kVA ratings correspond to full load conditions.

By full load is meant the load (obviously, on secondary) which would make transformer transfer it's rated power from primary to secondary. For example when 50 kVA is on full load, it would be transferring 50 kVA from primary to secondary. If primary voltage on full load is 25000 V (as in your case) then primary current should be (50*10^3 / 25000) ampere.

Now coming to your primary question,

if there isn't such currents

means that there isn't such load on secondary that can fully load transformer.

If transformer is under-loaded, primary (and hence secondary) currents would be less than full load values. For e.g. if transformer is on half-load (means load connected at secondary is half the full load), then it would drive exactly half current than full load current. Say your full load current is 'I' ampere then you half load current would be 'I/2' ampere. If transformation ration is 'k', then half-load secondary current would be 'I/2k' ampere.

If transformer is over-loaded, then it may damage winding and/or insulation. However, if we neglect such effect (and you shouldn't neglect it in practice) then, transformer which is on double-load will be drawing twice the full load current in both, primary and secondary winding.

Imagine you wound an inductor but instead of one thicker wire you used two thinner wires (this is commonplace by the way). An AC voltage is applied on both wires - lets call the AC voltage AC live and AC neutral and lets say neutral is connected to 0V for reference.

No problem so far?

Now, imagine you only applied AC live to one of those wires and left the other one open circuit (AC neutrals both at 0V still). What would be the induced voltage in the unconnected wire? Would it be: -

- In phase
- Out of phase by 180degrees
- Same amplitude
- Different amplitude

If the induced voltage on the unconnected wire were antiphase then joining it up to the AC live would create a seriously big problem and there would be sparks and breakers tripping. If the induced voltage on the unconnected wire were significantly different to the AC live and you joined them up then same problem as before - sparks and breaker trips.

This means if you wind a transformer with two wires in the same way there will be no phase shift but you can always swap the primary wires or secondary wires over and get 180 degrees.

Inherently there is no 180 degrees phase shift - you only get 180 degrees when you get into a muddle with the wires.

## Best Answer

Your suspicions as to correctness of the statement \$I_1 = 0A\$ are justified.

It is obvious that once you open the secondary loop at points \$c\$ and \$d\$, there will be no current through this loop, and for all practical purposes (unless it is some special kind of a system) the presence of the secondary winding of the transformer may be completely ignored. In this case what you are left with is a simple RL circuit - you can easily calculate the current there.

Well, the conclusion then is that this text is incorrect, right? Not at all. The writer just assumed that the reader will be very comfortable with the concept of Thevenin (or Norton) equivalent. You're not that comfortable with it, therefore I'll try to provide a basic insight.

The crucial point to understand about Thevenin (Norton) equivalent is the following:

this equivalent is not an alternative description of the circuit. This means that you can't (in general) replace the circuit's schematic with its Thevenin equivalent while analyzing the circuit. Think of it: you're replacing a (potentially) complex circuit with a simple circuit of a single voltage source in series with a single impedance. There is some information about the circuit which is lost in this conversion, and you can't restore it (unless you saved the original schematics).Why then Thevenin is used at all? Well, imagine for a second that you're asked to develop some electrical component which should get its energy from a DC power supply. The internal implementation of the power supply isn't your concern at all, right? The only parameters that you take into account when connecting to it is its voltage and the internal resistance. But how come such a complex device as power supply may be adequately represented by just two parameters? Well, this is the strength of Thevenin theorem (and I hope you have already realized that the representation of the power supply as a voltage source in series with internal resistance is just a Thevenin equivalent).

When you use Thevenin theorem, you implicitly make the following assumption:

I have a single place in the circuit which is of interest to me. When this is the case, you don't mind to loose information about all other nodes in your circuits, as long as the description of the "node of interest" is correct. It means that once you replaced your circuit with its Thevenin equivalent, you can no longer analyze the internals of the circuit - the only place where the behavior of Thevenin equivalent is guaranteed to match the behavior of the original circuit is the "node of interest".In the example you provided the Tevenin theorem had been used to analyze a transformer. The currents calculated for Thevenin equivalent's internals are not the real currents that will flow in your circuit. However, the currents calculated for the "node of interest" (which is the load connected between points \$c\$ and \$d\$ in this example) will be completely correct (who guarantees this?).

EDIT:It may still be unclear why the author claims that \$I_1=0\$ and all the voltage is on primary winding. Think about it: the winding is an inductor and there is a voltage drop on it only when there is changing current through it. It means that the above two statement of the author are contradictory. Well, in this case I'd say that the author provided very poor explanation and made a life of the reader very hard.

I'll try to give alternative description:

When we disconnect the load there will be no more current in the secondary (\$I_2=0\$). It means that the following boundary condition must be satisfied by the Thevenin equivalent:

`when there is no load, the current through the equivalent is zero`

(this is trivial boundary condition satisfied by each Thevenin equivalent). Furthermore, we know that the maximal voltage on secondary is \$\frac{V_s}{a}\$ (see equations of an ideal transformer) and it is obtained when there is no load on secondary. It means that this boundary condition must also be satisfied:`when there is no load, the open circuit voltage is V_s/a`

. Combining the above two boundary conditions we conclude that the Thevenin voltage source is of magnitude \$\frac{V_s}{a}\$ and its polarity is inverted due to initial transformer's orientation.When we short the points \$c\$ and \$d\$, the current trough secondary will be \$I_s=aI_p\$ (see the equations of ideal transformer). This means that Thevenin equivalent must satisfy this boundary condition:

`when the output of Thevenin is shorted, the current equals to a*I_p`

. Given that we derived the Thevenin voltage source to be of magnitude \$\frac{V_s}{a}\$, the above boundary condition on the short-circuit current can be satisfied only if Thevenin's equivalent impedance equals \$\frac{Z_1}{a^2}\$.In summary: the same results as in your book, but no contradictory statements.

Hope this helps.