1. ECE 3183 – EE Systems Chapter 2 – Part A Parallel, Series and General Resistive Circuits

2. ECE 3183 – Chapter 2 – Part ACHAPTER 2 A-2 Chapter 2 Resistive Circuits 1.Solve circuits (i.e., find currents and voltages of interest) by combining resistances in series and parallel. 2.Apply the voltage-division and current-division principles. 3.Solve circuits by the node-voltage technique. 4.Solve circuits by the mesh-current technique. 5.Find Thévenin and Norton equivalents. 6.Apply the superposition principle.

3. ECE 3183 – Chapter 2 – Part ACHAPTER 2 A-3 FIRST GENERALIZATION: MULTIPLE SOURCES i(t) KVL Voltage sources in series can be algebraically added to form an equivalent source. We select the reference direction to move along the path. Voltage drops are subtracted from rises

4. ECE 3183 – Chapter 2 – Part ACHAPTER 2 A-4 FIRST GENERALIZATION: MULTIPLE SOURCES v eq = ?

5. ECE 3183 – Chapter 2 – Part ACHAPTER 2 A-5 SECOND GENERALIZATION: MULTIPLE RESISTORS APPLY KVL TO THIS LOOP VOLTAGE DIVISION FOR MULTIPLE RESISTORS

6. ECE 3183 – Chapter 2 – Part ACHAPTER 2 A-6 SECOND GENERALIZATION: MULTIPLE RESISTORS APPLY KVL TO THIS LOOP

7. ECE 3183 – Chapter 2 – Part ACHAPTER 2 A-7 SECOND GENERALIZATION: MULTIPLE RESISTORS

8. ECE 3183 – Chapter 2 – Part ACHAPTER 2 A-8

9. ECE 3183 – Chapter 2 – Part ACHAPTER 2 A-9 THE CONCEPT OF EQUIVALENT CIRCUITS THIS CONCEPT WILL OFTEN BE USED TO SIMPLFY THE ANALYSIS OF CIRCUITS. WE INTRODUCE IT HERE WITH A VERY SIMPLE VOLTAGE DIVIDER + - 1 R 2 R S v i 21 RR v i S   + - S v 21 RR  i AS FAR AS THE CURRENT IS CONCERNED BOTH CIRCUITS ARE EQUIVALENT. THE ONE ON THE RIGHT HAS ONLY ONE RESISTOR

10. ECE 3183 – Chapter 2 – Part ACHAPTER 2 A-10 THE DIFFERENCE BETWEEN ELECTRIC CONNECTION AND PHYSICAL LAYOUT SOMETIMES, FOR PRACTICAL CONSTRUCTION REASONS, COMPONENTS THAT ARE ELECTRICALLY CONNECTED MAY BE PHYSICALLY FAR APART IN ALL CASES THE RESISTORS ARE CONNECTED IN SERIES

11. ECE 3183 – Chapter 2 – Part ACHAPTER 2 A-11 SUMMARY OF BASIC VOLTAGE DIVIDER V R1 = ? V R2 = ?

12. ECE 3183 – Chapter 2 – Part ACHAPTER 2 A-12 SUMMARY OF BASIC VOLTAGE DIVIDER VOLUME CONTROL?

13. ECE 3183 – Chapter 2 – Part ACHAPTER 2 A-13 SUMMARY OF BASIC VOLTAGE DIVIDER A “PRACTICAL” POWER APPLICATION HOW CAN ONE REDUCE THE LOSSES?

14. ECE 3183 – Chapter 2 – Part ACHAPTER 2 A-14 THE “INVERSE” VOLTAGE DIVIDER

15. ECE 3183 – Chapter 2 – Part ACHAPTER 2 A-15 THE “INVERSE” VOLTAGE DIVIDER

16. ECE 3183 – Chapter 2 – Part ACHAPTER 2 A-16

17. ECE 3183 – Chapter 2 – Part ACHAPTER 2 A-17 Current Division

18. ECE 3183 – Chapter 2 – Part ACHAPTER 2 A-18 FIND V x, i 3

19. ECE 3183 – Chapter 2 – Part ACHAPTER 2 A-19 Find i 1, i 2, and i 3

20. ECE 3183 – Chapter 2 – Part ACHAPTER 2 A-20 SERIES AND PARALLEL RESISTOR COMBINATIONS UP TO NOW WE HAVE STUDIED CIRCUITS THAT CAN BE ANALYZED WITH ONE APPLICATION OF KVL(SINGLE LOOP) OR KCL(SINGLE NODE-PAIR) WE HAVE ALSO SEEN THAT IN SOME SITUATIONS IT IS ADVANTAGEOUS TO COMBINE RESISTORS TO SIMPLIFY THE ANALYSIS OF A CIRCUIT NOW WE EXAMINE SOME MORE COMPLEX CIRCUITS WHERE WE CAN SIMPLIFY THE ANALYSIS USING THE TECHNIQUE OF COMBINING RESISTORS… … PLUS THE USE OF OHM’S LAW

21. ECE 3183 – Chapter 2 – Part ACHAPTER 2 A-21 SERIES AND PARALLEL RESISTOR COMBINATIONS SERIES COMBINATIONS PARALLEL COMBINATION

22. ECE 3183 – Chapter 2 – Part ACHAPTER 2 A-22

23. ECE 3183 – Chapter 2 – Part ACHAPTER 2 A-23

24. ECE 3183 – Chapter 2 – Part ACHAPTER 2 A-24 FIRST WE PRACTICE COMBINING RESISTORS FIND R AB

25. ECE 3183 – Chapter 2 – Part ACHAPTER 2 A-25 FIRST WE PRACTICE COMBINING RESISTORS 6k||3k (10K,2K)SERIES = 12K SERIES R AB = 5K  (12K||6K) = 4K (4K,2K)SERIES = 6K We need to re-draw!

26. ECE 3183 – Chapter 2 – Part ACHAPTER 2 A-26 EXAMPLES COMBINATION SERIES-PARALLEL AN EXAMPLE WITHOUT REDRAWING

27. ECE 3183 – Chapter 2 – Part ACHAPTER 2 A-27 EXAMPLES COMBINATION SERIES-PARALLEL RESISTORS ARE IN SERIES IF THEY CARRY EXACTLY THE SAME CURRENT (SHARE ONE COMMON NODE) RESISTORS ARE IN PARALLEL IF THEY HAVE THE SAME VOLTAGE ACROSS THEM AND ARE CONNECTED EXACTLY BETWEEN THE SAME TWO NODES

28. ECE 3183 – Chapter 2 – Part ACHAPTER 2 A-28 Strategy for analyzing circuits with series and parallel combinations of resistors: Systematically reduce the resistive network so that the resistance seen by the source is represented by a single resistor. Determine the source current for a voltage source or the source voltage for a current source. Expand the network, apply Ohm’s law, KVL, KCL, voltage division, and current division to determine all currents and voltages in the network

29. ECE 3183 – Chapter 2 – Part ACHAPTER 2 A-29

30. ECE 3183 – Chapter 2 – Part ACHAPTER 2 A-30 Circuit analysis example Find I o in the circuit shown.

31. ECE 3183 – Chapter 2 – Part ACHAPTER 2 A-31 THIS IS AN INVERSE PROBLEM WHAT CAN BE COMPUTED?

32. ECE 3183 – Chapter 2 – Part ACHAPTER 2 A-32 Circuits with dependent sources When writing the KVL and/or KCL equations for the network, treat the dependent source as though it were an independent source. Write the equations that specify the relationship of the dependent source to the controlling parameters. Solve the equations for the unknowns. Be sure that the number of linearly independent equations matches the number of the unknowns.

33. ECE 3183 – Chapter 2 – Part ACHAPTER 2 A-33 A PLAN: IF V_s IS KNOWN V_0 CAN BE DETERMINED USING VOLTAGE DIVIDER. TO FIND V_s WE HAVE A SINGLE NODE-PAIR CIRCUIT ALGEBRAICALLY, THERE ARE TWO UNKNOWNS AND JUST ONE EQUATION THE EQUATION FOR THE CONTROLLING VARIABLE PROVIDES THE ADDITIONAL EQUATION SUBSTITUTION OF I_0 YIELDSVOLTAGE DIVIDER KCL TO THIS NODE. THE DEPENDENT SOURCE IS JUST ANOTHER SOURCE

34. ECE 3183 – Chapter 2 – Part ACHAPTER 2 A-34 Problem solving strategy: Circuit with dependent sources For the network shown, what is the resulting ratio, V o /V s ?

35. ECE 3183 – Chapter 2 – Part ACHAPTER 2 A-35 Problem solving strategy: Circuit with dependent sources For the network shown, what is the resulting ratio, V o /V s ? (voltage divider for resistors in series)

36. ECE 3183 – Chapter 2 – Part ACHAPTER 2 A-36 Problem solving strategy: Circuit with dependent sources For the network shown, what is the resulting ratio, V o /V s ? (voltage divider for resistors in series) (voltage divider for resistors in series)

37. ECE 3183 – Chapter 2 – Part ACHAPTER 2 A-37 Problem solving strategy: Circuit with dependent sources For the network shown, what is the resulting ratio, V o /V s ? (voltage divider for resistors in series) (voltage divider for resistors in series)