very interesting question.. 1

[bulancak]

we all know that the flux of a transformer is constant in normal operation because our main voltage is constant. So lets think our transfomer no-load the only current on the circuit is excitation current, and if we draw any current from secondary that is from the primary why doesnt our primary winding produce a bigger voltage according to v=Ldi/dt??? also how can the primary inductance can be constant all he time under a current remembering L= d fi/dt
if fi = flux is constant unless we change the voltage and the current is different than excitation current so we shall depict that L increases...so?

 

[bulancak]

excuse me the second formule is L= Nfi /i and and as the i increases and we assume that fi= constant so L decreases...

 

[bulancak]

if flux is proportional to the current why our no-load flux doesnt increase as we load the transformer? THESE ARE TR?CKY OR AM ? ?GNORANT???

 

[2dye4]

Flux in the core is proportional to the NET current surrounding the core. Current in the load side is offset by current in the source side.

The circuit must be analyzed in terms of MUTUAL inductance.

 

[bulancak]

primary current flows from primary winding directly and in rough logic i need to accept that primary winding is a coil and according to V=Ldi/dt formule V primary should be greater than main voltage but it is impossible because they are almost the same and must be so all the time, so if physics law is true so how an it be??

 

[electricpete]

http://pemclab.cn.nctu.edu.tw/w3elemac/W3slide/ch2.xformer/sld030.htm

Hit the right arrow to step thru the derivation of equivalent circuit for a power transformer including leakage reactance and exciting reactance.

On the simplest level to answer your question you need only an ideal transformer
v1 = N1 * d/dt(Phi)
v2 = N2 * d/dt (Phi)
Combining above two we have v1/N1 = v2/N2

Neglecting leakage reactance effects (assumed 0) an dmagnetizing reactance affects (assumed inifinite magnetizing reactance), there is no energy stored in the transformer and then the apparent power in equals apparent power out:
v1 * i1 = v2 * i2
i2/i1 = v1/v2 = N1/N2
Current is drawn on the primary to match the amp turns of current drawn on the secondary.

The mental picture of this idealized transformer is that it always has exactly the right amount of flux to balance the applied voltage. If you draw current on the secondary which would tend to disrupt this flux pattern, an equal/opposite current is drawn on the primary to restore the flux pattern.

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[ScottyUK]

Are you familiar with the model of the transformer shown on transformer model. This model is the normal basis for analysing transformer behaviour.

The Wikipedia entry is actually quite good.

http://en.wikipedia.org/wiki/Transformer

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If we learn from our mistakes I'm getting a great education!

 

[7anoter4]

I agree with electricpete explanations and taking into consideration his recommendation of "mental picture of this idealized transformer" I shall endeavor to create this "mental picture".
The magnetic flux which produces the EMF [both primary and secondary] is
produced by wp*Io [MMF] according relation : wp*Ip+ws*Is=wp*Io
wp is the primary winding no.of turns and ws is the secondary winding no. of turns.
If Is=0 then Ip=Io.
The leakage flux is produced by wp*Ip in primary winding and ws*Is in secondary. The leakage flux determines an EMF lagging Is by 90 degrees, the same as the leakage flux produced by ws*Is does. It is a behavior of a reactance creating a voltage drop- beside the primary circuit resistance [DeltaVp]-from supply voltage Vp up to primary EMF [Ep].
The magnetic flux which produces the EMF [both primary and secondary] it is the useful flux-transferring the power from primary to secondary winding and, since it flows in the laminate core, it is sensible to the core saturation. That means Xm will change with Ep change- due to change in supply voltage or voltage drop-due to Ip change.
It is possible to change the scheme from fig. 1 to fig.2 if we put E's=Ep
E's = the secondary EMF related to primary.
It is possible also to add- instead Ep- Rfe and Xm and we can split the no-load current Io in two others: Ife [produces the core losses] and Im [produces Ep through Xm].
From the iron losses [FeLosses], one may extract Rfe and further Ife, Im and Xm.
Rfe=Ep^2/FeLosses Ife=FeLosses/Ep Im=Io-Ife and Xm=Ep/Im