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Power factor correction with two loads

  • Thread starter jean28
  • Start date
  • #1
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Homework Statement



http://i1226.photobucket.com/albums/ee410/jean28x/8b1ad092d9dd25a8f10f1672c0073d97.jpg

Vs = 480<0 rms.
P1 105 W; pf1 0.7 lagging; P2 5 × 104 W; pf2 0.95 leading.

Homework Equations



S = sqrt(P^2 + Q^2), = Vrms* Irms(conjugate)


The Attempt at a Solution



I know how to calculate the capacitance with one load, but how do I work with the two loads? Do I find the capacitance of both and then add them up?
 

Answers and Replies

  • #2
NascentOxygen
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Homework Statement



http://i1226.photobucket.com/albums/ee410/jean28x/8b1ad092d9dd25a8f10f1672c0073d97.jpg

Vs = 480<0 rms.
P1 105 W; pf1 0.7 lagging; P2 5 × 104 W; pf2 0.95 leading.
Where did you get the detail that one is lagging and the other leading?
I know how to calculate the capacitance with one load, but how do I work with the two loads? Do I find the capacitance of both and then add them up?
If they both needed added capacitance, then because the loads are in parallel you can add the two capacitances. But if you are correct in saying one load has leading pf, then that load doesn't require added parallel capacitance.

First step: add the reactive currents of the two loads. Is it overall capacitive or inductive?
 
  • #3
rude man
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Power factor is an even function of the phase angle, meaning the phase itself can be either + or -, requiring different reactive components to bring the phase to zero. So you need to have this defined for you first.
 

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