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Transcript
1 ET115 DC Electronics Unit Four: Energy and Power John Elberfeld [email protected] WWW.J-Elberfeld.com 2 Schedule Unit Topic Chpt Labs 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 1 2 3 3 4 5 6 6 6 7 Quantities, Units, Safety Voltage, Current, Resistance Ohm’s Law Energy and Power Series Circuits Exam I Parallel Circuits Series-Parallel Circuits Thevenin’s, Power Exam 2 Superposition Theorem Magnetism & Magnetic Devices Course Review and Final Exam 2 (13) 3 + 16 5 (35) 6 (41) 7 (49) 9 (65) 10 (75) 19 (133) 11 (81) Lab Final 3 Unit 4 Objectives - I • Define energy and power. • State the common units of energy and power. • Perform energy and power calculations. • List factors that affect the power rating of resistors. • Explain energy conversion and voltage drop in a resistance. 4 Unit 4 Objectives – II • Construct basic DC circuits on a protoboard. • Use a digital multimeter (DMM) to measure a predetermined low voltage on a power supply. • Measure resistances and voltages in a DC circuit using a DMM. 5 Reading Assignment • Read and study • Chapter 3, Energy and Power: Pages 81-98 6 Lab Assignment • Lab Experiment 6: • Power in DC Circuits Pages 41-38 • Complete all measurements, graphs, and questions and turn in your lab before leaving the room • Use the special handout to organize your data 7 Written Assignments • Do the Unit 4 Homework handout. • Be prepared for a UNIT EXAM similar to the homeworks and quizzes • If there are any calculations, you must show ALL your work for credit: – Write down the formula – Show numbers in the formula – Circle answer with the proper units 8 Energy • Energy is the ability to do work • Energy can be measured in calories (for food, heat, etc.) or in Joules for physics and electricity • The more Joules you have available, the more work you can do – Work occurs when you use a force to move an object in the direction of the force 9 Time • Time is an important consideration when you have a job to do • It takes the same energy to lift one brick at a time from the floor to a table until 1000 bricks have been moved, compared to a forklift lifting the entire pallet of bricks to the table in a matter of seconds • The work done, and the energy used to do the work is the same 10 Power • Power is how much work you do per second. • Power is measured in Watts • P = Work / time = Joules / second • Power = P = W/t = (Watts) 11 Ideas • If you lift 1 thousand bricks, one brick at a time, from the floor to a table in 5 hours, you do the same work as a forklift that lifts the bricks all at once in 20 seconds. • The forklift is more powerful than you are because it does more work per second than you did. 12 General Power Formula • P = work/time • For example, if you do 350 J of work in 1 s, your power is: • P = 350 J / 1 s = 350 J/s = 350 Watts • Power is measured in Watts • 1 Watt = 1 Joule / second 13 Power Calculations • What is the power in watts when 7,500 J of energy is used in 5 hours? • P=W/t 14 Power Calculations • What is the power in watts when 7,500 J of energy is used in 5 hours? • P = W / t = 7,500 J/ 5 h = 1500 J/h • Watt is a Joule/second, so we have to convert: 7,500 j 1h .417W 417mW 5h 3600s 15 Unit Conversions • The easiest way to convert watts to kilowatts, or microwatts, or back, is to use the ENG key on your calculator • Enter the value as it is given, and hit ENG or SHIFT-ENG until you get the desired exponent • Get 103 of kilowatts or 10-6 for microwatts, for example 16 Convert Original value 1000 W 3750 W 160 W 50,000 W 1,000,000 W 3 x 106 W 15 x 107 W 8.700 kW Conversion kW kW kW kW MW MW MW MW 17 Convert Original value 1W 0.4 W 0.002 W 0.0125 W 1.5 kW 0.5 MW 350 mW 9,000 μW Conversion mW mW mW mW W W W W 18 Energy • You buy ENERGY, not power, when you pay your electric bill • Energy is sold in Kilowatt-hours – Energy =Power x time • For example, National Grid charges $0.08 per kWh for the electric energy, plus another $0.04 per kWh to deliver the electricity to my house – Plus a ton of other fees 19 Energy Consumption • On the average, my house uses 1.03 kW of power at all times • A month has 30 days and 24 hours in a day, for a total of 720 hours • If P = W/t the W (Energy) = P • T • Energy = 1.03 kW • 720 hours • Energy = 742 kWh (kiloWatt hours) • At $0.12/kWh, my bill is $89.04 – Plus another $34 in fees and taxes 20 Energy in Joules • Doing unit conversions: J 3600s 742kWh 742 x10 2.67GJ s h 3 Humans are much more comfortable with numbers like 742 compared to 2.67 x 109 or Giga anythings 21 Power Formula • Power = Work / time, but in electric terms: •P=VI • The product of voltage and current gives the electric work done per second, or power. 22 Logic • • • • 1 Volt = 1 Joule / Coulomb 1 Ampere = 1 Coulomb / Second P = V•I Unit are Power = volts x amps = J/s = watts 1J 1C 1J 1VA 1W C s s 23 Modify Power Formula • Power: P = V I • Ohms Law: V = I R • P=VI – Substitute V = I R • P = (I R) I = I2R •P= 2 IR 24 Modify Power Formula • Power: P = V I • Ohms Law: V = I R • P=VI – Substitute I = V / R • P = V (V / R) = V2 / R •P= 2 V / R 25 Power • You must memorize: P=VI • You can use algebra to find the other methods to calculate power • P = I2 R • P = V2 / R 26 Power Example • Fact: P = V I • What power is produced by 3 mA of current with a 5.5 V drop? 27 Power Example • Fact: P = V I • What power is produced by 3 mA of current with a 5.5 V drop? • P=VI • P = 5.5 V • 3 ma • P 16.5 mW 28 Power Example • What is the power produced by 3 A with a 115 V drop? 29 Power Example • Fact: P = V I • What is the power produced by 3 A with a 115 V drop? • P=VI • P = 115 V • 3 A • P = 245 W 30 Power Example • Fact: P = V I • What power is produced by 2 A of current with a 3 V drop? • What is the power produced by 6 μA with a 3 kV drop? 31 Power Example • Fact: P = V I • What power is produced by 2 A of current with a 3 V drop? • P = V I = 3 V x 2 A = 6 VA = 6 W • What is the power produced by 6 μA with a 3 kV drop? • P = V I = 6 μA x 3 kV = 18mW 32 Power • Power: P = VI or P = I2R or P = V2 / R • What is the power when there is 500 mA of current through a 4.7 kΩ resistor? 33 Power • Power: P = VI or P = I2R or P = V2 / R • What is the power when there is 500 mA of current through a 4.7 kΩ resistor? • V = I R = 500 mA • 4.7 kΩ = 2.35 kV • P = V I = 2.35 kV • 500 mA = 1.18 kW • OR • P = I2R = (500 ma)2 • 4.7 kΩ = 1.18 kW 34 Power • Power: P = VI or P = I2R or P = V2 / R What is the power when there is 100 μA of current through a 10 kΩ resistor? 35 Power • Power: P = VI or P = I2R or P = V2 / R What is the power when there is 100 μA of current through a 10 kΩ resistor? • V = I R = 100 μA • 10 kΩ = 1 V • P = V I = 1 V • 100 μA = 100 μW • OR • P = I2R = (100 μa)2 • 10 kΩ = 100 μW • Which is easier? 36 Power • Power: P = VI or P = I2R or P = V2 / R What is the power when there is 60 V across a 620 Ω resistor? 37 Power • Power: P = VI or P = I2R or P = V2 / R What is the power when there is 60 V across a 620 Ω resistor? • I = V / R = 60 V / 620 Ω = 96.8 mA • P = V I = 60 V • 96.8 mA = 5.8 W • OR • P = V2 / R = (60 V)2 / 620 Ω = 5.8 W • Which method is easier? 38 Power • Power: P = VI or P = I2R or P = V2 / R • What is the power when there is 1.5 V across a 56 Ω resistor? 39 Power • Power: P = VI or P = I2R or P = V2 / R • What is the power when there is 1.5 V across a 56 Ω resistor? • I = V / R = 1.5 V / 56 Ω = 26.8 mA • P = V I = 1.5 V • 26.8 mA = 40.2 mW • OR • P = V2 / R = (1.5 V)2 / 56 Ω = 40.2 mW • Which method is easier? 40 Power • Power: P = VI or P = I2R or P = V2 / R • If power is 100 W, and current is 2 A, find the resistance. 41 Power • Power: P = VI or P = I2R • If power is 100 W, and current is 2 A, find the resistance. • P = VI, so V = P / I = 100 W / 2 A = 50 V • V = I R so R = V / I = 50 V / 2 A = 25 Ω • OR • P = I2R, so R = P / I2 = 100 W / (2 A)2 = • R = 25 Ω 42 Power • Power: P = VI or P = I2R or P = V2 / R • If power is 75 W, and voltage is 120 V, find the resistance. 43 Power • Power: P = VI or P = V2 / R • If power is 75 W, and voltage is 120 V, find the resistance. • P = VI, so I = P / V = 75 W / 120 V = I = 625 mA • V = I R, R = V / I = 120 V / 625 mA = 192 Ω • OR • P = V2 / R, so R = V2 / P = (120 V)2 / 75 W = • R = 192 Ω 44 Practical Applications • A resistor is rated as ½ W. It has a value of 1.2 kΩ. • A. What is the maximum current? • B. What is the maximum voltage? • C. What is the product of maximum current time maximum voltage equal to? 45 Practical Applications • A resistor is rated as ½ W. It has a value of 1.2 kΩ. • A. What is the maximum current? • P = I2 R and P = V2 / R I P .5W 20.4ma R 1.2k V PR .5W 1.2k 24.5V P V I 24.5V 20.4ma 0.5W 46 Power Supplies • A battery is rated at 650 mAh. It has to last 48 hours. What is the maximum current you can take out of the battery? • 650 mAh = Current x hours = I x 48 h • I = 650 mAh / 48 h = 13.5 mA 47 Lab 6 1. Locate a 2.7 kΩ Resistor: 2.7 k Ω = 2 7 00 Ω = _____ ____ ____ (If you don’t have 2.7 kΩ, use one close in value, like 2.6 kΩ or 2.8 kΩ) 2. Wire the resistor in series with a 10 kΩ variable resistor as instructed in class so the current has to go through both the fixed and variable resistor. 48 HUGE Ideas • The current is the same through both resistors • The voltage drop across both resistors adds up to 12 volts • The total resistance is the sum of the two resistors 12 V 49 Big Ideas • The power given off by the battery is equal to the total power used up by the resistors • As you INCREASE the total resistance, the total current DECREASES, and the total power will DECREASE. – The distribution of power across the two resistors is interesting • In series resistors, the biggest resistor has the highest voltage drop and uses the most power. 50 New Chart • Instead of using the chart in the book, use the bigger chart on the handout so you see the complete picture of what is happening in the circuit. 51 Discover On Your Own • When is power in the fixed resistor (R1) a maximum? A minimum? • When is power in the variable resistor (R2) a maximum? A minimum? • Is there any connection between the two resistors and their power? 52 Unit 4 Summary • Define energy and power • Perform energy and power calculations • Calculate power in resistors