1. Lecture 32 Review: AC power analysis Power factor correction
Average power, complex power, power triangles
RMS values
Power factor
Power factor correction
Related 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:

4. Power factor (pf)
Power factor:
Load impedance:

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 impedance
2. find source current
3. complex power = VI*/2
4. 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 load
The current provided to the load is higher than necessary
Results in additional power dissipated during delivery
Power companies don’t like this!

10. 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

11. Power factor correction – continued
We have an inductive load with some power factor cos1:
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

14. Example 2 – Determine pf

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