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.
Best Answer
That is entirely correct and that current is called the primary magnetization current.
Correct, for an ideal transformer this is entirely true.
Yes, but, if the unloaded secondary was left in place you’d still get the same primary magnetization inductance.