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In an inductor, a current waveform 90° 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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In a transformer, the primary has two components of current: -

  • Magnetization current i.e. the open-secondary primary current.
  • Current that (ignoring small leakage inductances) finds it way (via the turns ratio) to the secondary.

Point 1 - the phase angle of the mag current is 90 degree lagging the primary voltage (just like an inductor). In fact with the secondary open circuit, the transformer is just an inductor.

Point 2 - the current in the secondary is in phase with the secondary voltage for a purely resistive load. The power delivered to the load is "X" watts and the power taken from the supply via the perfect transformer is "X" watts.

Primary current is mainly\$^1\$ in phase with primary voltage when full load is on the secondary. As the secondary loading reduces, the primary current starts to look more reactive and eventually it is 90 degrees out of phase.


\$^1\$ As Dave tweed points out, for many low power transformers (on full load) the primary magnetizing current (at 90 degrees) is still quite domainant and therefore the phase angle won't be "mainly in phase with the primary voltage". The example he gives is a transformer with 1 henry primary mag inductance. This will take a reactive current of 120 V/ XL.

XL is about 377 ohms therefore, current is 318 mA. Current attributed to the 12V load is 1A and due to the turns ratio produces a current of 100mA in the primary. In other words the primary current at full-load, for this type of transformer, is still mainly reactive and closer to 90 degrees than zero.

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  • \$\begingroup\$ "Full load" is a nebulous term. It would be more precise to say that you can think of the load connected to the secondary as appearing in parallel with the primary inductance as a "transformed" value -- adjusted by the square of the turns ratio. Suppose you have a 120V to 12V transformer with a 1A secondary. "Full load" would be a 12-ohm resistor. Suppose the primary inductance is 1H. You can calculate the magnitude and phase of the primary current by treating it as 1H in parallel with 1200 ohms. Even at "full load", it's still mostly reactive current. \$\endgroup\$ – Dave Tweed Jun 12 '15 at 15:37
  • \$\begingroup\$ @Andy aka ,I still have one more question: across the primary coil the voltage is always equal to the supply voltage if there is no resistance in primary coil. This voltage is equal to rate of change of magnetic flux. I wonder which component (or both?)of the current does this flux corresponding to? No matter whether there is a secondary load, primary supply voltage doesn't change, so we can only get the same flux and thus current ! It is strange. \$\endgroup\$ – Kelvin S Jun 12 '15 at 15:55
  • \$\begingroup\$ @KelvinS yes, the flux is rather constant in a transformer and this is due to the load's ampere-turns in the primary exactly cancelling the secondary ampere-turns. It's easy but a tad longwinded to prove of course. \$\endgroup\$ – Andy aka Jun 12 '15 at 16:26
  • \$\begingroup\$ @Andyaka Thanks. I have seen similar statement in textbooks that i1N1 = i2N2 since the counter mmf is opposing the mmf of the primary coil. I just wonder if they are exactly the same does it implies that flux will be canceled out? I know flux should not canceled out but the two mmfs are the same this make me confused. \$\endgroup\$ – Kelvin S Jun 12 '15 at 16:53
  • \$\begingroup\$ @KelvinS yes the two load fluxes exactly cancel out leaving just the residual flux due to magnetization current in the primary \$\endgroup\$ – Andy aka Jun 12 '15 at 17:24

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