ElectricalPower SystemsmediumFE ExamCourseworkPE / Advanced

Real Power in a Balanced Three-Phase Load

A balanced three-phase load draws a line current of $I_L = 20\ \text{A}$ at a line-to-line voltage of $V_L = 480\ \text{V}$ , with a power factor of $0.85$ lagging.

Find the total real power drawn by the load.

Given

VL	480 V	Line-to-line voltage
IL	20 A	Line current
pf	0.85 -	Power factor

Hint 1

There is a standard three-phase formula in terms of line voltage and line current.

Hint 2

P = √3·V_L·I_L·cosθ — valid for both wye and delta.

Hint 3

The coefficient is √3 ≈ 1.732, not 3.

Worked solution — try the problem first

For any balanced three-phase load, in terms of line quantities:

$P = \sqrt{3}\, V_L I_L \cos\theta$

This form holds for both wye and delta connections — that is its whole convenience.

$P = 1.732 \times 480 \times 20 \times 0.85$

$= 1.732 \times 480 = 831.4$ $831.4 \times 20 = 16{,}628$ $16{,}628 \times 0.85 = 14{,}134\ \text{W}$

$\boxed{P = 14.1\ \text{kW}}$

Where the √3 comes from. Per-phase power is $V_\phi I_\phi \cos\theta$ , and total power is three times that. In wye, $V_\phi = V_L/\sqrt{3}$ and $I_\phi = I_L$ , so $P = 3(V_L/\sqrt3)I_L\cos\theta = \sqrt3 V_L I_L\cos\theta$ . The delta case gives the same result by the reciprocal argument on currents.

Do not use $3V_LI_L\cos\theta$ — that mixes per-phase counting with line quantities and overstates power by $\sqrt3 = 1.73$ .

Concepts:Three-phase powerBalanced loadsPower factorLine vs phase quantities

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