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

€

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

€

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:

€

L = 20cm

Want:

€

R€

D = 4.0mm

€

A = πr2 = (3.14)(.0020m)2 =1.26 ×10−5 m2

€

1m100 cm( ) = 0.20m

€

1m1,000 mm( ) = .0040m

€

ρ =1×10−4 Ω ⋅m (from table)

Example

solve for R

€

R = ρ LA

€

R = (1×10−4 Ω ⋅ / m ) .20 / m 1.26 ×10−5 / m 2 ⎛ ⎝ ⎜

⎞ ⎠ ⎟

€

R = 1.59Ω

Current

rate of electric flow (of charge)

€

current = chargetime

⇒ I = qt

•q• t• I

º charge (C) º time (s)º current (A)

€

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

€

I = VR

€

also, V = IR

€

or R = VI

€

I ∝ 1R

Ohm’s LawWater Circuit AnalogyPotential Difference•amount of water (pressure)

Ohm’s LawWater Circuit Analogy

Switch• turns flow on and off


Resistance• thickness & length of hose


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

€

1Watt (W ) = 1Joulesecond

€

volt (V ) = JouleCoulomb

€

current (I) = Coulombsecond

and

€

VI = J/ C ⎛ ⎝ ⎜

⎞ ⎠ ⎟

/ C s ⎛ ⎝ ⎜

⎞ ⎠ ⎟= J

s

€

P = VIso,

Combining Equations

€

P = VI

€

⇒ P = V VR ⎛ ⎝ ⎜

⎞ ⎠ ⎟

€

⇒ P = V 2

R

€

P = VI

€

⇒ P = (IR)I

€

⇒ P = I2R

twinkle, twinkle…

Power and Energy

Energy Equations

€

E = Pt or E = VIt

€

E = Pt

€

⇒ E = V 2

Rt

€

⇒ 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:

€

t = 3min

Want:

€

R€

E = 48,000J

€

V = 120V€

60 sec1min( ) = 180 s

Example

pick an equation

€

E = V 2

Rt

€

R = (120V )2

(48,000 J)180 s

€

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:

€

t = 24 hr

Want:

€

cost€

P = 60W

€

3,600 sec1hr( ) = 86,400 s

Problem: what is a kW-hr?

Example Let’s take a close look at the kilowatt-hour:

€

1kW− hr

€

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

€

=3.6 ×106 J/ s / s = 3.6 ×106 J


• Let’s solve for energy

€

1kW − hr = 3.6 ×106 J

€

E = Pt

€

E = (60W )(86,400 s)

€

E = 5,184,000J

€

1kW − hr3.6 ×106 J ⎛ ⎝ ⎜

⎞ ⎠ ⎟=

€

1.44 kW− hr


€

1.44 kW− hr

€

$.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,

€

IT

€

=I1

€

=I2

€

=I3

Series CircuitResistance•The current is forced to pass through each resistor.

So,

€

RT

€

=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,

€

VT

€

=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,

€

VT

€

=V1

€

=V2

€

=V3

Parallel CircuitCurrent•The total flow is split into each branch-off loop

So,

€

IT

€

=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

€

VT

RT

= V1

R1

+ V2

R2

+ V3

R3

€

VRT

= VR1

+ VR2

+ VR3

Parallel Circuit

Resistance•Now divide each side by V

€

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.

€

1RT

= 1R1

+ 1R2

+ 1R3

€

1RT

= 1100Ω

+ 1200Ω

+ 1300Ω

€

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.

€

=11

600Ω

€

1RT

= 6600Ω

+ 3600Ω

+ 2600Ω

€

RT = 600Ω11

€

= 55Ω

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Circuits