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Transformer %Z

Transformer %Z

Transformer %Z

3% impedance is the leakage impedance and magnetising impedance is separate.

RE: Transformer %Z

During a fault the magnetising impedance is, to a first approximation, negligible.
  

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

RE: Transformer %Z

(OP)
Thanks for your input, would I be able to get a very top level explanation as to why the magnetizing impedance is negligible during a fault?  From a first glance, I would say that if a fault occurs on just one side of the transformer, the other side is not shorted, and thus the magnetizing current still couples with the other winding (?).

Thanks,

Daniel

RE: Transformer %Z

The magnetizing branch has a very high impedance compared to loads. Compared to faults the impedance of the magnetizing branch is so high that it's inclusion or exclusion results it differences in the third or fourth digit at most.  Since the calculation is good to more than 2 or 3 digits anyway, there's no need to be concerned with the magnetizing branch.  

RE: Transformer %Z

As davidbeach said the effective magnetic field reactance can be neglected when one states the transformer impedance.
If V1 is the primary voltage , Zp=Rp+j*Xp primary winding impedance [j=i=sqrt(-1)] , Xp it is the reactance of magnetic field losses of primary winding and Rp is the primary winding resistance, E1 is mmf [ magnetomotive force ] of primary winding, I1 is the primary current [ all parameter as complex number] then:
V1=I1*Zp+E1
The same remarks for secondary winding:
E2=I2*Zs+V2
E1/E2=w1/w1=kw where w1,w2 is number of turns of primary respective secondary windings.
If we'll multiply the second equation by kw we'll get:
E1=I'2*Z's+V'2 and substituting E1 in the first equation we'll get:
V1=I1*Zs+I'2*Z's+V'2
Then we can draw the following scheme:

From this we can state:
I1=Io+I'2
Io=E1/j*Xm +E1/Rfe [ Io=I1 when I'2=0]. Where Xm=effective magnetic field reactance and Rfe = magnetic core
active losses equivalent resistance. Io=E1 (Xm -j*Rfe)/(Rfe*Xm)=E1*Const
Usually Io=2-3%*Irated if V1=primary rated voltage. As |Zp*I1rated| is less than 5%of V1rated  E1>0.95*V1rated.
Vsc=short-circuit voltage for I1=I1rated  [V'2=0]
in order to find Vsc the secondary is short-circuited and the I1=Irated.
Vsc=Zp*I1rated+E1sc  Zp*I1rated is still approx.5% then E1sc=0-5%*V1rated.
As Io=E1*Const for full V1 is only 3% for E1sc will be [directly proportional] 3%*5%=1.5/1000*I1rated.That means may
be neglected. I1=Io+I'2 if Io~0 then I1=I'2 and Vsc=(Zp+Z's)*I1rated so Zsc=Vsc/Irated=Zp+Z's.
 

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