1. PHY 202 (Blum)1 Combinations of Resistors Series, Parallel and Kirchhoff

2. PHY 202 (Blum)2 Simplifying circuits using series and parallel equivalent resistances

3. PHY 202 (Blum)3 Analyzing a combination of resistors circuit Look for resistors which are in series (the current passing through one must pass through the other) and replace them with the equivalent resistance (R eq = R 1 + R 2 ). Look for resistors which are in parallel (both the tops and bottoms are connected by wire and only wire) and replace them with the equivalent resistance (1/R eq = 1/R 1 + 1/R 2 ). Repeat as much as possible.

4. PHY 202 (Blum)4 Look for series combinations R eq =3k  R eq =3.6 k 

5. PHY 202 (Blum)5 Look for parallel combinations R eq = 1.8947 k  R eq = 1.1244 k 

6. PHY 202 (Blum)6 Look for series combinations R eq = 6.0191 k 

7. PHY 202 (Blum)7 Look for parallel combinations R eq = 2.1314 k 

8. PHY 202 (Blum)8 Look for series combinations R eq = 5.1314 k 

9. PHY 202 (Blum)9 Equivalent Resistance I = V/R = (5 V)/(5.1314 k  ) = 0.9744 mA

10. PHY 202 (Blum)10 Backwards 1 V= (3)(.9744) = 2.9232 V= (2.1314)(.9744) = 2.0768

11. PHY 202 (Blum)11 Backwards 2 V = 2.0768=I (3.3) I=0.629mA V = 2.0768=I (6.0191) I=0.345mA

12. PHY 202 (Blum)12 Backwards 3 V=(.345)(1.1244)=0.388 V=(.345)(1.8947)=0.654 V=(.345)(3)=1.035

13. PHY 202 (Blum)13 Kirchhoff’s Rules When series and parallel combinations aren’t enough

14. 14PHY 202 (Blum) Some circuits have resistors which are neither in series nor parallel They can still be analyzed, but one uses Kirchhoff’s rules.

15. 15PHY 202 (Blum) Not in series The 1-k  resistor is not in series with the 2.2-k  since the some of the current that went through the 1-k  might go through the 3-k  instead of the 2.2-k  resistor.

16. 16PHY 202 (Blum) Not in parallel The 1-k  resistor is not in parallel with the 1.5-k  since their bottoms are not connected simply by wire, instead that 3-k  lies in between.

17. 17PHY 202 (Blum) Kirchhoff’s Node Rule A node is a point at which wires meet. “What goes in, must come out.” Recall currents have directions, some currents will point into the node, some away from it. The sum of the current(s) coming into a node must equal the sum of the current(s) leaving that node. I 1 + I 2 = I 3 I 1   I 2  I 3 The node rule is about currents!

18. 18PHY 202 (Blum) Kirchhoff’s Loop Rule 1 “If you go around in a circle, you get back to where you started.” If you trace through a circuit keeping track of the voltage level, it must return to its original value when you complete the circuit Sum of voltage gains = Sum of voltage losses

19. 19PHY 202 (Blum) Batteries (Gain or Loss) Whether a battery is a gain or a loss depends on the direction in which you are tracing through the circuit Gain Loss Loop direction

20. 20PHY 202 (Blum) Resistors (Gain or Loss) Whether a resistor is a gain or a loss depends on whether the trace direction and the current direction coincide or not. II LossGain Loop direction Current direction

21. PHY 202 (Blum)21 Branch version

22. 22PHY 202 (Blum) Neither Series Nor Parallel I 1  I 2.2  I 1.5  I 1.7  I 3  Assign current variables to each branch. Draw loops such that each current element is included in at least one loop.

23. 23PHY 202 (Blum) Apply Current (Node) Rule I 1  I 1 -I 3  I 1.5  I 1.5 +I 3  I 3  *Node rule applied. **

24. 24PHY 202 (Blum) Three Loops Voltage Gains = Voltage Losses 5 = 1 I 1 + 2.2 (I 1 – I 3 ) 1 I 1 + 3 I 3 = 1.5 I 1.5 2.2 (I 1 – I 3 ) = 3 I 3 + 1.7 (I 1.5 + I 3 ) Units: Voltages are in V, currents in mA, resistances in k 

25. 25PHY 202 (Blum) 5 = 1 I 1 + 2.2 (I 1 – I 3 ) I 1  I 1 -I 3  I 1.5  I 1.5 +I 3  I 3 

26. 26PHY 202 (Blum) 1 I 1 + 3 I 3 = 1.5 I 1.5 I 1  I 1 -I 3  I 1.5  I 1.5 +I 3  I 3 

27. 27PHY 202 (Blum) 2.2 (I 1 – I 3 ) = 3 I 3 + 1.7 (I 1.5 + I 3 ) I 1  I 1 -I 3  I 1.5  I 1.5 +I 3  I 3 

28. 28PHY 202 (Blum) Simplified Equations 5 = 3.2 I 1 - 2.2 I 3 I 1 = 1.5 I 1.5 - 3 I 3 0 = -2.2 I 1 + 1.7 I 1.5 + 6.9 I 3 Substitute middle equation into others 5 = 3.2 (1.5 I 1.5 - 3 I 3 ) - 2.2 I 3 0 = -2.2 (1.5 I 1.5 - 3 I 3 ) + 1.7 I 1.5 + 6.9 I 3 Multiply out parentheses and combine like terms.

29. 29PHY 202 (Blum) Solving for I 3 5 = 4.8 I 1.5 - 11.8 I 3 0 = - 1.6 I 1.5 + 13.5 I 3 Solve the second equation for I 1.5 and substitute that result into the first 5 = 4.8 (8.4375 I 3 ) - 11.8 I 3 5 = 28.7 I 3 I 3  0.174 mA

30. 30PHY 202 (Blum) Comparison with Simulation

31. 31PHY 202 (Blum) Other currents Return to substitution results to find other currents. I 1.5 = 8.4375 I 3 = 1.468 mA I 1 = 1.5 I 1.5 - 3 I 3 I 1 = 1.5 (1.468) - 3 (0.174) I 1 = 1.68 mA

32. PHY 202 (Blum)32 Loop version

33. 33PHY 202 (Blum) Neither Series Nor Parallel Draw loops such that each current element is included in at least one loop. Assign current variables to each loop. Current direction and lop direction are the same. JAJA JBJB JCJC

34. 34PHY 202 (Blum) Loop equations 5 = 1  (J A - J B ) + 2.2  (J A - J C ) 0 = 1  (J B - J A ) + 1.5  J B + 3  (J B - J C ) 0 = 2.2  (J C - J A ) + 3  (J C - J B ) + 1.7 J C “Distribute” the parentheses 5 = 3.2J A – 1 J B - 2.2J C 0 = -1J A + 5.5 J B – 3J C 0 = -2.2J A – 3 J B + 6.9J C

35. 35PHY 202 (Blum) Algebra J C = (2.2/6.9)J A + (3/6.9)J B J C = 0.3188 J A + 0.4348 J B 5 = 3.2 J A – 1 J B - 2.2 (0.3188 J A + 0.4348 J B ) 0 = -1 J A + 5.5 J B – 3 (0.3188 J A + 0.4348 J B ) 5 = 2.4986 J A – 1.9566 J B 0 = -1.9564 J A + 4.1956 J B

36. 36PHY 202 (Blum) More algebra J B = (1.9564/4.1956) J A J B = 0.4663 J A 5 = 2.4986 J A – 1.9566 (0.4663 J A ) 5 = 1.5862 J A J A = 3.1522 mA

37. 37PHY 202 (Blum) Other loop currents J B = 0.4663 J A = 0.4663 (3.1522 mA) J B = 1.4699 mA J C = 0.3188 J A + 0.4348 J B J C = 0.3188 (3.1522) + 0.4348 (1.4699) J C = 1.644 mA

38. 38PHY 202 (Blum) Branch Variables I 1  I 2.2  I 1.5  I 1.7  I 3  Assign current variables to each branch. Draw loops such that each current element is included in at least one loop.

39. 39PHY 202 (Blum) Loop Variables Draw loops such that each current element is included in at least one loop. Assign current variables to each loop. Current direction and lop direction are the same. JAJA JBJB JCJC

40. 40PHY 202 (Blum) Branch Currents from Loop currents I 1 = J A – J B = 3.1522 – 1.4699 = 1.6823 mA I 1.5 = JB = 1.4699 mA

41. PHY 202 (Blum)41 Matrix equation

42. 42PHY 202 (Blum) Loop equations as matrix equation 5 = 3.2J A – 1 J B - 2.2J C 0 = -1J A + 5.5 J B – 3J C 0 = -2.2J A – 3 J B + 6.9J C

43. 43PHY 202 (Blum) Enter matrix in Excel, highlight a region the same size as the matrix.

44. 44PHY 202 (Blum) In the formula bar, enter =MINVERSE(range) where range is the set of cells corresponding to the matrix (e.g. B1:D3). Then hit Crtl+Shift+Enter

45. 45PHY 202 (Blum) Result of matrix inversion

46. 46PHY 202 (Blum) Prepare the “voltage vector”, then highlight a range the same size as the vector and enter =MMULT(range1,range2) where range1 is the inverse matrix and range2 is the voltage vector. Then Ctrl-Shift-Enter. Voltage vector

47. 47PHY 202 (Blum) Results of Matrix Multiplication