(1) I have completed at least 50% of the reading and study-guide assignments associated with the lecture, as indicated on the course schedule.

a) True b) False

iClicker Quiz

In static equilibrium, E = 0 inside a conducting material.

Now consider non-equilibrium situations where E 0 and V 0 in a conductor.

A non-zero E-field inside a conductor drives a non-zero current density J, which requires a continous source of flowing electric charge (e.g. a battery or power supply).

When the electric current in a conductor is constant in time, we do not say that the charges present are in equilibrium, but instead say that they are in a “steady state”.

I = current or charge flux through a defined areascalar quantity (coulombs/sec)

J = current densityvector quantity (coulombs/sec/m2)

AE dΦAnalogous to E-field flux:

AJ dI

carrier density flux density(current density)

Flux (current)

Fluid flow kg/m3 kg/m2/sec kg/sec

Electricity flow Coulombs/m3 Coulombs/m2/sec(Amps/m2)

Coulombs/sec (Amps)

Energy flow Joules/m3 Joules/m2/sec(Watts/m2)

Joule/sec(Watts)

Electric fields (Faraday’s analogy)

Fluxons/m3

Electric fieldFluxons/m2/sec(N/C = Volts/m)

Fluxons/sec(Volt·m)

AJ

vJ

Electric-fields vs flow-related quantities

Fluxon (made it up myself) = kgm3/Csec The flow analogy requires imagination.

Rank the four regions in order of current magnitude.

Now define the positive direction to be to the right.

Rank the four regions in order of signed current.

dccdc

ccc

vqnvJ

qn

(carriers/m3)(charge/carrier) = charge/m3

(charge/m3)(m/sec) = charge/sec/area

2

2

1

cc

c

c

cccc

c

c

c

cd

qn

m

EEm

qnE

m

qqnJ

Em

qE

m

qav

Resistivity is determined entirely by microscopic parameters

including the average thermal collision rate (1/).

Ohm’s Law: linear response

Global

A

LR

RIA

LIV

JE Local

ALV

AE

JAI

)/( AJ

Analogous to Poiseuille’s Law of fluid flow (hyperphysics).

Ohmic non-Ohmic

current difference potential

GLOBAL

field electric densitycurrent

LOCAL

onaccelerati citydrift velo

CMICROSCOPI

IV

EJ

avd

resistance area

length GLOBAL

yresistivit tyconductivi

density charge LOCAL

collisionsbetween mean time masscarrier

densitycarrier chargecarrier

CMICROSCOPI

RA

L

mnq

c

c

c

c

The resistor The flow

Electric Current Flow Quantities

Assume that the charge (qc), mass (m), density of carriers (nc) and mean collision time () in a wire are known.

Which of the following quantities can be determined?

1) Resistivity ()2) Resistance (R)3) Current (I)4) Current density (J)5) Drift velocity (vd)6) Electric field (E)

If we are further given the potential difference (V) across the wire, which of these quantities can be determined?

If we are also given the shape of the wire (L and A), which of these quantities can be determined?

A

LR

IRV

JAAdI

JL

VE

EvJ

Em

qv

qn

m

dc

c

cd

cc

c

J

2

1

Quiz: If we double the current through a copper wire of length L and area A, the resistance will be:(1) quadrupled (2) doubled (3) unchanged (4) halved (5) quartered

A

L

I

VR

Quiz: If we double the length and the diameter of copper wire, its resistance will be:(1) quadrupled (2) doubled (3) unchanged (4) halved (5) quartered.

Quiz: If we stretch a copper wire to twice its original length at constant volume, its resistance will be:(1) quadrupled (2) doubled (3) unchanged (4) halved (5) quartered.

qVW

Work to move charge q through a potential difference V:

If the average kinetic energy of the carriers doesn’t increase, then the energy expended has been converted to heat via atomic collisions. Let power (P) be the rate at which energy is dissipated as heat.

R

V

R

VVRIIIR

VIdt

dqV

dt

dWP

22)(

R

VRIP

22

R

VRIP

A

L

I

VR

22

Quiz: If we double the resistance of an Ohmic wire of length L and area A, the power dissipated in the wire will be:(1) quadrupled (2) doubled (3) unchanged (4) halved (5) quartered.

Quiz: If we double the length of an Ohmic wire of area A and resistivity , while holding the potential difference V across the wire constant, the power dissipated in the wire will be:(1) quadrupled (2) doubled (3) unchanged (4) halved (5) quartered.

(Not enough information! We could fix either V or I?)

Non-Ohmic behavior

Diode

Cell Membrane

Varistor

http://www.wikipedia.comhttp://www.onsemi.com

The resistivity of a metal increases as it gets warmer.

The resistivity of an insulator or semiconductors decreases as it gets warmer.

Practical definition of metal/insulatorslope of ρ vs. T > 0 → metalslope of ρ vs. T < 0 → insulator

2cc

c

qn

m T

R

R

Superconductors ( = 0)

Apply 100 V to a 1 meter length of 12-gauge copper wire (r = 1 mm).

3

6

8

104.5)1014.3(

)0.1)(107.1(

A

LR

A105.18)104.5(

)100( 33

R

VI

296

3

A/m109.5)1014.3(

)105.18(

A

IJ

V/m100)109.5)(107.1( 98 JE

m/s433.0)10602.1)(105.8(

)109.5(1928

9

cccd qn

JJv

s106.24)100)(10602.1(

)433.0)(1011.9( 1519

31

Eq

vm

c

d

262 m1014.3)001.0( Am107.1:Cu 8

W1085.1)104.5()105.18( 63232 RIP

Apply 100 V to a 1 meter length of 12-gauge copper wire (r = 1 mm).

kJ133)kJ/mol 5.300)(mol 443.0(

kJ161)K13562840)(g 13.28)(J/g/K 85.3(

kJ82.5)kJ/mol 14.13)(mol 443.0(

kJ4.11)K3001356)(g 13.28)(J/g/K 85.3(

2

1

2

1

vapmolvap

mw

meltmolmelt

mw

HnE

TmcE

HnE

TmcE

ms72ms87ms2.3ms1.62

1

1

vapwmelt

ww ttt

E

Pt

K 28401356

kJ/mol 5.300kJ/mol 14.13J/g/K 85.3

g/cm96.8/5.63 3

boilmelt

vapmeltm

matomic

TT

HHc

molgm

mol 443.0g/mol) 5.63(

)g 28.13(/

g 28.13m) 1)(m1014.3)(g/cm96.8( 363

molarmol

m

mmn

LAm

So what do you call a length of 12-gauge copper wire connected across the terminals of a good 120 V power source?

A fuse!

Cost of power used for home lighting(50 bulbs, 25 Watts each, 50% use, $0.10 per kWhr)

W2250bulb

W25bulbs 50

P

Amp/bulb21.0V 120

W25

V

PI

month/81$month

hr 2430

hrkW

10.0$)kW 25.2%)(50(cost

High-Voltage Transmission Line(V = 120 V vs. V = 120 kV)

622

2

2

2

2

101000

1

')/(

)'/'(''

V

V

VP

VP

RI

RI

P

P

trans

trans

loss

loss

transtrans PIVIVP '''

:smittedPower tran

RIPloss2

:losspower Resistive

High-Voltage Transmission Line (aluminum)L = 100 km, r = 0.5 cm, Ptrans = 10 MW, V = 120 kV)

%3)101(

)103(V3280)36)(91(

7

5

trans

lossloss P

PIRV

/hr$30kW300)36()91( 22 RIPloss

36)005.0(

)10)(1082.2(2

58

A

LR

A91)10011(

)1010(3

6

V

PI trans

65 103$m)($30/m) 10(cost Cable

Increasing A and cable cost by factor of 3 (adds 2×106 $) will decrease R and Ploss by factor of 3 (saves 20 $/hr). The reduced losses pay for the more expensive cable after 11.4 years.

Effective resistance: two resistors in series

R1 R2

R1 and R2 experience the same current but different voltages. Largest R has largest V.

212121 RR

I

RIRI

I

VV

I

VReq

Req is larger than either R1 or R2.

R1

R2

Effective resistance: two resistors in parallel

R1 and R2 experience the same voltage but different currents. Smallest R has largest I.

1

212121

11

//

RRRVRV

V

II

V

I

VReq

Req is smaller than smallest of R1 and R2.

Reduction of a resistive network

R R

R R

R

R R

R R

R

2R

R

2R

R/2

Reduction of a resistive network

R RR RR

R

R

R

R Ra b

(7/3)R

R/3R R

ab

ab

Reduction of a resistive network

Apply 42 V between a and c.

What is I between a and c?

What is Vbc?

I = 3 A

What is I2?

Vbc = 6 V

I2 = 2 A

The switch is initially open. When the switch is closed, the current measured by the ammeter will:

(1) increase (2) decrease (3) stay the same (4) fall to zero.

Compare the brightness of the four identical bulbs in this circuit.

D is in parallel with a zero-resistance wire. The current will take the zero-resistance path and bypass D altogether.

A and B are in series. So they will burn equally bright. Together, they see the full battery voltage.

C experiences the full battery voltage, or twice the voltage experienced by A or B. So C is four times as bright.

R R

V

R

If R1 is removed, R2 will glow(1) more brightly.(2) less brightly.(3) same brightness as before.

If R1 is removed, R2 will glow(1) more brightly.(2) less brightly.(3) same brightness as before.

Household devices are wired to run in parallel!

http://www.ciphersbyritter.com/RADELECT/LITES/XMSLITES.HTM

Strings of 50 Christmas lights in series. Assume ~100 V source and 25 W power consumption.

What is the resistance of a single bulb?(1) 2 (2) 4 (3) 8 (4) 10

Resistance of an object with arbitrary shape

dzab )( 22

End-to-end:)(zA

dz

area

lengthddR

)()( 22022 ab

Ldz

abdRR

L

Inside-out:Lr

dr

2

)(rA

dr

area

lengthddR

a

b

Ldr

LdRR

b

aln

22

ab

L

dz

)()( 2 zr

dz

zA

dz

A

dLdRR

Resistance of an object with arbitrary shape

L

z

L

u

du

ab

L

Lzaba

dz0 20 2]/)([

ab

L

baab

L

11

• A battery or other constant-voltage device (e.g. power suppy) is usually the energy source in a direct-current (DC) circuit.

• The positive terminal of the battery has higher potential than the negative terminal.

• The electromotive force (emf, ) of a battery is the open-circuit voltage between its terminals when no current is flowing.

• Ideally, the battery has no internal resistance of its own.

• We often idealize the wires in a circuit to have zero resistance.

Batteries as circuit elements

+

Real batteries

• Chemical energy: an electrolyte solution allows negative ions to flow toward and react with the anode (), while positive ions flow toward and react with the cathode (+).

• Charge build-up prevents the reaction from proceeding unless an external circuit allows electrons accumulating at the anode to return to the cathode.

• Dead when the reactants are used up. Rechargeable if the anode/cathode reactions are reversible.

• Internal resistance tends to increase with age, use and multiple recharge cycles.

Internal resistance in non-ideal batteries

rIVV ab

RR

V

rRI

Terminal voltage:

Current:

Two AA batteries yield a combined = 3 V. You observe a terminal voltage of 2.7 V while delivering 300 mA of current to an ultra-bright flashlight.

What is the load resistance (R) ? (1) 1 (2) 3 (3) 9 (4) 10

What is the internal resistance of a single AA battery?(1) 0.5 (2) 1 (3) 2 (4) 3

Typical Alkaline (Zn/MnO2)1.5 to 1.6 V open circuit, 1.1 to 1.3 V closed circuit.163 W-hr/kg (590 J/g) – 400 kJ total for AA0.034 (122.4 C) to 15 A-hr (54 kC) of charge depending on size.85% Capacity after 4 years of non-use.

Lithium Ion3.2 V open circuit, 2.5 to 3.0 V closed circuit.230 W-hr/kg (828 J/g) – 460 kJ total for AA0.160 (576 C) to 1.4 Amp-hrs (5.04 kC) of charge95% capacity after 5 years of non-use

NiCd1.2 V open circuit50 W-hr/kg (180 J/g) – 140 kJ total for AA70% capacity after one month of non use (500-5000 cycles)

Zn-Air bus battery:200 W-hr/kg (720 J/g)320 kW-hr (1.15 GJ) of energy

Multi-loop circuits

Loop: A current path that begins and ends at the same circuit point, traversing one or more circuit branches, but without ever passing the same point twice.

Branch: An independent current path experiences only one current at a given moment. It may be a simple wire or may also contain one or more circuit elements connected in series.

Junction: A point where three or more circuit branches meet.

Multi-loop circuits

The circuit above has: 3 branches, 2 junctions, 3 loops.

To solve for 3 unknown branch currents, we need 3 equations.

To get these equations, use all but one (2 1 = 1) junction, and as many independent loops as needed ( 3 1 = 2).

Kirchoff’s current rule: 0 nI

021 III

Current rule: The total current flowing into a junction is zero.

Arrows define positive branch-current directions. A current later determined to flow opposite its arrow is “negative”.

021 III 021 III

Alpine loop elevation

Kirchoff’s Voltage Rule: 0 nV

V = + for a battery crossing from – to + terminal.

Voltage rule: The voltage changes around a loop sum to zero.

Arrows define positive branch-current directions.

Use V = –I R when crossing a resistor in the positive direction.

Use V = +I R when crossing a resistor in the negative direction.

011 RI

I

Single-loop circuit example

mA 35000/15

0)5000(15

)500(25)1500(

)1000(30)2000(20

I

I

II

II

V 19

)3000)(003.0(10)3000(10

)1000(30)2000(20

I

IIVV groundA

Multi-loop circuit example

0)2()1(5 32 II

0312 III

05)1()1(5 21 II

Bottom loop:I2 = I1

Substitute I2 in junction Eq:I3 = 2I1

Substitute I2 and I3 in top loop:I1 = 5/3

Solve for currents:I1 = 5/3, I2 = 5/3, I3 = 10/3

Multi-loop circuit example

0144106 21 II

65.1 12 II

06102 13 II

53 13 II

0321 III

0115.5

)53()65.1(

1

111

I

III

132 321 III