This is an old revision of the document!
Exercise E1 Dot convention and load current direction
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?
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.
Exercise E2 Mutual inductance and leakage from a magnetic path
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 \(k=\frac{M}{\sqrt{L_1L_2}}\)
The main-flux self-inductances are
\[ \begin{align*} L_{1{\rm H}} &= \frac{N_1^2}{R_{\rm mH}} = \frac{400^2}{1.6\cdot 10^6~1/{\rm H}} = 0.100~{\rm H}, \\[4pt] L_{2{\rm H}} &= \frac{N_2^2}{R_{\rm mH}} = \frac{100^2}{1.6\cdot 10^6~1/{\rm H}} = 0.00625~{\rm H} = 6.25~{\rm mH}. \end{align*} \]
The mutual inductance is
\[ \begin{align*} M = \frac{N_1N_2}{R_{\rm mH}} = \frac{400\cdot 100}{1.6\cdot 10^6~1/{\rm H}} = 0.025~{\rm H} = 25~{\rm mH}. \end{align*} \]
The total self-inductances are
\[ \begin{align*} L_1 &= L_{1{\rm H}}+L_{1\sigma} = 100~{\rm mH}+4.0~{\rm mH} = 104~{\rm mH}, \\[4pt] L_2 &= L_{2{\rm H}}+L_{2\sigma} = 6.25~{\rm mH}+0.30~{\rm mH} = 6.55~{\rm mH}. \end{align*} \]
Thus
\[ \begin{align*} k = \frac{M}{\sqrt{L_1L_2}} = \frac{0.025~{\rm H}}{\sqrt{0.104~{\rm H}\cdot 0.00655~{\rm H}}} \approx 0.96. \end{align*} \]
The coupling is strong, but not ideal, because leakage inductances are present.
Exercise E3 Short-circuit voltage from the transformer impedance
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 \(i_{\rm peak}\approx 2.54 I_{1{\rm k}}\)
The short-circuit impedance magnitude is
\[ \begin{align*} |\underline{Z}_{\rm k}| = \sqrt{R_{\rm k}^2+X_{\rm k}^2} = \sqrt{(1.5~\Omega)^2+(4.0~\Omega)^2} = 4.27~\Omega. \end{align*} \]
The primary voltage required to drive rated current through the short-circuited transformer is
\[ \begin{align*} U_{1{\rm k}} = |\underline{Z}_{\rm k}|I_{1{\rm N}} = 4.27~\Omega\cdot 2.0~{\rm A} = 8.54~{\rm V}. \end{align*} \]
The relative short-circuit voltage is
\[ \begin{align*} u_{\rm k} = \frac{U_{1{\rm k}}}{U_{1{\rm N}}}\cdot 100~\% = \frac{8.54~{\rm V}}{230~{\rm V}}\cdot 100~\% = 3.71~\%. \end{align*} \]
The prospective continuous short-circuit current is
\[ \begin{align*} I_{1{\rm k}} = I_{1{\rm N}}\cdot \frac{100~\%}{u_{\rm k}} = 2.0~{\rm A}\cdot \frac{100}{3.71} = 53.9~{\rm A}. \end{align*} \]
The approximate first peak current is
\[ \begin{align*} i_{\rm peak} \approx 2.54 I_{1{\rm k}} = 2.54\cdot 53.9~{\rm A} = 137~{\rm A}. \end{align*} \]
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.
Exercise E4 Voltage drop under load using the Kapp approximation
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
\[ \begin{align*} \Delta U_2\approx \frac{\Delta U_1}{n}. \end{align*} \]
The primary current magnitude is approximately
\[ \begin{align*} I_1 = \frac{I_2}{n} = \frac{4.0~{\rm A}}{10} = 0.40~{\rm A}. \end{align*} \]
For an inductive load with \(\cos\varphi=0.8\),
\[ \begin{align*} \sin\varphi = \sqrt{1-\cos^2\varphi} = \sqrt{1-0.8^2} = 0.6. \end{align*} \]
The primary-side voltage drop is
\[ \begin{align*} \Delta U_1 &\approx I_1\left(R_{\rm k}\cos\varphi+X_{\rm k}\sin\varphi\right) \\ &= 0.40~{\rm A} \left( 2.4~\Omega\cdot 0.8 + 3.6~\Omega\cdot 0.6 \right) \\ &= 0.40~{\rm A} \left( 1.92~\Omega+2.16~\Omega \right) \\ &= 1.63~{\rm V}. \end{align*} \]
Referred to the secondary side:
\[ \begin{align*} \Delta U_2 \approx \frac{\Delta U_1}{n} = \frac{1.63~{\rm V}}{10} = 0.163~{\rm V}. \end{align*} \]
The secondary voltage decreases by approximately \(0.16~{\rm V}\) for this operating point.
Exercise E5 Concept question: why the magnetizing branch can be neglected in the short-circuit test
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.
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.