1 / 37

ECE 3183 – EE Systems

ECE 3183 – EE Systems. Chapter 2 – Part A Parallel, Series and General Resistive Circuits. Chapter 2 Resistive Circuits. Solve circuits (i.e., find currents and voltages of interest) by combining resistances in series and parallel. Apply the voltage-division and current-division principles.

kirk-armstrong
Download Presentation

ECE 3183 – EE Systems

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

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

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

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

  4. FIRST GENERALIZATION: MULTIPLE SOURCES veq = ? ECE 3183 – Chapter 2 – Part A

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

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

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

  8. ECE 3183 – Chapter 2 – Part A

  9. R 1 + R R v v 1 2 S S R 2 v = i S + R R 1 2 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 i i + + - - AS FAR AS THE CURRENT IS CONCERNED BOTH CIRCUITS ARE EQUIVALENT. THE ONE ON THE RIGHT HAS ONLY ONE RESISTOR ECE 3183 – Chapter 2 – Part 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 ECE 3183 – Chapter 2 – Part A

  11. SUMMARY OF BASIC VOLTAGE DIVIDER VR1 = ? VR2 = ? ECE 3183 – Chapter 2 – Part A

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

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

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

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

  16. ECE 3183 – Chapter 2 – Part A

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

  18. FIND Vx, i3 ECE 3183 – Chapter 2 – Part A

  19. Find i1, i2, and i3 ECE 3183 – Chapter 2 – Part 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 ECE 3183 – Chapter 2 – Part A

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

  22. ECE 3183 – Chapter 2 – Part A

  23. ECE 3183 – Chapter 2 – Part A

  24. FIRST WE PRACTICE COMBINING RESISTORS FIND RAB ECE 3183 – Chapter 2 – Part A

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

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

  29. ECE 3183 – Chapter 2 – Part A

  30. Circuit analysis example Find Io in the circuit shown. ECE 3183 – Chapter 2 – Part A

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

  33. THE EQUATION FOR THE CONTROLLING VARIABLE PROVIDES THE ADDITIONAL EQUATION ALGEBRAICALLY, THERE ARE TWO UNKNOWNS AND JUST ONE EQUATION SUBSTITUTION OF I_0 YIELDS VOLTAGE DIVIDER KCL TO THIS NODE. THE DEPENDENT SOURCE IS JUST ANOTHER SOURCE 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 ECE 3183 – Chapter 2 – Part A

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

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

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

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

More Related