Circuits
description
Transcript of Circuits
It’s what we see…Circuits
Simple Circuits•A basic circuit contains 3 parts:
① A source of electric potential (voltage)
•gives charge an electric potential•ex. battery, solar panel
Simple Circuits② A resistance load• uses up energy•3 ways: converts to• light (ex. lightbulb)•heat (toaster)•motion (motor)
Simple Circuitsmost loads do more than one
•ex. lightbulbs also give off heat•ex. toaster wires glow (light)
Simple Circuits③ Conductors• provides path for
flow between source and load
•ex. wire
Resistance
•All materials give some resistance to flow of charge•exception. superconductors•Materials with high resistance• insulators•ex. rubber, plastic•Materials with low resistance•conductors•ex. metals
ResistanceAmount of resistance depends on:•material•length of wire•thickness of wire (cross-sectional area)•imagine a straw•easier to blow through:•shorter• thicker
Calculating Resistance
•R•L•A•ρ
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R = ρ LA
º resistance (Ω = Ohm) º length of wire (m)º cross sectional area of wire (m2)º resistivity (depends on material Ω•m)
Calculating Resistance
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R = ρ LA
semiconductor
ExampleHow much resistance is in the graphite of a 20. cm long pencil if the graphite has a diameter of 4.0 mm?Given:
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L = 20cm
Want:
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R€
D = 4.0mm
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A = πr2 = (3.14)(.0020m)2 =1.26 ×10−5 m2
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1m100 cm( ) = 0.20m
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1m1,000 mm( ) = .0040m
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ρ =1×10−4 Ω ⋅m (from table)
Example
solve for R
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R = ρ LA
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R = (1×10−4 Ω ⋅ / m ) .20 / m 1.26 ×10−5 / m 2 ⎛ ⎝ ⎜
⎞ ⎠ ⎟
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R = 1.59Ω
Current
rate of electric flow (of charge)
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current = chargetime
⇒ I = qt
•q• t• I
º charge (C) º time (s)º current (A)
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1 Ampere (A) = 1Coulombsecond
Current Flow•Conventional Current is the direction positive charges move
•Actual electron flow is the other way!
Ohm’s Law• Current depends on two things:
1. the potential difference, V
2. the resistance, R
Together, they make Ohm’s Law:€
I ∝ V
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I = VR
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also, V = IR
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or R = VI
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I ∝ 1R
Ohm’s LawWater Circuit AnalogyPotential Difference•amount of water (pressure)
Ohm’s LawWater Circuit Analogy
Switch• turns flow on and off
Ohm’s LawWater Circuit Analogy
Resistance• thickness & length of hose
Ohm’s LawWater Circuit Analogy
Current• rate of flow
Electric Shocks•1 mA •10 mA •100 mA
= pain= release current= death
Inside a Lightbulb•Each wire of filament goes to a different part•This allows for the current to flow through
Power and Energy
•Power = rate of energy useage•P = Energy/timeNew unit Watts
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1Watt (W ) = 1Joulesecond
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volt (V ) = JouleCoulomb
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current (I) = Coulombsecond
and
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VI = J/ C ⎛ ⎝ ⎜
⎞ ⎠ ⎟
/ C s ⎛ ⎝ ⎜
⎞ ⎠ ⎟= J
s
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P = VIso,
Combining Equations
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P = VI
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⇒ P = V VR ⎛ ⎝ ⎜
⎞ ⎠ ⎟
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⇒ P = V 2
R
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P = VI
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⇒ P = (IR)I
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⇒ P = I2R
twinkle, twinkle…
Power and Energy
Energy Equations
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E = Pt or E = VIt
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E = Pt
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⇒ E = V 2
Rt
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⇒ E = I2Rt
Units and Variables
Variable UnitqtVRIEP
coulombs (C)seconds (s)
volts (V)ohms (Ω)
amperes (A)joules (J)Watts (W)
ExampleIn 3.0 minutes an electric pot delivers 48,000 J of energy to the water inside it. The coffee pot is connected to a standard 120-volt source. What is the resistance of the coffee pot?Given:
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t = 3min
Want:
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R€
E = 48,000J
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V = 120V€
60 sec1min( ) = 180 s
Example
pick an equation
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E = V 2
Rt
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R = (120V )2
(48,000 J)180 s
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R = 54Ω€
⇒ R = V 2
Et
Example1 kW-hr costs $.50:Let’s find the cost of running a 60 W lightbulb for 24 hours
Given:
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t = 24 hr
Want:
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cost€
P = 60W
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3,600 sec1hr( ) = 86,400 s
Problem: what is a kW-hr?
Example Let’s take a close look at the kilowatt-hour:
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1kW− hr
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1,000W1 / k / W ( )
3,600 sec1 / h r( ) = 3.6 ×106 W ⋅ s
• kW = P•hr = t• Pt = E• kW-hr is a unit of energy
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=3.6 ×106 J/ s / s = 3.6 ×106 J
Example1 kW-hr costs $.50:Let’s find the cost of running a 60 W lightbulb for 24 hours
• Let’s solve for energy
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1kW − hr = 3.6 ×106 J
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E = Pt
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E = (60W )(86,400 s)
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E = 5,184,000J
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1kW − hr3.6 ×106 J ⎛ ⎝ ⎜
⎞ ⎠ ⎟=
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1.44 kW− hr
Example1 kW-hr costs $.50:Let’s find the cost of running a 60 W lightbulb for 24 hours
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1.44 kW− hr
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$.501kW − hr
⎛ ⎝ ⎜
⎞ ⎠ ⎟= .$72
Time to Practice
go to pg. 260
Measuring Resistance
•Symbol• Isolate from circuit•Place leads on each side• Ohmmeter
Measuring Voltage
•Symbol•Voltage source must be connected to circuit•Place leads on each side• Voltmeter
Measuring Current
•Need to break open circuit•Force current through meter• ammeter
Electrical Circuits Lab
Time for lab!Go to page 264
Series CircuitLet’s start with Current• In order for all of the current to make it around the loop, they all have to flow at the same rate!
So,
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IT
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=I1
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=I2
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=I3
Series CircuitResistance•The current is forced to pass through each resistor.
So,
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RT
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=R1 + R2 + R3
Series CircuitVoltage•The voltage drops a constant amount•The voltage drops are spread out proportionately between each resistor so they add up to the total
So,
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VT
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=V1 + V2 + V3
Parallel CircuitLet’s start with Voltage• Imagine parallel as separate serial circuits connected•Circuit 1•Circuit 2•Circuit 3•Each uses the whole voltageSo,
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VT
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=V1
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=V2
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=V3
Parallel CircuitCurrent•The total flow is split into each branch-off loop
So,
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IT
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=I1 + I2 + I3
Parallel CircuitResistance• Let’s start with the current
•now substitute I=V/R
• since VT = V1 = V2 = V3€
IT = I1 + I2 + I3
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VT
RT
= V1
R1
+ V2
R2
+ V3
R3
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VRT
= VR1
+ VR2
+ VR3
Parallel Circuit
Resistance•Now divide each side by V
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1RT
= 1R1
+ 1R2
+ 1R3
Parallel CircuitLooks weird you say?• let’s try an example:• Find total resistance if a 100 Ω, 200 Ω and a 300 Ω
resistor are all in parallel.
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1RT
= 1R1
+ 1R2
+ 1R3
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1RT
= 1100Ω
+ 1200Ω
+ 1300Ω
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1RT
= 6600Ω
+ 3600Ω
+ 2600Ω
Parallel CircuitLooks weird you say?• let’s try an example:• Find total resistance if a 100 Ω, 200 Ω and a 300 Ω
resistor are all in parallel.
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=11
600Ω
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1RT
= 6600Ω
+ 3600Ω
+ 2600Ω
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RT = 600Ω11
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= 55Ω