30
1 Monday July 14 • Lecture 25 – Capacitance demo slide 19 – Capacitors in series and parallel slide 33 – Elmo example • Lecture 26 Currents and Resistance • Lecture 29 Circuits • Wear Microphone

Monday July 14

  • Upload
    elaina

  • View
    31

  • Download
    0

Embed Size (px)

DESCRIPTION

Monday July 14. Lecture 25 Capacitance demo slide 19 Capacitors in series and parallel slide 33 Elmo example Lecture 26 Currents and Resistance Lecture 29 Circuits Wear Microphone. Lecture 6 Current and Resistance Ch. 26 We are now leaving electrostatics. - PowerPoint PPT Presentation

Citation preview

Page 1: Monday July 14

1

Monday July 14

• Lecture 25– Capacitance demo slide 19– Capacitors in series and parallel slide 33– Elmo example

• Lecture 26 Currents and Resistance• Lecture 29 Circuits• Wear Microphone

Page 2: Monday July 14

2

Page 3: Monday July 14

3

Page 4: Monday July 14

4

Lecture 6 Current and Resistance Ch. 26We are now leaving electrostatics.

• Cartoon -Invention of the battery and Voltaic Cell• Topics

– What is current?– Current density– Conservation of Current– Resistance– Temperature dependence– Ohms Law– Batteries, terminal voltage, impedance matching– Power dissipation– Combination of resistors

• Demos– Ohms Law demo on overhead projector– T dependence of resistance– Three 100 Watt light bulbs

• Elmo– Puzzle - Resistor network figure out equivalent resistance

Page 5: Monday July 14

5

Loop of copper wire

Nothing moving;

electrostatic equilibrium

Now battery voltage forces charge through the conductor and we have a field in the wire.

0=E

0≠E

Page 6: Monday July 14

6

What is Current I?It is the amount of positive charge that moves past a certain point per unit time.

Ampond

CoulombtQI ==ΔΔ

=sec

Copper wire with voltage across it

+ +

+ +

+ +

+

+

I

ΔLALtv Δ=Δ

Drift velocity of charge

Density of electrons 1.6 x 10-19 C

Divide both sides by Δt. nqAvtQI =ΔΔ

=

tnqAv QtAvnq

volume volume unit per charge

Δ=ΔΔ×=

×=ΔQ

Page 7: Monday July 14

7

Drift velocity

vd = InqA

Μ=

oNn ρ = 8.4x1022 atoms/cm3

21922 )0815(.106.1104.81

3 cmCampv

cmatomsd π××××

= −

sm

dv5105.3 −×= The higher the density

the smaller the drift velocity

I = 1 Amp

q = 1.6x10-19 C

A = π(.0815 cm)2

= 8.9 grams/cm 3

No = 6x1023 atoms/mole

M = 63.5 grams/mole

Example: What is the drift velocity for 1 Amp of current flowing through a 14 gauge copper wire of radius 0.815 mm?

What is the drift velocity vd ?

Page 8: Monday July 14

8

What is the current density J ?Directions of current i is defined as the direction of positive charge. I is not a vector.

(Note positive charge moves in direction of E) electron flow is opposite E.

i =nAqvdJ =nqvd

i =J⋅d

A∫

J =

iA when

J and

A ae πaallel

Page 9: Monday July 14

9

Currents: Steady motion of charge and conservation of current

Current is the same throughout all sections in the diagram below; it is continuous.

Current density J does vary

i0 =i1 + i2 (Kichoff's 2nd Rule)

Higher J

Page 10: Monday July 14

10

Question: How does the drift speed compare to the instantaneous speed?

Instantaneous speed 106 m/s

(This tiny ratio is why Ohm’s Law works so well for metals.)

Question: At this drift speed 3.5x10-5 m/s, it would take an electron 8 hours to go 1 meter. So why does the light come on immediately when you turn on the light switch?

It’s like when the hose is full of water and you turn the faucet on, it immediately comes out the ends. The charge in the wire is like the water. A wave of electric field travels very rapidly down the wire, causing the free charges to begin drifting.

tinsd vv tan11105.3 −×≈

sm

dv5105.3 −×=

Page 11: Monday July 14

11

Example: Recall typical TV tube,or CRT. The electron beam has a speed 5x107 m/s. If the current is I = 100 microamps, what is n?

n =10−4A

1.6 ×10−19C⋅10−6m 2 ⋅5 ×107 ms

For CRT

For Copper

The lower the density the higher the speed.

Take A =1 mm 2

=(10−3m )2

=10−6m 2

37

313

cmelectrons102.1

melectrons102.1n ×=×=

322

cmelectrons105.8n ×=

n =I

qAv

Page 12: Monday July 14

12

Ohm’s Law

Want to emphasize here that as long as we have current (charge moving) due to an applied potential, the electric field is no longer zero inside the conductor.

I

• •A B

ΔL

Potential difference

True for many materials – not all. Note that Ohms Law is an experimental observation and is not a true law.

Constant of proportionality between V and I is known as the resistance. The SI unit for resistance is called the ohm.

Demo: Show Ohm’s Law

Best conductors

Silver

Copper – oxidizes

Gold – pretty inert

Non-ohmic materials

Diodes

Superconductors

constant. is E where,LEVV AB Δ=−

law) s(Ohm' current LEI Δ∝=

V =RI R =VI amp

VoltOhm =

V =RI R =VI

Page 13: Monday July 14

13

A test of whether or not a material satisfies Ohm’s Law

satisfied islaw sOhm'

constant1Slope ==

=

=

R

RVI

IRVHere the slope depends onthe potential difference.Ohm's Law is violated for a pn junction diode.

Page 14: Monday July 14

14

What is Resistance?The collisions between the electrons and the atoms is the cause of resistance and the cause for a very slow drift velocity of the electrons. The higher the density, the more collisions you have.

The dashed lines represent the straight line tracks of electrons in between collisions----------Electric field is off.

----------Electric field is on. When the field is on, the electron drifted further to B’...

e-

field off

field on

extra distance electron traveled

Page 15: Monday July 14

15

• Depends on shape, material, temperature.

• Most metals: R increases with increasing T

• Semi-conductors: R decreases with increasing T

Define a new constant which characterizes resistance of materials.

ResistivityLAR= A L

ALR =

Demo: Show temperature dependence of resistance

For materials = 10-8 to 1015 ohms-meters

Example: What is the resistance of a 14 gauge Cu wire? Find the resistance per unit length.

mcu mAL

R Ω−−

×≅×Ω×== 3

23

8

108)10815(.14.3

107.1

Build circuits with copper wire. We can neglect the resistance of the wire. For short wires 1-2 m, this is a good approximation.

Resistance: More on Resistance- Define Resistivity

Note Conductivity = 1/Resistivity s = 1

Page 16: Monday July 14

16

Example Temperature variation of resistivity.

ALR =

can be positive or negative

Consider two examples of materials at T = 20oC.

20 (Ω-m) a(C-1) L Area R (20oC)

Fe 10 -7 0.005 6x106 m 1mm2(10-

6m2)0.1 Ω

Si 640 - 0.075 1 m 1 m2 640 Ω

Fe – conductor - a long 6x106 m wire.

Si – insulator - a cube of Si 1 m on each side

Question: You might ask is there a temperature where a conductor and insulator are one and the same?

=20 1 + α (T − 20)[ ] Demo: Show temperature dependence of resistance

Page 17: Monday July 14

17

Condition: RFe = RSi at what temperature?

Use ALR =

RFe = 10-7 Ω-m [ 1 + .005 (T-20)] 26

6

10106

mm

×

RSi = 640 Ω-m [ 1 - .075 (T-20)]21

1mm

Now, set RFe = RSi and solve for T

T – 20 C = – 196 C

T = – 176 C or 97 K

(pretty low temperature)

R =20 1+a(T −20 C)[ ]LA

Page 18: Monday July 14

18

Resistance at Different Temperatures

T =293K T = 77K (Liquid Nitrogen)

Cu .1194 Ω .0152 Ω conductorNb .0235 Ω .0209 Ω impureC .0553 Ω .069 Ω semiconductor

Page 19: Monday July 14

19

Power dissipation resistorsI Potential energy decrease

)( VtQ

tU

−ΔΔ

=ΔΔ

(drop the minus sign)

Rate of potential energy decreases equals rate of thermal energy increases in resistor.

Called Joule heating

• good for stove and electric oven

• nuisance in a PC – need a fan to cool computer

Also since V = IR,

All are equivalent.

Example: How much power is dissipated when I = 2A flows through the Fe resistor of

R = 10,000 Ω. P = I2R = 22x104 Ω = 40,000 Watts

)( VQU −Δ=Δ

IVP =

RVRIP

22 or =

Page 20: Monday July 14

20

BatteriesA device that stores chemical energy and converts it to electrical energy.

Emf of a battery is the amount of increase of electrical potential of the charge when it flows from negative to positive in the battery. (Emf stands for electromotive force.)

Carbon-zinc = Emf = 1.5V

Lead-acid in car = Emf = 2V per cell

(large areas of cells give lots of current) Car battery has 6 cells or 12 volts.

Power of a battery = P

P = eI e is the Emf

Batteries are rated by their energy content. Normally they give an equivalent measure such as the charge content in mA-Hrs

milliamp-Hours

Charge = (coulomb/seconds) x secondsInternal Resistance

As the battery runs out of chemical energy the internal resistance increases.

Terminal Voltage decreases quickly.

How do you visualize this?What is terminal voltage?

Page 21: Monday July 14

21

What is the relationship between Emf, resistance, current, and terminal voltage?

Circuit model looks like this:

I

r

e

R Terminal voltage = V

V = IR (decrease in PE)

The terminal voltage decrease = e - Ir as the internal resistance r increases or when I increases.

R)(rI

)(

+===−

+=+=ee

ee

IRVI

RIIRI

Page 22: Monday July 14

22

Example: This is called impedance matching. The question is what value of load resistor R do you want to maximize power transfer from the battery to the load.

P = I2R = power dissipated in load

RRr

EP 2

2

)( +=

RrEI+

=

0=dRdP

Solve for R

R = r

?

P

R

You get max. power when load resistor equals internal resistance of battery.

(battery doesn’t last long)

=current from the battery

Page 23: Monday July 14

23

Combination of resistors

Resistors in series

Resistors in parallelVoltages are the same, currents add.

21

2121 )(RRR

IRRIRIRVeqiv +=

+=+=

21

21

RV

RV

RV

III

+=

+=

21

21

21

111,

RRRRR

RRRSo

equiv +=

+=

Current is the same in both the resistors

Page 24: Monday July 14

24

Equivalent Resistance

=

=

R2R2 )RR()RR(R eq

=++=

R4R4R

2

eq =

RR eq =

R2R )RR(RR eq

=+=

R3R2R

2

eq =

R32R eq =

Page 25: Monday July 14

25

Resistance cube

The figure above shows 12 identical resistors of value R attached to form a cube. Find the equivalent resistance of this network as measured across the body diagonal---that is, between points A and B. (Hint: Imagine a voltage V is applied between A and B, causing a total current I to flow. Use the symmetry arguments to determine the current that would flow in branches AD, DC, and CB.)

I

I

Page 26: Monday July 14

26

Resistance Cube cont.

IRV eq=

CBDCAD VVVV ++=

RI65IR

3IR6

IR3IRIR

eq

eq

=

++=

R65R eq =

Because the resistors are identical, the current divides uniformly at each junction.

I

3I

6I

3I

3I

6I

6I

6I

6I

6I

3I

3I

3I I

Page 27: Monday July 14

27

Chapter 26 Problem 21

A wire with a resistance of 3.0 is drawn out through a die so that its new length is three times its original length. Find the resistance of the longer wire, assuming that the resistivity and density of the material are unchanged.

Page 28: Monday July 14

28

Chapter 26 Problem 23

Two conductors are made of the same material and have the same length. Conductor A is a solid wire of diameter 0.5 mm. Conductor B is a hollow tube of outside diameter 3.0 mm and inside diameter 1.5 mm. What is the resistance ratio RA/RB, measured between their ends?

Page 29: Monday July 14

29

Chapter 26 Problem 25

A common flashlight bulb is rated at 0.30 A and 2.9 V (the values of the current and voltage under operating conditions). If the resistance of the bulb filament at room temperature (20°C) is 1.8 , what is the temperature of the filament when the bulb is on? The filament is made of tungsten.

Page 30: Monday July 14

30

Chapter 26 Problem 27

When 125 V is applied across a wire that is 12 m long and has a 0.30 mm radius, the current density is 1.2 multiplied by 104 A/m2. Find the resistivity of the wire.