1 / 35

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

jerry-massey
Download Presentation

Networks I for M.E. ECE 09.201 - 2

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. 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

More Related