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

With the secondary open circuit, only the primary winding has effects on results and therefore the current that flows is the magnetizing current. The magnetizing current produces losses and these are called iron losses and they are partly due to eddy currents circulating in the the iron laminates that make the core. Iron isn't the best conductor of electricity and so it's resistance dissipates power due to these eddy currents.

Hysteresis loss is also measured by this test and, strictly speaking the total "loss" with the secondary open circuit is eddy current loss plus hysteresis loss (Remanence effects in the iron material leading to energy loss every cycle of ac).

Shorting the secondary tests copper losses and this is usually done at a much lower applied voltage to the primary. Because of this, iron losses are negligible and can be ignored. In this test you are measuring the I^2*R losses of primary and secondary together. If you know the turns ratio you can make reasonable assumptions about the distribution of loss in primary and secondary.

Below is the equivalent circuit of a transformer with the secondary (copper losses) referred back into the primary. This referral is done by multiplying them with turns ratio squared: -

Note component Xm - this is the reactance of primary inductance of the transformer. Xp and Xs are leakage reactances (turns of copper that don't couple magnetically).

Measurements

Watt-meter measurements will indicate the losses on both tests.

Ammeter tests will tell you the current in the primary and/or the secondary - this can also help you gain knowledge about the turns ratio. Ammeter tests also tell you that you are at the limit of current for testing with the secondary shorted i.e. at the manufacturer's recommended limit. Don't go higher is the rule.

Voltmeter tests only apply to primary testing when the secondary is shorted and apply to both primary and secondary when the secondary is open circuit.

If you don't have a watt-meter, voltmeter and ammeter tests plus a little bit of common sense and knowledge can help you calculate powers that would be indicated by the watt-meter.

This answer applies to regular power transformers - it isn't intended to cover high frequency transformers although the principles are the same BUT dielectric loss and resonance effects can play a big part.

Why do these equations contradict each other?

First of all, you're using ideal transformer equations and that's fine as long as you apply them properly.

But you haven't applied them properly in this case.

Assuming an ideal transformer, if the secondary is loaded with a short-circuit, the voltage across the secondary is zero volts and thus, the primary voltage must be zero.

The fact is that the equations you provide must be satisfied simultaneously.

So, assuming the secondary is loaded with impedance \$Z\$, the transformer equation becomes

$$V_s = I_s \cdot Z = kV_p $$

but

$$I_s = \frac{I_p}{k} $$

thus

$$V_p = \frac{I_p}{k^2}\cdot Z$$

And there you have it, \$V_p\$ and \$I_p\$ are not independent of the secondary load \$Z\$.

In fact, you see that when \$Z = 0 \mathrm{\Omega}\$, the primary voltage must be zero for any finite primary current.

So, there's no contradiction.

I want to know one equation that describes all parametrs of transformer where is possible to see how primary and secondary voltages and currents change depending on impedences in transformer.

Then start with this model of a physical transformer

and solve for the primary and secondary voltages and currents.

If I recall correctly,

• R1 models the primary winding resistance
• X1 models the primary leakage inductance
• Rm models core loss due to hysteresis
• Xm models the finite permeability of the core
• X2 models the secondary leakage inductance
• R2 models the secondary winding resistance