I have a list of basic problems. Let's suppose a textbook problem. A transformer has an efficiency of 93%. We know the primary voltage and current. We know the numbers of turns and the efficiency. The problems asks for the voltage and current output of the transformer.

Of course we can immediately write: $$ P_{sec} = P_{pri} \times efficiency $$

But, does that decrease in power is seen by measuring a lower-than-ideal voltage, lower-than-ideal current or both? If the dependance is complicated, I can't get why is this a basic problem, or what assumption do I have to make.

Edit: more information. This is a step-down transformer. The primary is working at a 12500 V and 13.4 A, so the power input is about 170kW. The turn ratio n1/n2 is 93.3. We intend to provide for a circuit that is aprox. 125 V 1250A, according to the solution. The solution states with no explanation that N1/N2=I2/I1, even in a non-ideal transformer. So the voltage is lower than ideal.

  • \$\begingroup\$ We know the primary voltage and intensity. What is "intensity" in this context? \$\endgroup\$ Commented Mar 3 at 20:05
  • \$\begingroup\$ Sorry, error in translation. We know the voltage and current. \$\endgroup\$ Commented Mar 3 at 20:20
  • \$\begingroup\$ The dependence is more complex if you want a realistic answer. Have you seen the equivalent circuit of a transformer? \$\endgroup\$
    – Andy aka
    Commented Mar 3 at 20:28
  • \$\begingroup\$ I have seen models, and have not found a simple answer to a text book problem. In the solution of the problem, they suggest that the the relation Nprim/Nsec=Isec/Iprim with no explanation, so the voltage is lower than the ideal case, but not the current. \$\endgroup\$ Commented Mar 3 at 20:49
  • \$\begingroup\$ @MartínMontotoFernández please fix the title of your post, it still says intensity \$\endgroup\$
    – jsotola
    Commented Mar 3 at 21:27

1 Answer 1


You can use the following equivalent circuit to get some insights: enter image description here There are many components there. I would highlight the following:

  • \$R_1\$ and \$R_2\$ are the resistance of the primary and secondary windings, respectively. Those components will cause the losses [\$W\$] and, of course, voltage drop. As you are mentioning efficiency, the losses generated in these resistances will be used to compute the efficiency. Higher the efficiency, lesser is the voltage drop at those resistances.
  • \$X_L1\$ and \$_XL2\$ refer to the reactance of the transformer (split in two just for convenience). At those inductors there is no loss, but there will be voltage drop.

both resistance and reactance together form the impedance of the transformer. Typically, large power transformer has \$X_L>>R\$, distribution transformers has \$X_L\sim R\$, and small control transformers has \$X_L<R\$.

Now regarding the currents, as typically \$i_0<<i_1\$, we can assume that the currents are not 'lost' from one side of the transformers to another (it's good to comply with Ampere's Law).

Finally, answering to your question, the power will be dissipated in form of Watts [\$W\$] in the resistances and generate voltage drop, so that \$U_2<U_1 N_2/N_1\$ (which is what you mention as 'lower-than-ideal voltage'). But worth to note that not all voltage drop in the transformer will be due to those losses.

  • \$\begingroup\$ I edited the question. It is indeed a ~170kW transformer. Thanks for the through explanation. \$\endgroup\$ Commented Mar 4 at 15:45

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