Power Systems Question - Power Factors


Joined Dec 29, 2008
hi ...
... suggest drawing a power triangle ... It is much easier to visualize.

The apparent power, in volt-amps, is always the hypotenuse. The real power, in watts, is on the real axis, and the reactive power, in vars, is perpendicular to watts, parallel to the imaginary axis. Vars can be either inductive, +, or capacitive, -.


Joined Mar 31, 2012
so is the total apparent power 7000 x 2 = 14000VA?
That yields the correct value in this case by coincidence. In general this approach is not valid.

That you can't just add the apparent powers becomes obvious if you consider the whole notion of power factor correction -- take something that has a high apparent power relative to its real power and add another load in parallel with it that has little to no real power but a high reactive power (and hence high apparent power) but of the opposite kind of reactive power and, when combined, the total apparent power is close to just the real power of the first load.

For loads in parallel (almost always the case in these kinds of problems), real powers adds and reactive powers add, but apparent powers do not.

Consider the following two inductive loads:

P1 = 3 kW, pf = 0.600
P2 = 12 kW, pf = 0.923

S1 = 3kW / 0.600 = 5 kVA
S2 = 12 kW/0.923 = 13 kVA

But while the total real power is 15 kW, the total apparent power is NOT 18 kVA. We need to find the reactive power for each load:

We know that

S² = P² + Q²

So solve for both of the reactive powers, noting that they are both positive since inductive loads have positive reactive power.

Q1 = 4 kVAR
Q2 = 8 kVAR

The total reactive power is therefore 12 kVAR. The total apparent power is then

S = √((15 kW)² + (12 kVAR)²) = 19.21 kVA

The combined power factor is

pf = 15 kW / 19.21 kVA = 0.781