Power factor of ideal transformer

In summary, an ideal transformer has zero resistance and inductance, but its power factor depends on the load it's connected to.
  • #1
Mr Genius
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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¿
 
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  • #2
Mr Genius said:
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?
 
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  • #3
1487956092576.jpeg
 
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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:
PF_PF.jpeg
 
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  • #5
Mr Genius said:
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.
 
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  • #6
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.
 
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  • #7
Aha !
@Mr Genius
expanding on Baluncore's accurate statement above which is the heart of the matter,,,
Mr Genius said:
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.
 
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  • #8
Mr Genius said:
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 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.
 
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Related to Power factor of ideal transformer

1. What is the power factor of an ideal transformer?

The power factor of an ideal transformer is defined as the ratio of the real power (measured in watts) to the apparent power (measured in volt-amperes) in an electrical circuit.

2. How is the power factor of an ideal transformer calculated?

The power factor of an ideal transformer can be calculated by dividing the cosine of the phase angle between the voltage and current by the ratio of the real power to the apparent power.

3. Does the power factor of an ideal transformer affect its efficiency?

Yes, the power factor of an ideal transformer can affect its efficiency. A lower power factor means that more reactive power is required to deliver the same amount of real power, resulting in lower efficiency.

4. Can the power factor of an ideal transformer be improved?

Yes, the power factor of an ideal transformer can be improved by adding power factor correction devices, such as capacitors, to the circuit. These devices help to reduce the amount of reactive power and improve the overall power factor.

5. How does the power factor of an ideal transformer impact the overall power system?

The power factor of an ideal transformer is an important factor in power system analysis as it affects the efficiency and capacity of the system. A low power factor can also result in increased losses and voltage drops in the system.

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