Lecture11a Phy

8/13/2019 Lecture11a Phy

http://slidepdf.com/reader/full/lecture11a-phy 1/43

Today’s lecture is brought to you by…

PhysicsMan



Not to be confused with…

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



Today’s 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.



In previous lecture (the first capacitors lecture) I suggestedusing conservation of energy to ―show‖ that the voltage dropacross circuit components in series is the sum of the individualvoltage drops:

circuit components in series

C1

C2

+ -

C3

a b

V2V

1

Vab

V

Vab = V = V1 + V2 + V3

V3



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 may use this in tomorrow’s homework. It ―is‖ * on yourstarting equations sheet, and is a consequence of conservation

of energy. Use this in combination with Ohm’s Law, V=IR.

I “derived” thisin lecture 7.

Here’s what your text means by Vab: Vab=Va-Vb=Vba

* V = 0 around closed loop

http://localhost/var/www/apps/conversion/tmp/scratch_8/10-01-08.ppt



















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

- IR 1 - IR 2 - IR 3 + V = 0

I



R3

R2

R1

+ -

VA

V1

V3

V2

a b

- V1 - V2 - V3 + V A - VB = 0

+-

VB

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

- IR 1 - IR 2 - IR 3 + V A - VB = 0

I



5

+ -

9 V

a

b

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

I

10



5

+ -

9 V

a

b

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

I

10

+-

6 V



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 be ―close enough‖ to give you ―a feel‖for the physics.

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

The mathematical analysis is more complex. We have other

things to explore this semester.



We have been making calculations with voltages from batteries without asking detailed questions about the batteries. Now it’stime to look inside the batteries.

We introduce a new term – emf – in this section.

Any device which transforms a form of energy into electricenergy is called a ―source of emf.‖

―emf‖ is an abbreviation for ―electromotive 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://phet.colorado.edu/en/simulation/battery-voltage

http://phet.colorado.edu/en/simulation/battery-resistor-circuit

http://www.energizer.com/learning-center/Pages/how-batteries-work.aspx

http://www.energizer.com/learning-center/Pages/how-batteries-work.aspx

http://phet.colorado.edu/en/simulation/battery-resistor-circuit

http://phet.colorado.edu/en/simulation/battery-voltage



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

Here’s a battery with an emf. All batteries have an ―internalresistance:‖

+ -

a bThe ―battery‖ is everything

inside the green box.

Hook up a voltmeter to measure the emf:

+ -

a b

The ―battery‖ is everythinginside the green box.

Getting ready to connect thevoltmeter (it’s not hooked upyet).

emf

emf is the zero-current potential difference



You can’t measure voltage without some (howeversmall) current flowing, so you can’t measure emfdirectly.

Measuring the emf???

+ -

a bThe ―battery‖ is 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 .



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

+ -

a bThe ―battery‖ is 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, Vab is related to the emf, , by

ab V = ε ± I r .

r

Vab



Why the sign? If the battery is delivering current, the V itdelivers is less than the emf, so the – sign is necessary.

If the battery is being charged, you have to ―force‖ the currentthrough the battery, and the V to ―force‖ the current through isgreater than the emf, so the + sign is necessary.

This will become clear as you work (and understand) problems.See, for example, ―Power Input to a Source,‖ page 865 of yourtext.

Your text writes and expects you to put thecorrect sign (+ or -) on the I. An endangered (about to goextinct) starting equation is

V = ε - I r

ab V = ε - I r .

Not recommended for use by childrenunder 6. Do not continue use if youexperience dizziness, shortness of

breath, or trouble sleeping.



So… to model a battery, simply include an extra resistor torepresent the internal resistance, and label the voltage source*

as an emf instead of V (units are still volts):

+ -

r

*Remember, all sources of emf —not just batteries —have an internal resistance.

If the internal resistance is negligible, just don’t include it!



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 voltmeter’s resistance is solarge that approximately zero

current flows through the voltmeter

emf

terminal voltage Vabinternal resistance r

R=1 IBecause the voltmeter draws

―no‖ current, r and R are in

series with a current I flowingthrough both.

ε - Ir - IR=0

IR, the potential drop across

the resistor, is just thepotential difference Vab.

ab V = IR



+ -

a b

emf

R=1 I

ε-Ir - IR= 0 ab V = IR

Ir = ε- IR

ε - IR r =

I

εr = - R

I

ab V

I=R

ab

εR r = - R

V

ab

εr = R - 1 V

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

3

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



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

ab V

I=R

3 I= = 3A 1

I’ll do a demo on this in a bit.



Today’s agenda:


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




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

We’d 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 doesn’t appear on your equation sheet, butit should appear in your brain.



And one more thing… the negative sign means energy is being

―lost.‖ So everybody writes

Let’s 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 P<0 means energy out, and P>0 meansenergy in.

dW dqP V.dt dt

P IV.

P IV



Also, using Ohm’s ―law‖ V=IR, we can write P = I2R = V2 /R.I can’t believe it, but I got soft and put P = I2R = V2 /R onyour starting equation sheet.

Truth in Advertising II. Your powercompany doesn’t 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.



Today’s agenda:


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




Which example do you want to see worked next?Not enough time to work them all.

1. An electric heater: power usage and cost. 2. How much energy is a kWh?

3. How much energy did the electric heater use?

4. Charge, current, and power in a lightning bolt.

5. Power input and output in a circuit.

6. Home power use.



What’s the meaning of this assumption aboutthe current direction?

The current in your household wiring doesn’t flow in onedirection, but because we haven’t talked about current otherthan a steady flow of charge, we’ll make the assumption. Ourcalculation will be a reasonable approximation to reality.

Example: an electric heater draws 15.0 A on a 120 V line. Howmuch power does it use and how much does it cost per 30 daymonth if it operates 3.0 h per day and the electric company

charges 10.5 cents per kWh. For simplicity assume the currentflows steadily in one direction.



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

Back to examples page.



How much energy is a kilowatt hour (kWh)?

So a kWh is a ―funny‖ unit 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




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



Energy Transformed 583,200,000 Joules used

That’s a ton of joules! Good bargain for $17. That’s about34,000,000 joules per dollar (or 0.0000029¢/joule).

OK, ―used‖ is not an SI unit, but I stuck it in there to help me

understand. And joules don’t come by the ton.

One last quibble. You know from energy conservation that youdon’t ―use up‖ energy. You just transform it from one form to

another.




learn about

lightning athowstuffworks What 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 are ―obvious:‖

Example: a typical lightning bolt cantransfer 109 J of energy across apotential difference of perhaps 5x107 V

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 “old” text. Actual current is likely far more.

http://www.howstuffworks.com/lightning.htm

http://www.howstuffworks.com/lightning.htm



We could calculate power right now, but let’s do this in theorder requested. Besides, we can’t 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 on both

plates; that’s untrue here.

U q V



Etransferred= Qtransferred Vif

Continuing with our energy transformation idea:

Qtransferred = Etransferred / Vif

Qtransferred = 109 J / 5x107 V

That’s a lot of charge (remember, typical charges are 10-6 C.

Qtransferred = 20 C

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

ΔQI = Δt

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 .

Average power is just the total energy transferred divided by

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




Average power is just the total energy transferred divided bythe total time.

F

F

WP =

t

transferredEP = t

910 J

P =

0.2 s

9P = 5×10 W

P = 5 GW

Holy ****, Batman. That’s the power output of five enormouspower plants!


The numbers in this calculation differ substantiallythan the numbers in a homework problem(notnecessarily assigned this semester). ―This‖lightning bolt carries relatively low current for along time through a high potential difference, andtransports a lot of energy.

In reality, there is no such thing as a universally-typical lightning bolt, so expect different results fordifferent bolts. See here.





(a) Rate of energy conversion.

R=4

+ -

I = 12 V

The total resistance in thecircuit is 6 , so

V = I R total

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

I = / R total = 12 V / 6 = 2 A



(b) Power dissipated internally in the battery.

R=4

+ -

I=

2A

= 12 V r=2

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

(c) Power output of the battery.

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



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

R=4

+ -

I=2A = 12 V r=2

I=2A

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

Poutput = Presistor = I2 R = (2 A)2 (4 ) = 16W.




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…



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

VT

VHI

R

P = I V = 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 V – 116.7 V)

2

/ (0.03

) = 363 W

Three different waysto solve; all will givethe correct answer.



(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 the ―incoming‖ and ―outgoing‖power lines, and the ―outgoing‖ is at ground (0 V in thiscase)…except…

…because the ―outgoing‖ power line is at 0 V, you can ―accidentally‖ get this correct if you simply multiply the currentby the voltage at the point where the power wire enters yourhouse.

( ) h d h



P = I V

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

VT

VHI

R

P = (110 A) (116.7 V – 0 V)

P = 12840 W

You don’t want to use the P=I2R=V2 /R equations because youdon’t know the effective resistance of your house (although youcould calculate it).

P = (110 A) (120 V) – (110 A)(3.3 V) isBack to examples page.

Lecture11a Phy

Documents

Transcript of Lecture11a Phy