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

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:

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

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

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!

• You seem to be calculating the mutual inductance by considering the open circuit secondary but I can't see how this can be done without knowing both the coupling, the output voltage and the secondary inductance. Maybe I missed something? Also, the primary inductance (L11) does not involve U2 although I get the same answer - maybe you typoed. – Andy aka Nov 28 '18 at 18:34
• Hi! It can be done applying the induction law as I did! Because I obtain it with the relation between u2 and i1. Because the current i2 is zero... Yes it was a typo, thanks! How would you calculate the coupling and the secondary inductanc? – Granger Obliviate Nov 28 '18 at 19:42
• On the short-circuit test, the primary voltage is lowered so there's the rated secondary current drawn at secondary short circuit. – Janka Nov 28 '18 at 19:48
• Janka I don't understand what you said and how it helps me – Granger Obliviate Nov 28 '18 at 19:51
• Anyone please?? – Granger Obliviate Nov 28 '18 at 20:43

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.

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

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

• How the hell do I work with this equations? I need to pass to phasor form but then how do I take the module? Because I will have the sum of phasor I1 and phasor I2 I have no idea on how to take absolute values. Can you be more specific than that? – Granger Obliviate Nov 28 '18 at 21:25
• Can you help me please? – Granger Obliviate Nov 29 '18 at 0:25
• can you help me please? – Granger Obliviate Nov 29 '18 at 18:04