1. L14 § 4.5 Thevenin’s Theorem A common situation: Most part of the circuit is fixed, only one element is variable, e.g., a load: i0i0 +  +  The fixed part, a two terminal circuit, can be very complex. How the variable part affects v o, i o, or the power? By Thevenin’s theorem, the complex two terminal circuit can be replaced with a simple circuit. Making the analysis very easy Thevenin’s theorem is the most important result in Circuit Theory. It will be a main tool for Chapters 7, 8.

2. L14 Thevenin’s theorem: Every linear two terminal circuit can be made equivalent to the series connection of a voltage source and a resistor. Same v  i relationship i +  +  i ++ V th R th V th : Open circuit voltage across the two terminals i=0 +  +  Also called V oc R th : equivalent resistance with respect to the two terminals when all independent sources are turned off Case 1: No dependent source. R th =R eq (as in Chapter 2) Case 2: With dependent source. Need to supply an external independent source at the two terminal. Then R th is the ratio between the voltage and current.

3. L14 Example: What is the relationship between R L and i? ++ 32V 2A 4Ω4Ω 12Ω 1Ω1Ω b a RLRL i Replacing the part to the left of “a”, “b” by Thevenin’s equivalent ++ V th R th b a RLRL i Clear relationship: Key point: V th =?, R th =? Solution: V th is the open circuit voltage. ++ 32V 2A 4Ω4Ω 12Ω 1Ω1Ω a i 1 Assign mesh current i 1 KVL along the mesh: 4i 1 +12(i 1 +2)=32 i 1 =0.5A V th =12(i 1 +2)=30V 3 For R th, turn off all independent sources: 4Ω4Ω 12Ω 1Ω1Ω R th =1+4//12 =4  i=0 + 0V 

4. L14 ++ 32V 2A 4Ω4Ω 12Ω 1Ω1Ω b a 4 10A 3Ω3Ω 1Ω1Ω a b 8A 2A 4Ω4Ω 12Ω 1Ω1Ω a b 30V 3Ω3Ω 1Ω1Ω a b ++ ++ b a

5. L14 Case 2: There are dependent sources V th : The same as case 1, the open circuit voltage For R th : 1.Turn off all independent sources. +  i0i0 +  ++ i0i0 v0v0 or, Give any number for i 0, e.g., 1A, 2A, 10A, Solve for v 0 Give any number for v 0, e.g., 1V, 2V, 20V, Solve for i 0 2. Connect the two terminals to a current/voltage source 3. Solve for the voltage/current of the source

6. L14 Explanation: as the independent sources are turned off, the two terminal circuit behaves like a single resistor. i0i0 i0i0 ++ R th

7. L14 Example: Find R th, V th for the two terminal circuit ++ 30V 6Ω6Ω3Ω3Ω a 2v x b For V th : a V th = v x ++ 30V 6Ω6Ω3Ω3Ω 2v x Use nodal analysis Pick ground. v x is the node voltage vx vx b KCL at node v x : v x = 2V V th =2V For R th : turn off 30V with short circuit. Approach 1: connect to voltage source v 0 =1V, 6Ω6Ω3Ω3Ω a 2v x vx vx ++ v 0 =1V i2i2 i0i0 i1i1 KCL at node v x : i 0 = i 1 +i 2 +2v x Given v x =1V

8. L14 Approach 2: connect to current source i 0 =1A, KCL at node v 0 : i 1 +i 2 +2v x =1 6Ω6Ω3Ω3Ω a 2v x v0 v0 i 0 =1A i2i2 i1i1 v 0 = v x v 0 =0.4 Two approaches yield the same R th.

9. 24V       + v x  0.5v x 24V       + v x  0.5v x 0A By voltage div

10. 24V       + v x  0.5v x 1A 2A     + v x  0.5v x -1A

11. R14 Practice 1: Find R th, V th for the two terminal circuit ++ 5V 5A 3Ω3Ω 4Ω4Ω 2Ω2Ω 4Ω4Ω a b

12. R14 Practice 2: Find R th, V th for the two terminal circuit 4Ω4Ω 2Ω2Ω a 2i x b ixix

13. R14 5Ω5Ω 2Ω2Ω a 2v x Practice 3: Find R th, V th for the two terminal circuit 10V 3Ω3Ω 2Ω2Ω b ++

14. L15 § 4.6 Norton’s Theorem Norton’s theorem: Every linear two terminal circuit can be made equivalent to the parallel connection of a current source and a resistor. I N : Short circuit current from “a” to “b”, R N : same as R th i +  +  i ININ RNRN a b b a i ++ V th R th Since the same circuit is also equivalent to Thevenin’s By source transformation, R N =R th, V th =R th I N, I N =V th /R th You can get Thevenin’s equivalent first, then use source transformation to get Norton’s equivalent, or vice vesa

15. L15 Example: Find the Norton equivalent ++ 12V 2A 4Ω4Ω 8Ω8Ω 5Ω5Ω b 8Ω8Ω a For I N, short circuit “a”, “b” Be careful: a resistor in parallel with short circuit does not take any current. It can be discarded. ++ 12V 2A 4Ω4Ω 8Ω8Ω 5Ω5Ω b IN IN 8Ω8Ω a The mesh current is I N KVL along center mesh: 8I N +8I N -12+4(I N -2)=0 I N =1A For R N, turn off 2A and 12V sources: 4Ω4Ω 8Ω8Ω 5Ω5Ω b 8Ω8Ω a R N =5//(8+8+4)=4  IN IN

16. L15 § 4.8 Maximum power transfer i +  +  a b RLRL Linear circuit connected to a variable load R L Power consumed by load: Main tool: Thevenin’s and Norton’s theorem

17. L15 By Thevenin’s theorem, the two terminal circuit can be replaced with a voltage source in series with a resistor: i ++ V th R th RLRL Clear relationship:

18. L15 A pure math problem: find R L so that p is maximized. p RLRL p max R0R0 At the maximal point, For dp/dR L = 0, we must have R L =R th At R L =R th,

19. L15 Conclusion: Maximal power is transferred to the load when R L =R th, and the maximal power is When the Norton’s equivalent is used, similar result: Conclusion: Maximal power is transferred to the load when R L =R N, and the maximal power is

20. L15 Example: Determine R so that maximal power is delivered. Also find the maximal power. ++ R 20V 4Ω4Ω 0.5v x 15Ω 6Ω6Ω 10Ω Solution: Need to find V th and R th with respect two the two terminals of R. For V th, disconnect R Since i=0, v x = 0. 10  and 15  in series By voltage division ++ 20V 4Ω4Ω 0.5v x 15Ω 6Ω6Ω 10Ω i=0 Also since i=0, 0.5v x =0 flows through 4Ω. Thus v 2 = 0V By KVL, V th =v x +v 1 -v 2 =0+12+0=12V

21. L15 For R th, turn off 20V, but keep 0.5v x dependent source. Because of the dependent source, supply an external current source i 0 = 1A. (you may also try a voltage source and compare) Need to find v 0 across the 1A current source (Active sign convention) Assign current i 2 for 4Ω By KCL, i 2 =0.5v x -1. Thus v 2 = 4(0.5v x -1) = 4(3-1)=8V v x =6i 0 =6V 4Ω4Ω 0.5v x 15Ω 6Ω6Ω 10Ω i 0 =1A 1A i 2 21 Finally, The maximal power is delivered when R=R th =4Ω. The maximal power is

22. L15 Question: Any relationship between the following circuits? i R ++ VsVs IsIs i R Norton’s equivalent:Thevenin’s equivalent: IsIs i i ++ VsVs 0 Any resistor in series with current source can be discarded Any resistor in parallel with voltage source can be discarded

23. R15 Practice 4: Find the Norton’s equivalent ++ 40V 3A 20Ω 40Ω 10Ω a b

24. R15 RLRL ixix Practice 5: Find the value of R L so that maximum power is delivered. Also find the maximum power. ++ 4V 2Ω2Ω 3i x 2Ω2Ω 1Ω1Ω ++

25. R15 60Ω Practice 6: Find the unknown R L so that maximum power is delivered. Also find the maximum power. 40V 80Ω RLRL 20Ω 40Ω + 