Electrical – Modeling A Current Transformer in FEMM

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

When modeling an ideal transformer, why do we not consider the inductance of the primary and secondary winding of the transformer.

For an ideal transformer, the primary and secondary inductances are arbitrarily large ('infinite'). This must be so since, for an ideal transformer, there is no frequency dependence.

To see this, consider the equations (in the phasor domain) for ideally coupled ideal inductors:

$$V_1 = j\omega L_1I_1 - j\omega M I_2$$

$$V_2 = j \omega M I_1 - j \omega L_2 I_2$$

where

$$M = \sqrt{L_1L_2}$$

Solving for \$V_2\$ yields

$$V_2 = \left(\sqrt{\frac{L_2}{L_1}}\right)V_1 = \frac{N_2}{N_1}V_1$$

Now, assume the primary is driven by a voltage source and that there is an impedance \$Z_2\$ connected to the secondary such that

$$V_2 = I_2 Z_2$$

It follows that

$$I_2 = \frac{j \omega M}{Z_2 + j \omega L_2}I_1 = \left(\sqrt{\frac{L_1}{L_2}}\cdot\frac{1}{1 + \frac{Z_2}{j\omega L_2}}\right)I_1 = \left(\frac{N_1}{N_2}\cdot\frac{1}{1 + \frac{Z_2}{j\omega L_2}}\right)I_1$$

This is certainly not the behaviour of an ideal transformer where we expect

$$I_2 = \frac{N_1}{N_2}I_1$$

But notice that in the case that \$j\omega L_2 \gg Z_2\$ we have

$$I_2 \approx \frac{N_1}{N_2}I_1$$

which is exact in the limit that \$\frac{Z_2}{j \omega L_2} \rightarrow 0\$

Thus, we recover the ideal transformer equations from the ideally coupled ideal inductors in the limit that \$L_1, L_2\$ go to infinity (keeping their ratio constant).

In summary, we don't consider the inductances for the ideal transformer since, as shown above, the ideal transformer equations hold only in the limit of arbitrarily large primary and secondary inductances.