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Transformer primary current in GreatScott videp

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I14R10

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Transformer primary current in GreatScott video

I was watching this video from GreatScott https://youtu.be/2cxcP5lY7K4?t=283
And I was wondering about something that he said in the video at 4:43 (the link will take you to that time). He said something like this.
Impedance of the primary is 1605 ohms, that gives the current I=143mA. In reality we get about 40mA current. "The reason is that magnetic flux also induces a voltage into a primary coil which opposes the mains voltage and lowers primary voltage and current".

I have trouble understanding that. Does he actually mean that the coil has self inductance that generates voltage (which opposes the current that created it) when the current through the coil changes? But I don't see how that can explain how the transformer in his video draws 41mA instead of 143mA. Inductor in transformer primary should only shift the voltage out of phase by 90 degrees. If the secondary is open than the transformer should draw current equal to V/Z_primary plus some current for core magnetization. When the secondary is connected the impedance is transformed according to ratio of the turns of secondary and primary. At least that's what I understand.

I don't really see how magnetic flux from primary can induce current in primary itself. Maybe somebody can help me understand this.

Can you actually explain what he really meant to say in that video?
 
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His explanation is bollocks - more likely he did the calc wrong or measured Lpri wrong, or his gear cannot measure 143mA very well (most likely ) or the mains is very flat topped or a combination of these things. At low current you can ignore the R of the pri coil, then Zpri = XLpri = 2.pi.F.L
 
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