1 unit six: parallel circuits john elberfeld [email protected] et115 dc electronics
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Unit Six:Parallel Circuits
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 6 Objectives - IUnit 6 Objectives - I
• Identify a parallel resistive circuit.Identify a parallel resistive circuit.• Determine total resistance, current, and Determine total resistance, current, and
power in a parallel resistive circuit.power in a parallel resistive circuit.• State Kirchhoff’s current law.State Kirchhoff’s current law.• Solve for an unknown current in a circuit Solve for an unknown current in a circuit
using Kirchhoff’s current law (KCL).using Kirchhoff’s current law (KCL).• State the general current-divider formula.State the general current-divider formula.• State the current-divider formula for two State the current-divider formula for two
branches.branches.
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Unit 6 Objectives – IIUnit 6 Objectives – II
• Apply the appropriate current-divider Apply the appropriate current-divider formula to circuits to find an unknownformula to circuits to find an unknown
• current.current.• 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.
Unit 6 Objectives – IIIUnit 6 Objectives – III
• Measure resistances and voltages in Measure resistances and voltages in a DC circuit using a DMM.a DC circuit using a DMM.
• Apply Ohm’s Law, Thevenin’s Apply Ohm’s Law, Thevenin’s theorem, KVL and KCL to practical theorem, KVL and KCL to practical circuits.circuits.
• Test circuits by connecting Test circuits by connecting simulated instruments in Multisim.simulated instruments in Multisim.
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Reading AssignmentReading Assignment
• Read and study Read and study
• Chapter 5: Parallel Circuits:Chapter 5: Parallel Circuits:Pages 163-198Pages 163-198
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Lab AssignmentLab Assignment
• Experiment 9, “Parallel Circuits,”Experiment 9, “Parallel Circuits,”page 65 of page 65 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 6 Homework sheetComplete the Unit 6 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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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.
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Series Circuit ReviewSeries Circuit Review
• There is just one path for the currentThere is just one path for the current• The current is the same in every The current is the same in every
component in a series circuitcomponent in a series circuit• The voltage of the battery is equal to the The voltage of the battery is equal to the
sum of voltage drops in the resistorssum of voltage drops in the resistors• Total resistance is the sum Total resistance is the sum
of the resistorsof the resistors
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Parallel CircuitsParallel Circuits• There are two or more paths for current There are two or more paths for current
flow.flow.• The voltage is the same across all parallel The voltage is the same across all parallel
branches.branches.• The total current entering any point is The total current entering any point is
equal to the current leaving the pointequal to the current leaving the point
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Parallel Combination of ResistancesParallel Combination of Resistances
• When all the resistors in a circuit are When all the resistors in a circuit are connected in parallel, the circuit is called a connected in parallel, the circuit is called a parallel circuit.parallel circuit.
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Determine Parallel ResistorsDetermine Parallel Resistors
• Resistors are parallel to each other if Resistors are parallel to each other if both pairs of ends are connected both pairs of ends are connected directly together with no electrical directly together with no electrical resistance between the endsresistance between the ends
• Parallel has NOTHING to do with Parallel has NOTHING to do with geometry or layout of the resistorsgeometry or layout of the resistors
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Which are Parallel?Which are Parallel?
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Parallel CircuitParallel Circuit
• When a circuit is connected in parallel, the When a circuit is connected in parallel, the potential difference in all the circuit potential difference in all the circuit elements remains the same and the elements remains the same and the current is distributed proportionally current is distributed proportionally among all the circuit elements.among all the circuit elements.
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• The voltage drop across each component The voltage drop across each component in a parallel in a parallel circuit is the same, since all the circuit is the same, since all the resistances are connected in parallel. Thus, resistances are connected in parallel. Thus, the voltage across each connection is the voltage across each connection is equal to the supply voltage.equal to the supply voltage.
• The voltage drop in a parallel circuit is:The voltage drop in a parallel circuit is: where, V is the supply where, V is the supply voltage.voltage.
VVVV 321
Voltage Drop in Parallel Resistive CircuitVoltage Drop in Parallel Resistive Circuit
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A Practical ExampleA Practical Example
• Car circuits (and house circuits) are Car circuits (and house circuits) are all parallelall parallel– You can run the radio without the horn You can run the radio without the horn
being turned on – an advantagebeing turned on – an advantage
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Find Equivalent ResistanceFind Equivalent Resistance
Voltage = 25 VVoltage = 25 V
Find the total Current_______Find the total Current_______
Find the equivalent Resistance ______Find the equivalent Resistance ______
V = 25V
R1 = 25Ω
I1 = ?
R2 = 50Ω
I2 = ?
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CalculationsCalculations
• II11 = V = V11/R/R11 = 25V/25 = 25V/25ΩΩ = 1 A = 1 A
• II22 = V = V22/R/R22 = 25V/50 = 25V/50ΩΩ = 500 mA = 500 mA
• IITT = I = I11+I+I22 = 1 A + 500 mA = 1.5 A = 1 A + 500 mA = 1.5 A
• RRT T = V= VTT/I/ITT = = 25V/ 25V/ 1.5 A = 16.7 1.5 A = 16.7 ΩΩ
• If both resistors are replaced by a single If both resistors are replaced by a single 16.7 16.7 ΩΩ resistor, the voltage and current resistor, the voltage and current between the two end points is unchangedbetween the two end points is unchanged
• Notice that the total resistance is LESS Notice that the total resistance is LESS than any one resistor in the parallel than any one resistor in the parallel circuit.circuit.
V = 25VRT = 16.7Ω
IT = 1.5A
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ReasoningReasoning
• Adding a resistor in parallel provides Adding a resistor in parallel provides another path for current, making it another path for current, making it easier for current to floweasier for current to flow
• This is the same as reducing the total This is the same as reducing the total resistance for the circuitresistance for the circuit
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Effective, Total ResistanceEffective, Total Resistance
• Ohm’s Ohm’s law law
method:method:
T
TT I
VR
R1 = 10kΩ R2 = 10kΩ
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Parallel Circuit NodesParallel Circuit Nodes
• Two types of nodes or Two types of nodes or connections:connections:– Dividing NodeDividing Node: A junction where : A junction where
current enters by one connection but current enters by one connection but leaves by two or more connectionsleaves by two or more connections
– Summing connectionSumming connection: A junction where : A junction where current enters a junction by two or more current enters a junction by two or more connections but leaves via oneconnections but leaves via one
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Parallel Circuit NodesParallel Circuit Nodes• Junctions A and B have current leaving Junctions A and B have current leaving
(Dividing Node), while junctions D and C (Dividing Node), while junctions D and C have current entering (Summing Node).have current entering (Summing Node).– Look at it from the Look at it from the electronselectrons point of view point of view
Summing
Dividing
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Parallel Circuit CurrentParallel Circuit Current
• Current leaving the power supply’sCurrent leaving the power supply’s (–) terminal is the same current (–) terminal is the same current entering the (+) terminal.entering the (+) terminal.
• This is referred to as total current (This is referred to as total current (IITT).).
• Individual electrons may follow a Individual electrons may follow a variety of paths, but every electron variety of paths, but every electron that leaves the battery is balanced by that leaves the battery is balanced by an electron entering the batteryan electron entering the battery
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Parallel Circuit CurrentParallel Circuit Current
• Since the total current in the Since the total current in the individual resistors is equal to individual resistors is equal to the current supplied by the the current supplied by the source, the total current can be source, the total current can be stated as:stated as:
IITT = = IIR R 11 + + IIR R 22 … + … + IIR nR n
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Total CurrentTotal Current
• IITT = I = I11 + I + I22 + I + I33
I1 I2 I3IT
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Kirchhoff’s Current LawKirchhoff’s Current Law
• Kirchhoff’s current law states Kirchhoff’s current law states that any current entering a that any current entering a junction must be equal to the junction must be equal to the current leaving the junction.current leaving the junction.
IIinin = = IIoutout
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• By definition of KCL, By definition of KCL, the algebraic sum of the the algebraic sum of the currents meeting at a point or junction is zero.currents meeting at a point or junction is zero.
• In +In +
• Out -Out -
• For the given circuit, For the given circuit,
0654321 IIIIII
Kirchhoff’s Current Law (KCL)Kirchhoff’s Current Law (KCL)
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Power In Parallel CircuitsPower In Parallel Circuits
1. Summation method 1. Summation method
PPTT = = PPR R 11 + + PPR R 22 … + … + PPR nR n
2. Formulas for Total Power2. Formulas for Total Power
TT
TT
TTT
R
VP
RIPVIP
2
2
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Sample Problems
Comp. Resistance Voltage Current Power
R1 2kΩ
R2 4kΩ
R3 6kΩ
Total
VT = 36V
R1=2kΩ
R3=6kΩ
R2=4kΩ
+
-
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Sample Problems
Comp. Resistance Voltage Current Power
R1 2kΩ 36V 18ma 648mW
R2 4kΩ 36V 9ma 324mW
R3 6kΩ 36V 6ma 216mW
Total 1.09 kΩ 36V 33ma 1188mW
VT = 36V
R1=2kΩ
R3=6kΩ
R2=4kΩ
+
-
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Check your workCheck your work
• Things look good, but..Things look good, but..• P = IP = I22RR• P = (33ma)P = (33ma)22 1.09 k 1.09 kΩΩ = 1187 mW = 1187 mW• The values for I and R work together The values for I and R work together
to give you the accepted value, so to give you the accepted value, so they are probably correct.they are probably correct.
• P = VP = V2/2/RR• P = (36V)P = (36V)22 / 1.09 k / 1.09 kΩΩ = 1189 mW = 1189 mW
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Sample Problems
Comp. Resistance Voltage Current Power
R1 1kΩ
R2 2kΩ
R3 3kΩ
Total
VT = 24V
R1=1kΩ
R3=3kΩ
R2=2kΩ
+
-A B C
Current A Current B Current C
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Sample Problems
Comp. Resistance Voltage Current Power
R1 1kΩ 24V 24ma 576mW
R2 2kΩ 24V 12ma 288mW
R3 3kΩ 24V 8ma 192mW
Total 545 Ω 24V 44ma 1056mW
VT = 24V
R1=1kΩ
R3=3kΩ
R2=2kΩ
+
-A B C
Current A Current B Current C
44ma 20ma 8ma
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Finding Total Finding Total SeriesSeries Resistance Resistance
• Calculating the total resistance for Calculating the total resistance for resistors in resistors in SERIESSERIES is easy is easy
• In SERIES, just add up the all the In SERIES, just add up the all the resistorsresistors
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Finding Total Finding Total SeriesSeries Resistance Resistance
• Ohm’s Law (V = I R) works for the total Ohm’s Law (V = I R) works for the total circuit as well as each element in the circuit as well as each element in the circuitcircuit
• Using Formulas:Using Formulas:
• VVT T = I= IT T RRT T and Rand RTT = V = VT T / I/ ITT
• Find VFind VTT/I/ITT and you have R and you have RTT
• VVTT = V = V11 + V + V22 +V +V33 +V +V44……
• SoSo
• VVT T = I= I11RR11 + I + I22RR2 2 + I+ I33RR3 3 + I+ I44RR4 .. 4 .. - Series- Series
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DerivationDerivation
• VVT T = I= I11RR11 + I + I22RR2 2 + I+ I33RR3 3 + I+ I44RR4 .. 4 .. - Series- Series
• But, in SERIES, IBut, in SERIES, ITT = I = I11 = I = I22 = I = I33 = I = I44 = … = …
• VVT T = I= ITTRR11 + I + ITTRR2 2 + I+ ITTRR3 3 + I+ ITTRR4 ..4 ..
• VVT T = I= IT T (R(R11 + R + R2 2 + R+ R3 3 + R+ R4 ..4 ..))
• VVT T / I/ IT T = (R= (R11 + R + R2 2 + R+ R3 3 + R+ R4 ..4 ..))
• RRT T = V= VT T / I/ IT T = (R= (R11 + R + R2 2 + R+ R3 3 + R+ R4 ..4 ..))
• RRT T = (R= (R11 + R + R2 2 + R+ R3 3 + R+ R4 ..4 ..) – Series!) – Series!
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Finding Total Finding Total ParallelParallel Resistance Resistance
• Every resistor you add in parallel Every resistor you add in parallel actually REDUCES the equivalent actually REDUCES the equivalent resistance in the circuitresistance in the circuit– Each resistor you add in parallel Each resistor you add in parallel
increases the current, which makes the increases the current, which makes the resistance appear lowerresistance appear lower
• Simple addition does NOT work for Simple addition does NOT work for resistors in parallelresistors in parallel
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Finding Total Finding Total ParallelParallel Resistance Resistance
• VVT T = I= IT T RRT T and Rand RTT = V = VTT / I / ITT and I and ITT = V = VTT/R/RTT
• and and 1/R1/RTT = I = ITT / V / VTT
• We will find 1/RWe will find 1/RTT in this math in this math
• IITT = I = I11 + I + I22 + I + I33 + I + I44 + … + … - Parallel- Parallel
• IITT = V = V11/R/R11 + V + V22/R/R22 + V + V33/R/R33 + V + V44/R/R44 + … + …
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DerivationDerivation
• But, in PARALLEL, But, in PARALLEL, • VVTT = V = V11 = V = V22 = V = V33 = V = V44 + … + …• IITT = V = VTT/R/R11 + V + VTT/R/R22 + V + VTT/R/R33 + V + VTT/R/R44 + … + …• IITT = V = VTT(1/R(1/R11 + 1/R + 1/R22 + 1/R + 1/R33 + 1/R + 1/R44 + …) + …)• IITT / V / VT T = (1/R= (1/R11 + 1/R + 1/R22 + 1/R + 1/R33 + 1/R + 1/R44 + …) + …)• 1/R1/RTT = I = ITT / V / VT T = (1/R= (1/R11 + 1/R + 1/R22 + 1/R + 1/R33 + +
1/R 1/R44 + …) + …)• 1/R1/RTT = 1/R = 1/R11 + 1/R + 1/R22 + 1/R + 1/R33 + 1/R + 1/R44 + … + …• Parallel FormulaParallel Formula
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EquationEquation
• The reciprocal of the total resistance is The reciprocal of the total resistance is equal to the sum of the reciprocals of the equal to the sum of the reciprocals of the resistors in parallelresistors in parallel
• 1/R1/RTT = 1/1k = 1/1kΩΩ + + 1/2k1/2kΩΩ + + 1/3k1/3kΩΩ
VT = 24V
R1=1kΩ
R3=3kΩ
R2=2kΩ
+
-A B C
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Using FractionsUsing Fractions
• 1/R1/RTT = 1/1k = 1/1kΩΩ + + 1/2k1/2kΩΩ + + 1/3k1/3kΩΩ
• LCD is 6KLCD is 6KΩΩ
• 1/R1/RTT = 6/6k = 6/6kΩΩ + + 3/6k3/6kΩΩ + + 2/6k2/6kΩΩ
• 1/R1/RTT = 11/6k = 11/6kΩΩ
• RRTT = = 6k6kΩΩ // 1111
• RRTT = 545 = 545 ΩΩ
• (Hard to do with 2.2K(Hard to do with 2.2KΩΩ, 470, 470ΩΩ, 5.6M, 5.6MΩΩ))
VT = 24V
R1=1kΩ
R3=3kΩ
R2=2kΩ
+
-A B C
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CalculationsCalculations
• 1/R1/RTT = 1/1k = 1/1kΩΩ + + 1/2k1/2kΩΩ + + 1/3k1/3kΩΩ
• 1 EXP 3 [INV] + 2 EXP 3 [INV] + 3 EXP 3 1 EXP 3 [INV] + 2 EXP 3 [INV] + 3 EXP 3 [INV] EQUALS [INV] [INV] EQUALS [INV]
• RRTT = 545 = 545 ΩΩ
• Learn how YOUR calculator works!Learn how YOUR calculator works!
VT = 24V
R1=1kΩ
R3=3kΩ
R2=2kΩ
+
-A B C
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CheckCheck
• II11 = 24V/1k = 24V/1kΩΩ = 24 ma = 24 ma
• II22 = 24V/2k = 24V/2kΩΩ = 12 ma = 12 ma
• II33 = 24V/3k = 24V/3kΩΩ = 8 ma = 8 ma
• IITT = 44 ma = 44 ma
• RRTT = V = VTT/I/ITT = 24V/44ma = 545 = 24V/44ma = 545 ΩΩ
VT = 24V
R1=1kΩ
R3=3kΩ
R2=2kΩ
+
-A B C
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Own your own!Own your own!
• Find RFind RTT using the inverse functions using the inverse functions
on your calculatoron your calculator
VT = 36V
R1=2kΩ
R3=6kΩ
R2=4kΩ
+
-
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CheckCheck
• II11 = 36V/2k = 36V/2kΩΩ = 18 ma = 18 ma
• II22 = 36V/4k = 36V/4kΩΩ = 9 ma = 9 ma
• II33 = 36V/6k = 36V/6kΩΩ = 6 ma = 6 ma
• IITT = 33 ma = 33 ma
• RRTT = V = VTT/I/ITT = 36V/33ma = 1.09k = 36V/33ma = 1.09kΩΩ
VT = 36V
R1=2kΩ
R3=6kΩ
R2=4kΩ
+
-
Problem Problem
• The following resistors are The following resistors are connected in parallel:connected in parallel:
• 1 M1 MΩΩ, 2.2 M, 2.2 MΩΩ, 4.7 M, 4.7 MΩΩ, 12 M, 12 MΩΩ, 22 M, 22 MΩΩ
• Use your calculator to determine the Use your calculator to determine the total resistance.total resistance.
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RRTT = 556 k = 556 k Ω Ω
ProblemProblem
• The total resistance of a three branch The total resistance of a three branch parallel circuit is 819 parallel circuit is 819 ΩΩ..
• If two of the resistors are 1.5 kIf two of the resistors are 1.5 kΩΩ and and 10 k10 kΩΩ, find the third resistor., find the third resistor.
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1 2 3
1 1 1 1
TR R R R
ProblemProblem50
1 2 3
3
3
3
1 1 1 1
1 1 1 1
819 1.5 10
1 1 1 1
819 1.5 10
2.2
TR R R R
k k R
R k k
R k
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POS : Product over the SumPOS : Product over the Sum
• If you have just TWO Resistors in If you have just TWO Resistors in parallel, the equation reduces to:parallel, the equation reduces to:
• 1/R1/RTT = 1/R = 1/R11 + 1/R + 1/R22
• Multiply every term by (RMultiply every term by (RTTRR11RR22))• (R(RTTRR11RR22)/R)/RTT = (R = (RTTRR11RR22)/R)/R11 + (R + (RTTRR11RR22)/R)/R22
• RR11RR22 = R = RTTRR22 + R + RTTRR1 1 Factor out RFactor out RTT
• RR11RR22 = R = RTT(R(R22 + R + R11))• RRTT = R = R11RR22 / (R / (R22 + R + R11))• Product over the sum!!Product over the sum!!
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Practice Practice
• RRTT = R = R11RR22/(R/(R11+R+R22))
• RRTT = (2k = (2kΩΩ 4k 4kΩΩ) /(2k) /(2kΩΩ + 4k + 4kΩΩ) = 1.333k ) = 1.333k ΩΩ
• 2 EXP 3 x 4 EXP 3 2 EXP 3 x 4 EXP 3 ∕ ∕ (( 2 EXP 3 + 4 EXP 3 2 EXP 3 + 4 EXP 3 )) = =
• Do NOT omit the Do NOT omit the ( )( )
VT = 36V
R1=2kΩ
R2=4kΩ
+
-
53
More PracticeMore Practice
• RRTT = R = R11RR22/(R/(R11+R+R22))
• RRTT = 1.33k = 1.33kΩΩ 6k 6kΩΩ /(1.33k /(1.33kΩΩ + 6k + 6kΩΩ) = 1.09k ) = 1.09k ΩΩ
• Do NOT omit the ( )Do NOT omit the ( )
VT = 36V
R1=1.33kΩ
R2=6kΩ
+
-
54
Practice Practice
• You can evaluate a series of parallel You can evaluate a series of parallel resistors by doing product over sum resistors by doing product over sum for pairs until you get only one left – for pairs until you get only one left – which is Rwhich is RTT..
VT = 36V
R1=2kΩ
R3=6kΩ
R2=4kΩ
+
-
55
CalculationsCalculations
• Combining 6kCombining 6kΩΩ and 4k and 4kΩΩ using POS using POS
• Combining 2kCombining 2kΩΩ and 2.4k and 2.4kΩΩ
VT = 36V
R1=2kΩ
R3=6kΩ
R2=4kΩ
+
-
VT = 36V
R1=2kΩ
R23=2.4kΩ
+
-
VT = 36V
R1=1.09kΩ
+
-
POS – Parallel ResistorsPOS – Parallel Resistors
RR11 RR22 RRTT
Y V Bk GdY V Bk Gd Gn Be Bk SiGn Be Bk Si
Gn Be Br GdGn Be Br Gd Br Bk R SiBr Bk R Si
4.7 k4.7 kΩΩ 2.2 k2.2 kΩΩ
27 27 ΩΩ 56 56 ΩΩ
1.5 k1.5 kΩΩ 2.2 k2.2 kΩΩ
10 k10 kΩΩ 10 k10 kΩΩ
6 M6 MΩΩ 6 M6 MΩΩ
56
POSPOS
RR11 RR22 RRTT
Y V Bk GdY V Bk Gd Gn Be Bk SiGn Be Bk Si 25.6 25.6 ΩΩ
Gn Be Br GdGn Be Br Gd Br Bk R SiBr Bk R Si 359 359 ΩΩ
4.7 k4.7 kΩΩ 2.2 k2.2 kΩΩ 1.5 k1.5 kΩΩ
27 27 ΩΩ 56 56 ΩΩ 18.2 18.2 ΩΩ
1.5 k1.5 kΩΩ 2.2 k2.2 kΩΩ 892 892 ΩΩ
10 k10 kΩΩ 10 k10 kΩΩ 5 k5 kΩΩ
6 M6 MΩΩ 6 M6 MΩΩ 3 M3 MΩΩ
57
58 TroubleshootingTroubleshootingOpens in a Parallel CircuitOpens in a Parallel Circuit
• There are two types of opens in a parallel There are two types of opens in a parallel circuit:circuit:
– Open in a main line Open in a main line
– Open in a branch lineOpen in a branch line
• A main line connects the power source to A main line connects the power source to the rest of the circuit.the rest of the circuit.
• A branch line is one of the different A branch line is one of the different branches in a parallel circuit.branches in a parallel circuit.
59
Opens in a Parallel CircuitOpens in a Parallel Circuit• Open in a main line:Open in a main line:
• Open in a branch line:Open in a branch line:
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Main line OpensMain line Opens
• An open in the main An open in the main line stops all current line stops all current flow through all elementsflow through all elements
• The voltage across the open is equal The voltage across the open is equal to the supply voltageto the supply voltage
• There is no current to cause a voltage There is no current to cause a voltage drop across any element in the circuitdrop across any element in the circuit
• There is no power used in the circuit There is no power used in the circuit since there is no current flowingsince there is no current flowing
+
-
61
Effects Of a Branch OpenEffects Of a Branch Open
1. Branch opens makes the current go 1. Branch opens makes the current go to 0 A just in that branchto 0 A just in that branch
2. If current decreases to 0, resistance 2. If current decreases to 0, resistance must have increased to infinity must have increased to infinity (definition of an open)(definition of an open)
3. Branch voltage remains the same, 3. Branch voltage remains the same, both across the open branch and both across the open branch and the other branchesthe other branches
An Open ResistorAn Open Resistor• If a resistor becomes If a resistor becomes
“open” in a pure “open” in a pure parallel circuit:parallel circuit:
1. Current in that branch goes to zero1. Current in that branch goes to zero
2. Power in that branch goes to zero2. Power in that branch goes to zero
3. Total resistance in the circuit INCREASES 3. Total resistance in the circuit INCREASES because a path is gonebecause a path is gone
4. Total current in the circuit DECREASES 4. Total current in the circuit DECREASES because a path is gonebecause a path is gone
5. Total power in the circuit DECREASES 5. Total power in the circuit DECREASES because a path is gonebecause a path is gone
62
+
-
Adding a ResistorAdding a Resistor
• If a new resistor isIf a new resistor is added to a pure added to a pure parallel circuit:parallel circuit:
1. Total resistance in the circuit 1. Total resistance in the circuit DECREASES because a path is addedDECREASES because a path is added
2. Total current in the circuit 2. Total current in the circuit INCREASES because a path is addedINCREASES because a path is added
3. Total power in the circuit INCREASES 3. Total power in the circuit INCREASES because a path is addedbecause a path is added
63
+
-
64
A Practical ExampleA Practical Example
• II11 = I = I22 = I = I33 = I = I44 = 100V/10k = 100V/10k = 10ma each = 10ma each
65
Shorts in a Parallel CircuitShorts in a Parallel Circuit
• A short in a parallel circuit occurs if both A short in a parallel circuit occurs if both the terminals of any branch are short-the terminals of any branch are short-circuited.circuited.
• A short draws almost infinite current A short draws almost infinite current because the resistance of the connecting because the resistance of the connecting wires is zero.wires is zero.
66
Shorts in a Parallel CircuitShorts in a Parallel Circuit
• To protect any branch of the circuit To protect any branch of the circuit from the short-circuited branch, a from the short-circuited branch, a resistance (or fuse) is placed in resistance (or fuse) is placed in between the branch and power between the branch and power supplysupply
67
Effects of a ShortEffects of a Short
• This is a dramatic and serious event This is a dramatic and serious event because total resistance goes to zero because total resistance goes to zero Ohms.Ohms.
• The power supply attempts to put out The power supply attempts to put out
infinite current ( I = V / R = V / 0 = infinite current ( I = V / R = V / 0 = ∞∞ ) )• Since there is 0 Resistance across the Since there is 0 Resistance across the
branches, there is no voltage drop branches, there is no voltage drop developed, and current is the maximum developed, and current is the maximum the power supply can producethe power supply can produce
• A protective device (fuse, circuit breaker) A protective device (fuse, circuit breaker) is required.is required.
68
Contrasting Series And Parallel Circuits
SERIES• IT is constant
• KVL is used
• VT = sum of drops
• RT is sum of resistors
PARALLEL• IT is the sum of
IRn
• KCL is used
• VT is constant
• 1/RT is sum of reciprocals
69
Multiple SourcesMultiple Sources• Multiple sources of power are usually connected in Multiple sources of power are usually connected in
parallel to meet additional power requirement at parallel to meet additional power requirement at constant voltage.constant voltage.– Voltages must be the same!Voltages must be the same!
• The total current supplied by all the voltage sources is The total current supplied by all the voltage sources is given by: given by:
NT IIIII ......321
70
Voltage Sources In ParallelVoltage Sources In Parallel• Parallel sources are used to Parallel sources are used to
increase the amount of total increase the amount of total current available. current available.
• While While VVTT remains the same, the remains the same, the maximum maximum IITT available increases available increases by the amount of each sourceby the amount of each source
• Actual IActual ITT is still controlled by the is still controlled by the resistance in the circuitresistance in the circuit
71
• Current divider consists of a parallel resistive Current divider consists of a parallel resistive circuit dividing the current, provided by the circuit dividing the current, provided by the voltage source, into the parallel resistors.voltage source, into the parallel resistors.
• The total current, The total current, II, is divided between resistors , is divided between resistors RR11 and and RR22..
• Smaller resistance value will draw more current Smaller resistance value will draw more current and vice versa.and vice versa.
Current DividersCurrent Dividers
72
Sample Problems
Comp. Resistance Voltage Current Power
R1 2 kΩ
R2 4 kΩ 8 mA
Total
VT = ?V
R1=2 kΩ
R2=4 kΩ
+
-A B
Current A Current B
PREDICT: Size of RT
73
Sample Problems
Comp. Resistance Voltage Current Power
R1 2 kΩ 32 V 16 mA 512mW
R2 4 kΩ 32 V 8 mA 256mW
Total 1.33 kΩ 32 V 24 mA 768mW
VT = ?V
R1=2 kΩ
R2=4 kΩ
+
-A B
Current A Current B
24 mA 8 mA
74
FormulasFormulas
• How can we find the currents in one step?How can we find the currents in one step?• For TWO parallel resistors, the voltage is For TWO parallel resistors, the voltage is
the samethe same
• VVT T = I= I RR = I= I11 R R11 = = II22 R R22
• IITT = I = I11 + I + I22, so , so II22 = I = ITT – I – I11
• II11 R R11 = = II22 R R2 2 = = (I(ITT – I – I11) ) RR22
• II11 R R11 = (I = (ITT – I – I11) R) R2 2 = = RR22 I ITT – R – R22II11
• Adding RAdding R22II11 to each side: to each side:
• II11 R R1 1 ++ R R22 I I11 = R = R22 I ITT = I = I11 (R (R1 1 + R+ R22 ) = R ) = R22 I ITT
• II11 = R = R22 I ITT / / (R(R1 1 + R+ R22 ) )
+
-
Common Sense ChallengeCommon Sense Challenge
• Unlike the voltage divider formula, to Unlike the voltage divider formula, to find Ifind I11, you put R, you put R2 2 , the other resistor, , the other resistor,
at the top of the formula.at the top of the formula.
75
21
1 2
12
1 2
T
T
RI I
R R
RI I
R R
76
Sample ProblemsSample Problems
VT = ???
R1=8kΩ
R2=12kΩ
+
-A = 5ma B
IT = 5maI1 = R2 IT / (R2 +R1) = 12kΩ 5ma/ (8kΩ + 12kΩ )I1 = 3maI2 = R1 IT / (R2 +R1) = 8kΩ 5ma/ (8kΩ + 12kΩ )I2 = 2maVT = 24V
77
Current Dividers In A Two-branch CircuitCurrent Dividers In A Two-branch Circuit
78
• In the previous circuit, the current In the previous circuit, the current through each resistor is given as:through each resistor is given as:
• where, where, II11 is the current through resistor is the current through resistor RR11, and , and II22 is the current through resistor is the current through resistor RR22..
• Current division is similar to voltage Current division is similar to voltage division, but the TOP RESISTORS are division, but the TOP RESISTORS are switchedswitched
IRR
RI .
21
21 I
RR
RI
21
12
Current DividersCurrent Dividers
79
Sample Problems
Comp. Resistance Voltage Current Power
R1 8kΩ
R2 12kΩ
Total
VT = 24V
R1=8kΩ
R2=12kΩ
+
-A B
Current A Current B
PREDICT: Which R has more current
80
Sample Problems
Comp. Resistance Voltage Current Power
R1 8kΩ 24V 3ma 72mW
R2 12kΩ 24V 2ma 48mW
Total 4. 8kΩ 24V 5ma 120mW
VT = 24V
R1=8kΩ
R2=12kΩ
+
-A B
Current A Current B
5ma 2ma
81
Sample Problems
Comp. Resistance Voltage Current Power
R1 4 kΩ 30 mA
R2
Total
VT = ?V
R1=4 kΩ
R2=? kΩ
+
-A B
Current A Current B
50 mA
PREDICT: Size of R2
Size of RT
82
Sample Problems
Comp. Resistance Voltage Current Power
R1 4 kΩ 120 V 30 mA 3.6 W
R2 6 kΩ 120 V 20 mA 2.4 W
Total 2.4 kΩ 120 V 50 mA 6 W
VT = ?V
R1=4 kΩ
R2=? kΩ
+
-A B
Current A Current B
50 mA 30 mA
PREDICT: Size of R2
Size of RT
83
ConductanceConductance
• Conductance is the reciprocal of the Conductance is the reciprocal of the resistanceresistance
• G = 1 / RG = 1 / R
• R = 1 / GR = 1 / G
• G is measured in siemensG is measured in siemens
• R = 2.2 KR = 2.2 KΩΩ
• G = 1 / G = 1 / 2.2 K2.2 KΩΩ = 454 = 454μμSS
84
Sample Problems
Comp R V I G P
R1 7kΩ
R2 14kΩ
Total
VT = 28V
R1=7kΩ
R2=14kΩ
+
-A B
Current A Current B
85
Sample Problems
Comp R V I G P
R1 7kΩ 28V 4ma 142μS 112mW
R2 14kΩ 28V 2ma 71μS 56mW
Total 4.67kΩ 28V 6ma 213 μS 168mW
VT = 28V
R1=7kΩ
R2=14kΩ
+
-A B
Current A Current B
6ma 2ma
86
Always Check Your AnswersAlways Check Your Answers
• GGTT = 1/R = 1/RTT
• P = IP = ITT22RRTT = V = VTT
22/R/RTT
• RRTT = V = VTT/I/ITT
• Your answers must be consistent no Your answers must be consistent no matter what formula you use.matter what formula you use.
• Try it!Try it!
87
Sample Problems
Comp R V I G P
R1 4kΩ
R2 4kΩ
Total
VT = 12V
R1=4kΩ
R2=4kΩ
+
-A B
Current A Current B
88
Sample Problems
Comp R V I G P
R1 4kΩ 12V 3ma 250μS 36mW
R2 4kΩ 12V 3ma 250μS 36mW
Total 2kΩ 12V 3ma 500 μS 72mW
VT = 12V
R1=4kΩ
R2=4kΩ
+
-A B
Current A Current B
6ma 3ma
89
Sample Problems
Comp R V I G P
R1
R2
R3
Total
VT = 12V
R1=4kΩ
R2=4kΩ
+
-A B
Current A Current B Current C
R3=4kΩ
C
90
Sample Problems
Comp R V I G P
R1 4kΩ 12V 3ma 250μS 36mW
R2 4kΩ 12V 3ma 250μS 36mW
R3 4kΩ 12V 3ma 250μS 36mW
Total 1.33kΩ 12V 9ma 750μS 108mW
VT = 12V
R1=4kΩ
R2=4kΩ
+
-A B
Current A Current B Current C
9ma 6ma 3ma
R3=4kΩ
C
Multiple ResistorsMultiple Resistors
• If you have two resistors that are the If you have two resistors that are the same:same:
• Three and four identical resistorsThree and four identical resistors
91
1 1 1 2
2
T
T
R R R R
RR
1 1 1 1 3
3
T
T
R R R R R
RR
1 1 1 1 1 4
4
T
T
R R R R R R
RR
Identical Parallel ResistorsIdentical Parallel Resistors
• If you have “n” identical resistors If you have “n” identical resistors with R ohms each, the total with R ohms each, the total equivalent resistance is:equivalent resistance is:
• The more resistor you add in parallel, The more resistor you add in parallel, the smaller the total resistancethe smaller the total resistance
92
1 1 1 1 1...
T
T
n
R R R R R R
RR
n
93
Sample Problems
Comp R V I G P
R1 60W
R2 75W
Total
VT = 120V
R1=???
R2=????
+
-A B
Current A Current B
94
Sample Problems
Comp R V I G P
R1 240Ω 120V 500ma 4.14mS 60W
R2 192Ω 120V 625ma 5.21mS 75W
Total 107 Ω 120V 1.125A 9.35ms 135W
VT = 120V
R1=???
R2=????
+
-A B
Current A Current B
1.125A 625mA
95
Sample Problems
Comp R V I G P
R1 5ma 100μS
R2 5kΩ
R3
Total 1.5W
VT = ??V
R1=??kΩ
R2=??kΩ
+
-A B
Current A Current B Current C
R3=??kΩ
C
96
Sample Problems
Comp R V I G P
R1 10kΩ 50V 5ma 100μS 250mW
R2 5kΩ 50V 10ma 200μS 500mW
R3 3.33kΩ 50V 15ma 300μS 750mW
Total 1.66 kΩ 50V 30ma 600 μS 1.5W
VT = 12V
R1=??kΩ
R2=??kΩ
+
-A B
Current A Current B Current C
30ma 25ma 15ma
R3=??kΩ
C
97
Sample Problems
Comp R V I G P
R1
R2
R3
R4
Total
VT = 60V
R1=3kΩ
R2=6kΩ
+
-A B
Current A Current B Current C Current D
R3=9kΩ
C
R4=12kΩ
98
Sample Problems
Comp R V I G P
R1 3kΩ 60V 20ma 333μS 1.2W
R2 6kΩ 60V 10ma 167μS 600mW
R3 9kΩ 60V 6.67ma 111μS 400mW
R4 12kΩ 60V 5ma 83 μS 300mW
Total 1.44kΩ 60V 41.7ma 694 μS 2.5W
VT = 60V
R1=3kΩ
R2=6kΩ
+
-A B
Current A Current B Current C Current D
41.7ma 21.7ma 11.7ma 5ma
R3=9kΩ
C
R4=12kΩ
D
99
ObservationsObservations• When When voltage is constantvoltage is constant, like it is in , like it is in
parallel circuits:parallel circuits:• Current is inversely proportional to the Current is inversely proportional to the
value of the resistance value of the resistance (V constant!)(V constant!)• Power is inversely proportional to the Power is inversely proportional to the
value of the resistance value of the resistance (V constant!)(V constant!)• Conductance is inversely proportional to Conductance is inversely proportional to
the value of the resistance the value of the resistance (Always)(Always)– When resistance goes UP, all three other When resistance goes UP, all three other
values go DOWNvalues go DOWN
100
Unit 6 SummaryUnit 6 Summary
• Identifying parallel circuitsIdentifying parallel circuits
• Calculating total resistance in Calculating total resistance in parallelparallel
• Solving for unknown current using Solving for unknown current using KCLKCL
• Solving problems with the current-Solving problems with the current-divider formuladivider formula