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Unfortunately in mech. eng. we didn't really studied transformers that deeply.

However I was struck with a question after talking with an electrician friend that was repairing a failed transformer*:

Considering an ideal transformer (without losses) and an open circuit on the secondary, could the current on the primary be calculated as: \$I = V/(2\pi fL)\$?

I am under the (possibly wrong) impression that since there is no load on the secondary, the power dissipated on the primary should be 0, and as such no current would flow through it.

A different but related question: say I removed the secondary from some transformer and was left with just a piece of copper wire wound about an iron core, would the current flowing through this inductor be calculated the same way as if there was a secondary wound on the core?

\$*\$ The failed transformer had no load on the secondary but the primary was draining around 7 amps at 230V! I'm guessing the isolation of the secondary got toast and it short-circuited.

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  • \$\begingroup\$ No power dissipation is not the same thing as no current flowing (under your ideal transformer conditions). \$\endgroup\$
    – John D
    Mar 2, 2020 at 19:48
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    \$\begingroup\$ * I have a fully working 240v transformer that pulls 7A on the primary with no load on the secondary. It's a microwave oven transformer, and they are designed to run hard into saturation to minimise their use of materials, aka designed down to a cost. But for any transformer other than a MOT, 7A would be a bad sign. \$\endgroup\$
    – Neil_UK
    Mar 2, 2020 at 20:21
  • \$\begingroup\$ It was actually a transformer for powering the trajectory correction sensor PCB of a GMAW robot arm. Luckily the 1amp fuse did it's job! \$\endgroup\$ Mar 2, 2020 at 20:35

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Considering an ideal transformer (without losses) and an open circuit on the secondary, could the current on the primary be calculated as: 𝐼=𝑉/(2𝜋𝑓𝐿)

That is entirely correct and that current is called the primary magnetization current.

I am under the (possibly wrong) impression that since there is no load on the secondary, the power dissipated on the primary should be 0, and as such no current would flow through it.

Correct, for an ideal transformer this is entirely true.

say I removed the secondary from some transformer and was left with just a piece of copper wire wound about an iron core, would the current flowing through this inductor be calculated the same way as if there was a secondary wound on the core?

Yes, but, if the unloaded secondary was left in place you’d still get the same primary magnetization inductance.

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  • \$\begingroup\$ Thanks for the answers! However there is something I don't understand: if there is a primary magnetization current, how can the power dissipated on the primary be 0 and no current flows through it? Or is it that there is current on the primary (the so called primary magnetization current) but power is considered 0, because half the time the circuit draws power from the grid and the other half it returns power back to the grid? \$\endgroup\$ Mar 2, 2020 at 20:25
  • \$\begingroup\$ The current into the primary inductance is not in phase with the applied primary voltage. Think about that a little because, when current lags by 90 degrees, the instantaneous multiplication of v and i waveforms produces a power waveform that has an average value of zero. Hence, power (really energy) is taken and given back equally (as per your question). \$\endgroup\$
    – Andy aka
    Mar 2, 2020 at 20:29
  • \$\begingroup\$ Thanks! That finally clears it. Now, does it exist any power source or element connected between the power source and the transformer that could reject this power returned back from the transformer? I.e.: some type of element that only allowed power to flow from the source to the transformer? If so, the power returned from the transformer had to go somewhere right? Probably heat losses. Does this means that this hypothetical circuit cannot be modeled with pure ideal components? \$\endgroup\$ Mar 2, 2020 at 20:44
  • \$\begingroup\$ Yes, the energy (preferred way of calling an accumulation of power), has to either remain stored (as per the dc voltage on a capacitor) or cause a spark as in the energy stored in an inductor. Because we have capacitors, diodes, inductors and electronic switches we can harness the stored energy and reuse it cyclically to convert voltage A into voltage B hence, we have what is called a power converter (sometimes called a switching converter or switch mode power supply). \$\endgroup\$
    – Andy aka
    Mar 2, 2020 at 21:20

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