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Unit Seven Series-Parallel Circuits

John Elberfeld

JElberfeld@itt-tech.edu

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 7 Objectives - IUnit 7 Objectives - I

• Identify series-parallel relationships in a Identify series-parallel relationships in a combinational resistive circuit.combinational resistive circuit.

• Determine total resistance, current, and Determine total resistance, current, and power in a series-parallel resistivecircuit.power in a series-parallel resistivecircuit.

• Draw the schematic for the Wheatstone Draw the schematic for the Wheatstone bridge.bridge.

• Explain how to determine if the Explain how to determine if the Wheatstone bridge is balanced.Wheatstone bridge is balanced.

• Determine the effect of one or more Determine the effect of one or more resistive loads on a voltage dividerresistive loads on a voltage divider

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Unit 7 Objectives – IIUnit 7 Objectives – II

• Determine the bleeder current in a resistive Determine the bleeder current in a resistive voltage divider.voltage divider.

• Determine the loading effect of a voltmeter Determine the loading effect of a voltmeter on a circuit given the internal resistance of on a circuit given the internal resistance of the voltmeter.the voltmeter.

• Define bipolar voltage dividerDefine bipolar voltage divider• Use proper prototype board wiring and test Use proper prototype board wiring and test

procedures for DC resistive circuit procedures for DC resistive circuit components including using the digital components including using the digital multimeter.multimeter...

Unit 7 Objectives – IIIUnit 7 Objectives – III

• Construct basic DC circuits on a Construct basic DC circuits on a protoboard.protoboard.

• 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.

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Reading AssignmentReading Assignment

• Read and study Read and study

• Chapter 6: Series-Parallel Circuits:Chapter 6: Series-Parallel Circuits:Pages 212-237 (First half of chapter)Pages 212-237 (First half of chapter)

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Lab AssignmentLab Assignment

• Experiment 10, “Series-Parallel Experiment 10, “Series-Parallel Combination Circuits,” beginning on Combination Circuits,” beginning on page 75 of page 75 of DC Electronics: Lab DC Electronics: Lab Manual and MultiSim Guide.Manual and 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 7 Homework sheetComplete the Unit 7 Homework sheet

• Show all your work!Show all your work!

• Be prepared for a your second Be prepared for a your second MAJOR EXAM on questions similar MAJOR EXAM on questions similar to those on the homework.to those on the homework.

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Ohms LawOhms Law

• MEMORIZE: V = I RMEMORIZE: V = I R• Ohms LawOhms Law• If you increase the voltage, you If you increase the voltage, you

increase the current proportionallyincrease the current proportionally– 3 times the voltage gives you three 3 times the voltage gives you three

times the currenttimes the current– Resistance (ohms) is the proportionality Resistance (ohms) is the proportionality

constant and depends on the atomic constant and depends on the atomic structure of the material conducting the structure of the material conducting the currentcurrent

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Power FormulaPower Formula

• Power = Work / timePower = Work / time• P = V IP = V I• Voltage is the work done per Voltage is the work done per

coulomb, and current is the number coulomb, and current is the number of coulombs per second passing by of coulombs per second passing by a point.a point.

• The product of voltage and current The product of voltage and current gives the work done per second, or gives the work done per second, or power.power.

11Series-Parallel Combination of Series-Parallel Combination of

ResistancesResistances

• This type of circuit consists of a This type of circuit consists of a combination of resistances in series combination of resistances in series and parallel. and parallel.

• Series-parallel combination of Series-parallel combination of resistances can be classified as:resistances can be classified as:– Simple series-parallel circuitsSimple series-parallel circuits– Complex series-parallel circuitsComplex series-parallel circuits

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Parallel ResistorsParallel Resistors

• For resistors to be in parallel, both For resistors to be in parallel, both ends of each resistor must be in ends of each resistor must be in direct electrical contact with each direct electrical contact with each other – with no resistance between other – with no resistance between the endsthe ends

• ParallelParallel Not Parallel (blue)Not Parallel (blue)

R1||R2

R1 R2

13Series-Parallel Combination of Series-Parallel Combination of

ResistancesResistances

• To differentiate between a simple series-To differentiate between a simple series-parallel circuit and a complex circuit:parallel circuit and a complex circuit:– observe simple series and parallel observe simple series and parallel

combinations of resistances in the given combinations of resistances in the given circuit.circuit.

– replace the simple series and simple parallel replace the simple series and simple parallel combinations by their single equivalent combinations by their single equivalent resistor.resistor.

– repeat the simplification process to verify repeat the simplification process to verify whether the circuit is reduced to a single whether the circuit is reduced to a single resistance circuit.resistance circuit.

– if a single resistor can replace ALL the if a single resistor can replace ALL the resistors, it is a simple series-parallel circuitresistors, it is a simple series-parallel circuit

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Recognizing Series ComponentsRecognizing Series Components

RRTT= R= R11+R+R22+R+R33||R||R44||(R||(R55+R+R66))

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AnalysisAnalysis

• Analysis of the circuit also Analysis of the circuit also requires one to recognize the requires one to recognize the various paths for current flow.various paths for current flow.

• The ability to recognize the The ability to recognize the points where current branches points where current branches out and where current converges out and where current converges (sums) is vital.(sums) is vital.

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EXAMPLEEXAMPLE

Electron Flow →

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Series-Parallel CircuitsSeries-Parallel Circuits

• Pure series-parallel circuits can be Pure series-parallel circuits can be reduced to a single equivalent resistor:reduced to a single equivalent resistor:

• ExampleExample

RRTT= R= R11+R+R22||R||R33

RR22||R||R33

RR11

RR11

RR22 RR33

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ExampleExample

• Another simple series-parallel circuitAnother simple series-parallel circuit

RRTT= R= R11+R+R22||R||R33+RR44||R||R55

RR22||R||R33

RR11 RR11RR22

RR33

RR44

RR55

RR44||R||R55

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RRTT=R=R22||[||[RR11+R+R33||(R||(R44+R+R55)])]RR11

RR22

RR33

RR44

RR55

RR11

RR22

RR33 RR44+RR55

RR11

RR22

RR33||(R4+R5)

RR22

RR11+R+R33||(R4+R5) RRTT

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Complex CircuitComplex Circuit

• Unable to simplify to a single resistorUnable to simplify to a single resistor– Where the battery connects to the Where the battery connects to the

circuit is important!!!!circuit is important!!!!

• Complex Complex SimpleSimple

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Complex CircuitComplex Circuit

• Unable to simplify to a single resistorUnable to simplify to a single resistor

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PracticePractice

• Classify this circuit as series-parallel Classify this circuit as series-parallel or complexor complex

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PracticePractice

• Classify this circuit as simple series-Classify this circuit as simple series-parallel or complexparallel or complex

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• Classify this circuit as series-parallel or Classify this circuit as series-parallel or complexcomplex

• It is the electrical connections, not the It is the electrical connections, not the drawing, that determines the type of drawing, that determines the type of circuitcircuit

PracticePractice

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PracticePractice

• Classify this circuit as series-parallel Classify this circuit as series-parallel or complexor complex

26Voltage Drop in Series-Parallel Voltage Drop in Series-Parallel

Resistive CircuitResistive Circuit

• To find the equivalent resistance of a To find the equivalent resistance of a series-parallel resistive circuit: series-parallel resistive circuit:– Identify the resistance combinations as either Identify the resistance combinations as either

series or parallel.series or parallel.– Simplify, combine and recombine series and Simplify, combine and recombine series and

parallel combinationsparallel combinations– Find the value of the total resistance of the Find the value of the total resistance of the

circuit.circuit.

• Once you know the equivalent resistance, Once you know the equivalent resistance, you can calculate voltage or currentyou can calculate voltage or current

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Total ResistanceTotal Resistance

• One of the common approaches One of the common approaches is called Outside Toward the is called Outside Toward the Source approach.Source approach.

• To implement this method, one To implement this method, one begins farthest from the source begins farthest from the source and works back toward the and works back toward the source.source.

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Equivalent ResistanceEquivalent Resistance

• The analysis of the circuit uses The analysis of the circuit uses equivalent resistance as circuit equivalent resistance as circuit reductions are performed.reductions are performed.

• For instance, if a 6-kFor instance, if a 6-k and a 3-k and a 3-k resistor are in parallel, their resistor are in parallel, their equivalent equivalent seriesseries resistance is 2 resistance is 2 kk..

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Current In a Series-Parallel CircuitCurrent In a Series-Parallel Circuit

• The analysis of current in this The analysis of current in this circuit is a fundamental step.circuit is a fundamental step.

• Kirchhoff’s Current Law must be Kirchhoff’s Current Law must be followed to see how current followed to see how current divides and sums together.divides and sums together.

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Circuit Analysis ToolsCircuit Analysis Tools

• The circuit tools used to The circuit tools used to determine circuit parameters determine circuit parameters have already been covered. have already been covered. They include:They include:-Ohm’s law -Power equations -Ohm’s law -Power equations

-KCL and KVL -Voltage divider -KCL and KVL -Voltage divider rulerule

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Voltage In a Series-parallel CircuitVoltage In a Series-parallel Circuit

• Voltage distribution throughout Voltage distribution throughout the circuit follows the laws the circuit follows the laws appropriate to series and parallel appropriate to series and parallel connections.connections.

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Circuit AnalysisCircuit Analysis

• Combine the two parallel resistors to Combine the two parallel resistors to one equivalent resistorone equivalent resistor

• 1/R1/R2323 = 1/20 = 1/20ΩΩ + + 1/301/30ΩΩ = = 3/603/60ΩΩ + + 2/602/60ΩΩ

• 1/R1/R2323 = 5/60 = 5/60ΩΩ

• RR2323= 12 = 12 ΩΩ

V = 25V

R3 = 30Ω

R2 = 20Ω

R1 = 10Ω

R23 = 12Ω

R1 = 10Ω V = 25V

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Circuit AnalysisCircuit Analysis

• Combine the two parallel resistors to Combine the two parallel resistors to one equivalent resistor using POSone equivalent resistor using POS

• RR2323 = 20 = 20ΩΩxx3030ΩΩ / ( / (2020ΩΩ++3030ΩΩ) = 12 ) = 12 ΩΩ

– Same results!Same results!

V = 25V

R3 = 30Ω

R2 = 20Ω

R1 = 10Ω

R23 = 12Ω

R1 = 10Ω V = 25V

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AnalysisAnalysis

• Combine the series resistorsCombine the series resistors

• RR123123 = 10 = 10 ΩΩ + + 12 12 ΩΩ = 22 = 22 ΩΩ

V = 25V

R1 = 10Ω

R23 = 12Ω

R123 = 22Ω

V = 25V

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AnalysisAnalysis

• V = 25 V, RV = 25 V, RTT = 22 = 22ΩΩ, I, ITT = 1.14 A = 1.14 A

• VVR1R1 = 11.4V, V = 11.4V, VR2R2 = V = VR3R3 = 25V-11.4V=13.6V = 25V-11.4V=13.6V

• I I R2R2= V= VR2R2/20/20 ΩΩ = .68A = .68A

• I I R3R3= V= VR3R3/30/30 ΩΩ = .45A = .45A

• Check: ICheck: ITT = I = IR1R1 = I = IR2R2 + I + IR3R3 = 1.14 A = 1.14 A

• VVTT = V = VR1R1+V+VR2R2 = 25V = 25V

V = 25V

R2 = 20Ω

R1 = 10Ω

R3 = 30Ω

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Current in Parallel Resistive CircuitCurrent in Parallel Resistive Circuit

• In the circuit shown, the current flowing In the circuit shown, the current flowing through Rthrough R11, R, R22, R, R33, and R, and R44 has two separate has two separate

paths called branches.paths called branches.• Applying KCL, the current, I, is equal to Applying KCL, the current, I, is equal to

the sum of the branch currents, Ithe sum of the branch currents, I11 and I and I22..

←Conventional Flow

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Class ActivityClass Activity

• What is the equivalent resistance of What is the equivalent resistance of the following circuit?the following circuit?

V = 30V

R2 = 10kΩ

R1 = 5.6kΩ

R3 = 4.7kΩ

R4 = 2.2kΩ

R5 = 3.3kΩ

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DescriptionDescription

• What is the equivalent resistance of What is the equivalent resistance of the following circuit?the following circuit?

V = 30V

R2 = 10kΩ

R1 = 5.6kΩ

R3 = 4.7kΩ

R4 = 2.2kΩ

R5 = 3.3kΩ

Class ActivityClass Activity

• RR22 and R and R33 are…. are….

– In series with each other AND the series In series with each other AND the series combination is in parallel with Rcombination is in parallel with R55

• RR11 and R and R44 are in series with the are in series with the

combination of Rcombination of R55 in parallel with R in parallel with R22

and Rand R33

• RRTT= R= R11+R+R44+R+R55||(R||(R22+R+R33))

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V = 30V R2 =

10kΩ

R1 = 5.6kΩ

R3 = 4.7kΩ

R5 = 3.3kΩ

R4 = 2.2kΩ

Predict Total ResistancePredict Total Resistance

• The total resistance must be The total resistance must be between:between:

• 5.6k5.6k Ω Ω + 2.2 k + 2.2 kΩΩ = 7.8k = 7.8kΩΩ Minimum Minimum

• 5.6k5.6k Ω Ω + 2.2 k + 2.2 kΩΩ + 3.3 k + 3.3 kΩΩ = 11.1 k = 11.1 kΩΩ MaximumMaximum

• WHY???WHY???

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Current AnalysisCurrent Analysis

• Combine resistors in seriesCombine resistors in series

• RR2323 = 10k = 10kΩΩ + 4.7k + 4.7kΩΩ = 14.7k = 14.7kΩΩ

V = 30V

R5 = 3.3kΩ

R1 = 5.6kΩ

R23 = 14.7kΩ

R4 = 2.2kΩ

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Current AnalysisCurrent Analysis

• Combine parallel resistors (POS)Combine parallel resistors (POS)• RR235235 = 14.7 k = 14.7 kΩΩ x 3.3 x 3.3kkΩΩ / ( / (14.7k14.7kΩΩ + 3.3 + 3.3kkΩΩ) )

• RR235235 = 2.70 k = 2.70 kΩΩ

V = 30V

R235 = 2.7 kΩ

R1 = 5.6kΩ

R4 = 2.2kΩ

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Current AnalysisCurrent Analysis

• Combine series resistorsCombine series resistors• RR1234512345 = 5.6 k = 5.6 kΩΩ + 2.7 + 2.7kkΩΩ +2.2 +2.2 kkΩΩ =10.5 =10.5 kkΩΩ

V = 30V

R12345 = 10.5kΩ

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More AnalysisMore Analysis

• IITT = V = VTT/R/RTT = 30V / = 30V / 10.5 10.5 KKΩΩ = 2.85 mA = 2.85 mA

• VVR1R1 = 16V, V = 16V, VR235R235 = 7.7V, V = 7.7V, VR4R4 = 6.3V = 6.3V

V = 30V

R235 = 2.7 kΩ

R1 = 5.6kΩ

R4 = 2.2kΩ

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Current AnalysisCurrent Analysis

• VVR235R235 = 7.7V = 7.7V• IIR23R23 = 523 = 523μμA, IA, IR5R5 = 2.33mA = 2.33mA

V = 30V

R5 = 3.3kΩ

R1 = 5.6kΩ

R23 = 14.7kΩ

R4 = 2.2kΩ

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CurrentCurrent

• VVR235R235 = 7.7V, I = 7.7V, IR23R23 = 523 = 523μμAA• VVR2R2 = 5.23V, V = 5.23V, VR3R3 = 2.46V, = 2.46V,

V = 30V

R2 = 10kΩ

R1 = 5.6kΩ

R3 = 4.7kΩ

R4 = 2.2kΩ

R5 = 3.3kΩ

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Check You ResultsCheck You Results

• 30V =V30V =VR1R1+V+VR2R2+V+VR3R3+V+VR4 R4 == VVR1R1+V+VR5R5+V+VR4R4

• IITT = I = IR23R23 + I + IR5R5

V = 30V

R2 = 10kΩ

R1 = 5.6kΩ

R3 = 4.7kΩ

R4 = 2.2kΩ

R5 = 3.3kΩ

More AnalysisMore Analysis

• Which resistor has the largest voltage Which resistor has the largest voltage drop? drop? – Is it necessarily the biggest resistor?Is it necessarily the biggest resistor?

• Definition: IDefinition: I11 is the current through R is the current through R11, I, I22 the the

current through Rcurrent through R22, etc., etc.

• What must IWhat must I11 be equal to? be equal to?

• What must IWhat must I11 be great than? be great than?

• Which currents equal IWhich currents equal ITT – the total current – the total current

output of the power source?output of the power source?

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Power In A Series-parallel CircuitPower In A Series-parallel Circuit

• Analysis of power distribution in Analysis of power distribution in the circuit follows the same rules the circuit follows the same rules from pure series and pure from pure series and pure parallel circuits.parallel circuits.

• Total power is a sum of all Total power is a sum of all individual power levels in the individual power levels in the circuit components.circuit components.

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Power in Series-Parallel Resistive CircuitPower in Series-Parallel Resistive Circuit

• Total power generated by the source of a Total power generated by the source of a series parallel circuit is dissipated in series parallel circuit is dissipated in various branches of the circuit.various branches of the circuit.

where Pwhere Pdd = power dissipated in the circuit = power dissipated in the circuit

• PPcc = power consumed by the circuit, = power consumed by the circuit,

• PP11, P, P22 …..P= are the powers dissipated …..P= are the powers dissipated

in the components of the circuit in the components of the circuit

ncd PPPPPP .............)( 321

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Effects Of OpensEffects Of Opens

• As noted earlier, the location of As noted earlier, the location of the open has a great influence the open has a great influence on how the electrical parameters on how the electrical parameters change with the open.change with the open.

• Total current decreases as a Total current decreases as a result.result.

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Troubleshooting An OpenTroubleshooting An Open

• As the open becomes more As the open becomes more dramatic, the technician will note dramatic, the technician will note that current and power levels that current and power levels greatly decrease.greatly decrease.

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Current MeasurementsCurrent Measurements

• Measuring current in the parallel Measuring current in the parallel portion will indicate if an open portion will indicate if an open exists.exists.

• Measuring voltage in the series Measuring voltage in the series portion will indicate opens via portion will indicate opens via the absence of voltage drops.the absence of voltage drops.

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Effects Of ShortsEffects Of Shorts

• The location of the short The location of the short determines how dramatic the determines how dramatic the impact will be to the circuit.impact will be to the circuit.

• A general rule is that the closer A general rule is that the closer the short is to the source, the the short is to the source, the greater the impact will be to greater the impact will be to circuit operation.circuit operation.

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What IFWhat IF

• What is the current IF What is the current IF RR22 is OPEN is OPEN??

• RRTT = R = R11+R+R55+R+R44 = 11.1k = 11.1kΩΩ

• IITT= 2.70mA (was 2.85 mA - DEcrease)= 2.70mA (was 2.85 mA - DEcrease)

V = 30V

R2 = 10kΩ

R1 = 5.6kΩ

R3 = 4.7kΩ

R4 = 2.2kΩ

R5 = 3.3kΩ

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What IFWhat IF

• What is the current IF What is the current IF RR11 is OPEN is OPEN??

• RRTT = infinite = infinite

• IITT = 0 A = 0 A (was 2.85 mA - DEcrease)(was 2.85 mA - DEcrease)

V = 30V

R2 = 10kΩ

R1 = 5.6kΩ

R3 = 4.7kΩ

R4 = 2.2kΩ

R5 = 3.3kΩ

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What IFWhat IF

• What is the current IF What is the current IF RR55 is SHORTED is SHORTED??

• RRTT = R = R11+R+R44 = 7.8k = 7.8kΩΩ

• IITT = 3.9mA (was 2.85 mA - INcrease) = 3.9mA (was 2.85 mA - INcrease)

V = 30V

R2 = 10kΩ

R1 = 5.6kΩ

R3 = 4.7kΩ

R4 = 2.2kΩ

R5 = 0

58Application of Kirchhoff’s Current Law Application of Kirchhoff’s Current Law

(KCL)(KCL)

• The following circuit consists of four The following circuit consists of four resistances and one voltage source, resistances and one voltage source, along with the total and branch along with the total and branch currents.currents.

• Applying KCL Applying KCL at node B: at node B: 21 III

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A Completed ExampleA Completed Example

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Class ActivityClass Activity

• Find the current IFind the current I44 flowing through flowing through

the resistance Rthe resistance R44 in the following in the following

circuit.circuit.

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Plan of AttackPlan of Attack

• How would you tackle this problem?How would you tackle this problem?

62

Plan of AttackPlan of Attack

• How would you tackle this problem?How would you tackle this problem?

63

Plan of AttackPlan of Attack

• How would you tackle this problem?How would you tackle this problem?

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Loaded Voltage DividersLoaded Voltage Dividers

• The Voltage Divider is a common The Voltage Divider is a common series-parallel type circuit.series-parallel type circuit.

• As loads are placed on the As loads are placed on the circuit, the analysis becomes a circuit, the analysis becomes a bit more difficult.bit more difficult.

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Voltage Divider ReviewVoltage Divider Review

• Under ideal conditions, IUnder ideal conditions, I11 = I = I2 2 = I= I

• V=IR (Ohm’s Law!!!)V=IR (Ohm’s Law!!!)

• VVCCCC = I (R = I (R11 + R + R22))

• I = I = VVCCCC / (R / (R11 + R + R22))

• VVR2R2 = R = R2 2 II

• VVR2R2 = R = R2 2 VVCCCC / (R / (R11 + R + R22))

VR2

VCC

R2

R1

I1

I2

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Voltage Divider ReviewVoltage Divider Review

• Under ideal conditions, IUnder ideal conditions, I11 = I = I2 2 = I= I33 = I = I

• VVCCCC = I (R = I (R11 + R + R2 2 + R+ R33))

• I = VI = VCCCC / (R / (R11 + R + R2 2 + R+ R33))

• VVR3R3 = R = R3 3 II

• VVR3R3 = R = R3 3 VVCCCC / (R / (R11 + R + R2 2 + R+ R33))VRr

VCC

R2

R1I1

I2

R3I3

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Loaded Voltage DividersLoaded Voltage Dividers

• An unloaded voltage divider circuit has no An unloaded voltage divider circuit has no other components connected across it.other components connected across it.

• The circuit shows a loaded voltage divider The circuit shows a loaded voltage divider circuit, where resistor Rcircuit, where resistor R33 is connected is connected across the points, A and ground.across the points, A and ground.

• The resistors RThe resistors R22 and R and R33 are are in parallel. Therefore, their in parallel. Therefore, their voltages will be same and voltages will be same and currents through them will currents through them will be inversely proportional be inversely proportional to their resistances.to their resistances.

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AnalysisAnalysis

• If the applied voltage is If the applied voltage is 30V, and R30V, and R11= 4.7k= 4.7kΩΩ and R and R22

= 5.6k= 5.6kΩΩ, find V, find VR2R2..

VR2

VCC

R2

R1

I1

I2

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CalculationsCalculations

• VVR2R2 = 30V 5.6k = 30V 5.6kΩΩ / ( 4.7k / ( 4.7kΩΩ + 5.6k + 5.6kΩΩ))

• VVR2R2 = 16.3 V = 16.3 V

• IIR2R2 = 16.3 V/ 5.6k = 16.3 V/ 5.6kΩΩ = 2.91 mA = 2.91 mA

• Check:Check:

• IITT = = 30V / ( 4.7k30V / ( 4.7kΩΩ + 5.6k + 5.6kΩΩ))

• IITT = 2.91 mA = 2.91 mA

VR2

VCC

R2

R1

I1

I2

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AnalysisAnalysis

• If the applied voltage is If the applied voltage is 30V, and R30V, and R11= 4.7k= 4.7kΩΩ, R, R22 = =

5.6k5.6kΩΩ, and R, and R33 = 8.7k = 8.7kΩΩ, ,

findfind all currents and find all currents and find VVR2R2

VR2

VCC

R2

R1

I1

I2

R3

I3

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CalculationsCalculations

• RR22 is parallel to R is parallel to R3 3 (POS!)(POS!)• RR2323 = 5.6k = 5.6k ΩΩ 8.7k 8.7kΩΩ / (5.6k / (5.6kΩΩ + 8.7k + 8.7kΩΩ))• RR2323 = 3.4k = 3.4kΩΩ• VV2 2 = 30V3.4k = 30V3.4k ΩΩ /( /(3.4K3.4KΩΩ + 4.7 + 4.7kkΩΩ))• VV2 2 = 12.6 V= 12.6 V• II22 = 12.6V/5.6k = 12.6V/5.6k ΩΩ = 2.25mA = 2.25mA• II33 = 12.6V/8.7k = 1.45 mA = 12.6V/8.7k = 1.45 mA• IITT = 30V / ( = 30V / (3.4k3.4kΩΩ + 4.7 + 4.7kkΩΩ) ) • IITT = 3.70 mA = 3.70 mA• CHECK: ICHECK: ITT = I = I22 + I + I33

VR2

VCC

R2

R1

I1

I2

R3

I3

Voltmeters Voltmeters

• You measure voltage by placing a You measure voltage by placing a voltmeter across a resistor in a active voltmeter across a resistor in a active circuit.circuit.

• A perfect voltmeter has infinite A perfect voltmeter has infinite resistance and draws no current, but resistance and draws no current, but in reality, a voltmeter might have a in reality, a voltmeter might have a resistance of 10 Mresistance of 10 MΩΩ

• Any measuring device affects the Any measuring device affects the value being measuredvalue being measured

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AnalysisAnalysis

• If the applied voltage is 30V, If the applied voltage is 30V, and Rand R11= 200 k= 200 kΩΩ, R, R22 = 100 k = 100 kΩΩ, ,

and Rand R33 = 10 M = 10 MΩΩ (a voltmeter), (a voltmeter),

findfind all currents and find V all currents and find VR2R2

• Ideally, Ideally,

• VVR2R2 = 30V(100 k = 30V(100 kΩΩ)/(300)/(300 k kΩΩ))

• VVR2R2 = 10 V = 10 V

VR2

VCC

R2

R1

I1

I2

R3

I3

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CalculationsCalculations

• RR22 is parallel to R is parallel to R3 3 (POS!)(POS!)• RR2323 = 100k = 100kΩΩ 10 M 10 MΩΩ / (100k / (100kΩΩ + 10 M + 10 MΩΩ ) )• RR2323 = 99.01 k = 99.01 kΩΩ• VV2 2 = = 30V 30V 99.01 k99.01 kΩΩ =9.93 V =9.93 V

( (99.01k99.01kΩΩ+200k+200kΩΩ))• VV2 2 = 9.93 V = 9.93 V • The meter causes less than 1%The meter causes less than 1%

drop in measured voltage, butdrop in measured voltage, butthis increases as the size of thethis increases as the size of themeasured resistors increase measured resistors increase

VR2

VCC

R2

R1

I1

I2

R3

I3

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Special ConditionsSpecial Conditions

• If two If two equalequal resistors are in parallel, resistors are in parallel, the current AND voltage is the same the current AND voltage is the same for both resistorsfor both resistors

• If two If two equalequal resistors are in parallel, resistors are in parallel, the equivalent resistance is equal to the equivalent resistance is equal to just ½ of one of the resistorsjust ½ of one of the resistors

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ExampleExample

• IIR1R1= 25V/4.7k= 25V/4.7kΩΩ = 5.32 mA = 5.32 mA

• IIR2R2= 25V/4.7k= 25V/4.7kΩΩ = 5.32 mA = 5.32 mA

• IITT = I = IR1 R1 + I+ IR2 R2 = 5.32 mA + 5.32 mA = 10.6mA= 5.32 mA + 5.32 mA = 10.6mA

• RRTT = V = VTT/I/ITT = 25V / 10.6mA = 2.36 k = 25V / 10.6mA = 2.36 k ΩΩ

• Check: RCheck: RTT = 4.7k = 4.7kΩΩ 4.7k 4.7kΩΩ / (4.7k / (4.7kΩΩ 4.7k 4.7kΩΩ))

• RRTT = 2.35 k = 2.35 kΩΩ

• Currents same, RCurrents same, RTT = R = R11/2/2V = 25V

R1 = 4.7kΩ

R2 = 4.7kΩ

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ExampleExample

• All resistors are All resistors are 2020ΩΩ. .

• The current in RThe current in R33

is 10mA. is 10mA. • Find the voltage Find the voltage

drop across and drop across and the current the current through all the through all the other resistors.other resistors.

R2

R1

RL2

R3

RL3

RL1

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AnalysisAnalysis

• VVR3R3 = = 10mA 20 10mA 20 ΩΩ = .2 V = .2 V• VVL3 L3 == VVR3 R3 so Iso IL3L3 = = .2 V/ 20 .2 V/ 20 ΩΩ = 10 mA = 10 mA• IIR2R2 = = IIR3 R3 + I+ IL3 L3 = = 10 mA + 10 mA = 20 mA10 mA + 10 mA = 20 mA• VVR2R2 = = 20mA 20 20mA 20 ΩΩ = .4 V = .4 V• VVL2 L2 == VVR2R2 + V + VR3 R3 so Iso IL2L2 = = .6 V/ 20 .6 V/ 20 ΩΩ = 30 mA = 30 mA• IIR1R1 = = IIR2 R2 + I+ IL2 L2 = = 20 mA + 30 mA = 50 mA20 mA + 30 mA = 50 mA• VVR1R1 = = 50mA 20 50mA 20 ΩΩ = 1 V = 1 V• VVL1 L1 == VVR1+ R1+ VVR2R2 + V + VR3 R3 so so • IIL1L1 = = 1.6 V/ 20 1.6 V/ 20 ΩΩ = 80 mA = 80 mA• IITT = = IIR1 R1 + I+ IL1 L1 = = 50 mA + 80 mA = 130 mA50 mA + 80 mA = 130 mA

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ChecksChecks

• VVTT = V = VR1R1 + V + VR2R2 + V + VR3 R3 = .2V+.4V+1.0V = 1.6V= .2V+.4V+1.0V = 1.6V

• RRTT=V=VTT/I/IT T = 1.6V / 130mA = 12.3 = 1.6V / 130mA = 12.3 ΩΩ

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Check your work – Find RCheck your work – Find RTT

20Ω

??V

20Ω

20Ω

20Ω

20Ω

20Ω

20Ω

18V

20Ω

10Ω

20Ω

20Ω

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SimplifySimplify

20Ω

??V

20k

10 Ω

20k

20Ω

30 Ω

??V

20Ω 20Ω

20Ω

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SimplifySimplify

??V

20Ω 20Ω

12Ω

30 Ω

??V

20Ω 20Ω

20Ω

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SimplifySimplify

18V

20Ω 20Ω

12Ω 18V

20Ω

32Ω

84

Checks!!!!Checks!!!!

20Ω

32Ω

??V ??V

= 12.3 Ω

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Bridge CircuitsBridge Circuits• In a balanced bridge circuit, the current In a balanced bridge circuit, the current

through and the voltage drop across Rthrough and the voltage drop across R55 = 0: = 0:

• If any one of the outer resistances is If any one of the outer resistances is increased, the voltage drop across that increased, the voltage drop across that resistance will also increase. Therefore, it is resistance will also increase. Therefore, it is an unbalanced bridge circuit.an unbalanced bridge circuit.

4

2

3

1

R

R

R

R

A B

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The Wheatstone BridgeThe Wheatstone Bridge

• This circuit is a series-parallel This circuit is a series-parallel circuit that is very popular in circuit that is very popular in controls and industrial controls and industrial applications.applications.

• There are two states for the There are two states for the bridge:bridge:BalancedBalanced UnbalancedUnbalanced

VVAA = = VVBB VVAA ≠≠ VVBB

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Why??Why??

• In a balanced bridge, the voltage drop In a balanced bridge, the voltage drop across Racross R33 = voltage drop across R = voltage drop across R44

• There is NO current across the resistor There is NO current across the resistor connecting Rconnecting R3 3 to Rto R44

• VVR3 R3 = V= VInInRR33/(R/(R33+R+R11))

• VVR4 R4 = V= VInInRR44/(R/(R44+R+R22))

• VVR3 R3 = V= VR4R4A B

88

CalculationsCalculations

• VVInInRR33/(R/(R33+R+R11) = V) = VInInRR44/(R/(R44+R+R22))• Multiply both sides by (RMultiply both sides by (R33+R+R11)(R)(R44+R+R22) )

and divide both by Vand divide both by Vinin

• RR33 (R (R44+R+R22) =R) =R44 (R (R33+R+R11))• RR33 R R44+ R+ R33 R R22 =R =R44 R R33+ R+ R44 R R11

• RR33 R R22 =R =R44 R R11

• RR11 / R / R33 =R =R22 / R / R44

• In most problems, given 3, you find In most problems, given 3, you find the 4th valuethe 4th value

89

Class ActivityClass Activity

• For the following For the following balancedbalanced bridge bridge circuit, calculate the value of the circuit, calculate the value of the resistance, R.resistance, R.

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Lab ExperimentLab Experiment

• With a wire, resistance is With a wire, resistance is proportional to the length of the wireproportional to the length of the wire– The longer the wire, the more resistance The longer the wire, the more resistance

it hasit has

• A wire exactly 1 meter long is A wire exactly 1 meter long is connected to a batteryconnected to a battery– A slider that moves up and down the A slider that moves up and down the

wire makes an electrical connection to it wire makes an electrical connection to it

91

Lab SetupLab Setup

35 VA

73 cm

27 cm

XXX Ω

0 mA

1kΩ

L R

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Suppose you forgot?Suppose you forgot?

• VVLL = 35V 27 cm / (27cm + 73 cm) = 9.45 = 35V 27 cm / (27cm + 73 cm) = 9.45 V = VV = VRR

• II1k1k = 9.45 V / 1 k = 9.45 V / 1 kΩ = 9.45 mAΩ = 9.45 mA• VVXXX XXX = 35 V – 9.45 V = 25.55 V= 35 V – 9.45 V = 25.55 V• RRXXXXXX = V/I = 25.55 V / 9.45 ma = 2.70 kΩ = V/I = 25.55 V / 9.45 ma = 2.70 kΩ• RR11 / R / R33 =R =R22 / R / R44

• 73/27 = X/1k 73/27 = X/1k ΩΩ• X = 1k 73/ 27 = 2.70 X = 1k 73/ 27 = 2.70 kΩkΩ

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PracticePractice

20 VA

33 cm

67 cm

XXX Ω

0 mA

5kΩ

L R

Unit 7 SummaryUnit 7 Summary

1. Identifying series/parallel relationships in 1. Identifying series/parallel relationships in combinational circuitscombinational circuits

2. Calculating resistance, current or voltage 2. Calculating resistance, current or voltage in combinational resistive circuitsin combinational resistive circuits

3. Applying KVL and KCL to combinational 3. Applying KVL and KCL to combinational resistive circuitsresistive circuits

4. Wheatstone bridge4. Wheatstone bridge

5. Loaded voltage-dividers5. Loaded voltage-dividers

6. Meter loading6. Meter loading

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