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I'm doing some experiments with a transformer, measuring some physical quantities.

First I did an experiment with the transformer in open-circuit. I determined the voltages and currents on both primary and secondary winding.

I know the power factor measures the ratio between active power and apparent power (it was in fact this way that I computed it).

However, what does it exactly mean? Specially in open-circuit? Is it the percentage of power that doesn't get "lost"?

I know that if I had a load connected it would be the percentage of power delivered to the load. But this way I'm not sure.

Then I computed the parameters of the transformer G_m and B_m. I converted them to per-unit values. Does this have any physical significance? Because B_m is more close to 1 than G_m. Does just mean that the inductor parameter is more relevant than the resistor one?

Second I did an experiment with the transformer in short-circuit. I determined the voltages and currents on both primary and secondary winding.

I have the same question about the power factor, what does it mean here? Power delivered to a short-circuit?

Then I computed the parameters of the transformer R_t and X_t. Again, converted them to per-unit values. I have the exactly same questions as before.

Third, I experimented with a load, same thing as before.

Now the power factor must mean the percentage of power delivered to the load, right?

Thanks in advance!

EDIT: Transformer model (valid in pu values).

enter image description here

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  • \$\begingroup\$ Transformer...... then you talk about transistor. \$\endgroup\$ – Andy aka Mar 21 at 8:04
  • \$\begingroup\$ My mistake! I was tired lol. Thank you for the correction \$\endgroup\$ – Granger Obliviate Mar 21 at 10:09
  • \$\begingroup\$ What is "G_m" and "B_m"? What is "R_t" and "X_t". You should show the equivalent circuit of a transformer and refer to that circuit to avoid confusion: stades.co.uk/index_files/transformer%20equivalent%20circuit.PNG \$\endgroup\$ – Andy aka Mar 21 at 10:51
  • \$\begingroup\$ Added the model! \$\endgroup\$ – Granger Obliviate Mar 21 at 11:01
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enter image description here

This shows the Mutual Inductance which typically draws a VAR power 10% of rated VA so the input current is low P.F.

The short circuit current Isc is then the Rated VA/Zpu so a Zpu=10% of Rated V/I max at full load is 10% of Isc or

Isc = ( V * A )/Zpu

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