1. 8. 3 Power Factor Correction 8.1 Power in a Sinusoidal Steady-State Circuit 8.2 Maximum Average Power Transfer Chapter 8 AC Power Analysis 交流功率分析

2. 8.1 Power in a Sinusoidal Steady-State Circuit Let the voltage and current at the terminals of the circuit be The instantaneous power absorbed by the circuit is N0N0 v i + - 瞬时功率

3. The average power is the average of the instantaneous power over one period. cos  ： power factor  =  v -  i ： power factor angle average power is also called real power (W) reactive power (VAR ) apparent power (VA) i leads v : leading power factor i lags v : lagging power factor 功率因数 领先的功率因数 滞后的功率因数 功率因数角 有功功率 ( 瓦） 无功功率 ( 乏） 视在功率 ( 伏安）

4.  S P Q Power triangle 功率三角形 Similar triangle 相似三角形 Impedance triangle 阻抗三角形 |Z| R X zz Complex power (VA) 复功率 ( 伏安）  =  Z

5. Real Power and Reactive Power for R 、 L 、 C P R =VIcos  =VIcos0  =VI P L =VIcos  =VIcos90  =0 P C =VIcos  =VIcos( - 90  )=0 Q R =VIsin  =VIsin0  =0 =I2R=V2/R=I2R=V2/R Q L =VIsin  =VIsin90  =VI Q C =VIsin  =VIsin ( - 90  )= - VI

6. The complex, real,and reactive powers of the sources equal the respective sums of the complex, real,and reactive powers of the individual loads. but Conservation of AC Power 功率守恒

7. N0N0 v i + - Solution: Example 8.1 Find P, Q, S and in the circuit.

8. Example 8.2 Determine the power factor of the entire circuit as seen by the source. Calculate the average power delivered by the source. Solution: The total impedance is The power factor is The average power supplied by the source is or leading

9. Example 8.3 In the circuit, f=50Hz ， the readings of voltmeter, ammeter and wattmeter are 50V ， 1A and 30W, respectively. Calculate R and L. Solution: and

10. 8.2 Maximum Average Power Transfer The Thevenin impedance Z Th and the load impedance Z L are 最大有功功率传输

11. The maximum average power is We obtain That is rms value

12. Example 8.4 In the circuit, find the value of Z L that will absorb the maximum average power. Calculate that power. Solution: The maximum average power is By voltage division

13. The process of increasing the power factor without altering the voltage or current to the original load is known as power factor correction. 8.3 Power Factor Correction and 功率因数提高

14. Example 8.5 When connected to a 120V(rms), 60Hz power line, a load absorbs 4kW at a lagging power factor of 0.8. Find the value of capacitance necessary to raise the pf to 0.95(lagging). Solution: If the pf=0.8, then When the pf is raised to 0.95,

15. 部分电路图和内容参考了： 电路基础（第 3 版），清华大学出版社 电路（第 5 版），高等教育出版社 特此感谢！