Announcements 1 st Midterm and Homeworks #1 and #2 will be returned - - PowerPoint PPT Presentation

Announcements

• 1st Midterm and Homeworks #1 and #2 will be returned this

week in workshops

• Discussion of 1st exam results Monday, Oct 13th, after Prof.

y, , McFarland is back from Japan

• 2nd Midterm Exam is 23 October 8:00-9:20am Hubbell

2 Midterm Exam is 23 October, 8:00 9:20am, Hubbell

• HW#5 due Wednesday Oct 15th 11:30
• Preflights #10 and #11 will be posted later this week

– due Monday and Wednesday, 7am, respectively – everything with workshops also back on the normal schedule this week

Questions

• “AWESOME job using the chalk board to do examples.

Please keep doing this! keeps people awake and makes us follow along sicne we write it down”

– I will keep trying to do this. It’s a little uncomfortable for me b f h k d f h b d because of the awkwardness of the boards.

– “can you please not make HW sets due Tuesdays? The ChemE students in the class all have 2 other major problem t d T d d thi ld j t h th sets due Tuesdays and this would just push us over the edge”

– I will try to keep them on Wednesday, but I need better participation in the Wednesday preflights participation in the Wednesday preflights – If you think of other ways I can achieve that, please let me

• know. I’m open to other ideas.

– “You have to love baseball!!! Didn't your party school U – You have to love baseball!!! Didn t your party school U. Chicago teach you anything about loving the sport?”

– Nope. The Brewers’ spectacular collapse doesn’t help either.

Personal note from Prof McFarland: thank you so much for – Personal note from Prof. McFarland: thank you so much for not mentioning football in the comments this week. Oy!

Leftover: Lecture 8, Concept 2

• Two parallel plate capacitors are identical

(same A, same d) except that C1 has half of

κ

C E =? +Q

(same A, same d) except that C1 has half of the space between the plates filled with a material of dielectric constant κ as shown.

κ

C1 E1=? +Q

• Q

– If both capacitors are given the same

C2 E2=? +Q

• Q

amount of charge Q, what is the relation between E1, the electric field in the air of C1, and E2, the electric field in the air of C2

(1) E1 < E2 (2) E1 = E2 (3) E1 > E2

• The key here is to realize that the electric field in the air in C must be
• The key here is to realize that the electric field in the air in C1 must be

equal to the electric field in the dielectric in C1!!

• WHY?
• Plates are conductors; therefore they are equipotential surfaces.

F thi t h th h d it h l t b

• For this to happen, the charge density on each plane must be non-

uniform to create equal electric fields!!

• Another way (microscopic view): polarized dielectric attracts charge

from conductor, thus moving charge from air region of C1

• Consequently, E1 < E2.
• So V1 < V2, and thus C1 > C2. (A reasonable guess!)

Today…

• Current and How Current Works

Current and How Current Works

• Devices

– Batteries – Resistors

• Ohm’s Law
• Power in Resistors (

b bl t l t )

• Power in Resistors (probably next lecture)

Text Reference: Chapter 25:1-8 Examples: 25. 1,3,5,6,10

Current is charge in motion

C

• Charge, e.g. electrons, exists in conductors

with a density, ne (ne approx 1023 cm-3)

• “Somehow” put that charge in motion:
• Somehow put that charge in motion:

– effective picture – some net charge flows with an average velocity per charge carrier, vd – more on how this happens in a minute!

C t d it j i i b j

• Current density, j, is given by j = qenevd

– unit of j is C/m2sec or A/m2 (A = Ampere) and 1A = 1C/s – current, I, is j times cross sectional area, I = j πr2 – for 10 Amp in 1mm x 1mm area, j=10+7 A/m2 , and vd is about 10-3 m/s

Circuit Elements

• Capacitors:

Purpose is to store charge (energy).

• We have calculated the capacitance of a system
• We had to modify Gauss' Law to account for bulk

matter effects (dielectrics) … C = κC0

• We calculated effective capacitance of series or

p parallel combinations of capacitors

• Batteries (“Voltage sources”):

+

• Purpose is to provide a constant

potential difference between 2 points. +

OR

• Electrical ↔ chemical energy conversion
• For now, we will think of them as an ideal

source of voltage

V +

• Later: Non-ideal batteries will be dealt with

in terms of an "internal resistance".

Circuit Elements (Cont’d)

R i

• Resistors:

Purpose is to limit current drawn in a circuit. p

dQ I dt =

UNIT: Ampere = A = C/s Note:

• Resistance can be calculated from knowledge of the

dt

geometry of the resistor AND the “resistivity” of the material out of which it is made.

• The effective resistance of series and parallel

combinations of resistors will be calculated using Kirchhoff's Laws (which are really just statements about potential difference and current conservation).

• “Conductors” as we have considered them before,

have resistance!

Defining Resistance

• Resistance

R

Resistance Resistance is defined to be the ratio of the applied voltage to the current passing through.

I I R

the current passing through.

V

I V R ≡

UNIT: OHM = Ω

• Is this a good definition?

i.e., does the resistance belong only to the resistor? Recall the case of capacitance: (C=Q/V) depended on the geometry, not on Q or V individually g y, Q y Does R depend on V or I ? It seems as though it should, at first glance...

C id i i h i h l d i R

PF09 Question #2

Consider two circuits, each with a load resistor Ri driven by a battery with voltage Vi. V1=5 Volts, V2=20 Volts, R1=10 Ω, R2=50 Ω.

R V

2

,

1

,

2

Compare the currents through the two resistors. a) I1 > I2 b) I1 = I2 c) I1 < I2

95% 40% 60% 80% 100% 3%

3%

0% 20% 40%

What you said …

What I would have said

R V

What I would have said … I=V/R. I1=5V/10Ω, I2=20V/50Ω

R V

Ri h Right:

Ohm's law says that I=V/R. I1= 5/10 > I2= 20/50 For question 2 I = V/R and for question 4 P = IV. Equation roulette :) [ok, you win this time… care to try your luck again?]

Wrong:

Current is the flow of electric charge. Resistors reduce the flow of charge, and the battery provides the charge. Although there is less charge provided by V1, the ratio of Voltage/Resistance for V1/R1 is larger than V2/R2. So despite the fact that more charge is flowing from V2, R2 is reducing the charge much more. So I1>I2. [Careful not to confuse charge flow with voltage. Voltage is just a potential difference… tells you how much higher energy of charge leaving battery is than charge returning. #2 has smaller CURRENT at higher VOLTAGE] The current for circuit 1 is higher because the battery is working against a smaller resistance [Right answer, but… current is also proportional to V. More V means more force on charge carriers]

Resistance and Currents

• Let’s analyze the flow of charge

I I R

y g through a resistor…

– If there is a potential difference inside the resistor, then there is also…?

V

– A constant electric field

• Constant electric field implies

what force on charge carriers? g

–Constant, and therefore a constant acceleration!

• If acceleration of charge carriers occurs, then how does

g , current vary with time?

–Well, it should increase linearly with time, like velocity does under constant acceleration, right?

• This isn’t what we observe…

– We observe that current is proportional to voltage and doesn’t change over long period of time – Let’s understand this…

Resistance and Currents (cont’d)

• Charges are put in motion, but scatter in a very short

time from things that get in the way

– it’s crowded inside that metal – defects, vibrational motion of atoms (phonons), etc

1

• Typical scattering time τ = 10-14 sec
• charges accelerated for this time and then randomly scattered

charges accelerated for this time and then randomly scattered

• average velocity attained in this time is v = aτ = eEτ/m
• current density, j = enev, so current is proportional to E which is

proportional to Voltage proportional to Voltage

• OHM’s LAW j = (e2neτ/m)E (e2neτ/m=σ , called conductivity)

Lecture 9, Concept 1

• Which analogy would you use to explain flow of charge in a

resistor in response to an potential difference? Why?

(1) S f lli t id i d (1) Snow falling outside your window (2) A car rolling down a hill towards a wall (3) Molecules of a scent traveling through the air traveling to your nose

Lecture 9, Concept 1

• Which analogy would you use to explain flow of charge in a

resistor in response to an potential difference? Why?

(1) S f lli t id i d (1) Snow falling outside your window (2) A car rolling down a hill towards a wall (3) Molecules of a scent traveling through the air traveling to your nose

• The snow is falling at an apparently constant velocity despite the

constant acceleration due to gravity because of microscopic collisions with air molecules in its path collisions with air molecules in its path

• The car rolls down the hill in response to a constant acceleration,

and increases in speed until a catastrophic collision at the bottom.

• The scent molecules diffuse through the air, but they diffuse

g , y uniformly.

Resistance

• How do we calculate R??

I I R

How do we calculate R??

• To calculate R, must calculate current

which flows when voltage V is applied.

V

• Applying voltage V sets up an

electric field in the resistor. What determines the current?

V

I V R ≡

• Current is charge flowing past a point per unit time.

Need to know velocity of charges. I thi ff t d b fi ld?

• Is this affected by field?
• Force on charge carriers accelerates charge between

collisions collisions

• Empirically, we separate this into two parts

– Part of this we bury as the “resistivity” of the material – Part depends on the geometry of the resisting material

Resistivity

• Property of bulk matter related to

E

Property of bulk matter related to resistance of a sample is the resistivity ρ defined as:

L A E j

j E ≡ ρ

2

1

e

m e n ρ σ τ = =

where E = electric field and j = current density in conductor.

L

For uniform case:

I j

EL V

,

For uniform case:

A j =

EL V =

⇒

⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = = = = A ρL I L A I L EL V ρ ρj

⇒

L R ρ =

where

IR V =

So, in fact, we can compute the resistance if we know a bit about the material, and YES, the property belongs to the material!

⇒

⎟ ⎠ ⎜ ⎝ A A ρ ρj

A ρ

e.g., for a copper wire, ρ ~ 10-8 Ω-m, 1mm radius, 1 m long, then R ≈ .01Ω

Makes sense?

L

A E j

A L R ρ =

• Increase the Length, flow of electrons impeded

L A j

• Increase the cross sectional Area, flow facilitated
• The structure of this relation is identical to heat flow

through materials through materials

• think of window for an intuitive example

How thick? How big? What’s it made of?

• r

2

Two cylindrical resistors are made PF09 Question #5-6 y from the same material, and they are equal in length. The first resistor has diameter d and the resistor has diameter d, and the second resistor has diameter 2d.

100%

Compare the resistance of the two cylinders. ) R R b) R R ) R R

82% 40% 60% 80%

a) R1 > R2 b) R1 = R2 c) R1 < R2 If the same current flows through both resistors,

5%

13%

0% 20%

If the same current flows through both resistors, compare the average velocities of the electrons in the two resistors:

69% 40% 50% 60% 70% 80%

a) v1 > v2 b) v1 = v2 c) v1 < v2

7%

25%

0% 10% 20% 30%

What you said …

Right: R is proportional to (length)/(cross-secional area) so the same length and larger area in R2

means a lower resistance in R2. The current density is larger in R1 [for the same current] so the average drift velocities will also be larger.

Wrong:

R1=R2 because resistance is not determined by size but by Ohm's Law. . IR=V since R1 is greater the potential difference is greater therefore, the velocity is greater IR=V since R1 is greater the potential difference is greater therefore, the velocity is greater [oh no, it’s equation roulette! the “V” in Ohm’s Law is potential difference, not velocity] The current is the same flowing through the resistors and the length of the resistors is the same so charge will flow over the same distance with the same drift velocity. [current is flow of charges past a given point, so it doesn’t depend on length. the density of charge carriers is a property of the material, so for the same material what matters is the cross-sectional area… more carriers in parallel means they can go more slowly and carry the same current.] When A is bigger, drift velocity will be less. Intuitively, with a larger area, there will be less electron collisions and the velocity of electrons will be smaller at equilibrium [the answer is right, but it is because V is smaller (R smaller), so less acceleration. still as many targets to collider with.]

Lecture 9, Concept 2

• Two cylindrical resistors, R1 and R2, are made of identical material.

y

1 2

R2 has twice the length of R1 but half the radius of R1. – These resistors are then connected to a battery V as shown:

I1 I2

V

1

– What is the relation between I1, the current flowing in R1 , and I2 , the current flowing in R2?

(a) I1 < I2 (b) I1 = I2 (c) I1 > I2 ( )

1 2

( )

1 2

( )

1 2

Lecture 9, Concept 2

• Two cylindrical resistors, R1 and R2, are made of identical material.

y

1 2

R2 has twice the length of R1 but half the radius of R1. – These resistors are then connected to a battery V as shown:

I1 I2

V

1

– What is the relation between I1, the current flowing in R1 , and I2 , the current flowing in R2?

(a) I1 < I2 (b) I1 = I2 (c) I1 > I2

• The resistivity of both resistors is the same (ρ).
• Therefore the resistances are related as:

2 L L L

( )

1 2

( )

1 2

( )

1 2

1 1 1 1 1 2 2 2

8 8 ) 4 / ( 2 R A L A L A L R = = = = ρ ρ ρ

• The resistors have the same voltage across them; therefore

1 1 2 2

8 1 8 I R V R V I = = =

Ohm Example: Resistors in Series

a

R

I

The Voltage “drops”: b

R1

1

IR V V

b a

= −

2

IR V V

c b

= −

) (

2 1

R R I V V

c a

+ = − a c

R2

Whenever devices are in SERIES, the current is the same through both !

Reffective

c current is the same through both ! This reduces the circuit to: c

) (

2 1

R R Reffective + =

Hence:

Resistivity Example:

R1

Resistors in Series

Consider two cylindrical resistors with lengths L1 and L2

V

L1

L R

1

ρ =

V

L2

A R1 ρ = L R

2 2

ρ =

R2

A

2

ρ

Put them together, end to end to make a longer one...

2 1 2 1

R R A L L Reffective + = + = ρ

2 1

R R R + =

What about Work and Power?

• V=IR

– V is potential difference across which a charge is moved – How much charge? Current is the rate of flow of charge g g

• dQ/dt = I
• So for constant current, Q=I×time
• The work required to move a charge through a potential

The work required to move a charge through a potential difference is…?

– W=V×Q

• Where does this work go in a resistor?

Where does this work go in a resistor?

– Does it go into kinetic energy of the charge?

• No, charge scatters and slows

– So where does it go? So where does it go?

• Heat in a toaster element, for example. Or light in a light bulb.
• Remember Power from Physics 121?

– Power is Work / Time Power is Work / Time

Power

Batteries & Resistors Energy expended

chemical to electrical to heat

work

( )

s J

power time work =

Rate is:

Charges per time

Therefore:

VI P =

Potential difference per charge p g For Resistors:

( )

R I I IR P

2

= =

( )

R V R V V P

2

= =

• r you can write it as

Units okay?

Watt s J second Coulomb Coulomb Joule = = ×

C id i i h i h l d i R d i b b

PF09 Question #3

Consider two circuits, each with a load resistor Ri driven by a battery with voltage Vi. V1=5 Volts, V2=20 Volts, R1=10 Ω, R2=50 Ω. Compare the power dissipated in the two resistors.

R V

a) P1 > P2 b) P1 = P2 c) P1 < P2

R V

83%

80% 100%

83%

20% 40% 60% 12% 5% 0%

What you said …

What I would have said …

R V

Ri h

P=V2/R. I1=(5V)2/10Ω, I2=(20V)2/50Ω

R V

Right:

[Question 2] Using the equation V=IR, we can plug in the given values of V and R to find the current, I, for each circuit. The current in circuit 1 is .5 A, and the current in circuit 2 is .4 A, so the I1>I2. [Question 3] Using the equation P=IV we can plug in the given value of V with the value of I found [Question 3] Using the equation P=IV, we can plug in the given value of V with the value of I found in Question 2 to find the power, P, for each circuit. The power in circuit 1 is 2.5 W while the power in circuit 2 is 8 W, So P2>P1.

Wrong:

The power in the first circuit is larger than those in the other because current is larger [correct about current, but P=I^2R. High current but no resistance would result in no power lost]

Electric ovens, which must deliver a certain amount of power PF09 Question #8 , p to cook food, run on 240 Volts AC instead of the more common household voltage of 120 Volts AC. Which of the following might be a reason why? (You may choose more following might be a reason why? (You may choose more than one answer).

a) It may be difficult to make a very high resistance heating element. b) It is dangerous to carry too much current in household b) It is dangerous to carry too much current in household wiring. c) A low voltage oven would have to be supplied by thin i hi h ld b t f il t th h h wires which would be too fragile to run through a house. d) High voltage supply wires are safer in the event of a short in an oven.

80% 60% 80% 100% 22%

9% 25%

0% 20% 40%

What you said …

What I would have said …

P=V2/R. So the lower voltage (120V instead of 240V) would require ¼ the resistance for equal power. Twice the current. (a) 120V would require lower resistance (c) thin wires (high resistance) (a) 120V would require lower resistance, (c) thin wires (high resistance) would be bad with high current, (d) high voltage shorts are very bad… Right: Doubling the voltage requires that only half the current be drawn to achieve the same Right:

too much current in a wire is usually the cause of fires in homes. [high current & resistance in wires and connections means high power dissipated there, which could create dangerous amounts of heat] Doubling the voltage requires that only half the current be drawn to achieve the same amount of power. [true, so lower current for the 240V oven]

Wrong:

The heat cooking the food comes from many collisions inside the wire. An oven needs a lot of heat, and therefore the heating element needs a high resistance. If voltage is kept constant (at 120 V) and V=IR, resistance has to be much higher [good points, but hi h i t ill l t ( t fi d lt ) d th f l ] Danger is augmented by the increased resistance of thin wires, since thermal energy production is proportional to R [true: thin wires have high R. so low voltage oven would require THICK wires] higher resistance will mean lower current (at fixed voltage) and therefore lower power] Using high voltage reduces the magnitude of the current needed to generate a specific level of power P=IV, so high voltage wires reduce the dangers associated with a large current [true oven current is small, but a short means a low resistance to ground, so high voltage means higher current in short]

Lecture 9, Concept 3

• Consider the circuit at the right.

I

g – Which of the following modifications to the circuit will result in the maximum power dissipated in the resistor?

I R V

(a) Double R, leaving V unchanged

I R V

(b) Double V, leaving R unchanged (c) Double R and V

Lecture 9, Concept 3

• Consider the circuit at the right.

I

g – Which of the following modifications to the circuit will result in the maximum power dissipated in the resistor?

I R V

(a) Double R, leaving V unchanged

I R V

(b) Double V, leaving R unchanged (c) Double R and V

• Power = IV = V2/R
• Therefore, want to increase V and decrease R to

maximize the power dissipated in the resistor p p

Household Electrical Safety

• Most dangerous aspect of everyday

household current: ventricular fibrillation household current: ventricular fibrillation

• AC current of a few mA at 60Hz

through the heart can be fatal

• this translates to a current across the chest of ~100mA
• why? current density (through the heart) vs. current

y y ( g )

• I’ve been shocked from hand-to-hand by ~110 Volts twice in

my life (once doing a demo in Physics 122) my life (once doing a demo in Physics 122)

• so, why am I still alive?
• resistance of dry skin is ~10kΩ (wet skin is only ~1kΩ)
• resistance of dry skin is ~10kΩ (wet skin is only ~1kΩ)

so V=IR says current was only ~10mA

Household Electrical Safety

• Most safety tips focus on avoiding

current through the chest current through the chest

• Work on electrical panels with one

hand in your pocket, standing on insulated platforms to avoid any return path for current Th l

• Three-prong plugs
• Provide a separate

current path usually current path, usually connected to a metal case of an appliance, i f h t b t in case of a short between the “hot” wire and the case

Superconductivity…

• An amazing fact of nature…

some materials at low temperature have a resistance so low that it cannot have a resistance so low that it cannot be measured

• “Superconductors”

Superconductors

• What must the electric field be in a superconductor?
• Tremendous savings from high current applications

(like high field magnets… more on this later) l t d t i t t fl

• no power lost due to resistance as current flows, as

long as you can keep your conductor very, very cold

• another amazing feature

superconductors repel

• another amazing feature… superconductors repel

magnetic fields. Our first hint (in this course) that magnetism is related to electricity!

Summary…

• What Current Is

What Current Is

• Devices

– Batteries – Resistors

• Ohm’s Law

Nature of conduction and resistance in materials – Nature of conduction and resistance in materials

• Power

Next Time…

• Real Batteries
• Combining Circuit Elements

T t R f Ch t 25 9 10 26 1 3 6

• Combining Circuit Elements
• Kirchhoff’s Rules

Text Reference: Chapter 25:9-10, 26: 1-3, 6 Examples: 26: 1, 3, 6, 8