Understanding Inductor Behavior in Transformers: Voltage and Current Phasing

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In transformers, the phase relationship between voltage and current is crucial for understanding power dynamics. A perfect transformer with zero resistance and ideal core properties will exhibit a phase difference influenced by the reactive power consumed. When power is delivered to the secondary circuit, the primary coil's phase shift is determined by the ratio of reactive power to real power (Q/P). As real power approaches zero, the phase shift approaches 90 degrees, indicating maximum reactive power consumption. Conversely, as real power increases indefinitely, the phase shift approaches zero degrees, indicating minimal reactive power consumption.
kelvin490
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In an inductor, a current waveform 90o out of phase with the voltage waveform creates a condition where power is alternately absorbed and returned to the circuit by the inductor. If the inductor is perfect (no wire resistance, no magnetic core losses, etc.), it will dissipate zero power.

I wonder how this will change in the case of transformer. Suppose the primary and secondary coils are of zero resistance and the core is perfectly ferromagnetic, what should be the phase different between the voltage and current in the primary coil? Power is delivered to the secondary circuit so it should not be absorbed by the primary voltage source as in the case of pure inductor.
 
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kelvin490 said:
what should be the phase different between the voltage and current in the primary coil?

There are different models of a transformer. Here is one:

TX_equiv_circuit.png

If your transformer is perfect, Rw,p, Rc and Rw,s are zero. Say you load the transformer by a resistor, RL, the transformer will consume reactive power in Xl,p, Xm and Xl,s. So all in all the transformer will consume the power on the primary side:

Sp = PL + j( Ql,p + Qm + Ql,s ) = P + jQ

giving a phase shift, φ = arctan( Q / P ). So the phase shift depends on the relation Q/P.

If P→0, then φ→90°.

If P→∞, then φ→0°

If the load is not ohmic, you must add the reactive power consumed by the load, QL in the equation above.
 
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I am trying to understand how transferring electric from the powerplant to my house is more effective using high voltage. The suggested explanation that the current is equal to the power supply divided by the voltage, and hence higher voltage leads to lower current and as a result to a lower power loss on the conductives is very confusing me. I know that the current is determined by the voltage and the resistance, and not by a power capability - which defines a limit to the allowable...

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