But when I calculate X=2πfL it should have the same current or voltage.The important one for current calculation is L=1H.
The primary will have an impedance of X=2πfL, through which current will flow - the Magnetising current.
This will be in addition to the current that flows because of the load on the secondary.
Ok but why in the simulation where I have ideal parts the currents aren't equalA real transformer is an ideal transformer, with an inductor in parallel with the primary winding.
If it were an ideal transformer (the higher the inductance, the closer to ideal it gets) then the two currents would be equal, but for a real transformer the primary current will be higher by V/(2πfL).
(Even that is a simplification, because there is resistance, core losses etc.)
OooYou don’t.
An ideal transformer has infinite magnetising inductance, and yours has 1 Henry.
Transformers can be non-ideal in many different ways!Ooo
I thought that ideal transformer is when the coupling factor is equal 1
So may I ask you how to I calculate with 1H and with 100H ?
Because as I see for 1H the equation U1/U2 doesn't work, as well I2/I1 also doesn't work.
But for 100H U1/U2 and I2/I1 works.
I'm sorry to be very annoying but I've read that ideal transformer is when we don't have any lost energy (for example in heat. etc.) That the whole energy is changed into electricity.Transformers can be non-ideal in many different ways!
The coupling factor controls the leakage inductance.
The leakage inductance is the fraction of the primary inductance that doesn't couple with the secondary inductance.
The magnetising inductance depends on the type of transformer - toroids have much higher magnetising inductance than E-I laminated transformers, and smaller transformers have more magnetising inductance than larger ones.
From my notes, a 3VA pcb mount transformer has a magnetising inductance of 66H. A 1.6VA toroid is 1500H. A 7kVA dual bobbin transformer was 1.02H
These are 230V primaries - 115V primaries will have a quarter of the inductance.
Permeability is a characteristic of the material that the core is made out of, and its value is called μ
Permeance is a characteristic of the core of a particular transformer, and is equal to μA/l where l is the magnetic path length (the distance the flux covers to move around the core)
Inductance is equal to permeance multiplied by the square of the number of turns.
So infinite permeability would give infinite inductance, and infinite inductance would result in zero magnetising current, so the primary and secondary currents would be equal.
If you look back at @Danko 's calculations, you will see that when he used a large value of inductance the currents became closer to equal.
Just to complete the set, reluctance is the reciprocal of permeance.
Magnetomotive force is the product of current and number of turns
Flux is Magnetomotive force divided by reluctance.
Flux density is flux divided by cross sectional area.
If flux density gets too big, the core saturates and no longer works as a transformer. Yet another example of non-ideal behaviour!
This happens at a flux density of about 1.8 Tesla for iron cores and 0.3T for ferrite.
You could try just making the inductance ridiculously large (say 10000H). I presume the simulator would cope with it, even though practically it would be impossible to make.ahaaaaaaa, okey I think I get it now.
So the simulation gives me an option to change the value of the coil from 1H to 100H. The ideal tranformer if when primary and secondary coil are infinite.
Okej so that's how it works.
As I can understand when I don't have an ideal transformer (ideal here means that the coil value isn't high enough for example 100H etc. ) then I have to use different equations? What do I mean by that is I can't use U1/U2 = I2/I1 = N1/N2 in 1H coil example ?
You can not change inductance of primary coil,So may I ask you how to I calculate with 1H and with 100H ?
Power factor correction for a rather peculiar mains frequency!You can not change inductance of primary coil,
because in this clever problem, inductance 1 H and C1 capacitance 15 μF
are resonance circuit (tank), which oscillates on frequency 41.09363≈41.1 Hz.
On diagram below you can see currents in L1, C1, L2.
Note, all currents are in different phases.
So current in L2 exactly equals sum of currents L1 and C1.
View attachment 270363
I'm sorry I didn't understand the phrase with 1H coil. So the simulation I've shown where I2/I1 isn't equal 1 is because this transformer doesn't work like ideal transformer but with 100H it works very close like ideal. Am I right ?You could try just making the inductance ridiculously large (say 10000H). I presume the simulator would cope with it, even though practically it would be impossible to make.
The real transformer with 1H primary looks like an ideal transformer with a 1H inductor connected across the primary.
I know it is a resonance frequancy because I had a small problem understanding transformer.You can not change inductance of primary coil,
because in this clever problem, inductance 1 H and C1 capacitance 15 μF
are resonance circuit (tank), which oscillates on frequency 41.09363≈41.1 Hz.
On diagram below you can see currents in L1, C1, L2.
Note, all currents are in different phases.
So current in L2 exactly equals sum of currents L1 and C1.
View attachment 270363