Parameters of transformer: open-circuit test and short-circuit test

Question

Ok I have the following circuit and data (when the subscript is "ef" it means "rms" values):

I am asked to determine the paramenters of the transformer r1 L11, L22 and LM with the given experimental data.

I had no problem extracting data from the open circuit experiment. Using the fact that the active power is given by $$P=r_1 I_{rms}^2$$

I found $$r_1=10 \Omega$$

Then applying induction law in both primary and secondary leaves us with:

$$u_1(t)=r_1i_1(t)+L_{11}\frac{di_1(t)}{dt}$$ $$u_2(t)=-L_{M}\frac{di_1(t)}{dt}$$

Applying phasor notation and taking the rms values will lead us to obtain

$$L_M=\frac{U_{2_{rms}}}{\omega I_{1_{rms}} }=31.83 mH$$ $$L_{11}=\sqrt{(\frac{U_{1_{rms}}^2}{I_{1_{rms}}^2} - r_1^2) \frac{1}{\omega^2}}=55.13 mH$$

Ok and there is no more data we can extract form the open-circuit experiment.

Passing to the short-circuit experiment I will obtain from induction law again:

$$0=-L_{M}\frac{di_1(t)}{dt}-L_{22}\frac{di_2(t)}{dt}$$

Which leads to

$$L_{22}=\frac{L_M I_{1_{rms}}}{I_{2_{rms}} }$$

Problem now is I don't know the value of the root-mean square of current 2 and have no idea how to find it out. My guess is that I need to use the reactive power. But how? I know from Poynting complex theorem:

$$P_Q= 2\omega ((W_e)_{av} - (W_m)_{av})$$

But, and that is another question I have and would like to get ans answer on? How should I apply this formula. For the electrical energy, should I take the capacitor? But what's the voltage value? The same as the open-circuit experiment? And for the magnetic energy? What inductances should I consider? Do I need to calculate an equivalent circuit?

I'm really confused and would appreciate some help. Thanks!

Answer 1

The closed circuit experiment will yield two equations instead of one:

$$\begin{align} u_1(t) &= r_1i_1(t) + L_{11}\frac{di_1}{dt} - L_M\frac{di_2}{dt}\\ u_2(t) &= r_2i_2(t) - L_M\frac{di_1}{dt} + L_{22}\frac{di_2}{dt} \end{align}$$

(Note that the signs may need to be changed)

Meaning it will have 2 equations and 2 unknowns (\$i_1(t)\$ and \$i_2(t)\$), which can be solved.

[Edit]

For your short-circuit experiment and when using phasors, this would result into something like:

$$\begin{align} \underline{U_1} &= \left(r_1 + j\omega L_{11}\right) \underline{I_1} - j\omega L_M \underline{I_2}\\ 0 &= -j\omega L_M \underline{I_1} + \left(r_2 + j\omega L_{22}\right) \underline{I_2} \end{align}$$

The solution to this set of equations will be two complex values for \$\underline{I_1}\$ and \$\underline{I_2}\$. You can take the absolute value like with any phasor.

$$\underline{U_1} = U_{RMS}\cdot \sqrt{2}$$

$$I_{RMS} = \sqrt{\frac{\text{re}[\underline{I}]^2+\text{im}[\underline{I}]^2}{2}}$$

And you can calculate the complex power as well

$$\underline{P} = \underline{U_1}\cdot\underline{I_1}^*$$