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Unit Four:Unit Four:Energy and PowerEnergy and Power

John ElberfeldJohn Elberfeld

JElberfeld@itt-tech.edu

WWW.J-Elberfeld.comWWW.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 4 Objectives - IUnit 4 Objectives - I

• Define energy and power.Define energy and power.• State the common units of energy and State the common units of energy and

power.power.• Perform energy and power calculations.Perform energy and power calculations.• List factors that affect the power rating of List factors that affect the power rating of

resistors.resistors.• Explain energy conversion and voltage Explain energy conversion and voltage

drop in a resistance.drop in a resistance.

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Unit 4 Objectives – IIUnit 4 Objectives – II

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

• Use a digital multimeter (DMM) to measure Use a digital multimeter (DMM) to measure a predetermined low voltage on a power a predetermined low voltage on a power supply.supply.

• Measure resistances and voltages in a DC Measure resistances and voltages in a DC circuit using a DMM.circuit using a DMM.

Reading AssignmentReading Assignment

• Read and study Read and study

• Chapter 3, Energy and Power:Chapter 3, Energy and Power: Pages 81-98 Pages 81-98

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Lab AssignmentLab Assignment

• Lab Experiment 6:Lab Experiment 6:

• Power in DC Circuits Pages 41-38Power in DC Circuits Pages 41-38

• 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

• Use the special handout to organize Use the special handout to organize your datayour data

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Written AssignmentsWritten Assignments

• Do the Unit 4 Homework handout.Do the Unit 4 Homework handout.

• Be prepared for a UNIT EXAM similar Be prepared for a UNIT EXAM similar to the homeworks and quizzesto the homeworks and quizzes

• If there are any calculations, you If there are any calculations, you must show ALL your work for credit:must show ALL your work for credit:– Write down the formulaWrite down the formula– Show numbers in the formulaShow numbers in the formula– Circle answer with the proper unitsCircle answer with the proper units

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EnergyEnergy

• Energy is the ability to do workEnergy is the ability to do work

• Energy can be measured in calories Energy can be measured in calories (for food, heat, etc.) or in Joules for (for food, heat, etc.) or in Joules for physics and electricityphysics and electricity

• The more Joules you have available, The more Joules you have available, the more work you can dothe more work you can do– Work occurs when you use a force to Work occurs when you use a force to

move an object in the direction of the move an object in the direction of the forceforce

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TimeTime

• Time is an important consideration when Time is an important consideration when you have a job to doyou have a job to do

• It takes the same energy to lift one brick It takes the same energy to lift one brick at a time from the floor to a table until at a time from the floor to a table until 1000 bricks have been moved, compared 1000 bricks have been moved, compared to a forklift lifting the entire pallet of to a forklift lifting the entire pallet of bricks to the table in a matter of secondsbricks to the table in a matter of seconds

• The work done, and the energy used to The work done, and the energy used to do the work is the samedo the work is the same

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PowerPower

• Power is how much work you do per Power is how much work you do per second.second.

• Power is measured in WattsPower is measured in Watts

• P = Work / time = Joules / secondP = Work / time = Joules / second

• Power = P = W/t = (Watts)Power = P = W/t = (Watts)

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IdeasIdeas

• If you lift 1 thousand bricks, one If you lift 1 thousand bricks, one brick at a time, from the floor to a brick at a time, from the floor to a table in 5 hours, you do the same table in 5 hours, you do the same work as a forklift that lifts the bricks work as a forklift that lifts the bricks all at once in 20 seconds.all at once in 20 seconds.

• The forklift is more powerful than The forklift is more powerful than you are because it does more work you are because it does more work per second than you did.per second than you did.

General Power FormulaGeneral Power Formula

• P = work/timeP = work/time

• For example, if you do 350 J of work For example, if you do 350 J of work in 1 s, your power is:in 1 s, your power is:

• P = 350 J / 1 s = 350 J/s = 350 WattsP = 350 J / 1 s = 350 J/s = 350 Watts

• Power is measured in WattsPower is measured in Watts

• 1 Watt = 1 Joule / second1 Watt = 1 Joule / second

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Power CalculationsPower Calculations

• What is the power in watts when What is the power in watts when 7,500 J of energy is used in 5 hours?7,500 J of energy is used in 5 hours?

• P = W / tP = W / t

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Power CalculationsPower Calculations

• What is the power in watts when What is the power in watts when 7,500 J of energy is used in 5 hours?7,500 J of energy is used in 5 hours?

• P = W / t = 7,500 J/ 5 h = 1500 J/hP = W / t = 7,500 J/ 5 h = 1500 J/h

• Watt is a Joule/second, so we have Watt is a Joule/second, so we have to convert:to convert:

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7,500 1.417 417

5 3600

j hW mW

h s

Unit ConversionsUnit Conversions

• The easiest way to convert watts to The easiest way to convert watts to kilowatts, or microwatts, or back, is to kilowatts, or microwatts, or back, is to use the ENG key on your calculatoruse the ENG key on your calculator

• Enter the value as it is given, and hit Enter the value as it is given, and hit ENG or SHIFT-ENG until you get the ENG or SHIFT-ENG until you get the desired exponentdesired exponent

• Get 10Get 1033 of kilowatts or 10 of kilowatts or 10-6-6 for for microwatts, for examplemicrowatts, for example

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ConvertConvert

Original valueOriginal value Conversion Conversion

1000 W1000 W kWkW

3750 W3750 W kWkW

160 W160 W kWkW

50,000 W50,000 W kWkW

1,000,000 W1,000,000 W MWMW

3 x 103 x 106 6 WW MWMW

15 x 1015 x 1077 W W MWMW

8.700 kW8.700 kW MWMW

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ConvertConvert

Original valueOriginal value Conversion Conversion

1 W1 W mWmW

0.4 W0.4 W mWmW

0.002 W0.002 W mWmW

0.0125 W0.0125 W mWmW

1.5 kW1.5 kW WW

0.5 MW0.5 MW WW

350 mW350 mW WW

9,000 9,000 μμWW WW

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EnergyEnergy

• You buy ENERGY, not power, when You buy ENERGY, not power, when you pay your electric billyou pay your electric bill

• Energy is sold in Kilowatt-hoursEnergy is sold in Kilowatt-hours– Energy =Power x timeEnergy =Power x time

• For example, National Grid charges For example, National Grid charges $0.08 per kWh for the electric energy, $0.08 per kWh for the electric energy, plus another $0.04 per kWh to deliver plus another $0.04 per kWh to deliver the electricity to my housethe electricity to my house– Plus a ton of other feesPlus a ton of other fees

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Energy ConsumptionEnergy Consumption

• On the average, my house uses 1.03 kW On the average, my house uses 1.03 kW of power at all timesof power at all times

• A month has 30 days and 24 hours in a A month has 30 days and 24 hours in a day, for a total of 720 hoursday, for a total of 720 hours

• If P = W/t the W (Energy) = P • TIf P = W/t the W (Energy) = P • T• Energy = 1.03 kW • 720 hoursEnergy = 1.03 kW • 720 hours• Energy = 742 kWh (kiloWatt hours)Energy = 742 kWh (kiloWatt hours)• At $0.12/kWh, my bill is $89.04 At $0.12/kWh, my bill is $89.04 – Plus another $34 in fees and taxesPlus another $34 in fees and taxes

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Energy in Joules Energy in Joules

• Doing unit conversions:Doing unit conversions:

Humans are much more comfortable Humans are much more comfortable with numbers like 742 compared to with numbers like 742 compared to 2.67 x 102.67 x 109 9 or Giga anythingsor Giga anythings

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3 3600742 742 10 2.67

J skWh x GJ

s h

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Power FormulaPower Formula

• Power = Work / time, but in electric Power = Work / time, but in electric terms:terms:

• P = V IP = V I• The product of voltage and current The product of voltage and current

gives the electric work done per gives the electric work done per second, or power.second, or power.

LogicLogic

• 1 Volt = 1 Joule / Coulomb1 Volt = 1 Joule / Coulomb

• 1 Ampere = 1 Coulomb / Second1 Ampere = 1 Coulomb / Second

• P = V•I P = V•I

• Unit are Unit are Power = volts x amps = J/s = wattsPower = volts x amps = J/s = watts

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1 1 11 1

J C JVA W

C s s

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Modify Power FormulaModify Power Formula

• Power: P = V IPower: P = V I

• Ohms Law: V = I ROhms Law: V = I R

• P = P = VV I I– Substitute Substitute V V = I R= I R

• P = P = (I R) (I R) I = II = I22RR

• P = IP = I22RR

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Modify Power FormulaModify Power Formula

• Power: P = V IPower: P = V I

• Ohms Law: V = I ROhms Law: V = I R

• P = V P = V II– Substitute Substitute I I = V / R= V / R

• P = V P = V (V / R) (V / R) = V= V2 2 / R/ R

• P = VP = V2 2 / R/ R

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PowerPower

• You must memorize:You must memorize:

P = V IP = V I• You can use algebra to find the other You can use algebra to find the other

methods to calculate powermethods to calculate power

• P = IP = I22 R R

• P = VP = V22 / R / R

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Power ExamplePower Example

• Fact: P = V IFact: P = V I

• What power is produced by 3 mA of What power is produced by 3 mA of current with a 5.5 V drop?current with a 5.5 V drop?

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Power ExamplePower Example

• Fact: P = V IFact: P = V I

• What power is produced by 3 mA of What power is produced by 3 mA of current with a 5.5 V drop?current with a 5.5 V drop?

• P = V IP = V I

• P = 5.5 V • 3 maP = 5.5 V • 3 ma

• P 16.5 mWP 16.5 mW

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Power ExamplePower Example

• What is the power produced by 3 What is the power produced by 3 A A with a 115 V drop?with a 115 V drop?

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Power ExamplePower Example

• Fact: P = V IFact: P = V I

• What is the power produced by 3 What is the power produced by 3 A A with a 115 V drop?with a 115 V drop?

• P = V IP = V I

• P = 115 V • 3 AP = 115 V • 3 A

• P = 245 WP = 245 W

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Power ExamplePower Example

• Fact: P = V IFact: P = V I

• What power is produced by 2 A of What power is produced by 2 A of current with a 3 V drop?current with a 3 V drop?

• What is the power produced by 6 What is the power produced by 6 μμA A with a 3 kV drop?with a 3 kV drop?

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Power ExamplePower Example

• Fact: P = V IFact: P = V I

• What power is produced by 2 A of What power is produced by 2 A of current with a 3 V drop?current with a 3 V drop?

• P = V I = 3 V x 2 A = 6 VA = 6 WP = V I = 3 V x 2 A = 6 VA = 6 W

• What is the power produced by 6 What is the power produced by 6 μμA A with a 3 kV drop?with a 3 kV drop?

• P = V I = P = V I = 6 6 μμA x 3 kV = 18mW A x 3 kV = 18mW

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PowerPower

• Power: P = VI or P = IPower: P = VI or P = I22R or P = VR or P = V22 / R / R

• What is the power when there is 500 What is the power when there is 500 mA of current through a 4.7 kmA of current through a 4.7 kΩΩ resistor?resistor?

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PowerPower

• Power: P = VI or P = IPower: P = VI or P = I22R or P = VR or P = V22 / R / R

• What is the power when there is 500 What is the power when there is 500 mA of current through a 4.7 kmA of current through a 4.7 kΩΩ resistor?resistor?

• V = I R = 500 mA • 4.7 kV = I R = 500 mA • 4.7 kΩΩ = 2.35 kV = 2.35 kV

• P = V I = 2.35 kV • 500 mA = 1.18 kWP = V I = 2.35 kV • 500 mA = 1.18 kW

• OROR

• P = IP = I22R = (500 ma)R = (500 ma)2 • 2 • 4.7 k4.7 kΩΩ = 1.18 kW = 1.18 kW

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PowerPower

• Power: P = VI or P = IPower: P = VI or P = I22R or P = VR or P = V22 / R / R What is the power when there is What is the power when there is 100 100 μμA of current through a 10 kA of current through a 10 kΩΩ resistor?resistor?

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PowerPower

• Power: P = VI or P = IPower: P = VI or P = I22R or P = VR or P = V22 / R / R What is the power when there is What is the power when there is 100 100 μμA of current through a 10 kA of current through a 10 kΩΩ resistor?resistor?

• V = I R = 100 V = I R = 100 μμA • 10 kA • 10 kΩΩ = 1 V = 1 V• P = V I = 1 V • 100 P = V I = 1 V • 100 μμA = 100 A = 100 μμW W • OROR• P = IP = I22R = (100 R = (100 μμa)a)2 • 2 • 10 k10 kΩΩ = 100 = 100 μμW W • Which is easier?Which is easier?

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PowerPower

• Power: P = VI or P = IPower: P = VI or P = I22R or P = VR or P = V22 / R / R What is the power when there is What is the power when there is 60 V across a 620 60 V across a 620 ΩΩ resistor? resistor?

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PowerPower

• Power: P = VI or P = IPower: P = VI or P = I22R or P = VR or P = V22 / R / R What is the power when there is What is the power when there is 60 V across a 620 60 V across a 620 ΩΩ resistor? resistor?

• I = V / R = 60 V / 620 I = V / R = 60 V / 620 ΩΩ = 96.8 mA = 96.8 mA

• P = V I = 60 V • 96.8 mA = 5.8 WP = V I = 60 V • 96.8 mA = 5.8 W

• OROR

• P = VP = V2 2 / R = (60 V)/ R = (60 V)22 / 620 / 620 ΩΩ = 5.8 W = 5.8 W

• Which method is easier?Which method is easier?

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PowerPower

• Power: P = VI or P = IPower: P = VI or P = I22R or P = VR or P = V22 / R / R• What is the power when there is What is the power when there is

1.5 V across a 56 1.5 V across a 56 ΩΩ resistor? resistor?

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PowerPower

• Power: P = VI or P = IPower: P = VI or P = I22R or P = VR or P = V22 / R / R• What is the power when there is What is the power when there is

1.5 V across a 56 1.5 V across a 56 ΩΩ resistor? resistor?• I = V / R = 1.5 V / 56 I = V / R = 1.5 V / 56 ΩΩ = 26.8 mA = 26.8 mA• P = V I = 1.5 V • 26.8 mA = 40.2 mWP = V I = 1.5 V • 26.8 mA = 40.2 mW• OROR• P = VP = V2 2 / R = (1.5 V)/ R = (1.5 V)22 / 56 / 56 ΩΩ = 40.2 mW = 40.2 mW• Which method is easier?Which method is easier?

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PowerPower

• Power: P = VI or P = IPower: P = VI or P = I22R or P = VR or P = V22 / R / R

• If power is 100 W, and current is 2 A, If power is 100 W, and current is 2 A, find the resistance.find the resistance.

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PowerPower

• Power: P = VI or P = IPower: P = VI or P = I22RR• If power is 100 W, and current is 2 A, If power is 100 W, and current is 2 A,

find the resistance.find the resistance.• P = VI, so V = P / I = 100 W / 2 A = 50 VP = VI, so V = P / I = 100 W / 2 A = 50 V• V = I R so R = V / I = 50 V / 2 A = 25 V = I R so R = V / I = 50 V / 2 A = 25 ΩΩ• OROR• P = IP = I22R, so R = P / IR, so R = P / I22 = 100 W / (2 A) = 100 W / (2 A)22 = =• R = 25 R = 25 ΩΩ

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PowerPower

• Power: P = VI or P = IPower: P = VI or P = I22R or P = VR or P = V22 / R / R

• If power is 75 W, and voltage is 120 V, If power is 75 W, and voltage is 120 V, find the resistance.find the resistance.

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PowerPower

• Power: P = VI or P = VPower: P = VI or P = V2 2 // RR• If power is 75 W, and voltage is 120 V, find the If power is 75 W, and voltage is 120 V, find the

resistance.resistance.• P = VI, so I = P / V = 75 W / 120 V = P = VI, so I = P / V = 75 W / 120 V =

I = 625 mAI = 625 mA• V = I R, R = V / I = 120 V / 625 mA = 192 V = I R, R = V / I = 120 V / 625 mA = 192 ΩΩ• OROR• P = VP = V2 2 / R, so R = V/ R, so R = V2 2 / P = (120 V)/ P = (120 V)22 / 75 W = / 75 W =• R = 192 R = 192 ΩΩ

Practical ApplicationsPractical Applications

• A resistor is rated as ½ W. It has a A resistor is rated as ½ W. It has a value of 1.2 kvalue of 1.2 kΩΩ..

• A. What is the maximum current?A. What is the maximum current?

• B. What is the maximum voltage?B. What is the maximum voltage?

• C. What is the product of maximum C. What is the product of maximum current time maximum voltage equal current time maximum voltage equal to?to?

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Practical ApplicationsPractical Applications

• A resistor is rated as ½ W. It has a A resistor is rated as ½ W. It has a value of 1.2 kvalue of 1.2 kΩΩ..

• A. What is the maximum current?A. What is the maximum current?

• P = IP = I22 R and P = V R and P = V22 / R / R

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.520.4

1.2

.5 1.2 24.5

24.5 20.4 0.5

P WI ma

R k

V PR W k V

P VI V ma W

Power Supplies Power Supplies

• A battery is rated at 650 mAh. It has A battery is rated at 650 mAh. It has to last 48 hours. What is the to last 48 hours. What is the maximum current you can take out of maximum current you can take out of the battery?the battery?

• 650 mAh = Current x hours = I x 48 h650 mAh = Current x hours = I x 48 h

• I = 650 mAh / 48 h = 13.5 mAI = 650 mAh / 48 h = 13.5 mA

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Lab 6Lab 6

1.1. Locate a 2.7 kLocate a 2.7 kΩΩ Resistor: Resistor:

2.7 k2.7 k Ω Ω = 2 7 00 = 2 7 00 ΩΩ = _____ ____ ____ = _____ ____ ____

(If you don’t have 2.7 k(If you don’t have 2.7 kΩΩ, use one , use one close in value, like 2.6 kclose in value, like 2.6 kΩΩ or 2.8 k or 2.8 kΩΩ))

2. Wire the resistor in series with a 2. Wire the resistor in series with a 10 k10 kΩΩ variable resistor as instructed variable resistor as instructed in class so the current has to go in class so the current has to go through both the fixed and variable through both the fixed and variable resistor.resistor.

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HUGE IdeasHUGE Ideas

• The current is the same through both The current is the same through both resistorsresistors

• The voltage drop across both The voltage drop across both resistors adds up to 12 voltsresistors adds up to 12 volts

• The total resistance is the sum of the The total resistance is the sum of the two resistorstwo resistors

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12 V

Big IdeasBig Ideas

• The power given off by the battery is equal The power given off by the battery is equal to the total power used up by the resistorsto the total power used up by the resistors

• As you INCREASE the total resistance, the As you INCREASE the total resistance, the total current DECREASES, and the total total current DECREASES, and the total power will DECREASE.power will DECREASE.– The distribution of power across the two The distribution of power across the two

resistors is interestingresistors is interesting

• In series resistors, the biggest resistor has In series resistors, the biggest resistor has the highest voltage dropthe highest voltage drop and uses the and uses the most power.most power.

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New ChartNew Chart

• Instead of using the chart in the Instead of using the chart in the book, use the bigger chart on the book, use the bigger chart on the handout so you see the complete handout so you see the complete picture of what is happening in the picture of what is happening in the circuit.circuit.

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Discover On Your OwnDiscover On Your Own51

• When is power in the fixed resistor When is power in the fixed resistor (R(R11) a maximum? A minimum?) a maximum? A minimum?

• When is power in the variable When is power in the variable resistor (Rresistor (R22) a maximum? A ) a maximum? A

minimum?minimum?

• Is there any connection between the Is there any connection between the two resistors and their power?two resistors and their power?

Unit 4 SummaryUnit 4 Summary

• Define energy and powerDefine energy and power

• Perform energy and power Perform energy and power calculationscalculations

• Calculate power in resistorsCalculate power in resistors

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