There are other things to consider. Here's the equivalent circuit of the transformer: -

The extra components that need to be considered are: -

- Xp and Xs - these are leakage inductances and no-matter how hard you try to make a perfect transformer there will be some flux in the primary that just won't couple to the wires in the secondary - this is portrayed as an inductor in series with the primary winding and an inductor in series with the secondary winding. These limit the amount of current that can be taken from the secondary and can be compensated for by raising the primary voltage BUT that runs the risk of causing greater saturation losses.
- Rc is the iron loss in the laminations - the core is conductive and without laminations, the core would act like a shorted turn so, laminations are used and this greatly reduces eddy currents circulating in the core but there is a power loss due to iron not being a perfect conductor and this will heat the core up and the greater you make the primary voltage (to accommodate Xp and Xs) the greater the heat generated.

I see both the above as limiting factors for power transfer.

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

## Best Answer

'Copper loss' is I

^{2}R power loss in the windings due to current flowing through them. As this current increases with higher loading, so copper loss also increases as loading increases.Eddy current loss is power loss in the magnetic core due to current induced into it (each lamination in the core is effectively a shorted turn, but the silicon steel has relatively high resistance which keeps the current down). The induced voltage - and thus current and power loss - doesn't change with loading because the magnetic flux in the core doesn't change. This is also I

^{2}R loss, but it is not copper loss.To summarize, there are

twoplaces in the transformer where I^{2}R losses occur - in the magnetic core and in the windings. However, only the windings have a power loss proportional to load current. That loss can rightly be called 'copper loss' because only the windings are made of copper.