# Power factor of ideal transformer

1. Feb 24, 2017

### 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¿

2. Feb 24, 2017

### Staff: Mentor

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?

Last edited: Feb 24, 2017
3. Feb 24, 2017

### Mr Genius

4. Feb 24, 2017

### Staff: Mentor

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:

5. Feb 24, 2017

### Staff: Mentor

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.

6. Feb 24, 2017

### 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.

7. Feb 25, 2017

### jim hardy

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

Last edited: Feb 25, 2017
8. Feb 25, 2017

### Staff: Mentor

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 Pin Pout.

If the device introduces some resistive losses, then Pin = Pout + Ploss∠0°

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

Only where both of these introduced terms are zero can the equation Pin = Pout hold.

i.e., only if it's ideal will we have Pin = Pout × 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.