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General Physics (PHY 2140)

Lecture 5Lecture 5Electrodynamics

Electric current• temperature variation of resistance• electrical energy and power

Direct current circuits• emf• resistors in series

Chapter 17-18

http://www.physics.wayne.edu/~alan/2140Website/Main.htm

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Hours of operationHours of operation::

Monday and Tuesday, 10:00 AM to 5:00 PMWednesday and Thursday

10:00 AM to 4:00 PMFriday, Saturday and Sunday

Closed

Undergraduate students taking PHY1020, PHY2130, PHY2140, PHY2170/2175 and PHY2180/2185 will be able to get assistance in this Center with their homework, labwork

and other issues related to their physics course.

The Center will be open: Monday, May 21 to Thursday, August 2, 2007.

Department of Physics and Astronomyannounces the Spring-Summer 2007 opening of

The Physics Resource CenterThe Physics Resource Centeron Monday, May 21 in

Room 172Room 172 of Physics Research Building.of Physics Research Building.

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Having trouble with Physics???

Come to Supplemental Instruction (SI) It’s free, helpful, and has been proven to work!

Mondays: pm @ Rm. 37 State HallTuesdays: pm @ Rm. 27 State HallWednesdays: pm @ Rm. 37 State HallFridays: pm @ Rm. 37 State Hall

If these times are inconvenient for you, emailor call or come visit the:

Academic success center @ 1600 UGL(313) 577-3165

www.success.wayne.edu

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Exam 1 this Wednesday 5/23Exam 1 this Wednesday 5/23

1212--14 questions14 questions

You must show your work for credit.You must show your work for credit.

Closed book. Closed book.

You may bring an 8 You may bring an 8 ½½

x 11 inch page of notes.x 11 inch page of notes.

Bring a calculator and a pen/pencil (I donBring a calculator and a pen/pencil (I don’’t have extras).t have extras).

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JUNIOR YEARFall

Physics 5340/41: OpticsChemistry 2220/2230 or 2280/2290

Biology ElectiveCollege Foreign Lang. II

Winter

Physics 5620: ElectronicsPhysics 6Z00: Biomedical Seminar#

ElectiveCollege Foreign Lang. III

SENIOR YEARFall

College Group Req.Physics 6X00: Biological Physics

ElectiveElectiveGeneral Education Req.

Winter

Physics/Radiology 6Y00: Physics in MedicinePhysics 6W00: Biomedical Research#

College Group Req.General Education Req.#These will also be offered in the summer

For more info, please contact Prof. Peter Hoffmann at hoffmann@wayne.edu

Suggested Course Sequence

FRESHMAN YEARFall

Physics 2130/2131 or 2170/2171

Mathematics 1800English (BC)UGE 1000General Education Req.

Winter

Physics 2140/2141 or 2180/2181

Chemistry 1220/1230Mathematics 2010English (IC)

SOPHOMORE YEARFall

Physics 3X00: Application of Mathematics in Biomed.

Chemistry 1240/50Mathematics 2020Biology 1500

Winter

Physics 4X00: Introduction to Biomedical Physics

Biology 1510College Foreign Lang. IGeneral Education Req.

New Degree Program in New Degree Program in BIOMEDICAL PHYSICSBIOMEDICAL PHYSICS

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Lightning ReviewLightning Review

Last lecture:1.1. Capacitance and capacitorsCapacitance and capacitors

Capacitors with dielectrics (CCapacitors with dielectrics (C↑↑ if if κκ ↑↑))

2.2. Current and resistanceCurrent and resistance

Current and drift speedCurrent and drift speedResistance and OhmResistance and Ohm’’s laws law••

I is proportional to VI is proportional to VResistivity Resistivity ••

material propertymaterial property

Review Problem:

Consider two resistors wired one after another. If there is an electric current moving through the combination, the current in the second resistor is

a. equal tob. halfc. smaller, but not necessarily half

the current through

the first resistor.

R Al

ρ =

QIt

Δ=Δ

dI nqv A=

V IR=

a

b

c

R1

R2

I

0 0,AC C Cd

κε κ= =

221 1

2 2 2QU QV CVC

= = =

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17.5 Resistivity - Example(a) Calculate the resistance per unit length of a 22-gauge nichrome wire of radius 0.321 m.

Cross section: ( )22 3 7 20.321 10 3.24 10A r m mπ π − −= = × = ×

Resistivity (Table): 1.5 x 10−6

Ωm. 6

7 2

1.5 10 4.63.24 10 m

R ml A m

ρ −Ω

−

× Ω= = =

×Resistance/unit length:

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17.5 Resistivity - Example(b) If a potential difference of 10.0 V is maintained across a 1.0-m length of the nichrome wire, what is the current?

10.0 2.24.6

V VI ARΔ

= = =Ω

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17.6 Temperature Variation of Resistance - Intro• The resistivity of a metal depends on many

(environmental) factors.• The most important factor is the temperature.• For most metals, the resistivity increases with

increasing temperature.• The increased resistivity arises because of larger

friction caused by the more violent motion of the atoms of the metal.

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For most metals, resistivity increases For most metals, resistivity increases approx. linearly with temperature.approx. linearly with temperature.

( )1o oT Tρ ρ α⎡ ⎤= + −⎣ ⎦

• ρ

is the resistivity at temperature T (measured in Celsius).

• ρο

is the reference resistivity at the reference temperature

Tο

(usually taken to be

20 oC).• α

is a parameter called temperature coefficient of resistivity.

For a conductor with fixed cross section.For a conductor with fixed cross section.

( )1o oR R T Tα⎡ ⎤= + −⎣ ⎦

ρ

TMetallic Conductor

ρ

TSuperconductor

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17.6 Temperature Variation of Resistance 17.6 Temperature Variation of Resistance -- ExampleExample

Platinum Resistance ThermometerPlatinum Resistance ThermometerA resistance thermometer, which measures temperature by measurinA resistance thermometer, which measures temperature by measuring the g the change in the resistance of a conductor, is made of platinum andchange in the resistance of a conductor, is made of platinum and

has a has a resistance of 50.0 resistance of 50.0 ΩΩ

at 20at 20ooC. When the device is immersed in a vessel C. When the device is immersed in a vessel containing melting indium, its resistance increases to 76.8 containing melting indium, its resistance increases to 76.8 ΩΩ. Find the melting . Find the melting point of Indium.point of Indium.

Solution:Using α=3.92x10-3(oC)-1 from table 17.1.

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Platinum Resistance ThermometerPlatinum Resistance Thermometer A resistance thermometer, which measures temperature by measurinA resistance thermometer, which measures temperature by measuring the change in the g the change in the resistance of a conductor, is made of platinum and has a resistaresistance of a conductor, is made of platinum and has a resistance of 50.0 nce of 50.0 ΩΩ

at 20at 20ooC. C. When the device is immersed in a vessel containing melting indiuWhen the device is immersed in a vessel containing melting indium, its resistance m, its resistance increases to 76.8 increases to 76.8 ΩΩ. Find the melting point of Indium.. Find the melting point of Indium.

Solution:Solution:Using Using αα=3.92x10=3.92x10--33((ooC)C)--11

from table 17.1. from table 17.1. RRoo

=50.0 =50.0 ΩΩ..TToo

=20=20ooC.C.R=76.8 R=76.8 ΩΩ..

( ) [ ]13

76.8 50.0

3.92 10 50.0

137157

oo

oo

o

o

R RT TR C

CT C

α −−

− Ω− Ω− = =

⎡ ⎤× Ω⎢ ⎥⎣ ⎦=

=

( )1o oR R T Tα⎡ ⎤= + −⎣ ⎦

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17.717.7

SuperconductivitySuperconductivity

19111911: H. K. : H. K. OnnesOnnes, who had figured out how to , who had figured out how to make liquid helium, used it to cool mercury to 4.2 K make liquid helium, used it to cool mercury to 4.2 K and looked at its resistance:and looked at its resistance:

19571957: Bardeen (: Bardeen (UIUC!UIUC!), Cooper, and Schrieffer (), Cooper, and Schrieffer (““BCSBCS””) publish theoretical ) publish theoretical explanation, for which they get the Nobel prize in 1972.explanation, for which they get the Nobel prize in 1972.

It was BardeenIt was Bardeen’’s s secondsecond Nobel prize (1Nobel prize (1’’st: 1956 st: 1956 –– transistor)transistor)

•Current can flow, even if E=0.•Current in superconducting rings can flow for years with no

decrease!

At low temperatures the resistance of some At low temperatures the resistance of some metalsmetals 0, measured to be less than 0, measured to be less than 1010--1616••ρρconductorconductor(i.e., (i.e., ρρ<<1010--24 24 ΩΩmm)!)!

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17.8 Electrical energy and power17.8 Electrical energy and power

In any circuit, battery is used to induce electrical currentIn any circuit, battery is used to induce electrical currentchemical energychemical energy of the battery is transformed into of the battery is transformed into kinetic energykinetic energyof mobile charge carriers (electrical energy gain)of mobile charge carriers (electrical energy gain)

Any device that possesses resistance (resistor) present Any device that possesses resistance (resistor) present in the circuit will transform electrical energy into heatin the circuit will transform electrical energy into heat

kinetic energykinetic energy of charge carriers is transformed into of charge carriers is transformed into heatheat via via collisions with atoms in a conductor (electrical energy loss)collisions with atoms in a conductor (electrical energy loss)

I

V = IR

+ -

B A

C D

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Electrical energyElectrical energy

Consider circuit on the right in detailConsider circuit on the right in detailAB: charge gains electrical energy AB: charge gains electrical energy form the batteryform the battery

(battery looses chemical energy)(battery looses chemical energy)CD: electrical energy lost (transferred CD: electrical energy lost (transferred into heat)into heat)Back to A: same potential energy Back to A: same potential energy (zero) as before(zero) as beforeGained electrical energy = lost Gained electrical energy = lost electrical energy on the resistorelectrical energy on the resistor

A

B

D

CE Q VΔ = Δ ⋅Δ

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PowerPower

Compute rate of energy loss (power dissipated on the Compute rate of energy loss (power dissipated on the resistor)resistor)

Use OhmUse Ohm’’s laws law

Units of power: SI: watt Units of power: SI: watt delivered energy: kilowattdelivered energy: kilowatt--hours hours

E QP V I Vt t

Δ Δ= = Δ = ΔΔ Δ

( )22 V

P I V I RR

Δ= Δ = =

( )( )3 61 kWh 10 3600 3.60 10W s J= = ×

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ExampleExamplePower Transmission linePower Transmission line

A highA high--voltage transmission line with resistance of 0.31 voltage transmission line with resistance of 0.31 ΩΩ/km carries 1000A , /km carries 1000A , starting at 700 kV, for a distance of 160 km. What is the power starting at 700 kV, for a distance of 160 km. What is the power loss due to loss due to resistance in the wire? resistance in the wire?

Given:

V=700000 Vρ=0.31 Ω/km L=160 kmI=1000 A

Find:P=?

Observations: 1.

Given resistance/length, compute total resistance2.

Given resistance and current, compute power loss

( )( )0.31 160 49.6R L km kmρ= = Ω = Ω

( ) ( )22 61000 49.6 49.6 10P I R A W= = Ω = ×

Now compute power

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MiniMini--quizquiz

Why do the old light bulbs usually fail just after you turn Why do the old light bulbs usually fail just after you turn them on?them on?

When the light bulb is off, its filament is cold, and its resistance is R0

. Once the switch it thrown, current passes through the filament heating it up, thus increasing the resistance,

This leads to decreased amount of power delivered to the light bulb, as

Thus, there is a power spike

just after the switch is thrown, which burns the light bulb.

( )1o oR R T Tα⎡ ⎤= + −⎣ ⎦

2 /P V R=

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Chapter 18:Chapter 18: Direct Current CircuitsDirect Current Circuits

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Introduction: elements of electrical circuits Introduction: elements of electrical circuits A branchA branch::

A branch is a single electrical element or device (resistor, etA branch is a single electrical element or device (resistor, etc.).c.).

A junctionA junction::

A junction (or node) is a connection point between two or moreA junction (or node) is a connection point between two or more

branches.branches.

If we start at any point in a circuit (node), proceed through coIf we start at any point in a circuit (node), proceed through connected nnected electric devices back to the point (node) from which we startedelectric devices back to the point (node) from which we started, , without without crossing a node more than one timecrossing a node more than one time, we form a closed, we form a closed--path (or path (or looploop).).

b b b b b

A circuit with 5 branches.

•

•

•

b

bb

A circuit with 3 nodes.

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Steady current (constant in magnitude and direction)• requires a complete circuit• path cannot consist only of resistances:cannot have only potential drops in direction of current flow

Electromotive Force (EMF)• provides increase in potential E• converts some external form of energy into electrical energy

Single emf and a single resistor: emf can be thought of as a “charge pump”

I

V = IR

E

+ - V = IR = E

18.1 Sources of EMF18.1 Sources of EMF

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Important note: in our complete circuit, conservation of energy requires that as we travel around the circuit we should encounter a number of voltage drops and increases so that we get back to the same potential we started with!

V = IR = E

18.1 Sources of EMF 18.1 Sources of EMF (continued)(continued)

I

V = IR

E

+ -Start with 0V hereand move clockwise

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EMFEMF

Each real battery has some Each real battery has some internal resistanceinternal resistanceAB: potential increases by AB: potential increases by EE, , the source of EMF, then the source of EMF, then decreases by decreases by IrIr

(because of (because of the internal resistance)the internal resistance)Thus, terminal voltage on the Thus, terminal voltage on the battery battery ΔΔV isV is

Note: Note: EE

is the same as the is the same as the terminal voltage when the terminal voltage when the current is zero (open circuit)current is zero (open circuit)

E

r

R

A

B C

DV IrΔ = −E

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EMF (continued)EMF (continued)

Now add a load resistance RNow add a load resistance RSince it is connected by a Since it is connected by a conducting wire to the battery conducting wire to the battery →→

terminal voltage is the same as terminal voltage is the same as the potential difference across the the potential difference across the load resistanceload resistance

Thus, the current in the circuit isThus, the current in the circuit is

E

r

R

A

B C

D

IR r

=+E

,V Ir IR orIr IR

Δ = − == +

EE

Power output:

2 2I I r I R= +E

Note: we’ll assume r negligible unless otherwise is stated

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Voltmeters measure Potential Difference (or voltage) across a device by being placed in parallel with the device.

V

Ammeters measure current through a device by being placed in series with the device.

A

Measurements in electrical circuitsMeasurements in electrical circuits

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ab eq

abeq

V IR

VRI

=

≡

Req

Ia

b

Direct Current CircuitsDirect Current Circuits

Two Basic Principles:Conservation of ChargeConservation of Energy

Resistance Networks

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1 21 2

eq

eq

V IR

IR IRVR R RI I

Δ =

+Δ≡ = = +

18.2 Resistors in series18.2 Resistors in series

1 2I I I= =

1. Because of the charge conservation, all charges going through the resistor R2

will also go through resistor R1

. Thus, currents

in R1

and R2

are the same,

1 2V IR IRΔ = +

2. Because of the energy conservation, total potential drop

(between A and C) equals to the sum of potential drops between A and B and B and C,

By definition,

Thus, Req

would be

1 2eqR R R= +

R2

R1

v2

v1+ +

+

_

_

_v i1

A B

C

I

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Resistors in series: notesResistors in series: notes

Analogous formula is true for any number of resistors,Analogous formula is true for any number of resistors,

It follows that the equivalent resistance of a series It follows that the equivalent resistance of a series combination of resistors is greater than any of the combination of resistors is greater than any of the individual resistors.individual resistors.

1 2 3 ...eqR R R R= + + + (series combination)

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Resistors in series: exampleResistors in series: example

R2

R1

v2

v1+ +

+

_

_

_v i1

A B

C

I

In the electrical circuit below, find voltage across the resistoIn the electrical circuit below, find voltage across the resistor Rr R11

in terms of in terms of the resistances Rthe resistances R11

, R, R22

and potential difference between the batteryand potential difference between the battery’’s s terminals V. terminals V.

1 2V V V= +

Energy conservation implies:

with 1 1 2 2andV IR V IR= =

Then, ( )1 21 2

, so VV I R R IR R

= + =+

Thus,1

11 2

RV VR R

=+

This circuit is known as voltage divider.

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Resistors in seriesConservation of Charge

I = I1 = I2 = I3Conservation of Energy

Vab = V1 + V2 + V3

321

3

3

2

2

1

1321

321

RRRRIV

IV

IV

IV

IV

IV

IVVV

IVR

eq

abeq

++=

++=++=

++=≡

I

R1V1 =I1 R1

R2V2 =I2 R2

R3V3 =I3 R3

a

b

Voltage Divider:eqab R

RVV 11 =

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Resistors in parallelConservation of Charge

I = I1 + I2 + I3Conservation of Energy

Vab = V1 = V2 = V3

321

3

3

2

2

1

1321

321

1111

1

RRRR

VI

VI

VI

VI

VI

VI

VIII

VI

R

eq

ababab

ababeq

++=

++=++=

++=≡

aI

R1V1 =I1 R1

R3V3 =I3 R3

b

Current Divider:

1

1

RR

II eq=

R2V2 =I2 R2

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R1 =4Ω

R2 =3Ω R3 =6Ω

E=18V

Example: Determine the equivalent resistance of the circuit as shown.Determine the voltage across and current through each resistor.Determine the power dissipated in each resistorDetermine the power delivered by the battery

+