A three-phase 25-kVA, 480-V, 60-Hz alternator, operating under balanced steady-state conditions, supplies a line current of 20 A per phase at a 0.8 lagging power factor and at rated voltage.

Determine the power triangle for this operating condition.

The apparent power is S = 9.6 kVA;

The true power is P = 7.68 kW;

The reactive power is Q = 7.075 kVAr (inductive)

Explanation:

There are three types of electrical power the apparent power S (V*I), true power P (V*I*cos(theta)) and reactive power Q (V*I*sin(theta)). The power factor is the same as cos(theta) So we can find theta by applying acos to it:

theta = acos(0.8) = 36.87º

Now we can compute the power triangle:

S = V*I = 480*20 = 9.6 kVA

P = V*I*cos(theta) = 9.6*(0.8) = 7.68 kW

Q = V*I*sin(theta) = 9.6*|sin(36.87)|= 9.6|(-0.737)| = 7.075 kVAr (inductive)

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