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Networks I for M.E. ECE 09.201 - 2

Networks I for M.E. ECE 09.201 - 2. James K. Beard, Ph.D. Admin. 1 – week from tomorrow Test 1 Cruise course website Questions thus far?. Networks I. Today’s Learning Objectives – Analyze DC circuits with passive elements including: resistance -- DONE Learn about switches -- DONE

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Networks I for M.E. ECE 09.201 - 2

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  1. Networks I for M.E.ECE 09.201 - 2 James K. Beard, Ph.D.

  2. Admin • 1 – week from tomorrow Test 1 • Cruise course website • Questions thus far? Class Name

  3. Networks I • Today’s Learning Objectives – • Analyze DC circuits with passive elements including: resistance -- DONE • Learn about switches -- DONE • Introduce KVL and KCL • What is voltage and current division? • Parallel and series sources • Solve some more difficult problems Class Name

  4. chapter 2 - overview • engineering and linear models - done • active and passive circuit elements - done • resistors – Ohm’s Law - done • dependent sources – done • independent sources – done • transducers – done • switches – in progress Class Name

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

  6. Gustav Robert Kirchhoff • 1824-1887 • two profound scientific laws published in 1847 • how old was he? LC #1 Class Name

  7. Kirchhoff’s laws • Kirchhoff’s Current Law (KCL): • The algebraic sum of the currents into a node at any instant is zero. • Kirchhoff’s Voltage Law (KVL): • The algebraic sum of the voltages around any closed path in a circuit is zero for all time. Class Name

  8. R1=10 Node 1 Node 2 _ + + + R3= 5 I=5A R2= 20 _ _ Node 3 KCL Assume passive sign convention Class Name

  9. R1=10 i1 _ + v1=50v + + I I=5A R3= 5 i2 i3 R2= 20 _ _ 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 Class Name

  10. R1=10 i1 _ + v1=50v + + I I=5A R3= 5 i2 i3 R2= 20 _ _ Node 1 Node 2 v2 v3 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 Class Name

  11. Learning check #1 • what is relationship between v2 and v3 in previous example? • <, >, = Class Name

  12. 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) vR1 = iR1 = VR1 /(R1 + R2) vR2 = iR2 = VR2/(R1 + R2) Class Name

  13. 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) vR1 = iR1 = VR1 /(R1 + R2) vR2 = iR2 = VR2/(R1 + R2) VOLTAGE DIVIDER Class Name

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

  15. Learning check #2 • what is value of Req in previous example when the three resistors are replaced with the following 4 new resistors? • 1 k, 100, 10, and 1 Class Name

  16. R = 9 R = 1 PARALLEL RESISTORS • resistors attached in parallel can be simplified by adding their conductances (G) together to get an equivalent resistance (R=1/G). Geq = Gr1 + Gr2 + etc.. When you only have two: Req = (R1*R2)/(R1+R2) Class Name

  17. Learning checks #3 & #4 • 4. what is value of Req in previous example? • 5. what is the new value of Req when the two parallel resistors are replaced 2 new resistors shown below? • 10 and 40 Class Name

  18. 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 Class Name

  19. Learning check #5 • 6a. What is effective value of V for the series voltage sources in the example on board? • 6b. What is the power dissipated in the resistor of 30? Class Name

  20. Networks I • Today’s Learning Objectives – • Use KVL and KCL • What is voltage and current division? • Parallel and series sources Class Name

  21. 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 – in progress • parallel resistors and current division • series V-sources / parallel I-sources • resistive circuit analysis Class Name

  22. Kirchhoff’s laws • Kirchhoff’s Current Law (KCL): • The algebraic sum of the currents into a node at any instant is zero. • Kirchhoff’s Voltage Law (KVL): • The algebraic sum of the voltages around any closed path in a circuit is zero for all time. Class Name

  23. 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 Use KVL and Ohm’s Law VOLTAGE DIVIDER i = V/(R1 + R2) vR1 = iR1 = VR1 /(R1 + R2) vR2 = iR2 = VR2/(R1 + R2) Class Name

  24. 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) vR1 = iR1 = VR1 /(R1 + R2) vR2 = iR2 = VR2/(R1 + R2) VOLTAGE DIVIDER Class Name

  25. R1=10 i1 _ + v1=50v + + I I=5A R3= 5 i2 i3 R2= 20 _ _ KCL Node 1 Node 2 v2 v3 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 Class Name

  26. R = 9 R = 1 PARALLEL RESISTORS • resistors attached in parallel can be simplified by adding their conductances (G) together to get an equivalent resistance (R=1/G). Geq = Gr1 + Gr2 + etc.. When you only have two: Req = (R1*R2)/(R1+R2) Class Name

  27. Equivalent parallel resistors • Example 3 parallel resistors: 6, 9, 18  what is the equivalent resistance? Geq = Gr1 + Gr2 + etc.. • 1/6 + 1/9 + 1/18 = 6/18 = 1/3 • If Geq = 1/3 then R = ? Class Name

  28. Learning check #1 • What is effective resistance value of three parallel resistors with values of 4, 5, 20? • Hint: calculate Geq , then R Class Name

  29. 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 Class Name

  30. Learning check #2 • 2a. What is effective value of i for the example of parallel current sources on the board (5, 10, 7, 4)? • 2b. What is the power dissipated in the resistor of 6? Class Name

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

  32. 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. Class Name

  33. 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: Class Name

  34. 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: Class Name

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

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