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--- dummy [2026/05/17 10:49] – mexleadmin
+++ dummy [2026/05/26 13:48] (current) – removed mexleadmin
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-#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~.1  Dot convention and load current direction
-#@TaskText_HTML@#
-A transformer winding pair is marked with dots. A positive reference current \(i_1\) enters the dotted terminal of winding 1.
-Assume positive coupling and define \(u_2\) positive at the dotted terminal of winding 2.
-  * What is the polarity of the induced voltage on winding 2?
-  * If a resistor \(R_2\) is connected to winding 2, in which direction does the load current flow?
-  * Why can the transformer current reference \(i_2\) be opposite to the actual load current?
-#@ResultBegin_HTML~ExerciseDotConvention~@#
-For positive coupling:
-\[
-\begin{align*}
-i_1 \text{ enters the dotted terminal}
-\quad \Rightarrow \quad
-\text{the dotted terminal of winding 2 becomes positive.}
-\end{align*}
-\]
-Since \(u_2\) is defined positive at the dotted terminal, the induced voltage is aligned with \(u_2\).
-If only a resistor \(R_2\) is connected to the secondary side, this voltage drives current out of the dotted terminal into the load.
-However, in transformer equivalent circuits \(i_2\) is often drawn as a passive current entering the transformer port. Then
-\[
-\begin{align*}
-i_2=-i_{\rm load}.
-\end{align*}
-\]
-This is not negative coupling. It only means that the secondary side delivers power to the load.
-#@ResultEnd_HTML@#
-#@TaskEnd_HTML@#
-#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~.1  Mutual inductance and leakage from a magnetic path
-#@TaskText_HTML@#
-Two coils are wound on the same magnetic core. The shared main magnetic path has the reluctance
-\[
-\begin{align*}
-R_{\rm mH}=1.6\cdot 10^6~\frac{1}{\rm H}.
-\end{align*}
-\]
-The numbers of turns are
-\[
-\begin{align*}
-N_1=400,
-\qquad
-N_2=100.
-\end{align*}
-\]
-The leakage inductances are
-\[
-\begin{align*}
-L_{1\sigma}=4.0~{\rm mH},
-\qquad
-L_{2\sigma}=0.30~{\rm mH}.
-\end{align*}
-\]
-Calculate:
-  * \(L_{1{\rm H}}\)
-  * \(L_{2{\rm H}}\)
-  * \(M\)
-  * 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
-\[
-\begin{align*}
-L_{1{\rm H}}
-&=
-\frac{N_1^2}{R_{\rm mH}}
-=
-\frac{400^2}{1.6\cdot 10^6~1/{\rm H}}
-=
-.100~{\rm H},
-\\[4pt]
-L_{2{\rm H}}
-&=
-\frac{N_2^2}{R_{\rm mH}}
-=
-\frac{100^2}{1.6\cdot 10^6~1/{\rm H}}
-=
-.00625~{\rm H}
-=
-.25~{\rm mH}.
-\end{align*}
-\]
-The mutual inductance is
-\[
-\begin{align*}
-M
-=
-\frac{N_1N_2}{R_{\rm mH}}
-=
-\frac{400\cdot 100}{1.6\cdot 10^6~1/{\rm H}}
-=
-.025~{\rm H}
-=
-~{\rm mH}.
-\end{align*}
-\]
-The total self-inductances are
-\[
-\begin{align*}
-L_1
-&=
-L_{1{\rm H}}+L_{1\sigma}
-=
-~{\rm mH}+4.0~{\rm mH}
-=
-~{\rm mH},
-\\[4pt]
-L_2
-&=
-L_{2{\rm H}}+L_{2\sigma}
-=
-.25~{\rm mH}+0.30~{\rm mH}
-=
-.55~{\rm mH}.
-\end{align*}
-\]
-Thus
-\[
-\begin{align*}
-k
-=
-\frac{M}{\sqrt{L_1L_2}}
-=
-\frac{0.025~{\rm H}}{\sqrt{0.104~{\rm H}\cdot 0.00655~{\rm H}}}
-\approx
-.96.
-\end{align*}
-\]
-The coupling is strong, but not ideal, because leakage inductances are present.
-#@ResultEnd_HTML@#
-#@TaskEnd_HTML@#
-#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~.1  Short-circuit voltage from the transformer impedance
-#@TaskText_HTML@#
-A transformer has the rated primary data
-\[
-\begin{align*}
-U_{1{\rm N}}=230~{\rm V},
-\qquad
-I_{1{\rm N}}=2.0~{\rm A}.
-\end{align*}
-\]
-The short-circuit equivalent impedance referred to the primary side is
-\[
-\begin{align*}
-R_{\rm k}=1.5~\Omega,
-\qquad
-X_{\rm k}=4.0~\Omega.
-\end{align*}
-\]
-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~@#
-The short-circuit impedance magnitude is
-\[
-\begin{align*}
-|\underline{Z}_{\rm k}|
-=
-\sqrt{R_{\rm k}^2+X_{\rm k}^2}
-=
-\sqrt{(1.5~\Omega)^2+(4.0~\Omega)^2}
-=
-.27~\Omega.
-\end{align*}
-\]
-The primary voltage required to drive rated current through the short-circuited transformer is
-\[
-\begin{align*}
-U_{1{\rm k}}
-=
-|\underline{Z}_{\rm k}|I_{1{\rm N}}
-=
-.27~\Omega\cdot 2.0~{\rm A}
-=
-.54~{\rm V}.
-\end{align*}
-\]
-The relative short-circuit voltage is
-\[
-\begin{align*}
-u_{\rm k}
-=
-\frac{U_{1{\rm k}}}{U_{1{\rm N}}}\cdot 100~\%
-=
-\frac{8.54~{\rm V}}{230~{\rm V}}\cdot 100~\%
-=
-.71~\%.
-\end{align*}
-\]
-The prospective continuous short-circuit current is
-\[
-\begin{align*}
-I_{1{\rm k}}
-=
-I_{1{\rm N}}\cdot \frac{100~\%}{u_{\rm k}}
-=
-.0~{\rm A}\cdot \frac{100}{3.71}
-=
-.9~{\rm A}.
-\end{align*}
-\]
-The approximate first peak current is
-\[
-\begin{align*}
-i_{\rm peak}
-\approx
-.54 I_{1{\rm k}}
-=
-.54\cdot 53.9~{\rm A}
-=
-~{\rm A}.
-\end{align*}
-\]
-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.
-#@ResultEnd_HTML@#
-#@TaskEnd_HTML@#
-#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~.1  Voltage drop under load using the Kapp approximation
-#@TaskText_HTML@#
-A transformer has the turns ratio
-\[
-\begin{align*}
-n=10.
-\end{align*}
-\]
-The short-circuit equivalent parameters referred to the primary side are
-\[
-\begin{align*}
-R_{\rm k}=2.4~\Omega,
-\qquad
-X_{\rm k}=3.6~\Omega.
-\end{align*}
-\]
-The secondary load current is
-\[
-\begin{align*}
-I_2=4.0~{\rm A}.
-\end{align*}
-\]
-The load has the power factor
-\[
-\begin{align*}
-\cos\varphi=0.8
-\end{align*}
-\]
-and is inductive.
-Estimate the voltage drop on the secondary side using
-\[
-\begin{align*}
-\Delta U_1
-\approx
-I_1\left(R_{\rm k}\cos\varphi+X_{\rm k}\sin\varphi\right)
-\end{align*}
-\]
-and
-\[
-\begin{align*}
-\Delta U_2\approx \frac{\Delta U_1}{n}.
-\end{align*}
-\]
-#@ResultBegin_HTML~ExerciseKappVoltageDrop~@#
-The primary current magnitude is approximately
-\[
-\begin{align*}
-I_1
-=
-\frac{I_2}{n}
-=
-\frac{4.0~{\rm A}}{10}
-=
-.40~{\rm A}.
-\end{align*}
-\]
-For an inductive load with \(\cos\varphi=0.8\),
-\[
-\begin{align*}
-\sin\varphi
-=
-\sqrt{1-\cos^2\varphi}
-=
-\sqrt{1-0.8^2}
-=
-.6.
-\end{align*}
-\]
-The primary-side voltage drop is
-\[
-\begin{align*}
-\Delta U_1
-&\approx
-I_1\left(R_{\rm k}\cos\varphi+X_{\rm k}\sin\varphi\right)
-\\
-&=
-.40~{\rm A}
-\left(
-.4~\Omega\cdot 0.8
-+
-.6~\Omega\cdot 0.6
-\right)
-\\
-&=
-.40~{\rm A}
-\left(
-.92~\Omega+2.16~\Omega
-\right)
-\\
-&=
-.63~{\rm V}.
-\end{align*}
-\]
-Referred to the secondary side:
-\[
-\begin{align*}
-\Delta U_2
-\approx
-\frac{\Delta U_1}{n}
-=
-\frac{1.63~{\rm V}}{10}
-=
-.163~{\rm V}.
-\end{align*}
-\]
-The secondary voltage decreases by approximately \(0.16~{\rm V}\) for this operating point.
-#@ResultEnd_HTML@#
-#@TaskEnd_HTML@#
-#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~.1  Concept question: why the magnetizing branch can be neglected in the short-circuit test
-#@TaskText_HTML@#
-In the short-circuit test of a transformer, the secondary side is shorted. The primary voltage is then increased only until rated current flows.
-Explain in two or three sentences why the magnetizing branch \(R_{\rm Fe}\parallel jX_{1{\rm H}}\) can usually be neglected in this test.
-#@ResultBegin_HTML~ExerciseMagnetizingBranchShortCircuit~@#
-In the short-circuit test, only a small fraction of the rated primary voltage is needed to drive rated current through the short-circuited transformer. Since the main flux is approximately proportional to the applied voltage,
-\[
-\begin{align*}
-\underline{U}_1 \approx j\omega N_1\underline{\Phi}_{\rm H},
-\end{align*}
-\]
-the main flux is also small.
-Therefore the magnetizing current through \(jX_{1{\rm H}}\) and the iron-loss current through \(R_{\rm Fe}\) are small compared with the short-circuit current through \(R_{\rm k}+jX_{\rm k}\). For this reason, the magnetizing branch is usually neglected in the short-circuit equivalent circuit.
-#@ResultEnd_HTML@#
-#@TaskEnd_HTML@#