the no-load voltages. Here it is denoted by \(\ddot{u}\):+The technical voltage ratio is often defined from the no-load voltages. Here it is denoted by \(\ddot{\rm u}\):
  
 \[ \[
Line 783: Line 783:
 \[ \[
 \begin{align*} \begin{align*}
-\ddot{\ rm u}\neq n,+\ddot{\rm u}\neq n,
 \end{align*} \end{align*}
 \] \]
Line 1020: Line 1020:
 </panel> </panel>
 </WRAP> </WRAP>
- 
-===== Exercises ===== 
  
 ===== Exercises ===== ===== Exercises =====
Line 1665: Line 1663:
 #@TaskText_HTML@# #@TaskText_HTML@#
  
-A transformer has a rated primary current $I_{1{\rm N}}=10~{\rm A}$ and a short-circuit voltage $u_{\rm k}=5~\%$.+A transformer has a rated primary current $I_{1{\rm N}}=10~{\rm A}$ and a short-circuit voltage ${\rm u_k}=5~\%$.
  
 1. Calculate the continuous short-circuit current $I_{1{\rm k}}$ when rated primary voltage is applied. 1. Calculate the continuous short-circuit current $I_{1{\rm k}}$ when rated primary voltage is applied.
Line 1676: Line 1674:
 I_{1{\rm k}} I_{1{\rm k}}
 = =
-I_{1{\rm N}}\cdot \frac{100~\%}{u_{\rm k}}+I_{1{\rm N}}\cdot \frac{100~\%}{\rm u_k}
 \end{align*} \end{align*}
  
Line 2069: Line 2067:
 </WRAP> </WRAP>
  
-3. Calculate the relative short-circuit voltage $u_{\rm k}$.+3. Calculate the relative short-circuit voltage ${\rm u_k}$.
 <WRAP group> <WRAP group>
 <WRAP half column rightalign> <WRAP half column rightalign>
Line 2076: Line 2074:
 The relative short-circuit voltage is: The relative short-circuit voltage is:
 \begin{align*} \begin{align*}
-u_{\rm k}+{\rm u_k}
 = =
 \frac{U_{1{\rm k}}}{U_{1{\rm N}}}\cdot 100~\% \frac{U_{1{\rm k}}}{U_{1{\rm N}}}\cdot 100~\%
Line 2083: Line 2081:
 Insert the values: Insert the values:
 \begin{align*} \begin{align*}
-u_{\rm k}+{\rm u_k}
 &= &=
 \frac{8.54~{\rm V}}{230~{\rm V}}\cdot 100~\% \frac{8.54~{\rm V}}{230~{\rm V}}\cdot 100~\%
Line 2096: Line 2094:
 #@ResultBegin_HTML~2026~@# #@ResultBegin_HTML~2026~@#
 \begin{align*} \begin{align*}
-u_{\rm k}=3.71~\%+{\rm u_k}=3.71~\%
 \end{align*} \end{align*}
 #@ResultEnd_HTML@# #@ResultEnd_HTML@#
Line 2111: Line 2109:
 I_{1{\rm k}} I_{1{\rm k}}
 = =
-I_{1{\rm N}}\cdot \frac{100~\%}{u_{\rm k}}+I_{1{\rm N}}\cdot \frac{100~\%}{\rm u_k}
 \end{align*} \end{align*}
  
Line 2352: Line 2350:
  
 At rated voltage and no-load operation, the magnetizing current is approximately $I_{\rm m,N}=0.12~{\rm A}$. At rated voltage and no-load operation, the magnetizing current is approximately $I_{\rm m,N}=0.12~{\rm A}$.
-The short-circuit voltage is $u_{\rm k}=6.0~\%$.+The short-circuit voltage is ${\rm u_k}=6.0~\%$.
  
 Assume that the magnetizing current is approximately proportional to the applied voltage. Assume that the magnetizing current is approximately proportional to the applied voltage.
Line 2365: Line 2363:
 U_{1{\rm k}} U_{1{\rm k}}
 = =
-\frac{u_{\rm k}}{100~\%}\cdot U_{1{\rm N}}+\frac{\rm u_k}{100~\%}\cdot U_{1{\rm N}}
 \end{align*} \end{align*}
  
Line 2956: Line 2954:
   * **Mixing peak values and RMS values:** In AC power and transformer ratings, \(U\) and \(I\) usually mean RMS values. Time functions are written \(u(t)\), \(i(t)\). Instantaneous short-circuit peaks are written here as \(i_{\rm p}\).   * **Mixing peak values and RMS values:** In AC power and transformer ratings, \(U\) and \(I\) usually mean RMS values. Time functions are written \(u(t)\), \(i(t)\). Instantaneous short-circuit peaks are written here as \(i_{\rm p}\).
   * **Confusing reluctance and resistance:** Magnetic reluctance \(R_{\rm m}\) has the unit \(1/{\rm H}\), not \(\Omega\).   * **Confusing reluctance and resistance:** Magnetic reluctance \(R_{\rm m}\) has the unit \(1/{\rm H}\), not \(\Omega\).
-  * **Confusing \(n\) and the technical no-load voltage ratio \(\ddot{u}\):** The ideal ratio is \(n=\frac{N_1}{N_2}\). The measured no-load voltage ratio is close to \(n\), but not exactly equal for a real transformer.+  * **Confusing \(n\) and the technical no-load voltage ratio \(\ddot{\rm u}\):** The ideal ratio is \(n=\frac{N_1}{N_2}\). The measured no-load voltage ratio is close to \(n\), but not exactly equal for a real transformer.
   * **Forgetting the square when referring impedances:** Voltages transform with \(n\), currents with \(\frac{1}{n}\), but impedances transform with \(n^2\).   * **Forgetting the square when referring impedances:** Voltages transform with \(n\), currents with \(\frac{1}{n}\), but impedances transform with \(n^2\).
   * **Ignoring leakage reactance:** Leakage reactance is often the dominant part of short-circuit impedance. It strongly affects fault current and voltage drop.   * **Ignoring leakage reactance:** Leakage reactance is often the dominant part of short-circuit impedance. It strongly affects fault current and voltage drop.