Transcript Unit One: AC Electronics - Helderberg Hilltowns Association

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ET115 DC Electronics
Unit Four:
Energy and Power
John Elberfeld
[email protected]
WWW.J-Elberfeld.com
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Schedule
Unit Topic
Chpt Labs
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2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
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2
3
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4
5
6
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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
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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.
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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.
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Reading Assignment
• Read and study
• Chapter 3, Energy and Power:
Pages 81-98
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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
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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
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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
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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
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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)
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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.
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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
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Power Calculations
• What is the power in watts when
7,500 J of energy is used in 5 hours?
• P=W/t
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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
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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
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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
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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
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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
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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
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Energy in Joules
• Doing unit conversions:
J 3600s
742kWh  742 x10 
 2.67GJ
s
h
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Humans are much more comfortable
with numbers like 742 compared to
2.67 x 109 or Giga anythings
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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.
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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
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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
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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
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Power
• You must memorize:
P=VI
• You can use algebra to find the other
methods to calculate power
• P = I2 R
• P = V2 / R
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Power Example
• Fact: P = V I
• What power is produced by 3 mA of
current with a 5.5 V drop?
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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
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Power Example
• What is the power produced by 3 A
with a 115 V drop?
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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
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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?
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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
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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?
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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
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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?
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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?
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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?
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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?
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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?
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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?
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Power
• Power: P = VI or P = I2R or P = V2 / R
• If power is 100 W, and current is 2 A,
find the resistance.
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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
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Power
• Power: P = VI or P = I2R or P = V2 / R
• If power is 75 W, and voltage is 120 V,
find the resistance.
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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
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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?
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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
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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
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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.
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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
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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.
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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.
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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?
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Unit 4 Summary
• Define energy and power
• Perform energy and power
calculations
• Calculate power in resistors