Jawaban Soal Uts Kelistrikan

First, we need to calculate the apparent power in kVA. We can do this by multiplying load

voltage by load current:

S = IE

S = (9.615 A)(240 V)

S = 2.308 kVA

As we can see, 2.308 kVA is a much larger figure than 1.5 kW, which tells us that the power

factor in this circuit is rather poor (substantially less than 1). Now, we figure the power factor

of this load by dividing the true power by the apparent power:

Using this value for power factor, we can draw a power triangle, and from that determine

If this load is an electric motor, or most any other industrial AC load, it will have a lagging

(inductive) power factor, which means that we’ll have to correct for it with a capacitor of appro-

priate size, wired in parallel. Now that we know the amount of reactive power (1.754 kVAR),

we can calculate the size of capacitor needed to counteract its effects:

Jadi agar Power factornya mendekati 1 maka rangkaian ditambahkan capasitor sebesar 80,761 mikro farad

sumber : Tony R Kuphaldt 2009 pdf