unit five: series circuits

1

Unit Five:Series Circuits

John Elberfeld

[email protected]

WWW.J-Elberfeld.com

ET115 DC Electronics

mailto:[email protected]

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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3

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).

4

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.

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.

5

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

?

13

Find the CurrentFind the Current

Voltage = 15 VVoltage = 15 V

Resistance = 1 kResistance = 1 kII = _______ = _______

V = 15V

R = 1kΩ I = ?

14

Find the CurrentFind the Current


Resistance = 1 kResistance = 1 kII = _______ = _______

mA15Amps015.k1

V15

R

VI

V = 15V

R = 1kΩ I = ?

15

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


Resistance = _______Resistance = _______

V = 60V

R = ?Ω I = 4mA

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Find the ResistanceFind the Resistance

Current = 4 mACurrent = 4 mA


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

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

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 = ______________

24

Find the Current Find the Current

V = 120V

R1 = 82 Ω

R2

= 56 Ω

I = ?mAV = 120V R = ???? Ω

Replacement Resistor

25


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


26


RT = R1 + R2 = __________________

V = 120V

R1 = Gn Be Br

R2

= Br Bk R

R1 = ______ R2 = ______

27


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


V = 120V

R1 = Gn Be Br

R2

= Br Bk R

R1 = 560 Ω R2 = 1 kΩ

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RT = R1 + R2 = __________________

V = 120V

R1 = O O Bk

R2

= Y V Bk

R1 = ______ R2 = ______

29


I = ?mAV = 120V R = 80 Ω

RT = 33 Ω + 47 Ω = 80 Ω

IT = VT/RT = 120V / 80 Ω = 1.5 A


V = 120V

R1 = O O Bk

R2

= Y V Bk

R1 = 33 Ω R2 = 47 Ω

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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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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


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

37


• 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

+ -

- +

+

-

+

-

38

• 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

39


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

40


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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• 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

42

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

43

Kirchhoff’s Voltage LawKirchhoff’s Voltage Law

44

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+

-

45

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+

-

46

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

47

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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• 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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• 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

50

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+

-

51

Sample Problems - 2


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+

-

52

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.

53

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

54

Same polaritySame polarity

• Equivalent circuitsEquivalent circuits

+

-+

-

V = 9V

V = 5V

+

-V = 14V

electron flow

55

Opposite PolarityOpposite Polarity

• Equivalent circuitsEquivalent circuits

+

-+

-

V = -9V

V = 3V

+

-

V = -6V

electron flow

56

Complex SourcesComplex Sources

• Equivalent sourceEquivalent source

+

-

+

-

V = -9V

V = 3V

+

-V = 12V +

-V = +6V

electron flow

57

Complex SourcesComplex Sources

• Equivalent sourceEquivalent source

+

-+

-

V = 10 V

V = 8 V

+

-V = 5 V +

-V = + 23V

electron flow

58

Complex Sources - 2Complex Sources - 2

• Equivalent sourceEquivalent source+

-

+

-

V = -50 V

V = -10 V

+

-V = 25 V

+

-

V = - ??V

electron flow

59

Complex Sources - 3Complex Sources - 3

• Equivalent sourceEquivalent source+

-

V = -8 V

+

-V = 8 V

+

-

V = ??

electron flow ?

60

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

61

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

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?

62

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

64

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

65

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.

66

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

67

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+

-

+

-+

-+

-

68

Voltage Reference PointsVoltage Reference Points

69

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

70

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

71


• 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

+ -

- +

+

-

+

-


0V 2.62V

10.56V

15V

7.94V 4.45V

12.39V

-7.94 V

-12.39V

G A

BC

72


• I = _____ mA


V = 12 V

R1 = 470 ΩVCB = _____ V

R3 = 680 ΩVAG = _____ V

R2 = 1.0 kΩVBA = _____ V

G A

BC

73


• I = 5.58 mA


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

74


• I = 2 mA


V = 120V

R1 = 10kΩV = 20 V

R3 = 30kΩV = 60 V

R2 = 20kΩV = 40 V

G A

BC

75


• I = 2 mA


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

76

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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• 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

78

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

79

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

80

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 = ????

+

-+

-

81

ExampleExample


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

+

-+

-

82

ExampleExample


V = 24V

R2 = 5 kΩVR2=15v

R1 = 3 kΩVR1=9V

V= 0 (Ground!)

V=15V

V=24V+

-+

-

83

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

84

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

85

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

PotentiometersPotentiometers

• Rotating the control divides the total Rotating the control divides the total resistance into different proportionsresistance into different proportions

86


• 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=?

+

-+

-


• 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


• 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

+

-+

-

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

90

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

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?

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?

94

• 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

95

• 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

96

• 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…

97

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

98


• 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

99


100


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 ...

(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

102

Sample Problems - 1


R1 2kΩ

R2 4kΩ

R3 6kΩ

Total 36V

V = 36V

R1 = 2kΩ

R3 = 6kΩ

R2 = 4kΩ+

-

103

Sample Problems - 1


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+

-

104

Sample Problems – 2


R1 500μA

R2 4kΩ

R3 6kΩ

Total 10V

V = 10V

R1 = ? kΩ

R3 = 6 kΩ

R2 = 4 kΩ+

-

105

Sample Problems - 2


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+

-

106

Sample Problems - 3


R1

R2 5kΩ

R3 7kΩ 112mW

Total 60V

V = 60V

R1 = ? kΩ

R3 = 7 kΩ

R2 = 5 kΩ+

-

107

Sample Problems - 3


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+

-

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

108

unit five: series circuits

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