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electrical_engineering_and_electronics_2:block09 [2026/05/18 02:30] – mexleadminelectrical_engineering_and_electronics_2:block09 [2026/05/19 02:55] (current) – mexleadmin
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 \] \]
  
-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 1026: Line 1026:
 #@TaskText_HTML@# #@TaskText_HTML@#
  
-A transformer has \(N_1=1200\) turns and \(N_2=300\) turns.   +A transformer has $N_1=1200$ turns and $N_2=300$ turns.  \\ 
-The primary RMS voltage is \(U_1=230~{\rm V}\).   +The primary RMS voltage is $U_1=230~{\rm V}$.  \\ 
-The secondary side supplies a load current \(I_2=2.0~{\rm A}\).+The secondary side supplies a load current $I_2=2.0~{\rm A}$.
  
-  * Calculate the turns ratio \(n\). +1. Calculate the turns ratio $n$. 
-  * Calculate the ideal secondary voltage \(U_2\). +<WRAP group> 
-  * Calculate the magnitude of the ideal primary current \(I_1\). +<WRAP half column rightalign> 
-  * State whether this is a step-up or step-down transformer. +#@PathBegin_HTML~2001~@# 
- +<WRAP leftalign> 
-#@ResultBegin_HTML~Exercise1~@# +The turns ratio of an ideal transformer is defined as:
- +
-\[+
 \begin{align*} \begin{align*}
 n=\frac{N_1}{N_2} n=\frac{N_1}{N_2}
-=+\end{align*} 
 + 
 +Insert the given values: 
 +\begin{align*} 
 +n 
 +&=
 \frac{1200}{300} \frac{1200}{300}
-= +\\ 
-4.+&= 
 +4
 \end{align*} \end{align*}
-\]+</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 +<WRAP half column> 
 +#@ResultBegin_HTML~2001~@# 
 +\begin{align*} 
 +n=4 
 +\end{align*} 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP>
  
-The secondary voltage is+2. Calculate the ideal secondary voltage $U_2$. 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~2002~@# 
 +<WRAP leftalign> 
 +For an ideal transformer, the voltage ratio follows the turns ratio: 
 +\begin{align*} 
 +n=\frac{U_1}{U_2} 
 +\end{align*}
  
-\[+Therefore:
 \begin{align*} \begin{align*}
 U_2 U_2
-=+&=
 \frac{U_1}{n} \frac{U_1}{n}
-=+\\ 
 +&=
 \frac{230~{\rm V}}{4} \frac{230~{\rm V}}{4}
-= +\\ 
-57.5~{\rm V}.+&= 
 +57.5~{\rm V}
 \end{align*} \end{align*}
-\]+</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 +<WRAP half column> 
 +#@ResultBegin_HTML~2002~@# 
 +\begin{align*} 
 +U_2=57.5~{\rm V} 
 +\end{align*} 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP>
  
-The primary current magnitude is+3. Calculate the magnitude of the ideal primary current $I_1$. 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~2003~@# 
 +<WRAP leftalign> 
 +For the ideal transformer, the current ratio is inverse to the voltage ratio: 
 +\begin{align*} 
 +I_1=\frac{I_2}{n} 
 +\end{align*}
  
-\[+Insert the values:
 \begin{align*} \begin{align*}
 I_1 I_1
-= +&=
-\frac{I_2}{n} +
-=+
 \frac{2.0~{\rm A}}{4} \frac{2.0~{\rm A}}{4}
-= +\\ 
-0.50~{\rm A}.+&= 
 +0.50~{\rm A}
 \end{align*} \end{align*}
-\]+</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 +<WRAP half column> 
 +#@ResultBegin_HTML~2003~@# 
 +\begin{align*} 
 +I_1=0.50~{\rm A} 
 +\end{align*} 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP>
  
-Because \(U_2<U_1\), it is a step-down transformer.+4. State whether this is a step-up or step-down transformer. 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~2004~@# 
 +<WRAP leftalign> 
 +Compare the primary and secondary voltages: 
 +\begin{align*} 
 +U_1 &= 230~{\rm V} 
 +\\ 
 +U_2 &= 57.5~{\rm V} 
 +\end{align*}
  
 +Since
 +\begin{align*}
 +U_2<U_1
 +\end{align*}
 +
 +the transformer reduces the voltage.
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~2004~@#
 +The transformer is a step-down transformer.
 #@ResultEnd_HTML@# #@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 #@TaskEnd_HTML@# #@TaskEnd_HTML@#
 +
  
 #@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~  Quick check: mutual inductance from reluctance #@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~  Quick check: mutual inductance from reluctance
Line 1086: Line 1163:
 The main magnetic reluctance is The main magnetic reluctance is
  
-\[ 
 \begin{align*} \begin{align*}
 R_{\rm mH}=2.0\cdot 10^6~\frac{1}{\rm H}. R_{\rm mH}=2.0\cdot 10^6~\frac{1}{\rm H}.
 \end{align*} \end{align*}
-\] 
- 
-The number of turns is \(N_1=500\) and \(N_2=100\). 
- 
-  * Calculate \(L_{1{\rm H}}\). 
-  * Calculate \(L_{2{\rm H}}\). 
-  * Calculate \(M\). 
-  * Check whether the units are correct. 
  
-#@ResultBegin_HTML~Exercise2~@#+The number of turns is $N_1=500$ and $N_2=100$.
  
-\[+1. Calculate $L_{1{\rm H}}$. 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~2005~@# 
 +<WRAP leftalign> 
 +The main-flux inductance of coil 1 is:
 \begin{align*} \begin{align*}
 L_{1{\rm H}} L_{1{\rm H}}
 = =
 \frac{N_1^2}{R_{\rm mH}} \frac{N_1^2}{R_{\rm mH}}
-=+\end{align*} 
 + 
 +Insert the values: 
 +\begin{align*} 
 +L_{1{\rm H}} 
 +&=
 \frac{500^2}{2.0\cdot 10^6~1/{\rm H}} \frac{500^2}{2.0\cdot 10^6~1/{\rm H}}
-= +\\ 
-0.125~{\rm H}.+&= 
 +0.125~{\rm H}
 \end{align*} \end{align*}
-\]+</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 +<WRAP half column> 
 +#@ResultBegin_HTML~2005~@# 
 +\begin{align*} 
 +L_{1{\rm H}}=0.125~{\rm H} 
 +\end{align*} 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP>
  
-\[+2. Calculate $L_{2{\rm H}}$. 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~2006~@# 
 +<WRAP leftalign> 
 +The main-flux inductance of coil 2 is:
 \begin{align*} \begin{align*}
 L_{2{\rm H}} L_{2{\rm H}}
 = =
 \frac{N_2^2}{R_{\rm mH}} \frac{N_2^2}{R_{\rm mH}}
-=+\end{align*} 
 + 
 +Insert the values: 
 +\begin{align*} 
 +L_{2{\rm H}} 
 +&=
 \frac{100^2}{2.0\cdot 10^6~1/{\rm H}} \frac{100^2}{2.0\cdot 10^6~1/{\rm H}}
 +\\
 +&=
 +0.0050~{\rm H}
 +\\
 +&=
 +5.0~{\rm mH}
 +\end{align*}
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~2006~@#
 +\begin{align*}
 +L_{2{\rm H}}
 = =
 0.0050~{\rm H} 0.0050~{\rm H}
 = =
-5.0~{\rm mH}.+5.0~{\rm mH}
 \end{align*} \end{align*}
-\]+#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP>
  
-\[+3. Calculate $M$. 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~2007~@# 
 +<WRAP leftalign> 
 +The mutual inductance is:
 \begin{align*} \begin{align*}
 M M
 = =
 \frac{N_1N_2}{R_{\rm mH}} \frac{N_1N_2}{R_{\rm mH}}
-=+\end{align*} 
 + 
 +Insert the values: 
 +\begin{align*} 
 +M 
 +&=
 \frac{500\cdot 100}{2.0\cdot 10^6~1/{\rm H}} \frac{500\cdot 100}{2.0\cdot 10^6~1/{\rm H}}
 +\\
 +&=
 +0.025~{\rm H}
 +\\
 +&=
 +25~{\rm mH}
 +\end{align*}
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~2007~@#
 +\begin{align*}
 +M
 = =
 0.025~{\rm H} 0.025~{\rm H}
 = =
-25~{\rm mH}.+25~{\rm mH}
 \end{align*} \end{align*}
-\]+#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP>
  
-The unit is correct because \(1/(1/{\rm H})={\rm H}\).+4. Check whether the units are correct. 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~2008~@# 
 +<WRAP leftalign> 
 +The reluctance is given in: 
 +\begin{align*} 
 +[R_{\rm mH}]=\frac{1}{\rm H} 
 +\end{align*}
  
 +The number of turns is dimensionless. Therefore:
 +\begin{align*}
 +\left[
 +\frac{N^2}{R_{\rm mH}}
 +\right]
 +=
 +\frac{1}{1/{\rm H}}
 +=
 +{\rm H}
 +\end{align*}
 +
 +The same argument applies to the mutual inductance:
 +\begin{align*}
 +\left[
 +\frac{N_1N_2}{R_{\rm mH}}
 +\right]
 +=
 +{\rm H}
 +\end{align*}
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~2008~@#
 +The unit is correct because
 +\begin{align*}
 +\frac{1}{1/{\rm H}}={\rm H}
 +\end{align*}
 #@ResultEnd_HTML@# #@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 #@TaskEnd_HTML@# #@TaskEnd_HTML@#
 +
  
 #@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~  Mutual inductance and leakage from a magnetic path #@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~  Mutual inductance and leakage from a magnetic path
Line 1151: Line 1332:
 Two coils are wound on the same magnetic core. The shared main magnetic path has the reluctance Two coils are wound on the same magnetic core. The shared main magnetic path has the reluctance
  
-\[ 
 \begin{align*} \begin{align*}
 R_{\rm mH}=1.6\cdot 10^6~\frac{1}{\rm H}. R_{\rm mH}=1.6\cdot 10^6~\frac{1}{\rm H}.
 \end{align*} \end{align*}
-\] 
  
 The numbers of turns are The numbers of turns are
  
-\[ 
 \begin{align*} \begin{align*}
 N_1=400, N_1=400,
Line 1165: Line 1343:
 N_2=100. N_2=100.
 \end{align*} \end{align*}
-\] 
  
 The leakage inductances are The leakage inductances are
  
-\[ 
 \begin{align*} \begin{align*}
 L_{1\sigma}=4.0~{\rm mH}, L_{1\sigma}=4.0~{\rm mH},
Line 1175: Line 1351:
 L_{2\sigma}=0.30~{\rm mH}. L_{2\sigma}=0.30~{\rm mH}.
 \end{align*} \end{align*}
-\] 
  
-Calculate:+1. Calculate $L_{1{\rm H}}$. 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~2009~@# 
 +<WRAP leftalign> 
 +The main-flux self-inductance of coil 1 is: 
 +\begin{align*} 
 +L_{1{\rm H}} 
 += 
 +\frac{N_1^2}{R_{\rm mH}} 
 +\end{align*}
  
-  * \(L_{1{\rm H}}\) +Insert the values:
-  * \(L_{2{\rm H}}\) +
-  * \(M\) +
-  * the total self-inductances \(L_1\) and \(L_2\) +
-  * the coupling coefficient \(k=\frac{M}{\sqrt{L_1L_2}}\) +
- +
-#@ResultBegin_HTML~ExerciseMutualReluctance~@# +
- +
-The main-flux self-inductances are +
- +
-\[+
 \begin{align*} \begin{align*}
 L_{1{\rm H}} L_{1{\rm H}}
 &= &=
-\frac{N_1^2}{R_{\rm mH}} 
-= 
 \frac{400^2}{1.6\cdot 10^6~1/{\rm H}} \frac{400^2}{1.6\cdot 10^6~1/{\rm H}}
 +\\
 +&=
 +0.100~{\rm H}
 +\end{align*}
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~2009~@#
 +\begin{align*}
 +L_{1{\rm H}}=0.100~{\rm H}=100~{\rm mH}
 +\end{align*}
 +#@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 +2. Calculate $L_{2{\rm H}}$.
 +<WRAP group>
 +<WRAP half column rightalign>
 +#@PathBegin_HTML~2010~@#
 +<WRAP leftalign>
 +The main-flux self-inductance of coil 2 is:
 +\begin{align*}
 +L_{2{\rm H}}
 = =
-0.100~{\rm H}, +\frac{N_2^2}{R_{\rm mH}} 
-\\[4pt]+\end{align*} 
 + 
 +Insert the values: 
 +\begin{align*}
 L_{2{\rm H}} L_{2{\rm H}}
 &= &=
-\frac{N_2^2}{R_{\rm mH}} 
-= 
 \frac{100^2}{1.6\cdot 10^6~1/{\rm H}} \frac{100^2}{1.6\cdot 10^6~1/{\rm H}}
 +\\
 +&=
 +0.00625~{\rm H}
 +\\
 +&=
 +6.25~{\rm mH}
 +\end{align*}
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~2010~@#
 +\begin{align*}
 +L_{2{\rm H}}
 = =
 0.00625~{\rm H} 0.00625~{\rm H}
 = =
-6.25~{\rm mH}.+6.25~{\rm mH}
 \end{align*} \end{align*}
-\]+#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP>
  
-The mutual inductance is +3. Calculate $M$. 
- +<WRAP group> 
-\[+<WRAP half column rightalign> 
 +#@PathBegin_HTML~2011~@# 
 +<WRAP leftalign> 
 +The mutual inductance is:
 \begin{align*} \begin{align*}
 M M
 = =
 \frac{N_1N_2}{R_{\rm mH}} \frac{N_1N_2}{R_{\rm mH}}
-=+\end{align*} 
 + 
 +Insert the values: 
 +\begin{align*} 
 +M 
 +&=
 \frac{400\cdot 100}{1.6\cdot 10^6~1/{\rm H}} \frac{400\cdot 100}{1.6\cdot 10^6~1/{\rm H}}
 +\\
 +&=
 +0.025~{\rm H}
 +\\
 +&=
 +25~{\rm mH}
 +\end{align*}
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~2011~@#
 +\begin{align*}
 +M
 = =
 0.025~{\rm H} 0.025~{\rm H}
 = =
-25~{\rm mH}.+25~{\rm mH}
 \end{align*} \end{align*}
-\]+#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP>
  
-The total self-inductances are+4. Calculate the total self-inductances $L_1$ and $L_2$. 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~2012~@# 
 +<WRAP leftalign> 
 +The total self-inductance is the sum of main-flux inductance and leakage inductance.
  
-\[+For coil 1:
 \begin{align*} \begin{align*}
 L_1 L_1
 &= &=
 L_{1{\rm H}}+L_{1\sigma} L_{1{\rm H}}+L_{1\sigma}
-=+\\ 
 +&=
 100~{\rm mH}+4.0~{\rm mH} 100~{\rm mH}+4.0~{\rm mH}
-= +\\ 
-104~{\rm mH}, +&= 
-\\[4pt]+104~{\rm mH} 
 +\end{align*} 
 + 
 +For coil 2: 
 +\begin{align*}
 L_2 L_2
 &= &=
 L_{2{\rm H}}+L_{2\sigma} L_{2{\rm H}}+L_{2\sigma}
-=+\\ 
 +&=
 6.25~{\rm mH}+0.30~{\rm mH} 6.25~{\rm mH}+0.30~{\rm mH}
-= +\\ 
-6.55~{\rm mH}.+&= 
 +6.55~{\rm mH}
 \end{align*} \end{align*}
-\]+</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 +<WRAP half column> 
 +#@ResultBegin_HTML~2012~@# 
 +\begin{align*} 
 +L_1 &= 104~{\rm mH} 
 +\\ 
 +L_2 &= 6.55~{\rm mH} 
 +\end{align*} 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP>
  
-Thus +5. Calculate the coupling coefficient $k=\frac{M}{\sqrt{L_1L_2}}$. 
- +<WRAP group> 
-\[+<WRAP half column rightalign> 
 +#@PathBegin_HTML~2013~@# 
 +<WRAP leftalign> 
 +The coupling coefficient is:
 \begin{align*} \begin{align*}
 k k
 = =
 \frac{M}{\sqrt{L_1L_2}} \frac{M}{\sqrt{L_1L_2}}
-=+\end{align*} 
 + 
 +Insert the values in henry: 
 +\begin{align*} 
 +k 
 +&=
 \frac{0.025~{\rm H}}{\sqrt{0.104~{\rm H}\cdot 0.00655~{\rm H}}} \frac{0.025~{\rm H}}{\sqrt{0.104~{\rm H}\cdot 0.00655~{\rm H}}}
-\approx +\\ 
-0.96.+&\approx 
 +0.96
 \end{align*} \end{align*}
-\] 
  
 The coupling is strong, but not ideal, because leakage inductances are present. The coupling is strong, but not ideal, because leakage inductances are present.
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~2013~@#
 +\begin{align*}
 +k\approx 0.96
 +\end{align*}
  
 +The coupling is strong, but not ideal.
 #@ResultEnd_HTML@# #@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 #@TaskEnd_HTML@# #@TaskEnd_HTML@#
  
Line 1272: Line 1554:
 #@TaskText_HTML@# #@TaskText_HTML@#
  
-A transformer has \(n=5\).   +A transformer has $n=5$.   
-The secondary winding resistance is \(R_2=0.20~\Omega\) and the secondary leakage reactance is \(X_{2\sigma}=0.35~\Omega\).+The secondary winding resistance is $R_2=0.20~\Omega$ and the secondary leakage reactance is $X_{2\sigma}=0.35~\Omega$.
  
-Calculate the values \(R'_2\) and \(X'_{2\sigma}\) referred to the primary side.+Calculate the values $R'_2$ and $X'_{2\sigma}$ referred to the primary side.
  
-#@ResultBegin_HTML~Exercise3~@# +1. Calculate $R'_2$. 
- +<WRAP group> 
-\[+<WRAP half column rightalign> 
 +#@PathBegin_HTML~2014~@# 
 +<WRAP leftalign> 
 +When a resistance is referred from the secondary side to the primary side, it is multiplied by $n^2$:
 \begin{align*} \begin{align*}
 R'_2 R'_2
 = =
 n^2R_2 n^2R_2
-=+\end{align*} 
 + 
 +Insert the values: 
 +\begin{align*} 
 +R'_2 
 +&=
 5^2\cdot 0.20~\Omega 5^2\cdot 0.20~\Omega
-=+\\ 
 +&=
 25\cdot 0.20~\Omega 25\cdot 0.20~\Omega
-= +\\ 
-5.0~\Omega.+&= 
 +5.0~\Omega
 \end{align*} \end{align*}
-\]+</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 +<WRAP half column> 
 +#@ResultBegin_HTML~2014~@# 
 +\begin{align*} 
 +R'_2=5.0~\Omega 
 +\end{align*} 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP>
  
-\[+2. Calculate $X'_{2\sigma}$. 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~2015~@# 
 +<WRAP leftalign> 
 +The secondary leakage reactance is also referred to the primary side by multiplying with $n^2$:
 \begin{align*} \begin{align*}
 X'_{2\sigma} X'_{2\sigma}
 = =
 n^2X_{2\sigma} n^2X_{2\sigma}
-=+\end{align*} 
 + 
 +Insert the values: 
 +\begin{align*} 
 +X'_{2\sigma} 
 +&=
 5^2\cdot 0.35~\Omega 5^2\cdot 0.35~\Omega
-=+\\ 
 +&=
 25\cdot 0.35~\Omega 25\cdot 0.35~\Omega
-= +\\ 
-8.75~\Omega.+&= 
 +8.75~\Omega
 \end{align*} \end{align*}
-\]+</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 +<WRAP half column> 
 +#@ResultBegin_HTML~2015~@# 
 +\begin{align*} 
 +X'_{2\sigma}=8.75~\Omega 
 +\end{align*} 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP>
  
-The unit remains \(\Omega\), because \(n\) is dimensionless.+3. Check the unit. 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~2016~@# 
 +<WRAP leftalign> 
 +The turns ratio $n$ is dimensionless: 
 +\begin{align*} 
 +[n]=1 
 +\end{align*}
  
 +Therefore, multiplying by $n^2$ does not change the unit:
 +\begin{align*}
 +[R'_2] &= \Omega
 +\\
 +[X'_{2\sigma}] &= \Omega
 +\end{align*}
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~2016~@#
 +The unit remains $\Omega$, because $n$ is dimensionless.
 #@ResultEnd_HTML@# #@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 #@TaskEnd_HTML@# #@TaskEnd_HTML@#
 +
  
 #@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~  Quick check: short-circuit voltage and fault current #@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~  Quick check: short-circuit voltage and fault current
 #@TaskText_HTML@# #@TaskText_HTML@#
  
-A transformer has a rated primary current \(I_{1{\rm N}}=10~{\rm A}\) and a short-circuit voltage \({\rm u_k}=5~\%\). +A transformer has a rated primary current $I_{1{\rm N}}=10~{\rm A}$ and a short-circuit voltage ${\rm u_k}=5~\%$.
- +
-  * Calculate the continuous short-circuit current \(I_{1{\rm k}}\) when rated primary voltage is applied. +
-  * Estimate the initial peak short-circuit current \(i_{\rm p}\) using \(i_{\rm p}\approx 2.54 I_{1{\rm k}}\). +
- +
-#@ResultBegin_HTML~Exercise4~@#+
  
-\[+1. Calculate the continuous short-circuit current $I_{1{\rm k}}$ when rated primary voltage is applied. 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~2017~@# 
 +<WRAP leftalign> 
 +The short-circuit current can be estimated from the rated current and the relative short-circuit voltage:
 \begin{align*} \begin{align*}
 I_{1{\rm k}} I_{1{\rm k}}
 = =
 I_{1{\rm N}}\cdot \frac{100~\%}{\rm u_k} I_{1{\rm N}}\cdot \frac{100~\%}{\rm u_k}
-=+\end{align*} 
 + 
 +Insert the values: 
 +\begin{align*} 
 +I_{1{\rm k}} 
 +&=
 10~{\rm A}\cdot \frac{100~\%}{5~\%} 10~{\rm A}\cdot \frac{100~\%}{5~\%}
-= +\\ 
-200~{\rm A}.+&= 
 +200~{\rm A}
 \end{align*} \end{align*}
-\]+</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 +<WRAP half column> 
 +#@ResultBegin_HTML~2017~@# 
 +\begin{align*} 
 +I_{1{\rm k}}=200~{\rm A} 
 +\end{align*} 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP>
  
-\[+2. Estimate the initial peak short-circuit current $i_{\rm p}$ using $i_{\rm p}\approx 2.54 I_{1{\rm k}}$. 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~2018~@# 
 +<WRAP leftalign> 
 +The initial peak current is estimated by:
 \begin{align*} \begin{align*}
 i_{\rm p} i_{\rm p}
 \approx \approx
 2.54\cdot I_{1{\rm k}} 2.54\cdot I_{1{\rm k}}
-=+\end{align*} 
 + 
 +Insert the continuous short-circuit current: 
 +\begin{align*} 
 +i_{\rm p} 
 +&\approx
 2.54\cdot 200~{\rm A} 2.54\cdot 200~{\rm A}
-= +\\ 
-508~{\rm A}.+&= 
 +508~{\rm A}
 \end{align*} \end{align*}
-\] 
  
 The short-circuit current is much larger than the rated current. Protection devices must be selected accordingly. The short-circuit current is much larger than the rated current. Protection devices must be selected accordingly.
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~2018~@#
 +\begin{align*}
 +i_{\rm p}\approx 508~{\rm A}
 +\end{align*}
  
 +Protection devices must be selected accordingly.
 #@ResultEnd_HTML@# #@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 #@TaskEnd_HTML@# #@TaskEnd_HTML@#
 +
  
 #@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~  Longer exercise: transformer equivalent circuit for an actuator supply #@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~  Longer exercise: transformer equivalent circuit for an actuator supply
Line 1357: Line 1744:
 Rated data and equivalent circuit data are: Rated data and equivalent circuit data are:
  
-\[ 
 \begin{align*} \begin{align*}
 U_{1{\rm N}}&=230~{\rm V}, U_{1{\rm N}}&=230~{\rm V},
Line 1373: Line 1759:
 X_{2\sigma}&=0.018~\Omega. X_{2\sigma}&=0.018~\Omega.
 \end{align*} \end{align*}
-\] 
- 
-Assume \(n=\frac{U_{1{\rm N}}}{U_{2{\rm N}}}\). The magnetizing branch is neglected for the loaded operating point. 
- 
-  * Calculate \(n\). 
-  * Refer \(R_2\) and \(X_{2\sigma}\) to the primary side. 
-  * Calculate \(R_{\rm k}\) and \(X_{\rm k}\). 
-  * Calculate the primary rated current magnitude \(I_{1{\rm N}}\) using the ideal current ratio. 
-  * Estimate the magnitude of the internal voltage drop \(U_{\rm k}\approx |\underline{Z}_{\rm k}|I_{1{\rm N}}\). 
  
-#@ResultBegin_HTML~Exercise5~@#+Assume $n=\frac{U_{1{\rm N}}}{U_{2{\rm N}}}$. The magnetizing branch is neglected for the loaded operating point.
  
-The turns ratio is +1. Calculate $n$. 
- +<WRAP group> 
-\[+<WRAP half column rightalign> 
 +#@PathBegin_HTML~2019~@# 
 +<WRAP leftalign> 
 +The turns ratio is estimated from the rated voltages:
 \begin{align*} \begin{align*}
 n n
 = =
 \frac{U_{1{\rm N}}}{U_{2{\rm N}}} \frac{U_{1{\rm N}}}{U_{2{\rm N}}}
-=+\end{align*} 
 + 
 +Insert the values: 
 +\begin{align*} 
 +n 
 +&=
 \frac{230~{\rm V}}{23~{\rm V}} \frac{230~{\rm V}}{23~{\rm V}}
-= +\\ 
-10.+&= 
 +10
 \end{align*} \end{align*}
-\]+</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 +<WRAP half column> 
 +#@ResultBegin_HTML~2019~@# 
 +\begin{align*} 
 +n=10 
 +\end{align*} 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP>
  
-The secondary quantities referred to the primary side are+2. Refer $R_2$ and $X_{2\sigma}$ to the primary side. 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~2020~@# 
 +<WRAP leftalign> 
 +Secondary quantities are referred to the primary side by multiplying them with $n^2$: 
 +\begin{align*} 
 +R'_2 &= n^2R_2 
 +\\ 
 +X'_{2\sigma} &= n^2X_{2\sigma} 
 +\end{align*}
  
-\[+With $n=10$:
 \begin{align*} \begin{align*}
 R'_2 R'_2
 &= &=
-n^2R_2 
-= 
 10^2\cdot 0.012~\Omega 10^2\cdot 0.012~\Omega
-= 
-1.2~\Omega, 
 \\ \\
 +&=
 +1.2~\Omega
 +\\[4pt]
 X'_{2\sigma} X'_{2\sigma}
 &= &=
-n^2X_{2\sigma} 
-= 
 10^2\cdot 0.018~\Omega 10^2\cdot 0.018~\Omega
-= +\\ 
-1.8~\Omega.+&= 
 +1.8~\Omega
 \end{align*} \end{align*}
-\]+</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 +<WRAP half column> 
 +#@ResultBegin_HTML~2020~@# 
 +\begin{align*} 
 +R'_2 &= 1.2~\Omega 
 +\\ 
 +X'_{2\sigma} &= 1.8~\Omega 
 +\end{align*} 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP>
  
-Therefore +3. Calculate $R_{\rm k}$ and $X_{\rm k}$. 
- +<WRAP group> 
-\[+<WRAP half column rightalign> 
 +#@PathBegin_HTML~2021~@# 
 +<WRAP leftalign> 
 +The short-circuit equivalent values are the sums of the primary quantities and the referred secondary quantities:
 \begin{align*} \begin{align*}
 R_{\rm k} R_{\rm k}
 &= &=
 R_1+R'_2 R_1+R'_2
-= 
-1.2~\Omega+1.2~\Omega 
-= 
-2.4~\Omega, 
 \\ \\
 X_{\rm k} X_{\rm k}
 &= &=
 X_{1\sigma}+X'_{2\sigma} X_{1\sigma}+X'_{2\sigma}
-=+\end{align*} 
 + 
 +Insert the values: 
 +\begin{align*} 
 +R_{\rm k} 
 +&= 
 +1.2~\Omega+1.2~\Omega 
 +\\ 
 +&= 
 +2.4~\Omega 
 +\\[4pt] 
 +X_{\rm k} 
 +&=
 1.8~\Omega+1.8~\Omega 1.8~\Omega+1.8~\Omega
-= +\\ 
-3.6~\Omega.+&= 
 +3.6~\Omega
 \end{align*} \end{align*}
-\]+</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 +<WRAP half column> 
 +#@ResultBegin_HTML~2021~@# 
 +\begin{align*} 
 +R_{\rm k} &= 2.4~\Omega 
 +\\ 
 +X_{\rm k} &= 3.6~\Omega 
 +\end{align*} 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP>
  
-The primary current magnitude is +4. Calculate the primary rated current magnitude $I_{1{\rm N}}$ using the ideal current ratio. 
- +<WRAP group> 
-\[+<WRAP half column rightalign> 
 +#@PathBegin_HTML~2022~@# 
 +<WRAP leftalign> 
 +For an ideal transformer, the primary current magnitude is:
 \begin{align*} \begin{align*}
 I_{1{\rm N}} I_{1{\rm N}}
 = =
 \frac{I_{2{\rm N}}}{n} \frac{I_{2{\rm N}}}{n}
-=+\end{align*} 
 + 
 +Insert the values: 
 +\begin{align*} 
 +I_{1{\rm N}} 
 +&=
 \frac{5.0~{\rm A}}{10} \frac{5.0~{\rm A}}{10}
-= +\\ 
-0.50~{\rm A}.+&= 
 +0.50~{\rm A}
 \end{align*} \end{align*}
-\]+</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 +<WRAP half column> 
 +#@ResultBegin_HTML~2022~@# 
 +\begin{align*} 
 +I_{1{\rm N}}=0.50~{\rm A} 
 +\end{align*} 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP>
  
-The magnitude of the short-circuit impedance is +5. Estimate the magnitude of the internal voltage drop $U_{\rm k}\approx |\underline{Z}_{\rm k}|I_{1{\rm N}}$. 
- +<WRAP group> 
-\[+<WRAP half column rightalign> 
 +#@PathBegin_HTML~2023~@# 
 +<WRAP leftalign> 
 +First calculate the magnitude of the short-circuit impedance:
 \begin{align*} \begin{align*}
 |\underline{Z}_{\rm k}| |\underline{Z}_{\rm k}|
 = =
 \sqrt{R_{\rm k}^2+X_{\rm k}^2} \sqrt{R_{\rm k}^2+X_{\rm k}^2}
-= 
-\sqrt{(2.4~\Omega)^2+(3.6~\Omega)^2} 
-= 
-4.33~\Omega. 
 \end{align*} \end{align*}
-\] 
  
-Thus the internal voltage drop estimate is+Insert the values: 
 +\begin{align*} 
 +|\underline{Z}_{\rm k}| 
 +&= 
 +\sqrt{(2.4~\Omega)^2+(3.6~\Omega)^2} 
 +\\ 
 +&= 
 +4.33~\Omega 
 +\end{align*}
  
-\[+Now calculate the internal voltage drop:
 \begin{align*} \begin{align*}
 U_{\rm k} U_{\rm k}
-\approx+&\approx
 |\underline{Z}_{\rm k}|I_{1{\rm N}} |\underline{Z}_{\rm k}|I_{1{\rm N}}
-=+\\ 
 +&=
 4.33~\Omega\cdot 0.50~{\rm A} 4.33~\Omega\cdot 0.50~{\rm A}
 +\\
 +&=
 +2.17~{\rm V}
 +\end{align*}
 +
 +This is a primary-side voltage drop. On the secondary side:
 +\begin{align*}
 +\frac{2.17~{\rm V}}{10}
 = =
-2.17~{\rm V}.+0.217~{\rm V}
 \end{align*} \end{align*}
-\] 
  
-This is a primary-side voltage drop. On the secondary side it corresponds approximately to +For a $23~{\rm V}$ actuator supply this is small but not zero. 
- +</WRAP> 
-\[+#@PathEnd_HTML@# 
 +</WRAP> 
 +<WRAP half column> 
 +#@ResultBegin_HTML~2023~@#
 \begin{align*} \begin{align*}
-\frac{2.17~{\rm V}}{10}=0.217~{\rm V}.+|\underline{Z}_{\rm k}| &= 4.33~\Omega 
 +\\ 
 +U_{\rm k} &\approx 2.17~{\rm V}
 \end{align*} \end{align*}
-\] 
- 
-For a \(23~{\rm V}\) actuator supply this is small but not zero. 
  
 +Secondary-side equivalent:
 +\begin{align*}
 +U_{\rm k,2}\approx 0.217~{\rm V}
 +\end{align*}
 #@ResultEnd_HTML@# #@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 #@TaskEnd_HTML@# #@TaskEnd_HTML@#
 +
  
 #@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~  Short-circuit voltage from the transformer impedance #@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~  Short-circuit voltage from the transformer impedance
Line 1503: Line 1986:
 A transformer has the rated primary data A transformer has the rated primary data
  
-\[ 
 \begin{align*} \begin{align*}
 U_{1{\rm N}}=230~{\rm V}, U_{1{\rm N}}=230~{\rm V},
Line 1509: Line 1991:
 I_{1{\rm N}}=2.0~{\rm A}. I_{1{\rm N}}=2.0~{\rm A}.
 \end{align*} \end{align*}
-\] 
  
 The short-circuit equivalent impedance referred to the primary side is The short-circuit equivalent impedance referred to the primary side is
  
-\[ 
 \begin{align*} \begin{align*}
 R_{\rm k}=1.5~\Omega, R_{\rm k}=1.5~\Omega,
Line 1519: Line 1999:
 X_{\rm k}=4.0~\Omega. X_{\rm k}=4.0~\Omega.
 \end{align*} \end{align*}
-\] 
- 
-Calculate: 
  
-  * \(|\underline{Z}_{\rm k}|\) 
-  * the rated short-circuit voltage \(U_{1{\rm k}}\) 
-  * the relative short-circuit voltage \(u_{\rm k}\) 
-  * the prospective continuous short-circuit current \(I_{1{\rm k}}\) for rated primary voltage 
-  * the approximate first peak current \(i_{\rm peak}\approx 2.54 I_{1{\rm k}}\) 
  
-#@ResultBegin_HTML~ExerciseShortCircuitVoltage~@# +1. Calculate $|\underline{Z}_{\rm k}|$. 
- +<WRAP group> 
-The short-circuit impedance magnitude is +<WRAP half column rightalign> 
- +#@PathBegin_HTML~2024~@# 
-\[+<WRAP leftalign> 
 +The short-circuit impedance magnitude is:
 \begin{align*} \begin{align*}
 |\underline{Z}_{\rm k}| |\underline{Z}_{\rm k}|
 = =
 \sqrt{R_{\rm k}^2+X_{\rm k}^2} \sqrt{R_{\rm k}^2+X_{\rm k}^2}
-=+\end{align*} 
 + 
 +Insert the values: 
 +\begin{align*} 
 +|\underline{Z}_{\rm k}| 
 +&=
 \sqrt{(1.5~\Omega)^2+(4.0~\Omega)^2} \sqrt{(1.5~\Omega)^2+(4.0~\Omega)^2}
-= +\\ 
-4.27~\Omega.+&= 
 +4.27~\Omega
 \end{align*} \end{align*}
-\]+</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 +<WRAP half column> 
 +#@ResultBegin_HTML~2024~@# 
 +\begin{align*} 
 +|\underline{Z}_{\rm k}|=4.27~\Omega 
 +\end{align*} 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP>
  
-The primary voltage required to drive rated current through the short-circuited transformer is +2. Calculate the rated short-circuit voltage $U_{1{\rm k}}$. 
- +<WRAP group> 
-\[+<WRAP half column rightalign> 
 +#@PathBegin_HTML~2025~@# 
 +<WRAP leftalign> 
 +The primary voltage required to drive rated current through the short-circuited transformer is:
 \begin{align*} \begin{align*}
 U_{1{\rm k}} U_{1{\rm k}}
 = =
 |\underline{Z}_{\rm k}|I_{1{\rm N}} |\underline{Z}_{\rm k}|I_{1{\rm N}}
-=+\end{align*} 
 + 
 +Insert the values: 
 +\begin{align*} 
 +U_{1{\rm k}} 
 +&=
 4.27~\Omega\cdot 2.0~{\rm A} 4.27~\Omega\cdot 2.0~{\rm A}
-= +\\ 
-8.54~{\rm V}.+&= 
 +8.54~{\rm V}
 \end{align*} \end{align*}
-\]+</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 +<WRAP half column> 
 +#@ResultBegin_HTML~2025~@# 
 +\begin{align*} 
 +U_{1{\rm k}}=8.54~{\rm V} 
 +\end{align*} 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP>
  
-The relative short-circuit voltage is +3. Calculate the relative short-circuit voltage ${\rm u_k}$. 
- +<WRAP group> 
-\[+<WRAP half column rightalign> 
 +#@PathBegin_HTML~2026~@# 
 +<WRAP leftalign> 
 +The relative short-circuit voltage is:
 \begin{align*} \begin{align*}
-u_{\rm k}+{\rm u_k}
 = =
 \frac{U_{1{\rm k}}}{U_{1{\rm N}}}\cdot 100~\% \frac{U_{1{\rm k}}}{U_{1{\rm N}}}\cdot 100~\%
-=+\end{align*} 
 + 
 +Insert the values: 
 +\begin{align*} 
 +{\rm u_k} 
 +&=
 \frac{8.54~{\rm V}}{230~{\rm V}}\cdot 100~\% \frac{8.54~{\rm V}}{230~{\rm V}}\cdot 100~\%
-= +\\ 
-3.71~\%.+&= 
 +3.71~\%
 \end{align*} \end{align*}
-\]+</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 +<WRAP half column> 
 +#@ResultBegin_HTML~2026~@# 
 +\begin{align*} 
 +{\rm u_k}=3.71~\% 
 +\end{align*} 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP>
  
-The prospective continuous short-circuit current is +4. Calculate the prospective continuous short-circuit current $I_{1{\rm k}}$ for rated primary voltage. 
- +<WRAP group> 
-\[+<WRAP half column rightalign> 
 +#@PathBegin_HTML~2027~@# 
 +<WRAP leftalign> 
 +The prospective continuous short-circuit current is:
 \begin{align*} \begin{align*}
 I_{1{\rm k}} I_{1{\rm k}}
 = =
-I_{1{\rm N}}\cdot \frac{100~\%}{u_{\rm k}} +I_{1{\rm N}}\cdot \frac{100~\%}{\rm u_k} 
-=+\end{align*} 
 + 
 +Insert the values: 
 +\begin{align*} 
 +I_{1{\rm k}} 
 +&=
 2.0~{\rm A}\cdot \frac{100}{3.71} 2.0~{\rm A}\cdot \frac{100}{3.71}
-= +\\ 
-53.9~{\rm A}.+&= 
 +53.9~{\rm A}
 \end{align*} \end{align*}
-\]+</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 +<WRAP half column> 
 +#@ResultBegin_HTML~2027~@# 
 +\begin{align*} 
 +I_{1{\rm k}}=53.9~{\rm A} 
 +\end{align*} 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP>
  
-The approximate first peak current is +5. Calculate the approximate first peak current $i_{\rm peak}\approx 2.54 I_{1{\rm k}}$. 
- +<WRAP group> 
-\[+<WRAP half column rightalign> 
 +#@PathBegin_HTML~2028~@# 
 +<WRAP leftalign> 
 +The approximate first peak current is:
 \begin{align*} \begin{align*}
 i_{\rm peak} i_{\rm peak}
 \approx \approx
 2.54 I_{1{\rm k}} 2.54 I_{1{\rm k}}
-=+\end{align*} 
 + 
 +Insert the short-circuit current: 
 +\begin{align*} 
 +i_{\rm peak} 
 +&\approx
 2.54\cdot 53.9~{\rm A} 2.54\cdot 53.9~{\rm A}
-= +\\ 
-137~{\rm A}.+&= 
 +137~{\rm A}
 \end{align*} \end{align*}
-\] 
- 
-Even though the rated current is only \(2.0~{\rm A}\), a short-circuit fault could lead to a much larger current until protection reacts. 
  
 +Even though the rated current is only $2.0~{\rm A}$, a short-circuit fault could lead to a much larger current until protection reacts.
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~2028~@#
 +\begin{align*}
 +i_{\rm peak}\approx 137~{\rm A}
 +\end{align*}
 #@ResultEnd_HTML@# #@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 #@TaskEnd_HTML@# #@TaskEnd_HTML@#
 +
  
 #@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~  Voltage drop under load using the Kapp approximation #@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~  Voltage drop under load using the Kapp approximation
 #@TaskText_HTML@# #@TaskText_HTML@#
  
-A transformer has the turns ratio+A transformer has the turns ratio $n=10$. 
 + 
 +The short-circuit equivalent parameters referred to the primary side are $R_{\rm k}=2.4~\Omega$, $X_{\rm k}=3.6~\Omega$. \\ 
 +The secondary load current is $I_2=4.0~{\rm A}$. The load has the power factor $\cos\varphi=0.8$ and is inductive. 
 + 
 +Estimate the voltage drop on the secondary side using
  
-\[ 
 \begin{align*} \begin{align*}
-n=10.+\Delta U_1 
 +\approx 
 +I_1\left(R_{\rm k}\cos\varphi+X_{\rm k}\sin\varphi\right)
 \end{align*} \end{align*}
-\] 
  
-The short-circuit equivalent parameters referred to the primary side are+and
  
-\[ 
 \begin{align*} \begin{align*}
-R_{\rm k}=2.4~\Omega, +\Delta U_2\approx \frac{\Delta U_1}{n}.
-\qquad +
-X_{\rm k}=3.6~\Omega.+
 \end{align*} \end{align*}
-\] 
  
-The secondary load current is+1. Calculate the primary current magnitude $I_1$. 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~2029~@# 
 +<WRAP leftalign> 
 +The primary current magnitude is approximately: 
 +\begin{align*} 
 +I_1 
 += 
 +\frac{I_2}{n} 
 +\end{align*}
  
-\[+Insert the values:
 \begin{align*} \begin{align*}
-I_2=4.0~{\rm A}.+I_1 
 +&= 
 +\frac{4.0~{\rm A}}{10} 
 +\\ 
 +&= 
 +0.40~{\rm A}
 \end{align*} \end{align*}
-\]+</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 +<WRAP half column> 
 +#@ResultBegin_HTML~2029~@# 
 +\begin{align*} 
 +I_1=0.40~{\rm A} 
 +\end{align*} 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP>
  
-The load has the power factor+2. Determine $\sin\varphi$ for the inductive load. 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~2030~@# 
 +<WRAP leftalign> 
 +For an inductive load with $\cos\varphi=0.8$: 
 +\begin{align*} 
 +\sin\varphi 
 += 
 +\sqrt{1-\cos^2\varphi} 
 +\end{align*}
  
-\[+Insert the value:
 \begin{align*} \begin{align*}
-\cos\varphi=0.8+\sin\varphi 
 +&= 
 +\sqrt{1-0.8^2} 
 +\\ 
 +&= 
 +0.6
 \end{align*} \end{align*}
-\]+</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 +<WRAP half column> 
 +#@ResultBegin_HTML~2030~@# 
 +\begin{align*} 
 +\sin\varphi=0.6 
 +\end{align*} 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP>
  
-and is inductive.+3. Estimate the primary-side voltage drop $\Delta U_1$. 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~2031~@# 
 +<WRAP leftalign> 
 +Use the Kapp approximation: 
 +\begin{align*} 
 +\Delta U_1 
 +&\approx 
 +I_1\left(R_{\rm k}\cos\varphi+X_{\rm k}\sin\varphi\right) 
 +\end{align*}
  
-Estimate the voltage drop on the secondary side using +Insert the values:
- +
-\[+
 \begin{align*} \begin{align*}
 \Delta U_1 \Delta U_1
 +&\approx
 +0.40~{\rm A}
 +\left(
 +2.4~\Omega\cdot 0.8
 ++
 +3.6~\Omega\cdot 0.6
 +\right)
 +\\
 +&=
 +0.40~{\rm A}
 +\left(
 +1.92~\Omega+2.16~\Omega
 +\right)
 +\\
 +&=
 +1.63~{\rm V}
 +\end{align*}
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~2031~@#
 +\begin{align*}
 +\Delta U_1\approx 1.63~{\rm V}
 +\end{align*}
 +#@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 +4. Estimate the secondary-side voltage drop $\Delta U_2$.
 +<WRAP group>
 +<WRAP half column rightalign>
 +#@PathBegin_HTML~2032~@#
 +<WRAP leftalign>
 +The secondary-side voltage drop is:
 +\begin{align*}
 +\Delta U_2
 \approx \approx
-I_1\left(R_{\rm k}\cos\varphi+X_{\rm k}\sin\varphi\right)+\frac{\Delta U_1}{n}
 \end{align*} \end{align*}
-\] 
  
-and+Insert the values: 
 +\begin{align*} 
 +\Delta U_2 
 +&\approx 
 +\frac{1.63~{\rm V}}{10} 
 +\\ 
 +&= 
 +0.163~{\rm V} 
 +\end{align*}
  
-\[+The secondary voltage decreases by approximately $0.16~{\rm V}$ for this operating point. 
 +</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 +<WRAP half column> 
 +#@ResultBegin_HTML~2032~@#
 \begin{align*} \begin{align*}
-\Delta U_2\approx \frac{\Delta U_1}{n}.+\Delta U_2\approx 0.163~{\rm V}
 \end{align*} \end{align*}
-\] 
  
-#@ResultBegin_HTML~ExerciseKappVoltageDrop~@#+The secondary voltage decreases by approximately $0.16~{\rm V}$. 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP>
  
-The primary current magnitude is approximately+#@TaskEnd_HTML@#
  
-\[+ 
 +#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~  Why the magnetizing branch can be neglected in the short-circuit test 
 +#@TaskText_HTML@# 
 + 
 +A transformer has the rated primary voltage $U_{1{\rm N}}=230~{\rm V}$. 
 +Its rated primary current is $I_{1{\rm N}}=3.0~{\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 ${\rm u_k}=6.0~\%$. 
 + 
 +Assume that the magnetizing current is approximately proportional to the applied voltage. 
 +\\ \\ \\ 
 +1. Calculate the short-circuit test voltage $U_{1{\rm k}}$. 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~2033~@# 
 +<WRAP leftalign> 
 +The short-circuit test voltage is:
 \begin{align*} \begin{align*}
-I_1+U_{1{\rm k}}
 = =
-\frac{I_2}{n}+\frac{\rm u_k}{100~\%}\cdot U_{1{\rm N}} 
 +\end{align*} 
 + 
 +Insert the values: 
 +\begin{align*} 
 +U_{1{\rm k}} 
 +&= 
 +0.06\cdot 230~{\rm V} 
 +\\ 
 +&= 
 +13.8~{\rm V} 
 +\end{align*} 
 +</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 +<WRAP half column> 
 +#@ResultBegin_HTML~2033~@# 
 +\begin{align*} 
 +U_{1{\rm k}}=13.8~{\rm V} 
 +\end{align*} 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP> 
 + 
 +2. Estimate the magnetizing current $I_{\rm m,k}$ during the short-circuit test. 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~2034~@# 
 +<WRAP leftalign> 
 +The magnetizing current is assumed to be proportional to the voltage: 
 +\begin{align*} 
 +I_{\rm m,k}
 = =
-\frac{4.0~{\rm A}}{10}+I_{\rm m,N}\cdot \frac{U_{1{\rm k}}}{U_{1{\rm N}}} 
 +\end{align*} 
 + 
 +Insert the values: 
 +\begin{align*} 
 +I_{\rm m,k} 
 +&= 
 +0.12~{\rm A}\cdot \frac{13.8~{\rm V}}{230~{\rm V}} 
 +\\ 
 +&= 
 +0.0072~{\rm A} 
 +\\ 
 +&= 
 +7.2~{\rm mA} 
 +\end{align*} 
 +</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 +<WRAP half column> 
 +#@ResultBegin_HTML~2034~@# 
 +\begin{align*} 
 +I_{\rm m,k} 
 += 
 +0.0072~{\rm A}
 = =
-0.40~{\rm A}.+7.2~{\rm mA}
 \end{align*} \end{align*}
-\]+#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP>
  
-For an inductive load with \(\cos\varphi=0.8\),+3. Compare $I_{\rm m,k}$ with $I_{1{\rm N}}$. 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~2035~@# 
 +<WRAP leftalign> 
 +Compare the short-circuit magnetizing current with the rated current: 
 +\begin{align*} 
 +\frac{I_{\rm m,k}}{I_{1{\rm N}}} 
 += 
 +\frac{0.0072~{\rm A}}{3.0~{\rm A}} 
 +\end{align*}
  
-\[+Calculate the ratio:
 \begin{align*} \begin{align*}
-\sin\varphi+\frac{I_{\rm m,k}}{I_{1{\rm N}}} 
 +&= 
 +0.0024 
 +\\ 
 +&= 
 +0.24~\% 
 +\end{align*} 
 +</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 +<WRAP half column> 
 +#@ResultBegin_HTML~2035~@# 
 +\begin{align*} 
 +\frac{I_{\rm m,k}}{I_{1{\rm N}}}
 = =
-\sqrt{1-\cos^2\varphi}+0.24~\% 
 +\end{align*} 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP> 
 + 
 +4. Explain why the magnetizing branch can be neglected. 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~2036~@# 
 +<WRAP leftalign> 
 +During the short-circuit test, the applied voltage is much smaller than the rated voltage: 
 +\begin{align*} 
 +U_{1{\rm k}} \ll U_{1{\rm N}} 
 +\end{align*} 
 + 
 +Since the magnetizing current is assumed to be approximately proportional to the applied voltage, the magnetizing current is also very small: 
 +\begin{align*} 
 +I_{\rm m,k}=7.2~{\rm mA} 
 +\end{align*} 
 + 
 +Compared with the rated current: 
 +\begin{align*} 
 +I_{1{\rm N}}=3.0~{\rm A} 
 +\end{align*} 
 + 
 +the magnetizing current is only $0.24~\%$. 
 + 
 +So during the short-circuit test, almost all current flows through the short-circuit path consisting of $R_{\rm k}$ and $X_{\rm k}$. 
 +</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 +<WRAP half column> 
 +#@ResultBegin_HTML~2036~@# 
 +The magnetizing branch $R_{\rm Fe}\parallel jX_{1{\rm H}}$ can usually be neglected in the short-circuit test. 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP> 
 + 
 +#@TaskEnd_HTML@# 
 + 
 + 
 +#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~  Why the short-circuit equivalent circuit is often sufficient under load 
 +#@TaskText_HTML@# 
 + 
 +A transformer has the turns ratio $n=10$. The short-circuit equivalent parameters referred to the primary side are $R_{\rm k}=2.0~\Omega$, $X_{\rm k}=4.0~\Omega$. 
 + 
 +The secondary load current is $I_2=5.0~{\rm A}$. The load is ohmic-inductive with $\cos\varphi=0.8$. 
 + 
 +The no-load current on the primary side is approximately $I_{10}=0.03~{\rm A}$. 
 +\\ \\ \\ 
 +1. Calculate the load-related primary current $I'_2$. 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~2037~@# 
 +<WRAP leftalign> 
 +The load-related primary current is: 
 +\begin{align*} 
 +I'_2
 = =
-\sqrt{1-0.8^2} +\frac{I_2}{n}
-= +
-0.6.+
 \end{align*} \end{align*}
-\] 
  
-The primary-side voltage drop is+Insert the values: 
 +\begin{align*} 
 +I'_2 
 +&= 
 +\frac{5.0~{\rm A}}{10} 
 +\\ 
 +&= 
 +0.50~{\rm A} 
 +\end{align*} 
 +</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 +<WRAP half column> 
 +#@ResultBegin_HTML~2037~@# 
 +\begin{align*} 
 +I'_2=0.50~{\rm A} 
 +\end{align*} 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP>
  
-\[+2. Estimate the primary-side voltage drop. 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~2038~@# 
 +<WRAP leftalign> 
 +The voltage drop is estimated with:
 \begin{align*} \begin{align*}
 \Delta U_1 \Delta U_1
-&\approx +\approx 
-I_1\left(R_{\rm k}\cos\varphi+X_{\rm k}\sin\varphi\right)+I'_2 
 +\left( 
 +R_{\rm k}\cos\varphi 
 ++ 
 +X_{\rm k}\sin\varphi 
 +\right) 
 +\end{align*} 
 + 
 +For $\cos\varphi=0.8$: 
 +\begin{align*} 
 +\sin\varphi 
 +&= 
 +\sqrt{1-\cos^2\varphi}
 \\ \\
 &= &=
-0.40~{\rm A}+\sqrt{1-0.8^2} 
 +\\ 
 +&= 
 +0.6 
 +\end{align*} 
 + 
 +Insert the values: 
 +\begin{align*} 
 +\Delta U_1 
 +&\approx 
 +0.50~{\rm A}
 \left( \left(
-2.4~\Omega\cdot 0.8+2.0~\Omega\cdot 0.8
 + +
-3.6~\Omega\cdot 0.6+4.0~\Omega\cdot 0.6
 \right) \right)
 \\ \\
 &= &=
-0.40~{\rm A}+0.50~{\rm A}
 \left( \left(
-1.92~\Omega+2.16~\Omega+1.6~\Omega+2.4~\Omega
 \right) \right)
 \\ \\
 &= &=
-1.63~{\rm V}.+2.0~{\rm V}
 \end{align*} \end{align*}
-\]+</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 +<WRAP half column> 
 +#@ResultBegin_HTML~2038~@# 
 +\begin{align*} 
 +\Delta U_1\approx 2.0~{\rm V} 
 +\end{align*} 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP>
  
-Referred to the secondary side: +3. Calculate the secondary-side voltage drop $\Delta U_2\approx \frac{\Delta U_1}{n}$. 
- +<WRAP group> 
-\[+<WRAP half column rightalign> 
 +#@PathBegin_HTML~2039~@# 
 +<WRAP leftalign> 
 +The secondary-side voltage drop is:
 \begin{align*} \begin{align*}
 \Delta U_2 \Delta U_2
 \approx \approx
 \frac{\Delta U_1}{n} \frac{\Delta U_1}{n}
 +\end{align*}
 +
 +Insert the values:
 +\begin{align*}
 +\Delta U_2
 +&\approx
 +\frac{2.0~{\rm V}}{10}
 +\\
 +&=
 +0.20~{\rm V}
 +\end{align*}
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~2039~@#
 +\begin{align*}
 +\Delta U_2\approx 0.20~{\rm V}
 +\end{align*}
 +#@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 +4. Estimate an upper bound for the neglected voltage drop caused by $I_{10}$.
 +<WRAP group>
 +<WRAP half column rightalign>
 +#@PathBegin_HTML~2040~@#
 +<WRAP leftalign>
 +First calculate the magnitude of the short-circuit impedance:
 +\begin{align*}
 +|\underline{Z}_{\rm k}|
 = =
-\frac{1.63~{\rm V}}{10} +\sqrt{R_{\rm k}^2+X_{\rm k}^2}
-= +
-0.163~{\rm V}.+
 \end{align*} \end{align*}
-\] 
  
-The secondary voltage decreases by approximately \(0.16~{\rm V}\) for this operating point.+Insert the values: 
 +\begin{align*} 
 +|\underline{Z}_{\rm k}| 
 +&= 
 +\sqrt{(2.0~\Omega)^2+(4.0~\Omega)^2} 
 +\\ 
 +&= 
 +4.47~\Omega 
 +\end{align*}
  
 +The upper bound for the voltage drop caused by the no-load current is:
 +\begin{align*}
 +\Delta U_{1,10}
 +&\leq
 +|\underline{Z}_{\rm k}|I_{10}
 +\\
 +&=
 +4.47~\Omega\cdot 0.03~{\rm A}
 +\\
 +&=
 +0.134~{\rm V}
 +\end{align*}
 +
 +On the secondary side:
 +\begin{align*}
 +\Delta U_{2,10}
 +&\leq
 +\frac{0.134~{\rm V}}{10}
 +\\
 +&=
 +0.0134~{\rm V}
 +\end{align*}
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~2040~@#
 +\begin{align*}
 +|\underline{Z}_{\rm k}| &= 4.47~\Omega
 +\\
 +\Delta U_{1,10} &\leq 0.134~{\rm V}
 +\\
 +\Delta U_{2,10} &\leq 0.0134~{\rm V}
 +\end{align*}
 #@ResultEnd_HTML@# #@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 +5. Decide whether the short-circuit equivalent circuit is sufficient for this load estimate.
 +<WRAP group>
 +<WRAP half column rightalign>
 +#@PathBegin_HTML~2041~@#
 +<WRAP leftalign>
 +The load-related secondary-side voltage drop is:
 +\begin{align*}
 +\Delta U_2\approx 0.20~{\rm V}
 +\end{align*}
 +
 +The estimated neglected secondary-side voltage drop caused by the no-load current is at most:
 +\begin{align*}
 +\Delta U_{2,10}\leq 0.0134~{\rm V}
 +\end{align*}
 +
 +This is small compared with the load-related drop of $0.20~{\rm V}$.
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~2041~@#
 +For this engineering estimate, the short-circuit equivalent circuit is sufficient.
 +#@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 #@TaskEnd_HTML@# #@TaskEnd_HTML@#
  
 +
 +#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~  Ideal transformer versus real transformer
 +#@TaskText_HTML@#
 +
 +A transformer has the rated primary voltage of $U_1=230~{\rm V}$ and and the turns ratio $n=10$. A resistive load draws $I_2=4.0~{\rm A}$. \\
 +For the real transformer, the short-circuit equivalent parameters referred to the primary side are $R_{\rm k}=2.4~\Omega$, $X_{\rm k}=3.2~\Omega$. \\
 +The iron loss is approximately $P_{\rm Fe}=1.5~{\rm W}$. \\
 +Assume a resistive load with $\cos\varphi=1$.
 +\\ \\ 
 +1. Calculate the ideal secondary voltage $U_{2,\rm ideal}$.
 +<WRAP group>
 +<WRAP half column rightalign>
 +#@PathBegin_HTML~2042~@#
 +<WRAP leftalign>
 +For the ideal transformer:
 +\begin{align*}
 +U_{2,\rm ideal}
 +=
 +\frac{U_1}{n}
 +\end{align*}
 +
 +Insert the values:
 +\begin{align*}
 +U_{2,\rm ideal}
 +&=
 +\frac{230~{\rm V}}{10}
 +\\
 +&=
 +23.0~{\rm V}
 +\end{align*}
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~2042~@#
 +\begin{align*}
 +U_{2,\rm ideal}=23.0~{\rm V}
 +\end{align*}
 +#@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 +2. Calculate the load-related primary current $I'_2$.
 +<WRAP group>
 +<WRAP half column rightalign>
 +#@PathBegin_HTML~2043~@#
 +<WRAP leftalign>
 +The load-related primary current is:
 +\begin{align*}
 +I'_2
 +=
 +\frac{I_2}{n}
 +\end{align*}
 +
 +Insert the values:
 +\begin{align*}
 +I'_2
 +&=
 +\frac{4.0~{\rm A}}{10}
 +\\
 +&=
 +0.40~{\rm A}
 +\end{align*}
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~2043~@#
 +\begin{align*}
 +I'_2=0.40~{\rm A}
 +\end{align*}
 +#@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 +3. Estimate the real secondary voltage.
 +<WRAP group>
 +<WRAP half column rightalign>
 +#@PathBegin_HTML~2044~@#
 +<WRAP leftalign>
 +For a resistive load, the approximate primary-side voltage drop is:
 +\begin{align*}
 +\Delta U_1
 +\approx
 +I'_2R_{\rm k}
 +\end{align*}
 +
 +Insert the values:
 +\begin{align*}
 +\Delta U_1
 +&\approx
 +0.40~{\rm A}\cdot 2.4~\Omega
 +\\
 +&=
 +0.96~{\rm V}
 +\end{align*}
 +
 +The corresponding secondary-side voltage drop is:
 +\begin{align*}
 +\Delta U_2
 +&\approx
 +\frac{\Delta U_1}{n}
 +\\
 +&=
 +\frac{0.96~{\rm V}}{10}
 +\\
 +&=
 +0.096~{\rm V}
 +\end{align*}
 +
 +Thus the real secondary voltage is approximately:
 +\begin{align*}
 +U_{2,\rm real}
 +&\approx
 +23.0~{\rm V}-0.096~{\rm V}
 +\\
 +&=
 +22.90~{\rm V}
 +\end{align*}
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~2044~@#
 +\begin{align*}
 +\Delta U_1 &\approx 0.96~{\rm V}
 +\\
 +\Delta U_2 &\approx 0.096~{\rm V}
 +\\
 +U_{2,\rm real} &\approx 22.90~{\rm V}
 +\end{align*}
 +#@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 +4. Estimate the copper losses.
 +<WRAP group>
 +<WRAP half column rightalign>
 +#@PathBegin_HTML~2045~@#
 +<WRAP leftalign>
 +The copper losses are:
 +\begin{align*}
 +P_{\rm Cu}
 +\approx
 +R_{\rm k}(I'_2)^2
 +\end{align*}
 +
 +Insert the values:
 +\begin{align*}
 +P_{\rm Cu}
 +&\approx
 +2.4~\Omega\cdot (0.40~{\rm A})^2
 +\\
 +&=
 +0.384~{\rm W}
 +\end{align*}
 +
 +The real transformer also has iron losses:
 +\begin{align*}
 +P_{\rm Fe}=1.5~{\rm W}
 +\end{align*}
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~2045~@#
 +\begin{align*}
 +P_{\rm Cu}\approx 0.384~{\rm W}
 +\end{align*}
 +
 +Additionally:
 +\begin{align*}
 +P_{\rm Fe}=1.5~{\rm W}
 +\end{align*}
 +#@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 +5. Compare ideal and real transformer behavior.
 +<WRAP group>
 +<WRAP half column rightalign>
 +#@PathBegin_HTML~2046~@#
 +<WRAP leftalign>
 +For the ideal transformer:
 +\begin{align*}
 +U_{2,\rm ideal}=23.0~{\rm V}
 +\end{align*}
 +
 +For the real transformer:
 +\begin{align*}
 +U_{2,\rm real}\approx 22.90~{\rm V}
 +\end{align*}
 +
 +The real transformer has copper losses and iron losses:
 +\begin{align*}
 +P_{\rm Cu}&\approx 0.384~{\rm W}
 +\\
 +P_{\rm Fe}&=1.5~{\rm W}
 +\end{align*}
 +
 +So the main differences are:
 +
 +  * the ideal transformer has exactly $U_2=23.0~{\rm V}$, the real transformer has a slightly lower voltage,
 +  * the ideal transformer has no losses, the real transformer has copper and iron losses,
 +  * the ideal transformer has no leakage voltage drop, the real transformer has a load-dependent voltage drop.
 +
 +For this operating point the transformer is close to ideal, but not exactly ideal.
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~2046~@#
 +The real transformer differs from the ideal transformer by:
 +
 +  * a slightly lower secondary voltage,
 +  * copper and iron losses,
 +  * a load-dependent voltage drop.
 +
 +For this operating point it is close to ideal, but not exactly ideal.
 +#@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 +#@TaskEnd_HTML@#
  
 ===== Common pitfalls ===== ===== Common pitfalls =====
Line 1746: 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.