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

1. Engineering 43 Chp 8.[7-8]AC Analysis Bruce Mayer, PE Licensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu

2. KCL & KVL for AC Analysis • Simple-Circuit Analysis • AC Version of Ohm’s Law → V = IZ • Rules for Combining Z and/or Y • KCL & KVL • Current and/or Voltage Dividers • More Complex Circuits • Nodal Analysis • Loop or Mesh Analysis • SuperPosition

3. Methods of AC Analysis cont. • More Complex Circuits • Thevenin’s Theorem • Norton’s Theorem • Numerical Techniques • MATLAB • SPICE

4. For The Ckt At Right, Find VS if Example • Then I2 by Ohm • Solution Plan: GND at Bot, then Find in Order • I3 → V1 → I2 → I1 → VS • I3 First by Ohm • Then I1 by KCL • Then V1 by Ohm

5. Then VS by Ohm & KVL Example cont. • Then Zeq • Note That we haveI1 and VS • Thus can find the Circuit’s Equivalent Impedance

6. For The Ckt at Right Find IO Use Node Analysis Specifically a SuperNode that Encompasses The V-Src Nodal Analysis for AC Circuits • The Relation For IO • And the SuperNode Constraint • In SuperNode KCL Sub for V1

7. Solving for For V2 Nodal Analysis cont. • The Complex Arithmetic • Recall • Or

8. Same Ckt, But Different Approach to Find IO Note: IO = –I3 Constraint: I1 = –2A0 Loop Analysis for AC Circuits • The Loop Eqns • Simplify Loop2 & Loop3 • Solution is I3 = –IO • Recall I1 = –2A0 • Two Eqns In Two Unknowns

9. Isolating I3 Loop Analysis cont • Then The Solution • The Next Step is to Solve the 3 Eqns for I2 andI3 • So Then Note • Could also use a SuperMesh to Avoid the Current Source

10. = Recall Source SuperPosition + • Circuit With Current Source Set To Zero • OPEN Ckt • Circuit with Voltage Source set to Zero • SHORT Ckt • By Linearity

11. Same Ckt, But Use Source SuperPosition to Find IO Deactivate V-Source AC Ckt Source Superposition • The Reduced Ckt • Combine The Parallel Impedances

12. Find I-Src Contribution to IO by I-Divider AC Source Superposition cont • The V-Src Contribution by V-Divider • Now Deactivate the I-Source (open it)

13. Sub for Z” AC Source Superposition cont.2 • The Total Response • Finally SuperPose the Response Components

14. When Sources of Differing FREQUENCIES excite a ckt then we MUST use Superposition for every set of sources with NON-EQUAL FREQUENCIES An Example Multiple Frequencies • We Can Denote the Sources as Phasors • But canNOT COMBINE them due to DIFFERENT frequencies

15. Must Use SuperPosition for EACH Different ω V1 first (ω = 10 r/s) Multiple Frequencies cont.1 • V2 next (ω = 20 r/s) • The Frequency-1 Domain Phasor-Diagram

16. The Frequency-2 Domain Phasor-Diagram Multiple Frequencies cont.2 • Recover the Time Domain Currents • Finally Superpose • Note the MINUS sign from V-src Polarity

17. Source transformation is a good tool to reduce complexity in a circuit WHEN IT CAN BE APPLIED “ideal sources” are not good models for real behavior of sources A real battery does not produce infinite current when short-circuited Resistance → Impedance Analogy Source Transformation

18. Same Ckt, But Use Source Transformationto Find IO Start With I-Src Source Transformation • Then the Reduced Circuit • Next Combine the VoltageSources And Xform

19. The Reduced Ckt Source Transformation cont • Now Combine theSeries-ParallelImpedances • The Reduced Ckt • IO by I-Divider

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

21. 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

22. 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 ZTH by Deactivating the 8+j2 V-Source

23. Then ZTH AC Thevenin Analysis cont.2 • So The Thevenin Equivalent Circuit • IO is Now Simple

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

25. 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

26. 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 Find IO • See Next Slide

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

28. Use Nodal Analysis to Find VO The KCL at Node-1 Example • Subbing for V1 Yields • Now KCL At VO • Factoring for VO Yields • Isolating VO • Multiply Node-1 KCL by 2j to Obtain

29. This Time Use Thevenin to Find VO Take Cap & Res in Series as the Load Example Alternative • Find VOC by V-Divider • Deactivate V-Src to Find ZTH

30. Now Apply the Thevenin Equivalent Circuit to the Load Example Alternative cont. • Calculate VO By V-Divider • Same as Before

31. WhiteBoard Work • Let’s Work This Nice Problem to Find VO