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CHAPTER 3

CHAPTER 3. NETWORKS 1: 0909201-01 23 September 2002 – Lecture 3a ROWAN UNIVERSITY College of Engineering Professor Peter Mark Jansson, PP PE DEPARTMENT OF ELECTRICAL & COMPUTER ENGINEERING Autumn Semester 2002. networks I. Announcements – Homework 2 answers posted today

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CHAPTER 3

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  1. CHAPTER 3 NETWORKS 1: 0909201-01 23 September 2002 – Lecture 3a ROWAN UNIVERSITY College of Engineering Professor Peter Mark Jansson, PP PE DEPARTMENT OF ELECTRICAL & COMPUTER ENGINEERING Autumn Semester 2002

  2. networks I • Announcements – • Homework 2 answers posted today • Returned after test Tuesday • First Test Tomorrow  Ch. 3: 24 Sep • Lab 1 assignment is due • Sec 1: TODAY - 23 Sep • Sec 2: TOMORROW - 24 Sep

  3. networks I • Today’s Learning Objectives – • Define voltage/current divider circuits • Analyze series V-sources • Analyze parallel current sources • Reduce resistive circuits • Analyze DC circuits with passive and active elements including: resistance and power sources

  4. chapter 3 - overview • electric circuit applications - DONE • define: node, closed path, loop - DONE • Kirchoff’s Current Law - DONE • Kirchoff’s Voltage Law - DONE • a voltage divider circuit • parallel resistors and current division • series V-sources / parallel I-sources • resistive circuit analysis reduction

  5. R1=10 _ + + V= 5v + LOOP 1 _ _ R2= 20 Start +V - vR1 - vR2 = 0 iV = iR1 = iR2 = i +V = iR1 + iR2 V = i(R1 + R2) KVL i = V/(R1 + R2) vR! = iR1 = VR1 /(R1 + R2) vR2 = iR2 = VR2/(R1 + R2)

  6. R1=10 _ + + V= 5v + LOOP 1 _ _ R2= 20 Start +V - vR1 - vR2 = 0 iV = iR1 = iR2 = i +V = iR1 + iR2 V = i(R1 + R2) SERIES RESISTORS NOTE i = V/(R1 + R2) vR! = iR1 = VR1 /(R1 + R2) vR2 = iR2 = VR2/(R1 + R2) VOLTAGE DIVIDER

  7. R = 2 R = 3 R = 9 R = 4 SERIES RESISTORS • resistors attached in a “string” can be added together to get an equivalent resistance.

  8. R1=10 i1 _ + v1=50v + + I I=5A v2=20v R3= 5 i2 i3 R2= 20 v3=20v _ _ Node 1 Node 2 Node 3 Node 1 +I - i1 = 0 Node 2 +i1 - i2 - i3 = 0 Node 3 +i2 + i3 - I = 0 i2 = v2/R2 i3 = v3/R3 Use KCL and Ohm’s Law CURRENT DIVIDER

  9. series voltage sources • when connected in series, a group of voltage sources can be treated as one voltage source whose equivalent voltage =  all source voltages • unequal voltage sources are not to be connected in parallel

  10. parallel current sources • when connected in parallel, a group of current sources can be treated as one current source whose equivalent current =  all source currents • unequal current sources are not to be connected in series

  11. loop2 loop1 PROBLEM SOLVING METHOD va vb _ _ node3 node1 node2 + + Rb Ra ib ia + ivs + vc + ic vis vs Rc is _ _ _ node4

  12. steps taken • Apply P.S.C. to passive elements. • Show current direction at voltages sources. • Show voltage direction at current sources. • Name nodes and loops. • Name elements and sources. • Name currents and voltages.

  13. va vb _ _ node3 node1 node2 + + Rb Ra ib ia + ivs + vc + ic vis vs Rc is _ loop2 loop1 _ _ node4 WRITE THE KCL EQUATIONS node1: node3: node2: node4:

  14. va vb _ _ node3 node1 node2 + + Rb Ra ib ia + ivs + vc + ic vis vs Rc is _ loop2 loop1 _ _ node4 WRITE THE KVL EQUATIONS loop1: loop2:

  15. va vb _ _ node3 node1 node2 + + Rb Ra ib ia + ivs + vc + ic vis vs Rc is _ loop2 loop1 _ _ node4 WRITE SUPPLEMENTARYEQUATIONS

  16. 10 30 45 iT i1 + 15 90 50 _ 5v 5 100 CIRCUIT REDUCTION

  17. 10 30 45 iT i1 + 15 90 50 _ 5v 5 100 • Begin with loop on far right. • Combine the three resistors that are in series. • Req = 45+50+100 = 195

  18. 10 30 iT i1 + 15 90 195 _ 5v 5 • Again using the loop on the far right. • The 90  and 195  resistors are in parallel. • Req= (90)(195)/(90+195) = 61.58 

  19. 10 30 iT + 15 61.58 _ 5v 5 • Still working with the loop on the far right. • The 30  and the 61.58  resistors are in series. • Req = 30 + 61.58 = 91.58 

  20. 10 iT + 15 91.58 _ 5v 5 • Again, the far right loop. • The 15  and 91.58  resistors are in parallel. • Req = (15)(91.58)/(15+91.58) = 12.9 

  21. 10 iT + 12.9 _ 5v 5 • Now there is only one loop. • All the resistors are in series. • Req = 10+12.9+5 = 27.9 

  22. 10 0.179A iT + + 12.9 27.9 _ 5v _ 5v 5 a • Use Ohm’s Law to determine iT. • iT = 5/27.9 = 0.179A • iT flows in all three resistors, the 12.9  resistor is the equivalent resistance of the entire circuit beyond points a and b. b

  23. 10 0.179A + 15 91.58 _ 5v 5 a ix • iT divides at a to flow through the 15  and the 91.58  resistors (the 91.58  is an equivalent resistance for the rest of the circuit). • Use current divider: ix = (0.179)(15)/(15+91.58) = 0.0252A.

  24. 10 30 0.179A + 15 61.58 _ 5v 5 a 0.0252A • No calculations are required at this step because the 0.0252A is flowing through both resistors in the right loop. • This circuit must be drawn however, because the 61.58  resistor is an equivalent for the circuit to the right of a and b. b

  25. 10 30 0.179A i1 + 15 90 195 _ 5v 5 a 0.0252A • Use the current divider equation again to determine i1. • i1 = (0.0252)(195)/(90+195) = 0.01724A = 17.24mA. • The current through the 195  resistor is 0.0252 - 0.01724 = 7.96mA b

  26. One Minute Paper • please complete handout • no names • leave in box on leaving • thanks

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