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Lecture 32 Review: AC power analysis Average power, complex power, power triangles RMS values Power factor Power factor correction Related educational modules: –Section 2.9.1, 2.9.2

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AC power analysis Average power: Average power in terms of RMS (or effective) values: Complex power:

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Power triangles Complex power (rectangular form): Real (average) and reactive power: Presented graphically:

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Power factor (pf) Power factor: Load impedance:

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Example 1 For the circuit below, determine: (a) the complex power delivered by the source (b) the average power delivered by the source

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Outline problem on previous slide: 1. find equivalent impedance 2. find source current 3. complex power = VI*/2 4. Average power = (Vm*Im/2)*cos(thetav- thetai)

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(a) Determine the complex power delivered by the source

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(b) Determine the average power delivered by the source

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Effect of pf on power delivery If v - i 0, we have some reactive power that is not consumed by the load – The current provided to the load is higher than necessary – Results in additional power dissipated during delivery – Power companies dont like this!

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Power factor correction Power companies may require that users maintain a minimum power factor – e.g. pf > 0.9 Most large loads are inductive in nature – e.g. inductive motors Power factor correction may be necessary – The approach must be inexpensive & simple to implement Adding a capacitor in parallel with the inductive load will increase the power factor

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Power factor correction – continued We have an inductive load with some power factor cos 1 : The power triangle is shown below:

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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:

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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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Example 2 – Determine pf

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Example 2 – Re-design circuit so that pf = 1

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