LESSON 27 AC THEVENIN MAXIMUM POWER TRANSFER OBJECTIVES

LESSON 27 AC THEVENIN & MAXIMUM POWER TRANSFER

OBJECTIVES • Apply Thevenin’s theorem to simplify AC circuits for analysis. • List the steps for determining the Thevenin Equivalent of an actual electronic circuit. • Using Thevenin’s theorem, analyze complex series -parallel circuits with multiple load values. • Explain under what conditions a source transfers maximum power to a load. • Determine the value of load impedance for which maximum power is transferred from the circuit. • Calculate the average power dissipated by a load.

Example 1 – Calculate the Thévenin Equivalent circuit. – Given: Vs = 25 V p 0 E; R 1 = 100 S ; R 2 = 25 S; RL = 1 k. S, XL = 80 S, f = 509 Hz – solution

Ex: Complex Conjugate • Find the complex conjugate of (polar & rectangular forms): 3 + j 7 = 7. 62 +66. 8° – 3 + j 7 CC: =3 - j 7 = 7. 62 -66. 8° – 5. 6 k -21° = 5. 23 k - j 2. 01 k 5. 6 k +21° = 5. 23 k +j 2. 01 k CC: = • Find the load which will result in maximum power transferred if ZTH = – 1. 8 k 42° – 4 k - j 5 k 1. 8 k 42° = 1. 34 k + j 1. 20 k CC: = 1. 8 k -42° = 1. 34 k -j 2. 01 k 4 k -j 5 k = 6. 40 k -51. 5° 4 k +j 5 = 6. 40 k +51. 5° CC: =

Example 2: – Find value of load impedance ZL to achieve Max Power Transfer: – Given: E = 60 V p 0 E , f = 500 Hz, L = 5 m. H, R 1 = 30 S solution

• Challenge Problem • Review Quiz

REVIEW QUIZ • Thévenin’s Theorem converts a circuit to an equivalent form consisting of ? • What is ETH ? • What is ZTH ? • What is the complex conjugate? • What type of impedance does the addition of an impedance and its complex conjugate produce? • When does Max Power Transfer occur?

Example 1: Solution Return

Example 2: Solution » Return

Challenge Problem • • Determine ZL required for maximum power transfer. What is the maximum power transferred to the load? What is the true and reactive power of the load? What components make up this load (and values)?

Challenge Problem Solution • ZL=*ZTH ZL = *ZTH = 311. 5 -29 = 271. 0 - j 153. 6

Example: continued • Max Power: Since XC = XL will cancel, all circuit power will be true. ZT=542 0 I P=VTH/(ZT) = 41. 9 ma -53. 2 Irms=29. 6 m. A Vrms = 16. 05 V Pa. L = I 2 rms | Z L| Pa. L = 273 (m. VA) Pr. L = Pasin(-29) = 132 m. VAR PTL = Pacos(-29) = 239 m. W

Example: continued • Since ZL(max power) = 332. 5 -20. 4 , can tell that the load is made up of a net resistance and capacitance. In rectangular form: • ZL = 311. 7 - j 116 where R = 311. 7 and XC = 116 and: return