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Unit Nine:Superposition Theorem
John Elberfeld
WWW.J-Elberfeld.com
ET115 DC Electronics
ScheduleSchedule
Unit Unit Topic Topic Chpt LabsChpt Labs1.1. Quantities, Units, SafetyQuantities, Units, Safety 11 2 (13)2 (13)2.2. Voltage, Current, ResistanceVoltage, Current, Resistance 22 3 + 163 + 163.3. Ohm’s LawOhm’s Law 33 5 (35)5 (35)4.4. Energy and PowerEnergy and Power 33 6 (41)6 (41)
5.5. Series CircuitsSeries Circuits Exam IExam I 44 7 (49)7 (49)
6.6. Parallel CircuitsParallel Circuits 55 9 (65)9 (65)
7.7. Series-Parallel CircuitsSeries-Parallel Circuits 66 10 (75)10 (75)
8.8. Thevenin’s, Power Thevenin’s, Power Exam 2Exam 2 66 19 (133)19 (133)
9.9. Superposition Theorem Superposition Theorem 66 11 (81)11 (81)
10.10. Magnetism & Magnetic DevicesMagnetism & Magnetic Devices 77 Lab Final Lab Final 11.11. Course Review and Course Review and Final ExamFinal Exam
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Unit 9 Objectives - IUnit 9 Objectives - I
• State the superposition theorem.State the superposition theorem.• List the steps required to apply the List the steps required to apply the
superposition theorem.superposition theorem.• Calculate the current and voltage in a Calculate the current and voltage in a
given resistor by applying the given resistor by applying the superposition theorem.superposition theorem.
• Calculate the effect of a load on a bipolar Calculate the effect of a load on a bipolar voltage divided by applying the voltage divided by applying the superposition theorem.superposition theorem.
• Construct basic DC circuits on a Construct basic DC circuits on a protoboard.protoboard.
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Unit 9 Objectives – IIUnit 9 Objectives – II
• Use a digital multimeter (DMM) to measure Use a digital multimeter (DMM) to measure a predetermined low voltage on a power a predetermined low voltage on a power supply.supply.
• Measure resistances and voltages in a DC Measure resistances and voltages in a DC circuit using a DMM.circuit using a DMM.
• Apply Ohm’s Law, Thevenin’s theorem, Apply Ohm’s Law, Thevenin’s theorem, KVL and KCL to practical circuits.KVL and KCL to practical circuits.
• Troubleshoot circuits constructed in Troubleshoot circuits constructed in Multisim exercises using simulated Multisim exercises using simulated instrumentsinstruments
Reading AssignmentReading Assignment
• Read and study Read and study
• Chapter 6: Superposition Theorem:Chapter 6: Superposition Theorem:Pages 247-256 (Last section of Pages 247-256 (Last section of chapter 6)chapter 6)
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Lab AssignmentLab Assignment
• Experiment 11, “Superposition Experiment 11, “Superposition Theorem,” beginning on page 81 of Theorem,” beginning on page 81 of DC Electronics: Lab Manual and DC Electronics: Lab Manual and MultiSim Guide.MultiSim Guide.
• Complete all measurements, graphs, Complete all measurements, graphs, and questions and turn in your lab and questions and turn in your lab before leaving the roombefore leaving the room
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Written AssignmentsWritten Assignments
• Complete the Unit 9 Homework sheetComplete the Unit 9 Homework sheet
• Show all your work!Show all your work!
• Be prepared for a quiz on questions Be prepared for a quiz on questions similar to those on the homework.similar to those on the homework.
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Superposition TheoremSuperposition Theorem
• The current through or voltage drop The current through or voltage drop across any component in a linear across any component in a linear circuit with multiple current or circuit with multiple current or voltage sources is equal to the voltage sources is equal to the algebraic sum of the currents or algebraic sum of the currents or voltages produce by each source voltages produce by each source considered independentlyconsidered independently
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Internal ResistancesInternal Resistances
• Actual batteries (Voltage sources) Actual batteries (Voltage sources) and current sources have an internal and current sources have an internal resistance that affects the circuit, in resistance that affects the circuit, in addition to voltage or current that addition to voltage or current that they add to the circuitthey add to the circuit
• Usually, we assume in internal Usually, we assume in internal resistance of a voltage source = 0, resistance of a voltage source = 0, and the internal resistance of a and the internal resistance of a current source is infinite (open)current source is infinite (open)
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Getting EquationsGetting Equations
• Replace all but one of the voltage Replace all but one of the voltage and current sources with internal and current sources with internal resistancesresistances
• Find the current and voltages for the Find the current and voltages for the circuit for the one remaining sourcecircuit for the one remaining source
• Repeat this process for each sourceRepeat this process for each source
• Find total voltages and currents by Find total voltages and currents by adding affects of each sourceadding affects of each source
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ExampleExample
• Familiar setup of three resistors and Familiar setup of three resistors and two power suppliestwo power supplies
• Note the orientation of the power Note the orientation of the power supplies!supplies!
VA2= 40V
R3 =50Ω
R2=50Ω
VA1=25V
R1=75Ω Vx
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ExampleExample
• Replace VReplace VA2A2 with its internal resistance (0 with its internal resistance (0ΩΩ))
• RR2||32||3 = 50 = 50ΩΩ5050ΩΩ/(50/(50ΩΩ+50+50ΩΩ) = 25 ) = 25 ΩΩ
• RR12||312||3 = 75 = 75 ΩΩ + 25 + 25 ΩΩ = 100 = 100ΩΩ
• IITT = 25V/100 = 25V/100 ΩΩ = 250ma = I = 250ma = IR1R1
• VVR1R1 = 250ma 75 = 250ma 75 ΩΩ = 18.75V = 18.75V
• VVR2||3R2||3=25V- 18.75V=6.25V=25V- 18.75V=6.25V
• IIR2R2=125ma=125ma
• IIR3R3=125ma=125ma VA2= 40V
R3 =50Ω
R2=50ΩVA1=
25V
R1=75Ω
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ExampleExample
• Replace VReplace VA1A1 with its internal resistance (0 with its internal resistance (0ΩΩ))
• RR1||21||2 = 75 = 75ΩΩ5050ΩΩ/(75/(75ΩΩ+50+50ΩΩ) = 30 ) = 30 ΩΩ
• RR1||231||23 = 50 = 50 ΩΩ + 30 + 30 ΩΩ = 80 = 80ΩΩ
• IITT = 40V/80 = 40V/80 ΩΩ = 500ma = I = 500ma = IR3R3
• VVR3R3 = 500ma 50 = 500ma 50 ΩΩ = 25V = 25V
• VVR1||2R1||2=40V- 25V=15V=40V- 25V=15V
• IIR1R1=200ma=200ma
• IIR2R2=300ma=300ma VA2= 40V
R3 =50Ω
R2=50Ω
VA1=25V
R1=75Ω
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Put It All TogetherPut It All Together
VA2= 40V
R3 =50Ω
R2=50Ω
VA1=25V
R1=75Ω
250ma, 18.75V500ma, 25V
125ma, 6.25V
200ma, 15V
125ma, 6.25V
300ma, 15V
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Put It All TogetherPut It All Together
VA2= 40V
R3 =50Ω
R2=50Ω
VA1=25V
R1=75Ω
50ma, 3.75V 375ma, 18.75V
425ma, 21.25V
KCL and KVLKCL and KVL• Check: 425 mA = 50 mA + 375 A – OKCheck: 425 mA = 50 mA + 375 A – OK• 25 V = 21.25 V + 3.75 V – OK25 V = 21.25 V + 3.75 V – OK• 25 V = 3.75 V – 18.75 V + 40 V – OK25 V = 3.75 V – 18.75 V + 40 V – OK• 40 V = 21.25 V + 18.75 V - OK40 V = 21.25 V + 18.75 V - OK
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VA2= 40V
R3 =50Ω
R2=50Ω
VA1=25V
R1=75Ω
50ma, 3.75V 375ma, 18.75V
425ma, 21.25V
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Superposition Theorem - AgainSuperposition Theorem - Again
• This theorem states that if a linear This theorem states that if a linear network contains several independent network contains several independent energy sources, the total response is the energy sources, the total response is the sum of all the responses, if each source sum of all the responses, if each source acted separately and all the other acted separately and all the other independent sources were replaced by independent sources were replaced by their internal resistances.their internal resistances.
• This theorem is used to determine This theorem is used to determine current in a network with multiple current in a network with multiple sources.sources.
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Another ExampleAnother Example
• In this example, the VIn this example, the VA2 A2 is flipped is flipped
compared to previous examples!compared to previous examples!
VA2= 10V
R3 =20Ω
R2=40Ω
VA1=20V
R1=10Ω
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ExampleExample
• Replace VReplace VA2A2 with its internal resistance (0 with its internal resistance (0ΩΩ))
• RR2||32||3 = 20 = 20ΩΩ4040ΩΩ/(20/(20ΩΩ+40+40ΩΩ) = 13.33 ) = 13.33 ΩΩ
• RR12||312||3 = 10 = 10 ΩΩ + 13.33 + 13.33 ΩΩ = 23.33 = 23.33ΩΩ
• IITT = 20V/23.33 = 20V/23.33 ΩΩ = 857ma = I = 857ma = IR1R1
• VVR1R1 = 857ma 10 = 857ma 10 ΩΩ = 8.57V = 8.57V
• VVR2||3R2||3=20V- 8.57V=11.43V=20V- 8.57V=11.43V
• IIR2R2=572ma=572ma
• IIR3R3=285ma=285ma
R3 =20Ω
R2=40Ω
VA1=20V
R1=10Ω
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ExampleExample
• Replace VReplace VA1A1 with its internal resistance (0 with its internal resistance (0ΩΩ))
• RR1||21||2 = 10 = 10ΩΩ4040ΩΩ/(10/(10ΩΩ+40+40ΩΩ) = 8 ) = 8 ΩΩ
• RR1||231||23 = 20 = 20 ΩΩ + 8 + 8 ΩΩ = 28 = 28ΩΩ
• IITT = 10V/28 = 10V/28 ΩΩ = 357ma = I = 357ma = IR3R3
• VVR3R3 = 357ma 20 = 357ma 20 ΩΩ = 7.14V = 7.14V
• VVR1||2R1||2=10V- 7.14V=2.86V=10V- 7.14V=2.86V
• IIR1R1=286ma=286ma
• IIR2R2=71ma=71ma VA2= 10V
R3 =20Ω
R2=40Ω
R1=10Ω
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Put It All TogetherPut It All Together
857ma, 8.57V357ma, 7.14V
285ma, 11.43V
286ma, 2.86V
572ma, 11.43V
71ma, 2.86V
VA2= 10V
R3 =20Ω
R2=40Ω
VA1=20V
R1=10Ω
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KCL and KVLKCL and KVL
1143ma, 11.43V
214ma, 8.57V
929ma, 18.57V
VA2= 10V
R3 =20Ω
R2=40Ω
VA1=20V
R1=10Ω
• 214 mA + 929 mA = 1143 mA – OK214 mA + 929 mA = 1143 mA – OK• 20 V = 11.43 V + 8.75 V – OK20 V = 11.43 V + 8.75 V – OK• 20 V = 11.43 V + 18.57 V – 10 V - OK20 V = 11.43 V + 18.57 V – 10 V - OK• 10 V = 18,57 V – 8.57 V - OK10 V = 18,57 V – 8.57 V - OK
On your own…On your own…23
VA2= 10V
R3 =2 kΩ
R2=2 kΩ
VA1=20V
R1=4 kΩ
Short VShort V11
• RR11||R||R22 = 4k = 4kΩΩ 2k 2kΩΩ/(4k/(4kΩΩ + 2k + 2kΩΩ) = 1.33k) = 1.33kΩΩ
• RRTT = R = R11||R||R22 + R + R33 = 2k = 2kΩΩ + 1.33k + 1.33kΩΩ = 3.33k = 3.33kΩΩ
• IITT = 10V/ 3.33k = 10V/ 3.33kΩΩ = 3 mA = 3 mA
• VVR3R3 = I = ITTRR33 = 3 mA 2k = 3 mA 2kΩΩ = 6 V = 6 V
• V V R1||R2 R1||R2 = 10 V – 6 V = 4V= 10 V – 6 V = 4V
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VA2= 10V
R3 =2 kΩ
R2=2 kΩ
R1=4 kΩ
Short VShort V11
• VVR1R1 = 4 V = 4 V IIR1R1 = 4 V/ 4 k = 4 V/ 4 kΩΩ = 1 mA = 1 mA
• VVR2R2 = 4 V = 4 V IIR2R2 = 4 V/ 2 k = 4 V/ 2 kΩΩ = 2 mA = 2 mA
• VVR3R3 = 6 V = 6 V I IR3R3 = 3 mA = 3 mA
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VA2= 10V
R3 =2 kΩ
R2=2 kΩ
R1=4 kΩ
Short VShort V22
• RR22||R||R33 = 2k = 2kΩΩ 2k 2kΩΩ/(2k/(2kΩΩ + 2k + 2kΩΩ) = 1 k) = 1 kΩΩ
• RRTT = R = R22||R||R33 + R + R22 = 2k = 2kΩΩ + 4 k + 4 kΩΩ = 5 k = 5 kΩΩ
• IITT = 20V/ 5 k = 20V/ 5 kΩΩ = 4 mA = 4 mA
• VVR1R1 = I = ITTRR1 1 = 4 mA 4 k= 4 mA 4 kΩΩ = 16 V = 16 V
• V V R1||R2 R1||R2 = 20 V – 26 V = 4V= 20 V – 26 V = 4V
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VA1= 20V
R3 =2 kΩ
R2=2 kΩ
R1=4 kΩ
Short VShort V11
• VVR1R1 = 4 V = 4 V IIR1R1 = 4 V/ 2 k = 4 V/ 2 kΩΩ = 2 mA = 2 mA
• VVR2R2 = 4 V = 4 V IIR2R2 = 4 V/ 2 k = 4 V/ 2 kΩΩ = 2 mA = 2 mA
• VVR1R1 = 16 V = 16 V IIR1R1 = 4 mA = 4 mA
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VA2= 10V
R3 =2 kΩ
R2=2 kΩ
R1=4 kΩ
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Put It All TogetherPut It All Together
4 mA, 16 V3 ma, 6 V
2 mA, 4 V
1 mA, 4 V
2 mA, 4 V
2 mA, 4 V
VA2= 10V
R3 =2kΩ
R2=2kΩ
VA1=20V
R1=4kΩ
• 4 mA = 3 mA + 1 mA4 mA = 3 mA + 1 mA• 20 V = 12 V + 8 V20 V = 12 V + 8 V• 20 V = 12 V – 2 V + 10 V20 V = 12 V – 2 V + 10 V• 10 V = 8 V + 2 V10 V = 8 V + 2 V
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KCL and KVLKCL and KVL
3 mA, 12 V
4 mA, 8 V
1 mA, 2 V
VA2= 10V
R3 =2 kΩ
R2=2 kΩ
VA1=20V
R1=4 kΩ
SummarySummary
• Steps for the superposition theoremSteps for the superposition theorem
• Calculating multisource resistive Calculating multisource resistive circuits using superpositioncircuits using superposition
• MultisimMultisim
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