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the way I understood a basic inductive circuit (voltage source + inductor) is that all the circuit current (Io) lags the voltage by 90 degrees. However, I am following this blog tutorial to refresh myself (https://www.electronics-tutorials.ws/transformer/transformer-loading.html), where it can be seen in the V-flux triangle posted that only a small component of current (magnetising current or Im) is what actually lags the source voltage whereas the other component of the current (Ie) is actually in-phase with the source voltage (Vp).

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In addition, it is stated that eddy currents and hysteresis losses are provided by this in-phase current which I thought was due to magnetic field growing and collapsing in the transformer windings. I guess it also makes sense that these losses are real power losses but how do they show up in the primary side? As the winding impedance being lower and thus drawing more current (P=IV)? Would this be dissipated as (greater) real power across the winding or as (greater) imaginary power VAr being required by the inductor?

I would really appreciate your help as I am really confused with this. Thank you in advance.

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the way I understood a basic inductive circuit (voltage source + inductor) is that all the circuit current (Io) lags the voltage by 90 degrees.

That's correct, for a pure inductance.

If the circuit has both inductive and resistive parts, then indeed the current will be at a more general angle.

A transformer winding, which ideally is all inductance, will in the real case have resistive parts due to the resistance of the wire, and the eddy current losses in the core. The power absorbed by these losses is due to the component of current in phase with the voltage.

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  • \$\begingroup\$ Thank you very much @Neil_UK but how can resistance affect the angle of current? Moreover, for hysteresis losses, would they also show up as part of the real power loss across an inductor? \$\endgroup\$ Nov 30, 2023 at 15:48
  • \$\begingroup\$ @BaldovínCadenaMejía Any loss shows up as a real part of the current. You might prefer to think of resistance altering the angle of the voltage, with a given current the voltage across a resistance is in phase, that across an inductor is in quadrature, and the two voltage vectors/phasors add. The result is the same though, the ratio of current to voltage is a general angle if both in phase and quadrature componetnts are non-zero. \$\endgroup\$
    – Neil_UK
    Nov 30, 2023 at 16:30
  • \$\begingroup\$ Thank you @Neil_UK but still forgive me for my difficulty in grasping this, how can resistance affect the angle of the voltage? As you said current and voltage are in phase (no angle change) across a resistor so how can resistance cause the current to be at an angle other than 90 degrees (lagging) with respect to voltage? \$\endgroup\$ Nov 30, 2023 at 18:16
  • \$\begingroup\$ @BaldovínCadenaMejía consider an AC current flowing through two impedances in series. One is a pure resistor, the other is a pure inductor. The resistor generates an in phase voltage VR across itself. The inductor generates a quadrature voltage VL across itself. The two voltages are at right angles. If they are equal magnitude, then the resultant will be at 45 degrees to the current. With different magnitudes, you get different angles. See the very diagram you have posted, but think if Im and Ie as VR and VL, change the ratio of their magnitudes and the angle with respect to current changes. \$\endgroup\$
    – Neil_UK
    Nov 30, 2023 at 21:02

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