1. Chapter 6 Series and Parallel Circuits 1 of 81 MECH1100 1 of 81 Chapter 6 Series and Parallel Circuits MECH1100 Topics Identifying Series- Parallel Relationships Analysis of Series-Parallel Resistive Circuits Voltage Dividers with Resistive Loads Loading Effect of a Voltmeter Wheatstone Bridge Thevenin’s Theorem Maximum Power Transfer Superposition

2. Chapter 6 Series and Parallel Circuits 2 of 81 MECH1100 Most practical circuits have combinations of series and parallel components. Identifying series-parallel relationships Components that are connected in series will share a common path. Components that are connected in parallel will be connected across the same two nodes. 1 2 From Chapters 4 and 5

3. Chapter 6 Series and Parallel Circuits 3 of 81 MECH1100 You can frequently simplify analysis by combining series and parallel components. Combination circuits Solve by forming the simplest equivalent circuit possible. 1.has characteristics that are electrically the same as another circuit 2.is generally simpler. Circuits containing both series and parallel circuits are called COMBINATION circuits An equivalent circuit is one that :

4. Chapter 6 Series and Parallel Circuits 4 of 81 MECH1100 Equivalent circuits is equivalent to There are no electrical measurements that can distinguish between the boxes.

5. Chapter 6 Series and Parallel Circuits 5 of 81 MECH1100 Another example: There are no electrical measurements that can distinguish between the boxes. is equivalent to Equivalent circuits

6. Chapter 6 Series and Parallel Circuits 6 of 81 MECH1100 is equivalent to There are no electrical measurements that can distinguish between the three boxes. Equivalent circuits

7. Chapter 6 Series and Parallel Circuits 7 of 81 MECH1100 What Do We Know 1.For Series Circuits: a.Current at all points is the I T = I R1 = I R2 b.Voltage across each resistor V T = V 1 + V 2 2.For Parallel Circuit a.Current I T = I R1 + I R2 b.Voltage at all nodes is the VT = V1 = V2 same drops divides across each resistor in the branch

8. Chapter 6 Series and Parallel Circuits 8 of 81 MECH1100 Seven Step Process for Solving a Combination Circuit 1.Simplify the circuit to a series circuit by finding the effective equivalent resistance (R EQ ) of each parallel section in the circuit. Redraw the circuit each time it is simplified. 2.Calculate the total resistance (R T ) of the circuit by adding all R EQ ’s to the other series resistances. 3.Calculate the total current (I T ) using R T in Ohm’s law. 4.Calculate the voltage drop across any series resistances or R EQ ’s using Ohm’s law. 5.Calculate the branch currents in all parallel sections of the circuit using the voltage drop across R EQ and Ohm’s law. 6.Use the branch currents and resistance values to calculate the voltage of the parallel resistances. 7.Make a summary of the voltage drops and currents for each resistance to make sure they total correctly.

9. Chapter 6 Series and Parallel Circuits 9 of 81 MECH1100 A simple series-parallel resistive circuit.

10. Chapter 6 Series and Parallel Circuits 10 of 81 MECH1100 R 4 is added to the circuit in series with R 1.

11. Chapter 6 Series and Parallel Circuits 11 of 81 MECH1100 R 5 is added to the circuit in series with R 2.

12. Chapter 6 Series and Parallel Circuits 12 of 81 MECH1100 R 6 is added to the circuit in parallel with the series combination of R 1 and R 4.

13. Chapter 6 Series and Parallel Circuits 13 of 81 MECH1100 Sketch the circuits for nodes A to B A to C B to C

14. Chapter 6 Series and Parallel Circuits 14 of 81 MECH1100 Sketch the circuit and write the equation for nodes A to B

15. Chapter 6 Series and Parallel Circuits 15 of 81 MECH1100 Sketch the circuit and write the equation for nodes A to C

16. Chapter 6 Series and Parallel Circuits 16 of 81 MECH1100 Sketch the circuit and write the equation for nodes B to C

17. Chapter 6 Series and Parallel Circuits 17 of 81 MECH1100 Draw the schematic for this circuit board and give the equation for the relationship of the resistor branches

18. Chapter 6 Series and Parallel Circuits 18 of 81 MECH1100

19. Chapter 6 Series and Parallel Circuits 19 of 81 MECH1100 Reduce the following circuit to R EQ

20. Chapter 6 Series and Parallel Circuits 20 of 81 MECH1100 6 Ω

21. Chapter 6 Series and Parallel Circuits 21 of 81 MECH1100 10 Ω 6 4

22. Chapter 6 Series and Parallel Circuits 22 of 81 MECH1100 5 Ω 10

23. Chapter 6 Series and Parallel Circuits 23 of 81 MECH1100 50 Ω I T =2 amps ITIT 5 15 30

24. Chapter 6 Series and Parallel Circuits 24 of 81 MECH1100 Review of voltage relationships

25. Chapter 6 Series and Parallel Circuits 25 of 81 MECH1100 Kirchoff’s current law What are the readings for node A?

26. Chapter 6 Series and Parallel Circuits 26 of 81 MECH1100 Loaded voltage divider 1.A voltage-divider with a resistive load forms a combination (parallel) circuit. The voltage-divider equation was developed for a series circuit. Recall that the output voltage is given by A + 2.The voltage divider is said to be LOADED. 3.The loading reduces the total resistance from node A to ground.

27. Chapter 6 Series and Parallel Circuits 27 of 81 MECH1100 Loaded voltage divider A What is the voltage across R 3 ? Form an equivalent series circuit by combining R 2 and R 3 ; then apply the voltage- divider formula to the equivalent circuit: 8.10 V

28. Chapter 6 Series and Parallel Circuits 28 of 81 MECH1100 Stiff voltage divider A stiff voltage-divider is one in which the loaded voltage is nearly the same as the no-load voltage. To accomplish this, the load current must be small compared to the bleeder current (or R L is large compared to the divider resistors).

29. Chapter 6 Series and Parallel Circuits 29 of 81 MECH1100 Stiff Voltage Divider

30. Chapter 6 Series and Parallel Circuits 30 of 81 MECH1100 Stiff voltage divider Given: R 1 = R 2 = 1.0 k , 1.What value of R L will make the divider a stiff voltage divider? 2.What fraction of the unloaded voltage is the loaded voltage? Rule of Thumb: RL = R2*10; Thus: R L should be 10 k  or greater. This is 95% of the unloaded voltage.

31. Chapter 6 Series and Parallel Circuits 31 of 81 MECH1100 Currents in a two-tap loaded voltage divider. I Bleeder = I T - I RL1 – I RL2

32. Chapter 6 Series and Parallel Circuits 32 of 81 MECH1100 The loading effect of a voltmeter

33. Chapter 6 Series and Parallel Circuits 33 of 81 MECH1100 Loading effect of a voltmeter 1.A voltmeter has internal resistance 2. Thus R INT can change the resistance of the circuit under test. 3.A 1 M  internal resistance of the meter accounts for the readings. Given: V S = 10 V and R 1 and R 2 are not defective but the meter reads only 4.04 V when it is across either R 1 or R 2. 10 V What is a possible explanation of the meter not displaying 10 volts? 4.04 V

34. Chapter 6 Series and Parallel Circuits 34 of 81 MECH1100 Wheatstone bridge Voltage Divider 1 Voltage Divider 2

35. Chapter 6 Series and Parallel Circuits 35 of 81 MECH1100 Wheatstone Bridge V 1 =V 2 I 1 =I 3 V 3 =V 4 I 2 =I 4

36. Chapter 6 Series and Parallel Circuits 36 of 81 MECH1100 Wheatstone Bridge The Wheatstone bridge consists of: 1. a dc voltage source and 2.four resistive arms forming two voltage dividers. The output is taken between the dividers. Frequently, one of the bridge resistors is adjustable. (R 2 ). 4.04 V

37. Chapter 6 Series and Parallel Circuits 37 of 81 MECH1100 Balanced Wheatstone Bridge When the bridge is balanced, the output voltage is zero The products of resistances in the opposite diagonal arms are equal.

38. Chapter 6 Series and Parallel Circuits 38 of 81 MECH1100 Wheatstone bridge What is the value of R 2 if the bridge is balanced? 470  330  270  12 V 385 

39. Chapter 6 Series and Parallel Circuits 39 of 81 MECH1100 Finding an Unknown Resistance using a Galvanometer Scale Factor

40. Chapter 6 Series and Parallel Circuits 40 of 81 MECH1100 Measuring a physical parameter using a transducer. Unbalanced Wheatstone Bridge Unbalance occurs when V OUT ≠ 0 Used to measure 1.Mechanical Strain 2.Temperature 3.Pressure V OUT is converted to a digital output indicating the value of the reading.

41. Chapter 6 Series and Parallel Circuits 41 of 81 MECH1100 VOUT = VA-VB = 0

42. Chapter 6 Series and Parallel Circuits 42 of 81 MECH1100 10 Using the Thermistor chart, what is V OUT when the temperature is 50 o C V A =8.8v V B =6.0vV A-B =2.8v

43. Chapter 6 Series and Parallel Circuits 43 of 81 MECH1100 Example of a load cell

44. Chapter 6 Series and Parallel Circuits 44 of 81 MECH1100 Load Cell or Strain Gauge

45. Chapter 6 Series and Parallel Circuits 45 of 81 MECH1100 Typical load cell (strain gauge) schematic

46. Chapter 6 Series and Parallel Circuits 46 of 81 MECH1100 Typical load cell (strain gauge) schematic

47. Chapter 6 Series and Parallel Circuits 47 of 81 MECH1100 Typical load cell (strain gauge) schematic

48. Chapter 6 Series and Parallel Circuits 48 of 81 MECH1100

49. Chapter 6 Series and Parallel Circuits 49 of 81 MECH1100 Thevenin’s theorem Any two-terminal, resistive circuit can be replaced with a simple equivalent circuit when viewed from two output terminals. Equivalent Voltage (V TH ) is the “No Load” voltage between the two output terminals. Equivalent Resistance (R TH ) is the total resistance appearing between two output terminals Terminal equivalency

50. Chapter 6 Series and Parallel Circuits 50 of 81 MECH1100

51. Chapter 6 Series and Parallel Circuits 51 of 81 MECH1100 Simplification of a circuit by Thevenin’s theorem.

52. Chapter 6 Series and Parallel Circuits 52 of 81 MECH1100 Thevenin’s theorem Output terminals What is the Thevenin voltage for the circuit?8.76 V Remember, the load resistor has no affect on the Thevenin parameters.

53. Chapter 6 Series and Parallel Circuits 53 of 81 MECH1100 Thevenin’s theorem Output terminals What is the Thevenin resistance for the circuit? 7.30 k  Remember, the load resistor has no affect on the Thevenin parameters. SHORTED

54. Chapter 6 Series and Parallel Circuits 54 of 81 MECH1100 Thevenin’s equivalent depends on the output terminals from which the circuit is viewed.

55. Chapter 6 Series and Parallel Circuits 55 of 81 MECH1100 Thevenin’s equivalent depends on the output terminals from which the circuit is viewed.

56. Chapter 6 Series and Parallel Circuits 56 of 81 MECH1100 Thevenin’s theorem and Wheatstone Bridge Thevenin’s theorem is useful for solving the Wheatstone bridge. Thevenize the bridge into two Thevenin circuits 1.A to ground 2.B to ground. Resistance between point A and ground is R 1 ||R 3 Resistance between B to ground is R 2 ||R 4. The voltage on each side of the bridge is found using the voltage divider rule.

57. Chapter 6 Series and Parallel Circuits 57 of 81 MECH1100 Simplifying a Wheatstone bridge with Thevenin’s theorem.

58. Chapter 6 Series and Parallel Circuits 58 of 81 MECH1100 1.Remove RL to make an open circuit between A & B 2.Calculate R 1 ||R 3 = 3.Calculate R 2 ||R 4 = 4.Calculate the voltage from A to ground 5.Calculate the voltage from B to ground 6.Draw the Thevenin circuits for each of the bridge 165  179  7.5 V 6.88 V Thevenin’s theorem and Wheatstone Bridge

59. Chapter 6 Series and Parallel Circuits 59 of 81 MECH1100 Thevenin’s theorem and Wheatstone Bridge Thevenin’s Equivalent Circuit for a Wheatstone Bridge

60. Chapter 6 Series and Parallel Circuits 60 of 81 MECH1100 The maximum power is transferred from a source to a load when: R L = R S(resistance of the voltage source) The maximum power transfer theorem assumes the source voltage and resistance are fixed. Maximum power transfer Theorem

61. Chapter 6 Series and Parallel Circuits 61 of 81 MECH1100 Maximum power transfer Theorem What is the power delivered to the load? The voltage to the load is The power delivered is: 5.0 V

62. Chapter 6 Series and Parallel Circuits 62 of 81 MECH1100 Superposition theorem A way to determine currents and voltages in a linear circuit that has multiple sources 1.Take one source at a time and 2.Algebraically sum the results.

63. Chapter 6 Series and Parallel Circuits 63 of 81 MECH1100 Superposition theorem Four Step Process 1.Leave one voltage or current source at a time in the circuit (circuit 1) and replace the other (circuit 2) with a short. 2.Calculate R EQ/Total and then calculate the voltage or current for the resistor(s). 3.Repeat steps 1 and 2 for the other circuit (circuit 2). 4.Algebraically add the results for all sources.

64. Chapter 6 Series and Parallel Circuits 64 of 81 MECH1100

65. Chapter 6 Series and Parallel Circuits 65 of 81 MECH1100 6.10 k  What does the ammeter read for I 2 ? 1.97 mA0.985 mA 8.73 k  2.06 mA0.58 mA 1.56 mA Source 1: R T(S1) = I 1 = I 2 = Source 2: R T(S2) = I 3 = I 2 = Both sources I 2 = Set up a table of pertinent information and solve for each quantity listed: The total current is the algebraic sum. 1.56 mA

66. Chapter 6 Series and Parallel Circuits 66 of 81 MECH1100 Putting the load on the Thevenin circuits and applying the superposition theorem allows you to calculate the load current. The load current is: 1.27 mA Thevenin’s theorem and Wheatstone Bridge.0152-.01391

67. Chapter 6 Series and Parallel Circuits 67 of 81 MECH1100 67 of 81 Chapter 6 Series and Parallel Circuits MECH1100

68. Chapter 6 Series and Parallel Circuits 68 of 81 MECH1100 Troubleshooting The effective troubleshooter must think logically about circuit operation. Understand normal circuit operation and find out the symptoms of the failure. Decide on a logical set of steps to find the fault. Following the steps in the plan, make measurements to isolate the problem. Modify the plan if necessary.

69. Chapter 6 Series and Parallel Circuits 69 of 81 MECH1100 Troubleshooting The output of the voltage- divider is 6.0 V. Describe how you would use analysis and planning in finding the fault. From an earlier calculation, V 3 should equal 8.10 V. A low voltage is most likely caused by a low source voltage or incorrect resistors (possibly R 1 and R 2 reversed). If the circuit is new, incorrect components are possible. Decide on a logical set of steps to locate the fault. You could decide to 1) check the source voltage, 2) disconnect the load and check the output voltage, and if it is correct, 3) check the load resistance. If R 3 is correct, check other resistors. A

70. Chapter 6 Series and Parallel Circuits 70 of 81 MECH1100 Loading Load current Bleeder current Wheatstone bridge The current left after the load current is subtracted from the total current into the circuit. The output current supplied to a load. The effect on a circuit when an element that draws current from the circuit is connected across the output terminals. A 4-legged type of bridge circuit with which an unknown resistance can be accurately measured using the balanced state. Deviations in resistance can be measured using the unbalanced state. Key Terms

71. Chapter 6 Series and Parallel Circuits 71 of 81 MECH1100 Thevenin’s theorem Maximum power transfer Superposition The condition, when the load resistance equals the source resistance, under which maximum power is transferred to the load. A method for analyzing circuits with two or more sources by examining the effects of each source by itself and then combining the effects. A circuit theorem that provides for reducing any two-terminal resistive circuit to a single equivalent voltage source in series with an equivalent resistance. Key Terms

72. Chapter 6 Series and Parallel Circuits 72 of 81 MECH1100 1. Two circuits that are equivalent have the same a. number of components b. response to an electrical stimulus c. internal power dissipation d. all of the above Quiz

73. Chapter 6 Series and Parallel Circuits 73 of 81 MECH1100 2. If a series equivalent circuit is drawn for a complex circuit, the equivalent circuit can be analyzed with a. the voltage divider theorem b. Kirchhoff’s voltage law c. both of the above d. none of the above Quiz

74. Chapter 6 Series and Parallel Circuits 74 of 81 MECH1100 3. For the circuit shown, a. R 1 is in series with R 2 ∥R 3 b. R 1 is in parallel with R 2 c. R 2 is in series with R 3 d. R 2 is in parallel with R 1 + R 3 Quiz

75. Chapter 6 Series and Parallel Circuits 75 of 81 MECH1100 4. For the circuit shown, a. R 1 is in series with R 2 b. R 4 is in parallel with R 1 c. R 2 is in parallel with R 3 d. none of the above Quiz

76. Chapter 6 Series and Parallel Circuits 76 of 81 MECH1100 5. A signal generator has an output voltage of 2.0 V with no load. When a 600  load is connected to it, the output drops to 1.0 V. The Thevenin resistance of the generator is a. 300  b. 600  c. 900  d. 1200 . Quiz

77. Chapter 6 Series and Parallel Circuits 77 of 81 MECH1100 6. For the circuit shown, Kirchhoff's voltage law a. applies only to the outside loop b. applies only to the A junction. c. can be applied to any closed path. d. does not apply. Quiz

78. Chapter 6 Series and Parallel Circuits 78 of 81 MECH1100 7. The effect of changing a measured quantity due to connecting an instrument to a circuit is called a. loading b. clipping c. distortion d. loss of precision Quiz

79. Chapter 6 Series and Parallel Circuits 79 of 81 MECH1100 8. An unbalanced Wheatstone bridge has the voltages shown. The voltage across R 4 is a. 4.0 V b. 5.0 V c. 6.0 V d. 7.0 V VR 2 =5V VR 2 – VR L =4V VR 3 = 8V Quiz

80. Chapter 6 Series and Parallel Circuits 80 of 81 MECH1100 9. Assume R 2 is adjusted until the Wheatstone bridge is balanced. At this point, the voltage across R 4 is measured and found to be 5.0 V. The voltage across R 1 will be a. 4.0 V b. 5.0 V c. 6.0 V d. 7.0 V Quiz

81. Chapter 6 Series and Parallel Circuits 81 of 81 MECH1100 10. Maximum power is transferred from a fixed source when a. the load resistor is ½ the source resistance b. the load resistor is equal to the source resistance c. the load resistor is twice the source resistance d. none of the above Quiz

82. Chapter 6 Series and Parallel Circuits 82 of 81 MECH1100 Quiz Answers: 1. b 2. c 3. d 4. d 5. b 6. c 7. a 8. a 9. d 10. b