Unit Four: Energy and Power

ET115 DC Electronics Unit Four:Energy and Power John Elberfeld JElberfeld@itt-tech.edu WWW.J-Elberfeld.com

Schedule Unit Topic Chpt Labs • Quantities, Units, Safety 1 2 (13) • Voltage, Current, Resistance 2 3 + 16 • Ohm’s Law 3 5 (35) • Energy and Power 3 6 (41) • Series Circuits Exam I 4 7 (49) • Parallel Circuits 5 9 (65) • Series-Parallel Circuits 6 10 (75) • Thevenin’s, Power Exam 2 6 19 (133) • Superposition Theorem 6 11 (81) • Magnetism & Magnetic Devices 7 Lab Final • Course Review and Final Exam

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.

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.

Reading Assignment • Read and study • Chapter 3, Energy and Power: Pages 81-98

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

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

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

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

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)

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.

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

Power Calculations • What is the power in watts when 7,500 J of energy is used in 5 hours? • P = W / t

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:

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

Convert

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

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

Energy in Joules • Doing unit conversions: Humans are much more comfortable with numbers like 742 compared to 2.67 x 109 or Giga anythings

Power Formula • Power = Work / time, but in electric terms: • P = V I • The product of voltage and current gives the electric work done per second, or power.

Logic • 1 Volt = 1 Joule / Coulomb • 1 Ampere = 1 Coulomb / Second • P = V•I • Unit are Power = volts x amps = J/s = watts

Modify Power Formula • Power: P = V I • Ohms Law: V = I R • P = V I • Substitute V = I R • P = (I R) I = I2R • P = I2R

Modify Power Formula • Power: P = V I • Ohms Law: V = I R • P = V I • Substitute I = V / R • P = V (V / R) = V2 / R • P = V2 / R

Power • You must memorize: P = V I • You can use algebra to find the other methods to calculate power • P = I2 R • P = V2 / R

Power Example • Fact: P = V I • What power is produced by 3 mA of current with a 5.5 V drop?

Power Example • Fact: P = V I • What power is produced by 3 mA of current with a 5.5 V drop? • P = V I • P = 5.5 V • 3 ma • P 16.5 mW

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

Power Example • Fact: P = V I • What is the power produced by 3 A with a 115 V drop? • P = V I • 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?

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

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?

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

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?

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?

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?

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?

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?

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?

Power • Power: P = VI or P = I2R or P = V2 / R • If power is 100 W, and current is 2 A, find the resistance.

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

Power • Power: P = VI or P = I2R or P = V2 / R • If power is 75 W, and voltage is 120 V, find the resistance.

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

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?

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

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

Lab 6 • 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.

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

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.

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.

Unit Four: Energy and Power