The rating of a real transformer is 15 kVA , 2300/230-V then can we say that the primary rms is 2300 V or can we say that the secondary rms is 230 V or both.

To clarify further, suppose to solve for certain characteristics of this transformer like primary current and secondary current i use the equivalent circuit equation of the transformer referred to the primary side as:

Vp = aVs + Req*Ip + jXeq*Ip

Then what should i use for Vp or Vs in the above equation looking at the rating of the transformer.

In the above equation looking at the rating data given above is Vp= 2300 or Vs = 230 ?As mentioned earlier the transformer is mentioned to be real.

  • \$\begingroup\$ I think you should add a circuit diagram to accompany your formula. \$\endgroup\$
    – jippie
    Aug 17, 2014 at 8:34
  • \$\begingroup\$ I think you should define your symbols. | Why would you think that the 2300 or 230 volt values were "unreal"? \$\endgroup\$
    – Russell McMahon
    Aug 17, 2014 at 13:50

1 Answer 1


From these specs you know:

  • Maximum apparent power is 15kVA. This means that the product of Voltage and Current must not exceed 15kVA.
  • The transfer ratio is 10:1, largely independent of input voltage but may vary some with load.
  • The high voltage side is rated for a maximum of 2300V(RMS)
  • The low voltage side is rated for a maximum of 230V(RMS)

And when it comes to RMS, the transformer is probably spec'd for sine wave voltages. This means the peak voltage on high side can be no larger than 1.41 × 2300 = 3250V and 325V for the low voltage side.


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