Dc Circuit Cont. (1)

8/10/2019 Dc Circuit Cont. (1)

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http://www.nearingzero.net(nz014.jpg)

This relates (sort of) to a demo I ll do later.
http://www.nearingzero.net/http://www.nearingzero.net/


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Todays lecture is brought to you by

PhysicsMan


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Not to be confused with

Electro-Man (http://www.thinkgeek.com).


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Brain Freezehelp needed (Friday, October 4):

10:10 am11:00 am

11:10 am12:00 noon

free lunch after, if you want it

also need help 1:15-2:30 pm; same idea, different group

If you email about this, Ill try to get back to you with more information by Friday.


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Todays agenda:

Emf, Terminal Voltage, and Internal Resistance.You must be able to incorporate all of these quantities in your circuit calculations.

Electric Power.You be able to calculate the electric power dissipated in circuit components, andincorporate electric power in work-energy problems.

ExamplesFor you to work through outside of lecture.


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In lecture 7 (the first capacitors lecture) I suggested usingconservation of energy to show that the voltage drop acrosscircuit components in series is the sum of the individual voltagedrops:

circuit components in series

C1

C2

+ -

C3

a b

V2V1

Vab

V

Vab= V = V1+ V2+ V3

V3


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In general, the voltage drop across resistors in series (or othercircuit components) is the sum of the individual voltage drops.

circuit components in series

Vab= V = V1+ V2+ V3

R3

R2

R1

+ -

V

V1

V3

V2

a b

You mayuse this in tomorrows homework. Itis*on yourstarting equations sheet, and is a consequence of conservation

of energy. Use this in combination with Ohms Law, V=IR.

I derivedthisin lecture 7.

Heres what your text means by Vab:Vab=Va-Vb=Vba

* V = 0 around closed loop
http://localhost/var/www/apps/conversion/tmp/scratch_7/10-01-08.ppthttp://localhost/var/www/apps/conversion/tmp/scratch_7/10-01-08.ppthttp://localhost/var/www/apps/conversion/tmp/scratch_7/10-01-08.ppthttp://localhost/var/www/apps/conversion/tmp/scratch_7/10-01-08.ppthttp://localhost/var/www/apps/conversion/tmp/scratch_7/10-01-08.ppthttp://localhost/var/www/apps/conversion/tmp/scratch_7/10-01-08.ppthttp://localhost/var/www/apps/conversion/tmp/scratch_7/10-01-08.ppthttp://localhost/var/www/apps/conversion/tmp/scratch_7/10-01-08.ppthttp://localhost/var/www/apps/conversion/tmp/scratch_7/10-01-08.ppthttp://localhost/var/www/apps/conversion/tmp/scratch_7/10-01-08.ppthttp://localhost/var/www/apps/conversion/tmp/scratch_7/10-01-08.ppthttp://localhost/var/www/apps/conversion/tmp/scratch_7/10-01-08.ppthttp://localhost/var/www/apps/conversion/tmp/scratch_7/10-01-08.ppthttp://localhost/var/www/apps/conversion/tmp/scratch_7/10-01-08.ppthttp://localhost/var/www/apps/conversion/tmp/scratch_7/10-01-08.ppthttp://localhost/var/www/apps/conversion/tmp/scratch_7/10-01-08.ppthttp://localhost/var/www/apps/conversion/tmp/scratch_7/10-01-08.ppthttp://localhost/var/www/apps/conversion/tmp/scratch_7/10-01-08.ppthttp://localhost/var/www/apps/conversion/tmp/scratch_7/10-01-08.ppthttp://localhost/var/www/apps/conversion/tmp/scratch_7/10-01-08.ppthttp://localhost/var/www/apps/conversion/tmp/scratch_7/10-01-08.ppthttp://localhost/var/www/apps/conversion/tmp/scratch_7/10-01-08.ppthttp://localhost/var/www/apps/conversion/tmp/scratch_7/10-01-08.ppt


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R3

R2

R1

+ -

V

V1

V3

V2

a b

Starting at point a and following the current in a clockwise patharound the circuit and back to point a

- V1- V2- V3+ V = 0

- IR1- IR2- IR3+ V = 0

I


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R3

R2

R1

+ -

VA

V1

V3

V2

a b

- V1- V2- V3+ VA- VB= 0

+-

VB

Again, starting at point a and following the current in aclockwise path around the circuit and back to point a

- IR1- IR2- IR3+ VA- VB= 0

I


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5

+ -

9 V

a

b

Example: calculate I, Vab,and Vba(to be worked at theblackboard).

I

10

+-

6 V

Also discuss what happens if you guess wrong current direction.


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In Physics 24, whenever you work with currents in circuits, youshould assume (unless told otherwise)direct current.

DC Currents

Current in a dc circuit flows in one direction, from + to -.

We will not encounter ac circuits much in this course.

For any calculations involving household current, which isac, assuming dc will beclose enoughto give youa feelfor the physics.

If you need to learn about ac circuits, youll have coursesdevoted to them.

The mathematical analysis is more complex. We have other

things to explore this semester.


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We have been making calculations with voltages from batterieswithout asking detailed questions about the batteries. Now itstime to look inside the batteries.

We introduce a new termemfin this section.

Any device which transforms a form of energy into electricenergy is called asource of emf.

emfis an abbreviation forelectromotive force,but emf isnot a force!

The emf of a source is the voltage it produces when no currentis flowing.

emf, terminal voltage, and internal resistance

http://www.energizer.com
http://phet.colorado.edu/en/simulation/battery-voltagehttp://phet.colorado.edu/en/simulation/battery-resistor-circuithttp://www.energizer.com/learning-center/Pages/how-batteries-work.aspxhttp://www.energizer.com/learning-center/Pages/how-batteries-work.aspxhttp://www.energizer.com/learning-center/Pages/how-batteries-work.aspxhttp://www.energizer.com/learning-center/Pages/how-batteries-work.aspxhttp://www.energizer.com/learning-center/Pages/how-batteries-work.aspxhttp://phet.colorado.edu/en/simulation/battery-resistor-circuithttp://phet.colorado.edu/en/simulation/battery-voltage


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The voltage you measure across the terminals of a battery (orany source of emf) is less than the emf because of internalresistance.

Heres a battery with an emf. All batteries have aninternalresistance:

+ -

a bThebatteryis everything

inside the green box.

Hook up a voltmeter to measure the emf:

+ -

a b

Thebatteryis everythinginside the green box.

Getting ready to connect thevoltmeter (its not hooked upyet).

emf

emf is the zero-current potential difference


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You cant measure voltage without some (howeversmall) current flowing, so you cant measure emfdirectly.

Measuring the emf???

+ -

a bThebatteryis everythinginside the green box.

As soon as you connect thevoltmeter, current flows.

You can only measure Vab.

I

(emf)

Homework hint: an ideal voltmeter would be able to measure .

For example, problem 5.33.


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We model a battery as producing an emf, , and having aninternal resistance r:

+ -

a bThebatteryis everythinginside the green box.

The terminal voltage, Vab, is the voltage you measure acrossthe battery terminals with current flowing. When a current Iflows through the battery, Vabis related to the emf by

abV = I r .

r

Vab

Not recommended for use by children under 6. Donot continue use if you experience dizziness,

shortness of breath, or trouble sleeping. Do not

operate heavy machinery after using.

An extinctstarting

equation.


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To model a battery, simply include an extra resistor to representthe internal resistance, and label the voltage source* as an emf

instead of V (units are still volts):

+ -

r

*Remember, all sources of emfnot just batterieshave an internal resistance.

If the internal resistance is negligible, simply dont include it!

If you are asked to calculate the terminal voltage, it is just Vab= VaVb, calculated using thetechniques I am showing you today.

a b


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Summary of procedures for tomorrows homework:

Potential increases by whencurrent passes through an emf in

the direction from the - to +terminal.

Draw the current in a circuit so that it flows fromto +through the battery. The sum of the potential changesaround a circuit loop is zero.

+

-

I

V is +

Potential decreases by IR whencurrent goes through a resistor.

I

loop

V is -

Treat a battery internal resistance like any other resistor. If Iflows through a battery + to -, potential decreases by .


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Example: a battery is known to have an emf of 9 volts. If a 1ohm resistor is connected to the battery, the terminal voltage ismeasured to be 3 volts. What is the internal resistance of the

battery?

+ -

a b

the voltmeters resistance is solarge that approximately zero

current flows through the voltmeter

emf

terminal voltage Vabinternal resistance r

R=1 IBecause the voltmeter draws

nocurrent, r and R are in

series with a current I flowingthrough both.

- Ir - IR=0

IR, the potential drop across

the resistor, is just thepotential difference Vab.

abV = IR


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+ -

a b

emf

R=1 I

-Ir - IR= 0 abV = IR

Ir = - IR

- IRr =

I

r = - R

I

abV

I=R

ab

Rr = - R

V

ab

r = R - 1V

9r = 1 - 1 = 3- 1 =2

3

A rather unrealistically large value forthe internal resistance of a 9V battery.


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By the way, the experiment described in the previous exampleis not a very good idea.

abV

I=R

3I= = 3A1

Ill do a demo on this in a bit.


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Todays agenda:


Electric Power.You must be able to calculate the electric power dissipated in circuit components, andincorporate electric power in work-energy problems.

ExamplesFor you to work through outside of lecture.


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Last semester you defined power in terms of the work done bya force.

Wed better use the same definition this semester! So we will.

We focus here on the interpretation that power is energytransformed per time, instead of work by a force per time.

Electric Power

FF

dWP

dt

energy transformedPtime

The above equation doesnt appear on your equation sheet, butit should appear in your brain.


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However, we begin with the work aspect. We know the workdone by the electric force in moving a charge q through apotential difference:

The instantaneous power, which is the work per time done by

the electric force, is

i f i f dW dq V .

i f i f dW dq VP .dt dt

The work done by the electric force in moving an infinitesimal

charge dq through a potential difference is:

i f i f i f W U q V .


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And one more thing the negative sign means energy is being

lost.So everybody writes

Lets get lazy and drop the in front of the V, but keep in theback of our heads the understanding that we are talking aboutpotential difference. Then

But wait! We defined I = dQ/dt. So

and understands that P0 meansenergy in.

dW dqP V.dt dt

P IV.

P IV


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Also, using OhmslawV=IR, we can write P = I2R = V2/R.I cant believe it, but I got soft and put P = I2R = V2/R onyour starting equation sheet.

Truth in Advertising II.Your powercompany doesnt sell you power. It sellsenergy. Energy is power times time, so akilowatt-hour (what you buy from yourenergy company) is an amount of energy.

Truth in Advertising I.The V in P=IV is a potentialdifference, or voltage drop. It is really a V.


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Todays agenda:


Electric Power.You must be able to calculate the electric power dissipated in circuit components, andincorporate electric power in work-energy problems.

ExamplesMostly for you to work through outside of lecture.


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(a) Rate of energy conversion.

R=4

+ -

I= 12 V

Start at negative terminal ofbattery

Energy is converted at the rate Pconverted=I=(2 A)(12 V)=24W.

Example: A 12 V battery with 2 internal resistance isconnected to a 4 resistor. Calculate (a) the rate at whichchemical energy is converted to electrical energy in the battery,

(b) the power dissipated internally in the battery, and (c) thepower output of the battery.

r=2

Rate of energy conversion example.

- I R2- I R

4= 0

I = / (R2+ R4) =12 V / 6 = 2 A

V 0 around any closed circuit loop


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(b) Power dissipated internally in the battery.

R=4

+ -

I=

2A

= 12 Vr=2

Pdissipated= I2r = (2 A)2 (2 ) = 8 W.

(c) Power output of the battery.

Poutput= Pconverted- Pdissipated= 24 W - 8 W = 16W.



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(c) Power output of the battery (double-check).

R=4

+ -

I=2A= 12 Vr=2

I=2A

The output power is delivered to (and dissipated by) theresistor:

Poutput= Presistor= I2R = (2 A)2(4 ) = 16W.



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An electric heater draws 15.0 A on a 120 V line. How muchpower does it use.

P IV

P 15 A 120 V 1800 W = 1.8 kW

How much does it cost per 30 day month if it operates 3.0 hper day and the electric company charges 10.5 cents per kWh.

3 h $0.105

cos t 1.8 kW 30 days day kWh

cost $17.00

Electric heater example.


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How much energy is a kilowatt hour (kWh)?

So a kWh is afunnyunit of energy. K (kilo) and h (hours) arelowercase, and W (James Watt) is uppercase.

1 kW 1 h 1000 W 3600 s

J

1000 3600 ss

6= 3.6 10 J



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How much energy did the electric heater use?

done by force

average

W Energy TransformedP

time time

averageEnergy Transformed P time

J 3 h used 3600 s

Energy Transformed 1800 30 dayss day h

Energy Transformed 583,200,000 Joules used



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learn about

lightning athowstuffworksWhat kind of a problem is this?

You are given energy transferred, potential difference, time.

You need to calculate charge transferred, current, and averagepower. Equations for current and power areobvious:

Example: a typical lightning bolt cantransfer 109J of energy across apotential difference of perhaps 5x107V

during a time interval of 0.2 s. Estimatethe total amount of charge transferred,the current and the average power overthe 0.2 s.

avg

WP

t

avg

QI

t

Numbers obtained from an oldtext. Actual current is likely far more.

Lightning bolt example.
http://www.howstuffworks.com/lightning.htmhttp://www.howstuffworks.com/lightning.htm


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We could calculate power right now, but lets do this in theorder requested. Besides, we cant get current without Q,charge transferred.

We need to think in terms of energy transformations rather

than work done by forces. The equation above tells us thatpotential energy stored in clouds can be transferred to theground (at a different potential) by moving charge from cloudto ground. We are given energy transferred and potentialdifference, so we can calculate q.

Could I think of the cloud-earth system as a giant capacitor which stores energy?

You could, except our capacitor equation U=QV/2 assumes the same charge onboth plates; thats untrue here.

U q V



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Etransferred= QtransferredVif

Continuing with our energy transformation idea:

Qtransferred= Etransferred/ Vif

Qtransferred= 109J / 5x107V

Thats a lot of charge (remember, typical charges are 10-6C.

Qtransferred= 20 C

Once we have the charge transferred, the current is easy.

QI =t

20 CI = = 100 A

0.2 s

This is probably less than the actualcurrent in a lightning bolt by a factor ofanywhere from 10 to 1000. See this link.

http://hypertextbook.com/facts/1997/BrookeHaramija.shtmlhttp://hypertextbook.com/facts/1997/BrookeHaramija.shtml


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VT

VHI

R

(a) Find the voltage at the point where the power wire entersyour house.

VHT= IR

VT-VH

= IR

VH= VT-IR

VH= (120 V)(110 A) (0.03) = 116.7 V

Example: the electric utility companysupplies your house with electricityfrom the main power lines at 120 V.

The wire from the pole to your househas a resistance of 0.03 . Supposeyour house is drawing 110 A ofcurrent

Household voltage example.


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(b) How much power is being dissipated in the wire from thepole to your house?

VT

VHI

R

P = IV = I2R = (V)2/R

P = I(VT-VH) = I2R = (VT-VH)

2/R

P = (110 A) (120 V -116.7 V) = 363 W

or P = (110 A)2(0.03) = 363 W

or P = (120 V116.7 V)2/ (0.03) = 363 W

Three different waysto solve; all will givethe correct answer.



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(c) How much power are you using inside your house?

VT

VHI

R

You need to understand that your household voltage representsthe potential difference between theincomingandoutgoingpower lines, and theoutgoingis at ground (0 V in thiscase)except

because the outgoingpower line is at 0 V, you canaccidentallyget this correct if you simply multiply the currentby the voltage at the point where the power wire enters yourhouse.


( ) h d h


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P = IV

(c) How much power are you using inside your house?

VT

VHI

R

P = (110 A) (116.7 V0 V)

P = 12840 W

You dont want to use the P=I2R=V2/R equations because youdont know the effective resistance of your house (although youcould calculate it).

P = (110 A) (120 V)(110 A)(3.3 V) is also a reasonable way to work this part.

Dc Circuit Cont. (1)

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