Transformer

Vader ! Please send lots of tinfoil ! Emergency !

There are two reasons.

1- Copper loss which depends on transformer current.
2- Core loss depends on transformer voltage

and both these losses are independent of the power factor and that's why the transformer is always rated in kVA and transformer never makes any changes in power factor.

Are you serious?

have you ever measured power factor at the input of transformer at no load?

Please Read through again what OP has written.

OP: The transformer does affect the power factor. However the effect is so small that it can be neglected.

It is the magnetising current Vs active current ratio. The load current is dependent on the flux linkages across the primary and secondary and do not contribute to the powerfactor. It is the basic ecitation to the magnetic circuit and the unlinked/self linked fluxes that are the culprits and are kept to the minimum in any good transformer, so that the flux linkages are nearly total, and that automatically keeps the voltages and currents in phase (pri and secondary even when individual V & I are not)

what a joke " Transformer does affect but neglected"

Consider the equivalent circuit of transformer, which is available in any text book, it has two components restive and reactive.

From where the reactive component has come in?

What is the voltage regulation and percentage impedance of transformer? think about this and go back to your text books and the joke will be clear.

BTW: earlier also I have mentioned - what is the no load current of a practical transformer (ie with secondary OC in case...) is it purely resistive?

Devices like transformers and motors require power to maintain magnetic fields to perform their function. This so called "reactive" (kVAR) power flows into and out of the device but is not really consumed to perform work.

The equivalent circuit model for the non-ideal transformer is shown in Figure 1. An ideal transformer with resistors and inductors in parallel and series replaces the non-ideal transformer. This model is called the high side equivalent circuit model because all parameters have been moved to the primary side of the ideal transformer. The series resistance, R_eq, is the resistance of the copper winding. The series inductance, X_eq, accounts for the flux leakage

BUT: When the transformer becomes a real transformer with resistive losses, eddy current losses in the iron core, etc. these losses make the load seen by the line a little less then pure resistive. If the transformer is well designed and the losses are very low compared with the power being delivered to the load, then the overall load seen by the line would be essentially resistive and the power factor would be very near to ONE.
you place a reactive load on a transformer that is still delivering a large amount of power to the load compared with the internal losses, the line would see a load that is reactive just like the load.
In other words, in the real world where the transformer in the circuit is well suited to the load place on it the losses inside the transformer will be small compared to the power being delivered through to the load and the transformer is essentially or nearly invisible. Especially this would be the case if the turns ratio of the primary winding to the secondary winding is unity.
It is always better to brush-up your text books.

So, guys let me comeback to this discussion again.

It is engineering and it is not joking forum. "This is for Joke Guest"

I dont know your names to call you. Anyway,

I read your topic and explaination. They are really useful and now see the power factor formula for a transformer. I think you are missing something. See below!

You are right in (16)

Here the point is the OP is asking whether Transformer contributes to the power-factor.

ie is

Load pf (ie transformer output pf) = supply end pf (ie transformer's input pf) ?

Let us leave the joking guest to continue with his non-contribution and no point in raising to his bait.

The point as I raised is - look at the equivalent circuit of the transformer from primary side and then put the load on secondary,

the shunt reactance is going to add its effect. Of course in a rare situation when the pf of the transformer exactly equals the load pf (ie the R-L circuit of it matches with the load R-L circuit) the pf will of course be unaffected, else there will be a minuscule effect on the pf as seen from primary - which will be negligible near the full load condition, but it will be appreciable on low/no-load.

What a joke again " Brush-up my text books"? You still dont know the difference between efficiency and power factor

but you brused-up internet too still you have a problem.

Thanks and i also accept that but can you say why inductive loads consumes reactive power and generator why produces ractive power

"why inductive loads consumes reactive power .."

This is more serious than I thought

I dont want to argue or discuss farther regarding this. It is just my comment for you. So, think and accept the best comment for your question.

I just want to add some more information :

Real power is required to perform work while reactive power is necessary to assist or support the work being done.
A transformer does no work, it just transforms electrical energy.
In a transformer, the power factor of the load connected to the winding is the same power factor reflected back through the primary winding to the source and it does not consume or obserb any real or reactive power.

If you still not agree, I think here are thousand of engineers in this forum.

Let others to comment you better.

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