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Unit Five:Series Circuits
John Elberfeld
WWW.J-Elberfeld.com
ET115 DC Electronics
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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 5 Objectives - IUnit 5 Objectives - I
• Identify a series resistive circuit.Identify a series resistive circuit.• Determine total resistance, current, and Determine total resistance, current, and
power in a series resistive circuit.power in a series resistive circuit.• State Kirchhoff’s voltage law.State Kirchhoff’s voltage law.• Solve for an unknown voltage in a circuit Solve for an unknown voltage in a circuit
using Kirchhoff’s voltage law (KVL).using Kirchhoff’s voltage law (KVL).• State the voltage divider formula.State the voltage divider formula.• Apply the voltage-divider formula to series Apply the voltage-divider formula to series
circuits to find an unknown quantity circuits to find an unknown quantity (either a voltage or resistance).(either a voltage or resistance).
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Unit 5 Objectives – IIUnit 5 Objectives – II
• Calculate the range of output voltages Calculate the range of output voltages when a potentiometer is used as a when a potentiometer is used as a voltage-divider.voltage-divider.
• 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.
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Unit 5 Objectives – IIIUnit 5 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.
• Troubleshoot circuits constructed in Troubleshoot circuits constructed in Multisim exercises using simulated Multisim exercises using simulated instruments.instruments.
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Reading AssignmentReading Assignment
• Read and study Read and study
• Chapter 4: Series Circuits:Chapter 4: Series Circuits:Pages 109-147Pages 109-147
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Lab AssignmentLab Assignment
• Lab Experiment 7:Lab Experiment 7:“Series Circuits,” page 49 “Series Circuits,” page 49
• 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 5 Homework sheetComplete the Unit 5 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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Series CircuitSeries Circuit
• A series circuit has one path for A series circuit has one path for current flow.current flow.
• The current is the same at any point The current is the same at any point in the circuit.in the circuit.
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SERIES CIRCUITSSERIES CIRCUITS
• Total resistance in a series Total resistance in a series circuit is the summation of the circuit is the summation of the individual resistor values:individual resistor values:
RRTT = = RR1 1 + + RR2 2 + + RR3 3 …. + …. + RRnn
Where Where nn = the = the number of number of resistorsresistors
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OHM’S LAWOHM’S LAW
• Ohm’s Law applies to the total circuitOhm’s Law applies to the total circuit
• Total resistance in a series circuit is Total resistance in a series circuit is equal to the total circuit voltage divided equal to the total circuit voltage divided by the total current by the total current
• IITT is the same at every point in the is the same at every point in the
circuitcircuit
T
TT I
VR
VT = ITRT
IT = VT / RT
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Single Resistor CircuitSingle Resistor Circuit
• Single resistor example (Simplest)Single resistor example (Simplest)
• V = I R or R = V/IV = I R or R = V/I
• R = 10V/2A = 5 R = 10V/2A = 5
?
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Find the CurrentFind the Current
Voltage = 15 VVoltage = 15 V
Resistance = 1 kResistance = 1 kII = _______ = _______
V = 15V
R = 1kΩ I = ?
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Find the CurrentFind the Current
Voltage = 15 VVoltage = 15 V
Resistance = 1 kResistance = 1 kII = _______ = _______
mA15Amps015.k1
V15
R
VI
V = 15V
R = 1kΩ I = ?
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Find the VoltageFind the Voltage
Resistance = 10 kResistance = 10 kCurrent = 4 mACurrent = 4 mA
Voltage = ______Voltage = ______
V = ? V
R = 10kΩ I = 4mA
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Find the VoltageFind the Voltage
Resistance = 10 kResistance = 10 kCurrent = 4 mACurrent = 4 mA
Voltage = ______Voltage = ______
Volts40k10mA4 IRV
V = ? V
R = 10kΩ I = 4mA
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Find the ResistanceFind the Resistance
Current = 4 mACurrent = 4 mA
Voltage = 60 VVoltage = 60 V
Resistance = _______Resistance = _______
V = 60V
R = ?Ω I = 4mA
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Find the ResistanceFind the Resistance
Current = 4 mACurrent = 4 mA
Voltage = 60 VVoltage = 60 V
Resistance = _______Resistance = _______
k15mA4
V60
I
VR
V = 60V
R = ?Ω I = 4mA
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• Series circuit comprised of resistors:Series circuit comprised of resistors:
• To calculate the total resistance we use the To calculate the total resistance we use the formula Rformula RT T = R= R11 + R + R22 + R + R33 + etc. + etc.
• The total resistance for the figure is: The total resistance for the figure is:
• RRTT= 10= 10ΩΩ + + 2020ΩΩ + + 3030ΩΩ = = 60 OHMS60 OHMS
Multiple Resistor ExampleMultiple Resistor Example
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Three ResistorsThree Resistors
• Brown Black Black = ___________Brown Black Black = ___________
• Brown Grey Black = ___________Brown Grey Black = ___________
• Red Red Black = ___________Red Red Black = ___________
• Total Resistance = ___________Total Resistance = ___________
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V = 120V
Br Bk Bk
R R Bk
Br Gy Bk
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Simplify the CircuitSimplify the Circuit
• To make life easier, you can replace many To make life easier, you can replace many resistors in series in the circuit with a resistors in series in the circuit with a single resistor that has the same effect on single resistor that has the same effect on the circuit values as all the resistors it the circuit values as all the resistors it replacedreplaced
• The value of the replacement resistor is The value of the replacement resistor is equal to the sum of all the series resistors equal to the sum of all the series resistors that it is replacingthat it is replacing
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The BIG IdeaThe BIG Idea
• In a pure series circuit, the current in In a pure series circuit, the current in every resistor is the same – the same every resistor is the same – the same as the current that leaves the batteryas the current that leaves the battery
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Find the Current Find the Current 23
V = 95 V
R1
R3
R2
mA
mAmA
mA
• If the current in the bottom left Ammeter is 5 If the current in the bottom left Ammeter is 5 mA, what is the current in the:mA, what is the current in the:
• Bottom Right = ______________Bottom Right = ______________• Upper Left = ______________Upper Left = ______________• Upper Right = ______________Upper Right = ______________
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Find the Current Find the Current
V = 120V
R1 = 82 Ω
R2
= 56 Ω
I = ?mAV = 120V R = ???? Ω
Replacement Resistor
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Find the Current Find the Current
V = 120V
R1 = 82 Ω
R2
= 56 Ω
I = ?mAV = 120V R = 138 Ω
RT = 82 Ω + 56 Ω = 138 Ω
IT = VT/RT = 120V / 138 Ω = 870 mA
Replacement Resistor
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Find the Current Find the Current
RT = R1 + R2 = __________________
V = 120V
R1 = Gn Be Br
R2
= Br Bk R
R1 = ______ R2 = ______
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Find the Current Find the Current
I = ?mAV = 120V R = 1.56 kΩ
RT = 560 Ω + 1 kΩ = 1.56 kΩ
IT = VT/RT = 120V / 1.56 kΩ = 76.8 mA
Replacement Resistor
V = 120V
R1 = Gn Be Br
R2
= Br Bk R
R1 = 560 Ω R2 = 1 kΩ
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Find the Current Find the Current
RT = R1 + R2 = __________________
V = 120V
R1 = O O Bk
R2
= Y V Bk
R1 = ______ R2 = ______
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Find the Current Find the Current
I = ?mAV = 120V R = 80 Ω
RT = 33 Ω + 47 Ω = 80 Ω
IT = VT/RT = 120V / 80 Ω = 1.5 A
Replacement Resistor
V = 120V
R1 = O O Bk
R2
= Y V Bk
R1 = 33 Ω R2 = 47 Ω
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Find the Current Find the Current
V = 95V
R1 = Br Gn R
R3 = Br Bk O
R2
= R R R
R1 = ______ R2 = ______ R3 = ______
RT = R1 + R2 + R3 = __________________
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Find the Current Find the Current
V = 95 V
R1 = 1.5 kΩ
R3 = 10 kΩ
R2
= 2.2 kΩ
I = ?mAV = 95 V R = 13.7 kΩ
RT = 1.5 kΩ + 2.2 kΩ + 10 kΩ = 13.7 kΩ
IT = VT/RT = 95V / 13.7 kΩ = 6.93 mA
Replacement Resistor
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Multiple ResistorsMultiple Resistors
• What is the total resistance of twelve What is the total resistance of twelve 5.6 k5.6 kΩΩ resistors in series? resistors in series?
• RRTT = 12 x 5.6 k = 12 x 5.6 kΩΩ = 67.2 k = 67.2 kΩΩ
• What is the total resistance of six 47What is the total resistance of six 47 ΩΩ resistors, eight 100 resistors, eight 100 ΩΩ resistors and resistors and two 22 two 22 ΩΩ resistors in series? resistors in series?
• _________________ _________________
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Multiple ResistorsMultiple Resistors
• What is the total resistance of twelve What is the total resistance of twelve 5.6 k5.6 kΩΩ resistors in series? resistors in series?
• RRTT = 12 x 5.6 k = 12 x 5.6 kΩΩ = 67.2 k = 67.2 kΩΩ
• What is the total resistance of six 47What is the total resistance of six 47 ΩΩ resistors, eight 100 resistors, eight 100 ΩΩ resistors and resistors and two 22 two 22 ΩΩ resistors in series? resistors in series?
• 1.126 k1.126 kΩΩ
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Missing ResistorMissing Resistor
• Five resistors in series have a total Five resistors in series have a total resistance of 20 kresistance of 20 kΩΩ. .
• The value of four of the resistors are The value of four of the resistors are 4.7 k4.7 kΩΩ, 1.0 k, 1.0 kΩΩ, 2.2 k, 2.2 kΩΩ, and 3.9 k, and 3.9 kΩΩ..
• What is the value of the fifth What is the value of the fifth resistor?resistor?
• RR55 = ________________ = ________________
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Missing ResistorMissing Resistor
• Five resistors in series have a total Five resistors in series have a total resistance of 20 kresistance of 20 kΩΩ. .
• The value of four of the resistors are The value of four of the resistors are 4.7 k4.7 kΩΩ, 1.0 k, 1.0 kΩΩ, 2.2 k, 2.2 kΩΩ, and 3.9 k, and 3.9 kΩΩ..
• What is the value of the fifth resistor?What is the value of the fifth resistor?
• 20 k20 kΩΩ = 4.7k = 4.7kΩΩ+1.0k+1.0kΩΩ+2.2k+2.2kΩΩ+3.9k+3.9kΩΩ+X+X
• 20 k20 kΩΩ = 11.8k = 11.8kΩΩ+X+X
• X = 20 kX = 20 kΩΩ - 11.8k - 11.8kΩΩ = 8.2k = 8.2kΩΩ
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Voltage DropVoltage Drop
• V=IR (Ohm’s Law) applies to each V=IR (Ohm’s Law) applies to each resistor in the circuitresistor in the circuit
• I = 2 mA = the same current in each I = 2 mA = the same current in each and every resistor in a series circuitand every resistor in a series circuit
V = 120V
R1 = 10kΩ
R3 = 30kΩ
R2
= 20kΩ
+
-
RT = 60kΩ
IT=VT/RT
IT = 120 V/ 60kΩ
IT = 2 ma
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Voltage DropVoltage Drop
• I = 2 mAI = 2 mA
• VVAA = V = V11 + V + V22 + V + V3 3 = 20V + 40V + 60V = 120 V= 20V + 40V + 60V = 120 V• The sum of the voltage drops across all The sum of the voltage drops across all
the resistors adds up to the input voltagethe resistors adds up to the input voltage
V = 120V
R1 = 10kΩV = 2 ma 10k Ω =20 V
R3 = 30kΩV = 2ma 30k Ω = 60 V
R2 = 20kΩV = 2ma 20k ΩV= 40 V
+ -
- +
+
-
+
-
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• Voltage drop in the series circuit is Voltage drop in the series circuit is given by the sum of the individual given by the sum of the individual voltage drops across all the voltage drops across all the resistors: resistors:
n
n
VVVV
IRIRIRIRV
.........
........
321
321
Voltage Drop in a Series Resistive Voltage Drop in a Series Resistive CircuitCircuit
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Voltage DropVoltage Drop
V = 5.5 V
R1 = 2.2 kΩV1 = ????
R3 = 1.0 kΩV3 = ??????
R2 = 5.6 kΩV2 = ?????
+ -
- +
+
-
+
-
RT = 8.8kΩ
IT=VT/RT
IT = 5.5 V/8.8kΩ
IT = 625 µa
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Voltage DropVoltage Drop
V = 5.5 V
R1 = 2.2 kΩV1 = 625 µa 2.2 kΩ =1.375 V
R3 = 1.0 kΩV3 = 625 µa 1.0 kΩ = 625 mV
What is the sum of the voltage drops in the resistors?
R2 = 5.6 kΩV2 = 625 µa 5.6 kΩ =3.5 V
+ -
- +
+
-
+
-
RT = 8.8kΩ
IT=VT/RT
IT = 5.5 V/8.8kΩ
IT = 625 µa
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Voltage DropVoltage Drop
• VVTT = V = V11 + V + V22 + V + V3 3 = 1.375V + 3.5V + 625m V = = 1.375V + 3.5V + 625m V =
5.5 V5.5 V• The sum of the voltage drops across all the The sum of the voltage drops across all the
resistors adds up to the input voltageresistors adds up to the input voltage
V = 5.5 V
R1 = 2.2 kΩV1 = 625 µa 2.2 kΩ =1.375 V
R3 = 1.0 kΩV3 = 625 µa 1.0 kΩ = 625 mV
R2 = 5.6 kΩV2 = 625 µa 5.6 kΩ =3.5 V
+ -
- +
+
-
+
-
RT = 8.8kΩ
IT=VT/RT
IT = 5.5 V/8.8kΩ
IT = 625 µa
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Kirchhoff’s Voltage LawKirchhoff’s Voltage Law
• The sum of all voltage drops in a The sum of all voltage drops in a closed loop is equal to the value of closed loop is equal to the value of the applied voltagethe applied voltage– This deals with magnitudesThis deals with magnitudes
• The algebraic sum of all the voltage The algebraic sum of all the voltage drops and all the voltage sources in drops and all the voltage sources in any closed loop equals 0any closed loop equals 0– This includes the sign of the voltageThis includes the sign of the voltage
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Kirchhoff’s Voltage LawKirchhoff’s Voltage Law
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Sample Problems - 1
Comp. Resistance Voltage Current
R1 1 MΩ
R2 2.2 MΩ
R3 560 kΩ
Total 16V
V = 16V
R1 = 1 MΩ
R3 = 560 kΩ
R2 = 2.2 MΩ+
-
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Sample Problems - 1
Comp. Resistance Voltage Current
R1 1 MΩ 4.26 V 4.26 μA
R2 2.2 MΩ 9.37 V 4.26 μA
R3 560 kΩ 2.39 V 4.26 μA
Total 3.76 MΩ 16V 4.26 μA
V = 16V
R1 = 1 MΩ
R3 = 560 kΩ
R2 = 2.2 MΩ+
-
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Power CalculationsPower Calculations
• Power = Voltage x Current = WattsPower = Voltage x Current = Watts
• P = VIP = VI
• The more electric power a motor The more electric power a motor uses, the more work it can do per uses, the more work it can do per secondsecond
• Combining with Ohm’s Law:Combining with Ohm’s Law:
• P = IP = I22RR P = VP = V22/R/R
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Power in a Series CircuitPower in a Series Circuit
• P = IP = I22RR• Because current is the same at every Because current is the same at every
point in a series circuit, the resistance point in a series circuit, the resistance with the smallest value will also with the smallest value will also dissipate the smallest power value.dissipate the smallest power value.
• The largest resistor in the circuit will The largest resistor in the circuit will dissipate the largest amount of power.dissipate the largest amount of power.
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Power in a Series CircuitPower in a Series Circuit
• Since the current is the same at any Since the current is the same at any point in a series circuit, the equation point in a series circuit, the equation
PP = = I I 2 2 RR is often the is often the best equation best equation to use when both to use when both II and and RR are known are known..
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Power in a Series CircuitPower in a Series Circuit
• The total power dissipated in a The total power dissipated in a series circuit is also the amount series circuit is also the amount of power the Power Source must of power the Power Source must deliver. deliver. – The power supplied is equal to the total The power supplied is equal to the total
power dissipated in the circuit elementspower dissipated in the circuit elements
• This may also be expressed as:This may also be expressed as:
PPTT = = PPR R 11 + + PPR R 22 … + … + PPR nR n
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Sample Problems - 2
Comp. Resistance Voltage Current Power
R1 4.7 kΩ
R2 4.7 kΩ
R3 4.7 kΩ
Total 48 V
V = 48 V
R1 = 4.7 kΩ
R3 = = 4.7 kΩ
R2 = 4.7 kΩ+
-
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Sample Problems - 2
Comp. Resistance Voltage Current Power
R1 4.7 kΩ 16 V 3.40 mA 54.3 mW
R2 4.7 kΩ 16 V 3.40 mA 54.3 mW
R3 4.7 kΩ 16 V 3.40 mA 54.3 mW
Total 14.1 kΩ 48 V 3.40 mA 163.4 mW
V = 48 V
R1 = 4.7 kΩ
R3 = = 4.7 kΩ
R2 = 4.7 kΩ+
-
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Why use multiple voltage sources?Why use multiple voltage sources?
• A single voltage source is not able to A single voltage source is not able to supply above its rated power (rated power supply above its rated power (rated power = rated voltage X current capacity).= rated voltage X current capacity).
• If more power is needed, multiple sources If more power is needed, multiple sources are connected in parallel at constant are connected in parallel at constant voltage. voltage.
• However, multiple sources may also be However, multiple sources may also be connected in series, where additional connected in series, where additional (voltage) power is required, with the (voltage) power is required, with the current remaining constant.current remaining constant.
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Series Voltage Sources Series Voltage Sources
• If two voltage sources are in series If two voltage sources are in series and have the same polarity, the and have the same polarity, the voltages addvoltages add
• If two voltage sources are in series If two voltage sources are in series and have opposite polarities, the and have opposite polarities, the voltages subtractvoltages subtract
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Same polaritySame polarity
• Equivalent circuitsEquivalent circuits
+
-+
-
V = 9V
V = 5V
+
-V = 14V
electron flow
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Opposite PolarityOpposite Polarity
• Equivalent circuitsEquivalent circuits
+
-+
-
V = -9V
V = 3V
+
-
V = -6V
electron flow
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Complex SourcesComplex Sources
• Equivalent sourceEquivalent source
+
-
+
-
V = -9V
V = 3V
+
-V = 12V +
-V = +6V
electron flow
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Complex SourcesComplex Sources
• Equivalent sourceEquivalent source
+
-+
-
V = 10 V
V = 8 V
+
-V = 5 V +
-V = + 23V
electron flow
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Complex Sources - 2Complex Sources - 2
• Equivalent sourceEquivalent source+
-
+
-
V = -50 V
V = -10 V
+
-V = 25 V
+
-
V = - ??V
electron flow
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Complex Sources - 3Complex Sources - 3
• Equivalent sourceEquivalent source+
-
V = -8 V
+
-V = 8 V
+
-
V = ??
electron flow ?
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Find the Input Voltage - 1Find the Input Voltage - 1
V = ??V
R1 = 2.75kΩV = ?????
R3 = 6.15kΩV = ????
Determine the input voltage
R2 = 4.1kΩV = ????
+ -
- +
+
-
+
-
I = 2 mAI = 2 mA
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Find the Input Voltage - 1Find the Input Voltage - 1
• VVAA = V = V11 + V + V22 + V + V3 3 = 5.5V + 8.2V + 1.23V = 26 V= 5.5V + 8.2V + 1.23V = 26 V
• The sum of the voltage drops across all The sum of the voltage drops across all the resistors adds up to the input voltagethe resistors adds up to the input voltage
V = ??V
R1 = 2.75kΩV = 5.5 V
R3 = 6.15kΩV = 12.3 V
R2 = 4.1kΩV = 8.2 V
+ -
- +
+
-
+
-
I = 2 mAI = 2 mA
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Find the Missing VoltageFind the Missing Voltage
• Five resistors are in series with a 20 Five resistors are in series with a 20 V source.V source.
• The voltage drops across four of the The voltage drops across four of the resistors are 1.5 V, 5.5 V, 3 V, and 6V.resistors are 1.5 V, 5.5 V, 3 V, and 6V.
• How much is the voltage drop across How much is the voltage drop across the fifth resistor? the fifth resistor?
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Find the Missing VoltageFind the Missing Voltage
• Five resistors are in series with a 20 Five resistors are in series with a 20 V source.V source.
• The voltage drops across four of the The voltage drops across four of the resistors are 1.5 V, 5.5 V, 3 V, and 6V.resistors are 1.5 V, 5.5 V, 3 V, and 6V.
• How much is the voltage drop across How much is the voltage drop across the fifth resistor? the fifth resistor?
63
20 V = 1.5 V + 5.5 V + 3V + 6 V + X20 V = 1.5 V + 5.5 V + 3V + 6 V + XX = 20 V - 1.5 V - 5.5 V - 3V - 6 V X = 20 V - 1.5 V - 5.5 V - 3V - 6 V X = 4 VX = 4 V
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Voltage Reference PointsVoltage Reference Points
• Voltages are always relativeVoltages are always relative– That is, one point on a circuit has a That is, one point on a circuit has a
voltage that is higher (+) or lower (-) voltage that is higher (+) or lower (-) with respect to another pointwith respect to another point
• Often voltages are written with Often voltages are written with respect to groundrespect to ground– Ground is often the negative terminal of Ground is often the negative terminal of
the power supplythe power supply
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What is a circuit ground?What is a circuit ground?
• In most electronic equipment, a large In most electronic equipment, a large conductive area on a printed circuit board conductive area on a printed circuit board is used as the common reference point. is used as the common reference point.
• This point is called the ground. This point is called the ground.
• This ground provides a convenient way of This ground provides a convenient way of connecting all common points within the connecting all common points within the circuit back to one side of the battery or circuit back to one side of the battery or the voltage source.the voltage source.
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Voltages with Respect to GroundVoltages with Respect to Ground
V = 18V
R1 = 1 kΩVR1=3V
R3 = 2 kΩVR3=6v
R2 = 3 kΩVR2=9V
V=0V (Ground!)
V=6V
V=15V
V=18V =VT+
-
+
-+
-+
-
I= 3mA
V=0V
15V = voltage across 5k replacement
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Another Reference PointAnother Reference Point
V = 18V
R1 = 1 kΩVR1=3V
R3 = 2 kΩVR3=6v
R2 = 3 kΩVR2=9V
V=-15V (Ground!)
V=-9V
V=0V (Black Lead)
V=+3V+
-
+
-+
-+
-
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Voltage Reference PointsVoltage Reference Points
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Decoding the SymbolsDecoding the Symbols
• VVC C = Voltage from point C to ground= Voltage from point C to ground– Single letter implies the second point is Single letter implies the second point is
groundground– Red lead on the letter, black on groundRed lead on the letter, black on ground
• VVDADA = Voltage from Point D to Point A = Voltage from Point D to Point A– Double letter implies the second point is Double letter implies the second point is
another location in the circuitanother location in the circuit– Red lead on the first letter, black on the Red lead on the first letter, black on the
second lettersecond letter
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Voltage Calculations
• I = 794 μA
V = 15V
R1 = 5.6 kΩV = 4.45 V
R3 = 3.3 kΩV = 2.62 V
R2 = 10 kΩV = 7.94 V
+ -
- +
+
-
+
-
VG VA VB VC VBA VCB VCA VAB VAC
0V 2.62V
10.56V
15V V V V
-7.94 V V
G A
BC
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Voltage Calculations
• I = 794 μA
V = 15V
R1 = 5.6 kΩV = 4.45 V
R3 = 3.3 kΩV = 2.62 V
R2 = 10 kΩV = 7.94 V
+ -
- +
+
-
+
-
VG VA VB VC VBA VCB VCA VAB VAC
0V 2.62V
10.56V
15V
7.94V 4.45V
12.39V
-7.94 V
-12.39V
G A
BC
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Voltage Calculations
• I = _____ mA
VG VA VB VC VBA VCB VCA VAB VAC
V = 12 V
R1 = 470 ΩVCB = _____ V
R3 = 680 ΩVAG = _____ V
R2 = 1.0 kΩVBA = _____ V
G A
BC
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Voltage Calculations
• I = 5.58 mA
VG VA VB VC VBA VCB VCA VAB VAC
0 3.79V
9.37V
12V 5.58V 2.62V 8.2V -5.58V -8.2V
V = 12 V
R1 = 470 ΩVCB = 2.62 V V
R3 = 680 ΩVAG = 3.79 V
R2 = 1.0 kΩVBA = 5.58 V
G A
BC
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Voltage Calculations
• I = 2 mA
VG VA VB VC VBA VCB VCA VAB VAC
V = 120V
R1 = 10kΩV = 20 V
R3 = 30kΩV = 60 V
R2 = 20kΩV = 40 V
G A
BC
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Voltage Calculations
• I = 2 mA
VG VA VB VC VBA VCB VCA VAB VAC
0V 60V 100V 120V 40V 20V 60V -40V -60V
V = 120V
R1 = 10kΩV = 20 V
R3 = 30kΩV = 60 V
R2 = 20kΩV = 40 V
G A
BC
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Voltage DividerVoltage Divider
• A voltage divider consists of several A voltage divider consists of several resistors connected in series across resistors connected in series across a voltage source. a voltage source.
• The voltages developed across each The voltages developed across each of these resistors can be used for of these resistors can be used for different applications.different applications.
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Voltage DividerVoltage Divider
• Resistive circuits are used to Resistive circuits are used to obtain some percentage of the obtain some percentage of the applied voltage source.applied voltage source.
RT = 30kΩ
IT=VT/RT
IT = 90 V/ 30kΩ
IT = 3 ma
VR1= 3ma 10kΩ
VR1= 30V
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ExampleExample
V = 30V
R1 = 2 kΩVR1=6V
R3 = 5 kΩVR3=15v
R2 = 3 kΩVR2=9V
V= 0 (Ground!)
V=15V
V=24V
V=30V+
-
+
-+
-+
-
RT = 10kΩ
IT=VT/RT
IT = 30 V/ 10kΩ
IT = 3 ma
VR3= 3ma 5kΩ
VR3= 15V
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Voltage Divider ShortcutsVoltage Divider Shortcuts
• Voltage dividers are used in so many Voltage dividers are used in so many circuits they deserve more studycircuits they deserve more study
• Voltage dividers are a way to access Voltage dividers are a way to access a smaller voltage from a large a smaller voltage from a large voltage sourcevoltage source
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ExampleExample
• Find the total current, and then find Find the total current, and then find the voltage drop across the resistorsthe voltage drop across the resistors
V = 24V
R2 = 5 kΩVR2=____V
R1 = 3 kΩVR1=____V
V= 0 (Ground!)
RT = 3 kΩ + 5 kΩ
RT = 8 kΩ
IT = VT/RT = ????
IT = ????
V1 = ITR1 = ?? 3 kΩ
V1 = ???
V2 = ITR2 = ???? 5 kΩ
V2 = ????
+
-+
-
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ExampleExample
• Find the total current, and then find Find the total current, and then find the voltage drop across the resistorsthe voltage drop across the resistors
V = 24V
R2 = 5 kΩVR2=____V
R1 = 3 kΩVR1=____V
V= 0 (Ground!)
RT = 3 kΩ + 5 kΩ
RT = 8 kΩ
IT = VT/RT = 24 V/8 kΩ
IT = 3 mA
VR1 = ITR1 = 3 mA 3 kΩ
VR1 = 9 V
VR2 = ITR2 = 3 mA 5 kΩ
VR2 = 15 V
+
-+
-
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ExampleExample
• Find the total current, and then find Find the total current, and then find the voltage drop across the resistorsthe voltage drop across the resistors
V = 24V
R2 = 5 kΩVR2=15v
R1 = 3 kΩVR1=9V
V= 0 (Ground!)
V=15V
V=24V+
-+
-
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Voltage Divider in 1 StepVoltage Divider in 1 Step• IITT = V = VTT / (R / (R11 + R + R22))
• VVRR22 = = IITT RR22
• VVRR22 = [ V = [ VTT / (R / (R11 + R + R22) ] x R) ] x R22
• VVRR22 = (V = (VTT R R2 2 ) / (R) / (R11 + R + R22))
• If you work with lots of voltage dividers, you will find If you work with lots of voltage dividers, you will find it easier to memorize the voltage divider equationit easier to memorize the voltage divider equation
22
1 2s
RV V
R R
VVTT = V = VSS = V = VSourceSource
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ExampleExample
V = 16V
R2 = 5 kΩ
R1 = 3 kΩ
V= 0 (Ground!)
VR2=?
+
-+
-
I = 16V/(3kΩ+5kΩ)=2mA
V2 = 2mA 5kΩ = 10 V2
21 2
2
516 10
3 5
s
RV V
R R
kV V V
k k
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Voltage Divider RuleVoltage Divider Rule
• The voltage across any resistor in a The voltage across any resistor in a series may be determined by the series may be determined by the following equation:following equation:
• This works with many resistors!This works with many resistors!
TT
XX V
R
RV
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PotentiometersPotentiometers
• Rotating the control divides the total Rotating the control divides the total resistance into different proportionsresistance into different proportions
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Voltage DividerVoltage Divider
• The potentiometer The potentiometer acts like a voltage acts like a voltage divider:divider:
87
V = 15V
R2 = 35 kΩ
R1 = 15 kΩ
V= 0 (Ground!)
Wiper
V2=?
+
-+
-
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Voltage DividerVoltage Divider
• The potentiometer The potentiometer acts like a voltage acts like a voltage divider:divider:
88
V = 15V
R2 = 35 kΩ
R1 = 15 kΩ
V= 0 (Ground!)
Wiper
V2=?
+
-+
-
22
1 2
2
3515 10.5
15 35
s
RV V
R R
kV V V
k k
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Voltage DividerVoltage Divider
• If a 60k potentiometer is used to If a 60k potentiometer is used to generate 5 V from a 16 V power generate 5 V from a 16 V power source, what is the resistance source, what is the resistance between the wiper and ground?between the wiper and ground?
89
V = 16V
R2 = ?? kΩ
R1 = ?? kΩ
V= 0 (Ground!)
Wiper
V2=5 V
+
-+
-
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Grind it out!Grind it out!
• RRTT = 60 k = 60 kΩΩ, V = 16 V, , V = 16 V,
• I = V / R so I = 267 I = V / R so I = 267 μμAA
• RR22 = V = V22 / I / I
• RR22 = 5 V / 267 = 5 V / 267 μμA = 18.7 kA = 18.7 kΩΩ
• Double Check: The other resistance isDouble Check: The other resistance is 60 k 60 kΩΩ - 18.7 k - 18.7 kΩΩ = 41.25 k = 41.25 kΩΩ
• The voltage drop is 41.25 kThe voltage drop is 41.25 kΩΩ • 267 • 267 μμAA
• VV11 = 11 V and V = 11 V and V11 + V + V22 = 11V + 5 V = 16 V, the source = 11V + 5 V = 16 V, the source
voltagevoltage
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Another Approach Another Approach
• RR11+R+R22= 60 k= 60 kΩΩ – the value of the – the value of the
potentiometerpotentiometer
91
22
1 2
5 1660
5 6018.75
16
s
RV V
R R
XV V
k
V kX k
V
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92 Trouble Shooting: Trouble Shooting: Open in a Series CircuitOpen in a Series Circuit
• A series circuit is open when no A series circuit is open when no current flows in the circuit even current flows in the circuit even though there is a voltage source in though there is a voltage source in the circuitthe circuit..
• Why does the meter read V Volts?Why does the meter read V Volts?
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93 Trouble Shooting: Trouble Shooting: Open in a Series CircuitOpen in a Series Circuit
• An opens blocks all currentAn opens blocks all current
• How much power is used in the How much power is used in the circuit if there is no current?circuit if there is no current?
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• A short in a series circuit represents A short in a series circuit represents zero resistance between the terminals zero resistance between the terminals that are shorted.that are shorted.
• Why does the meter read 0 V? Why does the meter read 0 V?
Shorts in a Series CircuitShorts in a Series Circuit
Shorted Resistor
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• A short reduces the resistance in a series A short reduces the resistance in a series circuit circuit
• A short INCREASES the currentA short INCREASES the current
• A short INCREASES the total power A short INCREASES the total power
• P = V I - V stays the same, I P = V I - V stays the same, I increasesincreases
Shorts in a Series CircuitShorts in a Series Circuit
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• A circuit consists of 10 electrical A circuit consists of 10 electrical bulbs in series. If one of the bulbs is bulbs in series. If one of the bulbs is removed from the circuit, what will removed from the circuit, what will happen when the circuit is switched happen when the circuit is switched on? Give reasons to support your on? Give reasons to support your answer. answer.
• What is the voltage across the open in What is the voltage across the open in the circuit?the circuit?
Think about it…Think about it…
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Troubleshooting a Series CircuitTroubleshooting a Series Circuit
• If an open occurs at any point in If an open occurs at any point in a simple series circuit, current a simple series circuit, current will decrease to 0 A.will decrease to 0 A.
• All voltage drops across all All voltage drops across all resistors will decrease to 0 Vresistors will decrease to 0 V
• All the applied voltage appears All the applied voltage appears across the break in the circuitacross the break in the circuit
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Troubleshooting a Series CircuitTroubleshooting a Series Circuit
• If a short occurs across (through) If a short occurs across (through) a resistor, current will increase a resistor, current will increase because resistance decreases.because resistance decreases.
• As current increases, the voltage As current increases, the voltage across the remaining resistors across the remaining resistors will increase.will increase.
• Total Power will increaseTotal Power will increase
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Troubleshooting a Series CircuitTroubleshooting a Series Circuit
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Troubleshooting a Series CircuitTroubleshooting a Series Circuit
If a total short occurs, If a total short occurs, RRTT = 0 = 0
• Current will attempt to increase Current will attempt to increase to unacceptable levelsto unacceptable levels
• A protective device reacts to A protective device reacts to open the circuit, or ...open the circuit, or ...
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(Top)(Top)
Br Bk R = _____Br Bk R = _____
(Bottom)(Bottom)
0 0 R = _____0 0 R = _____
(Diagonal)(Diagonal)
V R R = _____V R R = _____
(Vertical)(Vertical)
Ohm meter = _____Ohm meter = _____
101Ohm Meter
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Sample Problems - 1
Comp. Resistance Voltage Current Power
R1 2kΩ
R2 4kΩ
R3 6kΩ
Total 36V
V = 36V
R1 = 2kΩ
R3 = 6kΩ
R2 = 4kΩ+
-
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Sample Problems - 1
Comp. Resistance Voltage Current Power
R1 2kΩ 6V 3ma 18mW
R2 4kΩ 12V 3ma 36mW
R3 6kΩ 18V 3ma 54mW
Total 12kΩ 36V 3ma 108mW
V = 36V
R1 = 2kΩ
R3 = 6kΩ
R2 = 4kΩ+
-
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Sample Problems – 2
Comp. Resistance Voltage Current Power
R1 500μA
R2 4kΩ
R3 6kΩ
Total 10V
V = 10V
R1 = ? kΩ
R3 = 6 kΩ
R2 = 4 kΩ+
-
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Sample Problems - 2
Comp. Resistance Voltage Current Power
R1 10kΩ 5V 500μA 2.5mW
R2 4kΩ 2V 500μA 1mW
R3 6kΩ 3V 500μA 1.5mW
Total 20kΩ 10V 500μA 5mW
V = 10V
R1 = ? kΩ
R3 = 6 kΩ
R2 = 4 kΩ+
-
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Sample Problems - 3
Comp. Resistance Voltage Current Power
R1
R2 5kΩ
R3 7kΩ 112mW
Total 60V
V = 60V
R1 = ? kΩ
R3 = 7 kΩ
R2 = 5 kΩ+
-
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Sample Problems - 3
Comp. Resistance Voltage Current Power
R1 3kΩ 12V 4mA 48mW
R2 5kΩ 20V 4mA 80mw
R3 7kΩ 28V 4mA 112mW
Total 15kΩ 60V 4mA 240mW
V = 60V
R1 = ? kΩ
R3 = 7 kΩ
R2 = 5 kΩ+
-
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Summary Unit 5Summary Unit 5
• Identifying series circuitsIdentifying series circuits
• Calculating total resistance in seriesCalculating total resistance in series
• Solving for unknown voltage using Solving for unknown voltage using KVLKVL
• Solving problems with the voltage-Solving problems with the voltage-divider formuladivider formula
• Solving problems with a Solving problems with a potentiometer as a voltage-dividerpotentiometer as a voltage-divider
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