1. EGR 2201 Unit 12 Sinusoidal Steady-State Analysis  Read Alexander & Sadiku, Chapter 10.  Homework #12 and Lab #12 due next week.  Quiz next week.

2. Summary of Chapter 9

3. Steps to Analyze AC Circuits 1. Transform the circuit from the time domain to the phasor domain. 2. Solve the problem using circuit techniques (Ohm’s law, Kirchhoff’s laws, voltage-divider rule, etc.) 3. Transform the resulting phasor to the time domain.

4. What’s Next?  Now we’ll see that we can also apply these other familiar techniques in the phasor domain: Nodal analysis Mesh analysis Superposition Source transformation Thevenin’s theorem Norton’s theorem

5. Review from Unit 3: Steps in Performing Nodal Analysis on a Circuit with No Voltage Sources  Given a circuit with n nodes, without voltage sources, follow these steps: 1. Select a node as the reference node. Assign voltages v 1, v 2, …, v n-1 to the remaining n-1 nodes. These voltages are relative to the reference node. 2. Apply KCL to each of the n-1 non- reference nodes. Use Ohm’s law to express the branch currents in terms of node voltages. Then simplify the equations. 3. Solve the resulting n-1 simultaneous equations to obtain the unknown node voltages.

6.  As described above, our procedure applies only to circuits without voltage sources.  But it’s not hard to extend the procedure to circuits with voltage sources.  The way you handle a voltage source depends on whether the source is connected to the reference node…. Review from Unit 3: What About Circuits with Voltage Sources?

7. Review from Unit 3: Nodal Analysis with A Voltage Source That Is Connected to the Reference Node A voltage source connected to the reference node is easy to handle, because it immediately reveals the voltage at one of the non-reference nodes.  Example: In the circuit shown, we can immediately see that v 1 = 10 V.

8. Review from Unit 3: Nodal Analysis with a Voltage Source That Is Not Connected to the Reference Node A voltage source not connected to the reference node is trickier. To handle it, we treat the voltage source and its two nodes (along with any elements in parallel with the voltage source), as a supernode.

9. Review from Unit 3: How to Handle a Supernode We apply KCL and KVL to the supernode to get two equations. Example: In the circuit shown, KCL gives i 1 + i 4 = i 2 + i 3 And KVL gives v 2 = 5 + v 3

10. Nodal Analysis in the Phasor Domain  Everything that we just reviewed about nodal analysis of DC circuits also applies to sinusoidal AC circuits in the phasor domain.  The only difference is that we’ll use complex numbers throughout.

11. What’s Next?  We can also apply these other familiar techniques in the phasor domain: Nodal analysis Mesh analysis Superposition Source transformation Thevenin’s theorem Norton’s theorem

12. Review from Unit 4: Steps in Performing Mesh Analysis on a Circuit with No Current Sources  Given a circuit with n meshes, without current sources, follow these steps: 1. Assign mesh currents i 1, i 2, …, i n to the n meshes. 2. Apply KVL to each of the n meshes. Use Ohm’s law to express the voltages in terms of mesh currents. Then simplify the equations. 3. Solve the resulting n simultaneous equations to obtain the unknown mesh currents.

13.  As described above, our procedure applies only to circuits without current sources.  But it’s not hard to extend the procedure to circuits with current sources.  The way you handle a current source depends on whether the source is located in only one mesh or is shared by two meshes…. Review from Unit 4: What About Circuits with Current Sources?

14. Review from Unit 4: Mesh Analysis with a Current Source Located in Only One Mesh A current source located in only one mesh is easy to handle, because it immediately reveals the mesh current in that mesh.  Example: In the circuit shown, we can immediately see that i 2 = 5 A.

15. Review from Unit 4: Mesh Analysis with a Current Source Shared by Two Meshes A current source shared by two meshes is trickier. To handle it, we create a supermesh by excluding the current source and any elements in series with it.

16. Review from Unit 4: How to Handle a Supermesh We apply KCL and KVL to the super- mesh to get two equations. Example: In the circuit shown, KCL gives i 2 = i 1 +6 And KVL around the supermesh gives 20 = 6i 1 + 10i 2 + 4i 2

17. Mesh Analysis in the Phasor Domain  Everything that we just reviewed about mesh analysis of DC circuits also applies to sinusoidal AC circuits in the phasor domain.  The only difference is that we’ll use complex numbers throughout.

18. What’s Next?  We can also apply these other familiar techniques in the phasor domain: Nodal analysis Mesh analysis Superposition Source transformation Thevenin’s theorem Norton’s theorem

19. Review from Unit 5: Superposition Principle (1 of 2)  The superposition principle says that to find the total effect of two or more independent sources in a linear circuit, you can find the effect of each source acting alone, and then combine those effects.

20. Review from Unit 5: Superposition Principle (2 of 2)  To find the effect of an independent source acting alone, you must turn off all of the other independent sources. To turn off a voltage source, replace it by a short circuit. To turn off a current source, replace it by an open circuit.

21. Review from Unit 5: Steps in Using the Superposition Principle 1.Select one independent source and turn off the others, as explained above. Find the desired voltage or current due to this source using techniques from earlier chapters. 2.Repeat step 1 for each of the other independent sources. 3.Add the values obtained from the steps above, paying careful attention to the signs (+ or ) of each value.

22. Superposition in the Phasor Domain  Everything that we just reviewed about superposition in DC circuits also applies to sinusoidal AC circuits in the phasor domain.  The only difference is that we’ll use complex numbers throughout.

23. What’s Next?  We can also apply these other familiar techniques in the phasor domain: Nodal analysis Mesh analysis Superposition Source transformation Thevenin’s theorem Norton’s theorem

24. Review from Unit 5: Source Transformation  Sometimes when you're analyzing a circuit, it can be useful to substitute a voltage source (and a resistor in series with it) by a current source (and a resistor in parallel with it), or vice versa.  Either substitution is called a source transformation.  Two questions: 1. Why would this be useful? 2. Are we allowed to do this?

25. Review from Unit 5: Example of Why Source Transformation Can Be Useful  Suppose we wish to find v o in the circuit to the right. This is not difficult, but it requires a bit of work.  We may get the answer more easily if we can first replace the current-source-and- parallel-resistor with a voltage-source-and-series-resistor.

26. Review from Unit 5: Are We Allowed to Do This?

27. Source Transformation in the Phasor Domain  Everything that we just reviewed about source transformation in DC circuits also applies to sinusoidal AC circuits in the phasor domain.  The only difference is that we’ll use complex numbers throughout, and instead of having a source resistance R, we’ll have a source impedance Z.

28. What’s Next?  We can also apply these other familiar techniques in the phasor domain: Nodal analysis Mesh analysis Superposition Source transformation Thevenin’s theorem Norton’s theorem

29. Review from Unit 6: Thevenin’s Theorem  Thevenin’s theorem says that any linear two-terminal circuit can be replaced by an equivalent circuit consisting of an independent voltage source V Th in series with a resistor R Th.  There’s a standard procedure for finding the values of V Th and R Th. But first, what exactly does this mean and why is it useful?

30. Review from Unit 6: What Thevenin’s Theorem Means  Thevenin’s theorem lets us replace everything on one side of a pair of terminals by a very simple equivalent circuit consisting of just a voltage source and a resistor. Original CircuitEquivalent Circuit

31. Review from Unit 6: Why It’s Useful  This greatly simplifies computation when you wish to find values of voltage or current for several different possible values of a load resistance. Original CircuitEquivalent Circuit

32. Review from Unit 6: Steps in Finding V TH and R TH 1.Open the circuit at the two terminals where you wish to find the Thevenin-equivalent circuit. (In the circuit shown, this means removing the load.) 2.V TH is the voltage across the two open terminals. Pay attention to polarity! 3.R TH is the resistance looking into the open terminals with all independent sources turned off. (Recall that to turn off a voltage source we replace it by a short, and to turn off a current source we replace it by an open.)

33. Thevenin’s Theorem in the Phasor Domain  Everything that we just reviewed about Thevenin’s theorem in DC circuits also applies to sinusoidal AC circuits in the phasor domain.  The only difference is that we’ll use complex numbers throughout, and instead of having a Thevenin resistance R Th we’ll have a Thevenin impedance Z Th.

34. What’s Next?  We can also apply these other familiar techniques in the phasor domain: Nodal analysis Mesh analysis Superposition Source transformation Thevenin’s theorem Norton’s theorem

35. Review from Unit 6: Norton’s Theorem  Norton’s theorem says that any linear two-terminal circuit can be replaced by an equivalent circuit consisting of an independent current source I N in parallel with a resistor R N. Original CircuitNorton-Equivalent Circuit

36. Review from Unit 6: Finding I N and R N Norton-Equivalent CircuitThevenin-Equivalent Circuit

37. Norton’s Theorem in the Phasor Domain  Everything that we just reviewed about Norton’s theorem in DC circuits also applies to sinusoidal AC circuits in the phasor domain.  The only difference is that we’ll use complex numbers throughout, and instead of having a Norton resistance R N we’ll have a Norton impedance Z N.