If I am not mistaken, this is the inductance that appears in series with transformer winding and causes the secondary voltage to drop. So for example, if a single phase, 1kVA, 600:120V has impedance of 5%, that means transformers' total leakage inductance is 1.91mH? and voltage drop at full load is 6V (i.e. secondary voltage is 114v?
Transformer Leakage inductance
- Thread starter Rho_HV
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Rattus is right - you need to take account of resistance.
When I don't have a known X/R, I use typical figures and that gives me the L of about 1.85mH. Not greatly different.
A couple of points here.
First, you need to know if the 120V is an on-load voltage or an-off load voltage. My experience is that it is often the on-load voltage but I'm from the other side of the pond from the majority of posters here. In any case it gives you the starting point for applying voltage drop.
The second point to note is that impedance isn't the same as regulation. The 5% is sometimes referred to as impedance volts and described as that voltage you would need to get full load current to flow into a short circuit.
It's a determining factor in calculating the prospective fault current.
This isn't the same as the regulation i.e. how much the terminal voltage drops on load.
Load power factor makes a significant difference.
I did a few calcs and, for typical loads on your transformer, you might expect about 3% voltage regulation at rated transformer kVA.
I understand that "5%" per unit impedance is not the same as voltage regulation, and voltage regulation is a function of load. But, if you are running a 120V, "5%" transformer at half the full load,shouldn't you expect to see 2.5% x 120V drop at the secondary (3V drop)?
You might see that under certain circumstance but it is likely to be less.
I think you might still be confusing impedance and regulation.
The voltage across the transformer impedance is not likely to be in phase with the voltage across the load so you can't do a simple subtraction arithmetic subtraction.
Some numbers to maybe illustrate my previous point now that I have a little more time.
Take your 1kVA single phase transformer.
Consider it as an ideal voltage source with an inductive reactance and resistance in series with it.
The rated output current of the 1kVA transformer at 120V is 8.3A.
Assume the transformer has a resistive load of 30 ohms and its terminal voltage is 120V with this load. That gives 4A - about half load. Again, sticking with simple numbers.
The reactance is about 0.7 ohms which is how we both derived about 1.9mH. A current of 4A in that would result in 2.8V across it - not too far from your 3V.
The resistance from my simple ratio is about 0.18 ohms. Add that to the resistive load and, at 4A, you get 120.7V across the restive part and 2.7V across the inductive part.
The voltages are 90deg out of phase.
So the total voltage is then sqrt(120.7^2 +2.7^2)
It's a drop of less than 1V.
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