1. Rowan Hall 238A beard@rowan.edu http://rowan.jkbeard.com September 18, 2006 Networks I for M.E. ECE 09.201 - 2 James K. Beard, Ph.D.

2. Slide 2 Class Name September 18, 2006 Admin 1 – week from tomorrow Test 1 Cruise course website Questions thus far?

3. Slide 3 Class Name September 18, 2006 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

4. Slide 4 Class Name September 18, 2006 chapter 2 - overview engineering and linear models - done engineering and linear models - done active and passive circuit elements - done active and passive circuit elements - done resistors – Ohm’s Law - done resistors – Ohm’s Law - done done dependent sources – done done independent sources – done transducers – done switches – in progress

5. Slide 5 Class Name September 18, 2006 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

6. Slide 6 Class Name September 18, 2006 Gustav Robert Kirchhoff 1824-1887 two profound scientific laws published in 1847 how old was he? LC #1

7. Slide 7 Class Name September 18, 2006 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.

8. Slide 8 Class Name September 18, 2006 KCL Assume passive sign convention R 2 = 20  I=5A R 1 =10  R 3 = 5  + _ + _ + _ Node 1Node 2 Node 3

9. Slide 9 Class Name September 18, 2006 R 2 = 20  I=5A R 1 =10  R 3 = 5  + _ + _ v 1 =50v+ _ i1i1 i2i2 i3i3 I Node 1 +I - i 1 = 0 Node 2 +i 1 - i 2 - i 3 = 0 Node 3 +i 2 + i 3 - I = 0 i 2 = v 2 /R 2 i 3 = v 3 /R 3 Node 1Node 2 Node 3 Use KCL and Ohm’s Law

10. Slide 10 Class Name September 18, 2006 R 2 = 20  I=5A R 1 =10  R 3 = 5  + _ + _ v 1 =50v+ _ i1i1 i2i2 i3i3 I Node 1 +I - i 1 = 0 Node 2 +i 1 - i 2 - i 3 = 0 Node 3 +i 2 + i 3 - I = 0 i 2 = v 2 /R 2 i 3 = v 3 /R 3 Node 1Node 2 Node 3 Use KCL and Ohm’s Law CURRENT DIVIDER v2v2 v3v3

11. Slide 11 Class Name September 18, 2006 Learning check #1 what is relationship between v 2 and v 3 in previous example?, =

12. Slide 12 Class Name September 18, 2006 KVL +V - v R1 - v R2 = 0 i V = i R1 = i R2 = i +V = iR 1 + iR 2 V = i(R 1 + R 2 ) R 2 = 20  V= 5v R 1 =10  + _ + _ LOOP 1 + _ Start i = V/(R 1 + R 2 ) v R1 = iR 1 = VR 1 /(R 1 + R 2 ) v R2 = iR 2 = VR 2 /(R 1 + R 2 )

13. Slide 13 Class Name September 18, 2006 SERIES RESISTORS +V - v R1 - v R2 = 0 i V = i R1 = i R2 = i +V = iR 1 + iR 2 V = i(R 1 + R 2 ) R 2 = 20  V= 5v R 1 =10  + _ + _ LOOP 1 + _ Start i = V/(R 1 + R 2 ) v R1 = iR 1 = VR 1 /(R 1 + R 2 ) v R2 = iR 2 = VR 2 /(R 1 + R 2 ) VOLTAGE DIVIDER NOTE

14. Slide 14 Class Name September 18, 2006 SERIES RESISTORS resistors attached in a “string” can be added together to get an equivalent resistance. R = 2  R = 3  R = 4  R = 9 

15. Slide 15 Class Name September 18, 2006 Learning check #2 what is value of R eq in previous example when the three resistors are replaced with the following 4 new resistors? 1 k , 100 , 10 , and 1 

16. Slide 16 Class Name September 18, 2006 PARALLEL RESISTORS resistors attached in parallel can be simplified by adding their conductances (G) together to get an equivalent resistance (R=1/G). R = 9  R = 1  G eq = G r1 + G r2 + etc.. When you only have two: R eq = (R 1 *R 2 )/(R 1 +R 2 )

17. Slide 17 Class Name September 18, 2006 Learning checks #3 & #4 4. what is value of R eq in previous example? 5. what is the new value of R eq when the two parallel resistors are replaced 2 new resistors shown below? 10  and 40 

18. Slide 18 Class Name September 18, 2006 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

19. Slide 19 Class Name September 18, 2006 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 resistorof 30  ?

20. Slide 20 Class Name September 18, 2006 Networks I Today’s Learning Objectives –  Use KVL and KCL  What is voltage and current division?  Parallel and series sources

21. Slide 21 Class Name September 18, 2006 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

22. Slide 22 Class Name September 18, 2006 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.

23. Slide 23 Class Name September 18, 2006 KVL +V - v R1 - v R2 = 0 i V = i R1 = i R2 = i +V = iR 1 + iR 2 V = i(R 1 + R 2 ) R 2 = 20  V= 5v R 1 =10  + _ + _ LOOP 1 + _ Start i = V/(R 1 + R 2 ) v R1 = iR 1 = VR 1 /(R 1 + R 2 ) v R2 = iR 2 = VR 2 /(R 1 + R 2 ) Use KVL and Ohm’s Law VOLTAGE DIVIDER

24. Slide 24 Class Name September 18, 2006 SERIES RESISTORS +V - v R1 - v R2 = 0 i V = i R1 = i R2 = i +V = iR 1 + iR 2 V = i(R 1 + R 2 ) R 2 = 20  V= 5v R 1 =10  + _ + _ LOOP 1 + _ Start i = V/(R 1 + R 2 ) v R1 = iR 1 = VR 1 /(R 1 + R 2 ) v R2 = iR 2 = VR 2 /(R 1 + R 2 ) VOLTAGE DIVIDER NOTE

25. Slide 25 Class Name September 18, 2006 R 2 = 20  I=5A R 1 =10  R 3 = 5  + _ + _ v 1 =50v+ _ i1i1 i2i2 i3i3 I Node 1 +I - i 1 = 0 Node 2 +i 1 - i 2 - i 3 = 0 Node 3 +i 2 + i 3 - I = 0 i 2 = v 2 /R 2 i 3 = v 3 /R 3 Node 1Node 2 Node 3 Use KCL and Ohm’s Law CURRENT DIVIDER v2v2 v3v3 KCL

26. Slide 26 Class Name September 18, 2006 PARALLEL RESISTORS resistors attached in parallel can be simplified by adding their conductances (G) together to get an equivalent resistance (R=1/G). R = 9  R = 1  G eq = G r1 + G r2 + etc.. When you only have two: R eq = (R 1 *R 2 )/(R 1 +R 2 )

27. Slide 27 Class Name September 18, 2006 Equivalent parallel resistors Example 3 parallel resistors: 6 , 9 , 18   what is the equivalent resistance? G eq = Gr1 + Gr2 + etc.. 1/6 + 1/9 + 1/18 = 6/18 = 1/3 If G eq = 1/3 then R = ?

28. Slide 28 Class Name September 18, 2006 Learning check #1 What is effective resistance value of three parallel resistors with values of 4 , 5 , 20  ? Hint: calculate G eq, then R

29. Slide 29 Class Name September 18, 2006 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

30. Slide 30 Class Name September 18, 2006 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 resistorof 6  ?

31. Slide 31 Class Name September 18, 2006 PROBLEM SOLVING METHOD + _ + _ + + _ + _ _ node1node2 node3 node4 RaRa RbRb RcRc vsvs isis iaia ibib icic vava vbvb vcvc i vs v is loop1loop2

32. Slide 32 Class Name September 18, 2006 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.

33. Slide 33 Class Name September 18, 2006 WRITE THE KCL EQUATIONS node1: node2:node4: node3: + _ + _ + + _ + _ _ node1node2 node3 node4 RaRa RbRb RcRc vsvs isis iaia ibib icic vava vbvb vcvc i vs v is loop1 loop2

34. Slide 34 Class Name September 18, 2006 WRITE THE KVL EQUATIONS + _ + _ + + _ + _ _ node1node2 node3 node4 RaRa RbRb RcRc vsvs isis iaia ibib icic vava vbvb vcvc i vs v is loop1 loop2 loop1:loop2:

35. Slide 35 Class Name September 18, 2006 WRITE SUPPLEMENTARY EQUATIONS + _ + _ + + _ + _ _ node1node2 node3 node4 RaRa RbRb RcRc vsvs isis iaia ibib icic vava vbvb vcvc i vs v is loop1 loop2