Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu

1. Engineering 43 Thevenin/NortonAC Power Bruce Mayer, PE Licensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu

2. Thevenin Equivalent Circuit for PART A Thevenin’s Equivalence Theorem • Resistance to Impedance Analogy

3. Same Circuit, Find IO by Thevenin Take as “Part B” Load the 1Ω Resistor Thru Which IO Flows AC Thevenin Analysis • The Thevenin Ckt • Now Use Src-Xform

4. The XformedCircuit AnotherSourceXform AC Thevenin Analysis cont. • Now the Combined 8+j2 Source and all The Impedance are IN SERIES • Thus VOC Determined by V-Divider • Find ZTHby Zeroing the 8+j2V-Source

5. ReDraw the Voltage Divider • The V-Divider Formula:

6. Find ZTH by Zeroing the 8+j2 V-Source AC TheveninAnalysis • So The Thevenin Equivalent Circuit • IO is Now Simple

7. Norton Equivalent Circuit for PART A Norton’s Equivalence Theorem • Resistance to Impedance Analogy

8. Same Circuit, Find IO by Norton Take as the “Part B” Load the 1Ω Resistor Thru Which IO Flows AC Norton Analysis • Shorting the Load Yields Isc= IN • Possible techniques to Find ISC: • Loops or Nodes • Source Transformation • SuperPosition • Choose Nodes

9. Short-Ckt Node On The Norton Circuit AC Norton Analysis cont • As Before Deactivate Srcs to Find ZN=ZTH • Use ISC=IN, and ZN to FindIO • See Next Slide

10. The Norton Equivalent Circuitwith LoadReattached AC Norton Analysis cont.2 • Find IO By I-Divider • Same as all the Others

11. Outline – AC Steady State Power • Instantaneous Power Concept • For The Special Case Of Steady State Sinusoidal Signals • Average Power Concept • Power Absorbed Or Supplied During an Integer Number of Complete Cycles • Maximum Average Power Transfer • When The Circuit is in Sinusoidal Steady State

12. Outline – AC SS Power cont. • Power Factor • A Measure Of The Angle Between the CURRENT and VOLTAGE Phasors • Power Factor Correction • Improve Power Transfer To a Load By “Aligning” the I & VPhasors • Single Phase Three-Wire Circuits • Typical HouseHold Power Distribution

13. Consider in the Time Domain a Voltage Source Supplying Current to an Impedance Load Instantaneous Power • Now Recall From Chp1 The Eqn for Power • Then at any Instant for Time-Varying Sinusoidal Signals • Now Use Trig ID • In the General Case

14. Rewriting the Power Relation for Sinusoids • The First Term is a CONSTANT, or DC value; i.e. There is No Time Dependence • The Second Term is a Sinusoid of TWICE the Frequency of the Driving Source Instantaneous Power cont. • Examine the TWO Terms of the Power Equation

15. For the Single Loop Ckt Example • Use Phasors To find I • To Obtain the Time Domain current Take the Real Part of the Phasor Current • Assume

16. Thus for This case Example cont • In the Power Equation • Or • The Amplitudes and Phase Angles • See Next Slide for Plots • v(t) • i(t) • p(t) = v(t)•i(t)

17. Max-p = 7.46W p(t) Calculated by p(t)=v(t)•i(t) Avg-p = 3.46W NEGATIVE POWER – Inductive Load can Release stored Energy to the Circuit p = 0 if either i or v are zero

18. For ANY Periodic Function, It’s Average Value Can be calculated by Integrating Over at Least ONE COMPLETE PERIOD, T, and Then Dividing the Integrated Value by the Period Average Power • Then the Average POWER for Electrical Circuits With Sinusoidal Excitation • And Recall

19. Also For ANY Periodic Function, the Average value May be Calculated over any INTEGER number of periods. This is, in Fact, How most Electrical Power Values are MEASURED. Mathematically: Average Power cont • Now Sub Into the Average Power Integral The Simplified (n =1) Expression for Instantaneous Power

20. Examine the Two Terms from the Average Power Eqn Average Power cont.2 • Thus the First Term is a CONSTANT that depends on the Relative Phase Angle • Also by Trig: cos(−) = cos() • Thus for the First Term It Does NOT Matter if the Current LEADS or LAGs the Voltage

21. The Second Term from the Average Power Eqn Average Power cont.3 • As the The Integral of a sin or cos over an INTEGER number of periods is ZERO • Thus the Average Power is Described by the FIRST TERM ONLY

22. For a Purely RESISTIVE Circuit There is NO Imaginary Component of the Impedance and thus no Phase Shift Between i & v. So for sinusoids Resistive & Reactive Power • For A Purely REACTIVE Circuit i&v are ±90° out of phase • So then Avg Power eqn • Thus Purely reactive Impedances absorb NO Power on Average • They STORE Energy over one Half-Cycle, and then RELEASE it over the NEXT • Thus the Resistors Absorb, or Dissipate, Power (as Heat) only

23. For This Circuit Example  Avg Power • The Power Parameters • The Power Calculation • FIND: The Dissipated Power • Use Phasor Algebra to Find the Current

24. Since Only the Resistor Dissipates Power, Check the Previous Calc by using Pres=vres•i Example  Avg Power cont • Alternatively by OHM • For a Resistor the Current & Voltage WITHIN the Resistor are IN-PHASE; thus • Use Phasors for VR 

25. For This Circuit Example  Cap Circuit • Find the Total Impedance Across the V-Source • Find The Dissipated Power in Each Resistor • Start by Finding The Current In the Ckt Branches that Contain the Resistors • Then The total Current I • i(t) LEADS vS(t)

26. Example  Cap Circuit cont • Now Find I2 by Current Divider • So the 2Ω Resistor Power Dissipation • Then the Power Absorbed by the 4Ω R

27. Recall From the Study of Resistive Ckts The Criteria for Max Power Xfer to a Load Resistor Maximum Avg Power Transfer • Consider This General Thevenin Equivalent Ckt • Where RTH is the Thevenin Equivalent Resistance for the Driving Ckt • Now Try to Develop a Similar Relationship for Impedances • For This Ckt the Avg Pwr Delivered to the Load

28. Now by Phasors Max Avg Power Transfer cont • Now By Euler Rln Recall • Where

29. Then the Load Voltage & Current Magnitudes Max Avg Power Transfer cont.2 • Also by Euler • Similarly • Now a Useful Trig ID • And

30. ReArrange Trig ID to Find Max Avg Power Transfer cont.3 • Again the Power Eqn • Or • And • Substitute to Find

31. Finally The Power Eqn Restated Max Avg Power Transfer cont.4 • Now To Maximize the Power Transfer, Set the Partial Derivatives to 0 • at LAST The Optimized Load • The Complex Conjugate • And the Power Transferred at Optimum

32. Check Max Avg Power Transfer cont.5 • At the Max Condition 

33. Find ZL for the Maximum Power Xfer Example  Maximum Power Xfer • And The Max Pwr Xfer • Need to find • VOC = VTH • RTH • Recall The Max Power Criteria

34. Remove Load and Find VOC by Loop Current Example  Max PwrXfer cont. • And by Ohm’s Law in the Frequency Domain • Find ZTH by Source Deactivation (Zeroing) • Using KVL on the Loop

35. Then ZTH Example  Max PwrXfer cont.2 • or • Taking The Conjugate • Then The Power Transferred to this Load

36. Consider a Complex Current Thru a Complex Impedance Load Power Factor • By Ohm & Euler • The Current and Load-Voltage Phasors (Vectors) Can Be Plotted on the Complex Plane • in the Electrical Power Industry θZ is the Power Factor Angle, or Simply the Phase Angle

37. The Phase Angle Can Be Positive or Negative Depending on the Nature of the Load V is the BaseLine Power Factor cont • Typical Industrial Case is the INDUCTIVE Load • Large Electric Motors are Essentially Inductors • Now Recall The General Power Eqn • Measuring the Load with an AC DMM yields • Vrms • Irms

38. The Product of the DMM Measurements is the APPARENT Power Power Factor cont.2 • Now Define the Power Factor for the Load • The Apparent Power is NOT the Actual Power, and is thus NOT stated in Watts. • Apparent Power Units = VA or kVA • Some Load Types

39. Consider this case Vrms = 460 V Irms = 200A pf = 1.5% Then Papparent = 92kVA Pactual =1.4 kW  This Load requires The Same Power as a Hair Dryer or Toaster However, Despite the low power levels, The WIRES and CIRCUIT BREAKERS that feed this small Load must be Sized for 200A! The Wires would be nearly an INCH in Diameter pf – Why do We Care?

40. The Local Power Company Services this Large Industrial Load Example  Power Factor • Then the I2R Loses in the 100 mΩ line • Improving the pf to 94% Power company I lags V • Find Irms by Pwr Factor

41. For This Ckt The Effect of the Power Factor on Line Losses Example - Power Factor cont

42. Consider a general Ckt with an Impedance Ld Active Power Reactive Power Complex Power • Mathematically • Converting to Rectangular Notation • For this Situation Define the Complex Power for the Load:

43. Thus S in Shorthand Complex Power cont • Alternatively, Reconsider the General Sinusoidal Circuit • S & Q are NOT Actual Power, and Thus all Terms are given Non-Watt Units • S→ Volt-Amps (VA) • Q → Volt-Amps, Reactive (VAR) • P is Actual Power and hence has Units of W • First: U vs. Urms

44. Now in the General Ckt By Ohm’s Law Complex Power cont.2 • And Again by Ohm • In the Last Expression Equate the REAL and Imaginary Parts

45. And by Complex Power Definition Complex Power cont.3 • Similarly for Q • Using the Previous Results for P • So Finally the Alternative Expression for S

46. The Expressions for S Complex Power Triangle • Plotting S in the Complex Plane • From The Complex Power “Triangle” Observe • Note also That Complex Power is CONSERVED

47. For the Circuit At Right Zline =0.09 Ω + j0.3 Ω Pload = 20 kW Vload = 2200° pf = 80%, lagging f = 60 Hz → ω = 377s-1 inductive capacitive Example - Complex Power • From the Actual Power • Lagging pf → Inductive • Thus • And Q from Pwr Triangle

48. Then SL Example - Complex Power cont • Recall the S Mathematical Definition • Note also that [U*]* = U • In the S Definition, Isolating the Load Current and then Conjugating Both Sides • Alternatively Lagging

49. Now Determine VS Example - Complex Pwr cont.2 • Then VS • Then The Phase Angle • To find the Src Power Factor, Draw the I & V Phasor Diagram

50. For the Circuit AtRight, Determine Real & Reactive Power losses in the Line Real & Reactive Power at the Source inductive capacitive Example - Complex Power kVAR • From the Actual Power • Lagging pf → Inductive • Thus • And by S Definition