The relative short-circuit voltage is

\[ \begin{align*} u_{\rm k} = \frac{U_{1{\rm k}}}{U_{1{\rm N}}}\cdot 100~\% = \frac{35~{\rm V}}{230~{\rm V}}\cdot 100~\% = 15.2~\%. \end{align*} \]

The blocked-rotor current at rated voltage is approximately

\[ \begin{align*} I_{1{\rm block}} = I_{1{\rm N}}\cdot \frac{100~\%}{u_{\rm k}} = 4.0~{\rm A}\cdot \frac{100}{15.2} = 26.3~{\rm A}. \end{align*} \]

Thus

\[ \begin{align*} \frac{I_{1{\rm block}}}{I_{1{\rm N}}} = \frac{26.3~{\rm A}}{4.0~{\rm A}} \approx 6.6. \end{align*} \]

The blocked-rotor current is high because the rotor cage behaves like a short-circuited secondary winding. Only the internal impedance of the motor limits the current.

When the rotor starts rotating, the relative motion between the rotating magnetic field and the rotor becomes smaller. The induced rotor voltage becomes smaller, the rotor current decreases, and the motor moves away from the transformer short-circuit situation.

The short-circuit test voltage is

\[ \begin{align*} U_{1{\rm k}} = \frac{u_{\rm k}}{100~\%}\cdot U_{1{\rm N}} = 0.06\cdot 230~{\rm V} = 13.8~{\rm V}. \end{align*} \]

The magnetizing current is assumed to be proportional to the voltage:

\[ \begin{align*} I_{\rm m,k} &= I_{\rm m,N}\cdot \frac{U_{1{\rm k}}}{U_{1{\rm N}}} \\ &= 0.12~{\rm A}\cdot \frac{13.8~{\rm V}}{230~{\rm V}} \\ &= 0.0072~{\rm A} = 7.2~{\rm mA}. \end{align*} \]

Compared with the rated current:

\[ \begin{align*} \frac{I_{\rm m,k}}{I_{1{\rm N}}} = \frac{0.0072~{\rm A}}{3.0~{\rm A}} = 0.0024 = 0.24~\%. \end{align*} \]

So during the short-circuit test, the magnetizing current is much smaller than the rated current. Almost all current flows through the short-circuit path consisting of \(R_{\rm k}\) and \(X_{\rm k}\).

Therefore, the magnetizing branch \(R_{\rm Fe}\parallel jX_{1{\rm H}}\) can usually be neglected in the short-circuit test.

The load-related primary current is

\[ \begin{align*} I'_2 = \frac{I_2}{n} = \frac{5.0~{\rm A}}{10} = 0.50~{\rm A}. \end{align*} \]

For \(\cos\varphi=0.8\),

\[ \begin{align*} \sin\varphi = \sqrt{1-\cos^2\varphi} = \sqrt{1-0.8^2} = 0.6. \end{align*} \]

The primary-side voltage drop is

\[ \begin{align*} \Delta U_1 &\approx I'_2 \left( R_{\rm k}\cos\varphi + X_{\rm k}\sin\varphi \right) \\ &= 0.50~{\rm A} \left( 2.0~\Omega\cdot 0.8 + 4.0~\Omega\cdot 0.6 \right) \\ &= 0.50~{\rm A} \left( 1.6~\Omega+2.4~\Omega \right) \\ &= 2.0~{\rm V}. \end{align*} \]

The secondary-side voltage drop is approximately

\[ \begin{align*} \Delta U_2 \approx \frac{\Delta U_1}{n} = \frac{2.0~{\rm V}}{10} = 0.20~{\rm V}. \end{align*} \]

The short-circuit impedance magnitude is

\[ \begin{align*} |\underline{Z}_{\rm k}| = \sqrt{R_{\rm k}^2+X_{\rm k}^2} = \sqrt{(2.0~\Omega)^2+(4.0~\Omega)^2} = 4.47~\Omega. \end{align*} \]

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 this is only

\[ \begin{align*} \Delta U_{2,10} \leq \frac{0.134~{\rm V}}{10} = 0.0134~{\rm V}. \end{align*} \]

This is small compared with the load-related drop of \(0.20~{\rm V}\). Therefore, for this engineering estimate, the short-circuit equivalent circuit is sufficient.

For the ideal transformer:

\[ \begin{align*} U_{2,\rm ideal} = \frac{U_1}{n} = \frac{230~{\rm V}}{10} = 23.0~{\rm V}. \end{align*} \]

The load-related primary current is

\[ \begin{align*} I'_2 = \frac{I_2}{n} = \frac{4.0~{\rm A}}{10} = 0.40~{\rm A}. \end{align*} \]

For a resistive load, the approximate primary-side voltage drop is

\[ \begin{align*} \Delta U_1 \approx I'_2R_{\rm k} = 0.40~{\rm A}\cdot 2.4~\Omega = 0.96~{\rm V}. \end{align*} \]

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

\[ \begin{align*} U_{2,\rm real} \approx 23.0~{\rm V}-0.096~{\rm V} = 22.90~{\rm V}. \end{align*} \]

The copper losses are

\[ \begin{align*} P_{\rm Cu} \approx R_{\rm k}(I'_2)^2 = 2.4~\Omega\cdot (0.40~{\rm A})^2 = 0.384~{\rm W}. \end{align*} \]

The real transformer also has iron losses:

\[ \begin{align*} P_{\rm Fe}=1.5~{\rm W}. \end{align*} \]

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.