Power factor of ideal transformer

[Mr Genius]

Can anyone explain why the power factor of an ideal transformer cos (phi) is equal to 1??
A simple ideal transformer is formed of 2 ideal coils. The power factor of an ideal coil is 0 since phase difference between the current and the applied voltage is - π/2. Then how can the power factor of the transformer equal 1¿

 

[berkeman]

Can anyone explain why the power factor of an ideal transformer cos (phi) is equal to 1??
A simple ideal transformer is formed of 2 ideal coils. The power factor of an ideal coil is 0 since phase difference between the current and the applied voltage is - π/2. Then how can the power factor of the transformer equal 1¿

The Power Factor will depend on the load placed at the output of the transformer. Can you post a copy of what you are reading?

 

[Mr Genius]

 

[berkeman]

Okay, it looks like they are talking about the transformer as part of a system. If you assume a purely resistive load at its output, then any deviation from a unity Power Factor would come from the transformer losses and parasitics. Does that make sense? Certainly if you connect a load to the transformer output that has a PF<1, the PF of the system will be <1 as well.

Here is a rotated and cleaned up copy of your attachment:

 

[anorlunda]

A simple ideal transformer is formed of 2 ideal coils. T

That statement is adding to your confusion. A real transformer has coils and depends on magnetiic forces. An ideal transformer is an imaginary device that defines relationships between voltages and cuteness. The imaginary device makes no reference to magnetiic or coils. I could implement something close to an ideal transformer using only microprocessors and software, no coils needed.An ideal resistor had zero L and C, whereas real life resistors have both.

Stop trying to imagine real life properties to any ideal component. That is not helpful.

 

[Baluncore]

The primary of the transformer has a magnetising current due to the “leakage inductance” of the primary that creates flux in the core. That reactive magnetising current is proportional to primary voltage, so it is quite independent of secondary current or load. Without load, the power factor of a real transformer is zero. As more real power is transformed, the magnetising current becomes less significant and the power factor approaches closer to 1 but never gets there.

An imaginary ideal transformer need not have any magnetising current, so the power factor can always be thought of as 1.

 

[jim hardy]

Aha !
@Mr Genius
expanding on Baluncore's accurate statement above which is the heart of the matter,,,

A simple ideal transformer is formed of 2 ideal coils.
Think a little harder. It's an ideal inductor with another winding. An ideal inductor has specified inductance edit: so it draws magnetizing current , as you've already said.
An ideal transformer edit: differs from an ideal inductor in that it has infinite inductance so there is no magnetizing current.
With no magnetizing current and no load current either there's nothing to measure the phase angle of, so power factor becomes moot.
Only when load current commences does an angle appear to take the cosine of.
So power factor of an ideal transformer is exactly that of its load. The power factor of an ideal coil is 0 since phase difference between the current and the applied voltage is - π/2.
Fine for a finite inductor ,
but not for one with infinite inductance because there's no current to measure the phase of.
( i know, I know, a preposition is a bad word to end a sentence with)

Then how can the power factor of the transformer equal 1¿
Ideal transformer just hands over the power factor of its load.

No load, no power factor.

 

[NascentOxygen]

Can anyone explain why the power factor of an ideal transformer cos (phi) is equal to 1??

We usually think of a transformer as a device which magnifies voltage or current, though (obviously) not both together. But if we think in terms of power, then a transformer is a device where P_in ≈ P_out.

If the device introduces some resistive losses, then P_in = P_out + P_loss∠0°

If the device itself introduces a reactive component, then P_in = P_out + ∆ ∠-90°

Only where both of these introduced terms are zero can the equation P_in = P_out hold.

i.e., only if it's ideal will we have P_in = P_out × 1.000000∠0°

If the proportionality factor (in blue) is not of unity magnitude, or the cosine of its angle is not unity, then it cannot be describing an ideal device.