The no-load current of a transformer consists of two components:

  1. The Magnetization Current iM is the current required to produce the flux in the transformer core.
  2. The Core-loss Current ih+e is the current required to make up for hysteresis and eddy current losses.

This is what my book says for the no-load current in a real transformer. To me it sounded silly that the magnetization current, which is a part of the overall no-load current, is solely responsible for the entire flux generated in the core.

How is it that not the overall but a part of the no-load current is responsible for all the flux that is in the core?


1 Answer 1


This might help, the equivalent circuit of a low frequency power transformer: -

enter image description here

  • \$L_P\$ is the primary leakage inductance
  • \$R_P\$ is the primary copper loss
  • \$R_C\$ is the core losses due to eddy currents and hysteresis
  • \$L_M\$ is the magnetization inductance
  • \$L_S\$ is the secondary leakage inductance
  • \$R_S\$ is the secondary copper loss

As you should be able to see, the core loss components produce heat and not magnetization. Above image from here.

  • \$\begingroup\$ There is no flux due to core-loss components - they are represented by a resistor (Rc in the above diagram). Flux is produced by inductance i.e. is a current that is 90 degrees lagging the primary voltage. The "loss currents" are in phase with primary voltage because Rc is a resistor and that Rc represents those losses. On the other hand, if you argued that the eddy current losses are due to induction then that would be correct but, the primary sourced flux that induces those eddy currents is exactly opposed by the flux due to the circulating eddy currents; hence no net flux. \$\endgroup\$
    – Andy aka
    Commented Jun 28, 2019 at 11:51
  • \$\begingroup\$ In exactly the same way, secondary resistive load currents produce a flux that entirely cancels the flux caused by the extra current that flows in the primary due to the load. In other words, for a transformer, resistive load currents and eddy current losses behave the same - no net flux increase no matter how hard you try. You might have thought that as you drew load current from the secondary, flux increases? \$\endgroup\$
    – Andy aka
    Commented Jun 28, 2019 at 11:54
  • \$\begingroup\$ And my second comment immediately above yours - does that not do what you asked? \$\endgroup\$
    – Andy aka
    Commented Jun 29, 2019 at 0:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.