I'm trying to solve this question.

For a specified input voltage and frequency, if the equivalent radius of the core of a transformer is reduced by half, the factor by which the number of turns in the primary should change to maintain the same no load current I

If radius is halved, area becomes \$ 1/4\$ times. Hence to maintain the same no load current, there needs to be 4 times the number of turns we had earlier.

But the given answer is 2 times the number of turns. Why is that?


1 Answer 1


This answer assumes that "no load current" refers to the magnetization current in the primary of the transformer and that is defined by the magnetization inductance.

If the radius is halved then the inductance falls by 4 times because area is quartered. If the core is a regular shaped core with a decent value of permeability then you can assume that inductance is proportional to the square of turns hence, to restore the inductance, you need to double the turns to increase the inductance 4 times: -

enter image description here

Picture from here.

For a closed magnetic core \$\ell\$ becomes the mean length around the core.

  • 1
    \$\begingroup\$ Again, nice and direct. Really enjoy this aspect of your writing on magnetics. \$\endgroup\$
    – jonk
    Feb 5, 2021 at 11:11
  • \$\begingroup\$ That clarifies it, thank you. But why does a transformer with large core area has less no of turns than a transformer with smaller area for the same voltage rating? \$\endgroup\$
    Feb 5, 2021 at 11:23
  • 1
    \$\begingroup\$ @ASWINVENU it boils down to creating enough inductance so that the magnetization current (for a given voltage supply) isn't so high that the core saturates too much. \$\endgroup\$
    – Andy aka
    Feb 5, 2021 at 11:28

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