Originally Posted by

**rcwilson**
That follows from the per unit math. Approximate voltage drop through the transformer is per unit amps times per unit impedance ( V=IZ). At full load per unit amps = 100% = 1.0.

Voltage drop is then (%Z x 100) times 1.0 = .05 x 1 = 5%.

To find approximate voltage drop at any other load (neglecting power factor) just ratio the current as gar did.

Another way to look at it: convert the motor inrush current to per unit.

First select the per unit base: to make it easy, select the trasnformer base: MVA base = 1500 kVA = 1.5 MVA

Select the voltage level you are looking at, Vbase= 480V.

Calculate the base current: Ibase = (MVA base /(kV base x 1.732)) x 1,000 = 1.5 x 1000/(.480 x 1,732) = 1804 Amps.

Using the base current, determine your per unit motor starting current: Istart = 2800/1804 = 1.55

Voltage drop = Ipu x Zpu = 1.55 x .05 = 0.0775 = 7.75%. ( Either gar or I are confused by root 3).

Another way. Motor HP = KVA = 300 KVA. Per unit = 300/1500=0.20. I = Ibase x 0.2 = 360A. You said inrush current is 7.75 times FLA, so inrush MVApu = 7.75 x 0.2 = 1.55.

Vdrop = 1.55 x 5% = 7.75%. (If pu MVA through the transformer is 1.55 the current is 1.55 pu).

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