Power factor correction with two loads

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SUMMARY

This discussion focuses on power factor correction for two loads connected in parallel, with specific parameters: Vs = 480<0 rms, P1 = 105 W (pf1 = 0.7 lagging), and P2 = 5 × 104 W (pf2 = 0.95 leading). The key conclusion is that to correct the power factor, one must first calculate the reactive currents of both loads to determine if the overall system is capacitive or inductive. If one load has a leading power factor, it does not require additional capacitance, while the lagging load may require it. The total capacitance can be calculated by adding the individual capacitances of both loads if necessary.

PREREQUISITES
  • Understanding of power factor concepts and calculations
  • Knowledge of reactive power and its relationship to real power
  • Familiarity with complex power calculations (S = sqrt(P^2 + Q^2))
  • Ability to analyze circuits with parallel loads
NEXT STEPS
  • Learn how to calculate reactive power for multiple loads
  • Study the effects of power factor correction on overall system efficiency
  • Explore methods for measuring and adjusting power factor in electrical systems
  • Investigate the use of capacitors in power factor correction for industrial applications
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Electrical engineers, students studying power systems, and professionals involved in power factor correction and energy efficiency optimization.

jean28
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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?
 
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jean28 said:

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?
 
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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