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Assume transformers are 10/0.4 kV 500kVA transformers. Primary is high voltage side and secondary is the low voltage side.

Normally we do Z_secondary(LV) = Z_primary(HV) * (Vrated2(LV)/Vrated1(HV))^2 to refer impedances from primary to secondary in case of a single transformer. My question is, if instead, two identical transformers were connected in parallel, will the referring be the same? Or will there be any changes in the above formula?

Thanks.

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Theoretically, with ideal transformers with infinite magnetization (primary) inductance, the formula for impedance transformation is: -

$$Z_{primary} = \left[\dfrac{N_p}{N_s}\right]^2\cdot Z_{secondary}$$

But because there is magnetization inductance (\$X_L\$) the above formula becomes: -

$$Z_{primary} = X_L||\left[\dfrac{N_p}{N_s}\right]^2\cdot Z_{secondary}$$

And with two identical parallel transformers, the formula becomes: -

$$Z_{primary} = \dfrac{X_L}{2}||\left[\dfrac{N_p}{N_s}\right]^2\cdot Z_{secondary}$$

Just switch this formula around to solve for the secondary impedance.

And, just in case there is any doubt about what \$X_L\$ is, I include the equivalent circuit for a transformer (to avoid ambiguity): -

enter image description here

If you need to consider leakage inductance then \$L_P\$ and \$L_S\$ are shown above in their circuit positions.

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  • \$\begingroup\$ indeed, I am interested on the modelling of transformers in power distrubiton systems. We(in classes) make some simplifications in our modellings. We do replace a transformer only with a reactance which we calculate with uk(%) * Vrated(l-to-l)^2/Srated(3-ph) and call is Xs generally. I assume we omit the X_L you mentioned. I think our X_s is leakage inductance only. So, if you also think that we do not include your XL in our transformer modelling based on my writings, then parallel connection of transformers will not have any affect the referring formula, right? \$\endgroup\$
    – muyustan
    Jun 4 '20 at 9:39
  • \$\begingroup\$ I have added the equivalent circuit of a transformer showing where the leakage inductances are - you should be able to see that they add (in complex math) to the load for LS. \$\endgroup\$
    – Andy aka
    Jun 4 '20 at 9:48
  • \$\begingroup\$ ok, guess we are not including Lm in our model, in such case, the formula for referring an external impedance connected to the primary to secondary will not be dependent on whether there are more than one identical transformers in parallel or only one of them. \$\endgroup\$
    – muyustan
    Jun 4 '20 at 10:06
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    \$\begingroup\$ That is correct, yes. \$\endgroup\$
    – Andy aka
    Jun 4 '20 at 11:34
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    \$\begingroup\$ yes it was clear already, I am totally ok that your answer had created no confusions regarding reactance impedance etc. I am just after to most open(clear, understandable) form of it can be. Anyway, this gone far much than needed, thanks again and have a nice day. \$\endgroup\$
    – muyustan
    Jun 4 '20 at 13:49

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