Electrical – Regulation of transformer with inductive load

Imagine you wound an inductor but instead of one thicker wire you used two thinner wires (this is commonplace by the way). An AC voltage is applied on both wires - lets call the AC voltage AC live and AC neutral and lets say neutral is connected to 0V for reference.

No problem so far?

Now, imagine you only applied AC live to one of those wires and left the other one open circuit (AC neutrals both at 0V still). What would be the induced voltage in the unconnected wire? Would it be: -

• In phase
• Out of phase by 180degrees
• Same amplitude
• Different amplitude

If the induced voltage on the unconnected wire were antiphase then joining it up to the AC live would create a seriously big problem and there would be sparks and breakers tripping. If the induced voltage on the unconnected wire were significantly different to the AC live and you joined them up then same problem as before - sparks and breaker trips.

This means if you wind a transformer with two wires in the same way there will be no phase shift but you can always swap the primary wires or secondary wires over and get 180 degrees.

Inherently there is no 180 degrees phase shift - you only get 180 degrees when you get into a muddle with the wires.

Torque applied is maintain at constant for both scenario. But during the load condition there is a significant drop in rotation speed.

If input torque is held constant then you are fashioning a constant current generator and not a constant voltage generator. Input mechanical speed would need to be held constant to produce a "near-enough" constant output voltage.

Mechanical power in = \$2\pi n T\$ where n is speed (rev/sec) and T is torque. You have kept T constant but allowed n to fall therefore the mechanical power into the generator has fallen just when you wanted power to increase!