Total power, reactive power and pf?

Hi.

A 1000 V rms 60 Hz voltage source delivers power to two loads that are in parallel. The first load is a 10 micro Farad capacitor and the second load absorbs an apparent power of 10kVA with an 80% lagging power factor.

Find :

a) Total power

b) Total reactive power

c) The power factor for the source

d) rms source current

I am stuck, I did try but didn´t get anywhere.

Thanks

2 Answers

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  • F
    Lv 6
    6 years ago
    Best Answer

    The first load is a 10F μcapacitor for which we have Z = 265,4 ohms@-90°. I = 1000/265,4 = 3,77A.

    Q = 1000 * 3,77* sin (-90°) = -3770 VAR

    Lagging angle means arccos (0,8) = 36,87° and since the unit is VA you have V*I = 10000 which means I = 10A since V = 10000.

    So 10000*10*cos(36,87°) = 8000W. Also Q = V*I*sin (angle) = 1000*10*sin (36,87°) = 6000 VAR

    Now for the source we have :

    Total reactive power is 6000-3770 = 2230 VAR. Total power is sqrt(P^2+Q^2) =8305 VA and I = P/V so I = 8,305A.

    pf = 8000/(1000*8,305) 96,327%

    Hope this helped.

  • 6 years ago

    Capacitor's impedance Xc= 1/(2×π×60×10×10⁻⁶ ) ≈ 265 ohms Reactive power drawn by capacitor = V²/X = 3.77 kVA.

    a) Total power (i.e. apparent power!!) = √{[10.sin(arc-cos(0.8)) – 3.77}² + 8²} = 8.31 kVA. Real power = 8 kW

    b) 10.sin(arc-cos(0.8)) – 3.77 = 2.23 kVA

    c) cos[arc-tan(2.23/8.31)] = 0.966

    d) 8.31 A

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