Presentation on theme: "Lecture 32 Review: AC power analysis Power factor correction"— Presentation transcript:
1 Lecture 32 Review: AC power analysis Power factor correction Average power, complex power, power trianglesRMS valuesPower factorPower factor correctionRelated educational modules:Section 2.9.1, 2.9.2
2 AC power analysis Average power: Average power in terms of RMS (or effective) values:Complex power:
3 Power triangles Complex power (rectangular form): Real (average) and reactive power:Presented graphically:
5 Example 1 For the circuit below, determine: (a) the complex power delivered by the source(b) the average power delivered by the source
6 Outline problem on previous slide: 1. find equivalent impedance2. find source current3. complex power = VI*/24. Average power = (Vm*Im/2)*cos(thetav-thetai)
7 (a) Determine the complex power delivered by the source
8 (b) Determine the average power delivered by the source
9 Effect of pf on power delivery If v - i 0, we have some reactive power that is not consumed by the loadThe current provided to the load is higher than necessaryResults in additional power dissipated during deliveryPower companies don’t like this!
10 Power factor correction Power companies may require that users maintain a minimum power factore.g. pf > 0.9Most large loads are inductive in naturee.g. inductive motorsPower factor correction may be necessaryThe approach must be inexpensive & simple to implementAdding a capacitor in parallel with the inductive load will increase the power factor
11 Power factor correction – continued We have an inductive load with some power factor cos1:The power triangle is shown below:
12 Power factor correction – continued again We can increase the power factor by adding a capacitor in parallel with the load:The power triangle then becomes:
13 Example 2 – power factor correction For the circuit below if(a) Determine the power factor(b) Re-design the circuit so that pf = 1
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