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

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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

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3

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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

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What is Current I?

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

Ampond

Coulomb

t

QI ==

ΔΔ

=sec

Copper wire with voltage across it

+ +

+ +

+ +

+

+

I

ΔLA

Ltv Δ=Δ

Drift velocity of charge

Density of electrons 1.6 x 10-19 C

Divide both sides by Δt.nqAv

t

QI =

ΔΔ

=

tnqAv QtAvnq

volume volume unit per charge

Δ=ΔΔ×=

×=ΔQ

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Drift velocity

€

vd =I

nqA

Μ=

oNn ρ = 8.4x1022 atoms/cm3

21922 )0815(.106.1104.81

3 cmCamp

vcmatomsd 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 ?

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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 =nAqvdrJ =nq

rvd

i =

rJ ⋅d

rA∫

J =

iA when

rJ and

rA are parallel

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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 (Kirchoff's 2nd Rule)

Higher J

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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.

€

vd ≈ 3.5 ×10−11v instant

sm

dv5105.3 −×=

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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 =10−4 A

1.6 ×10−19C ⋅10−6m2 ⋅5 ×107 ms

For CRT

For Copper

The lower the density the higher the speed.

Take A =1 mm2

=(10−3m)2

=10−6m2

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313

cmelectrons102.1

melectrons102.1n ×=×=

322

cmelectrons105.8n ×=

n =I

qAv

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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

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

satisfied islaw sOhm'

constant1Slope ==

=

=

R

RVI

IRVHere the slope depends on

the potential difference.

Ohm's Law is violated for a

pn junction diode.

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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

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• 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.

ResistivityL

AR= A L

A

LR =

Demo: Show temperature dependence of resistance

For materials ρ= 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

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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 σ1

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Example Temperature variation of resistivity.Calculate R at T= 20 C forFe – conductor - a long 6x106 m wire of area 1mm2.Si – insulator - a cube of Si 1 m on each side

A

LR =

can be positive or negative

Consider two examples of materials at T = 20o C, then

ρ20 (Ω -m) α (C-1) L Area R (20oC)

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

6m2)600 KΩ

Si 640 - 0.075 1 m 1 m2 0.640 KΩ

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

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Condition: RFe = RSi at what temperature?

Use A

LR =

RFe = 10-7 Ω-m [ 1 + .005 (T-20)] 26

6

10

106

m

m−

×

RSi = 640 Ω-m [ 1 - .075 (T-20)]21

1

m

m

Now, set RFe = RSi and solve for T

T – 20 C = – 170.50 C

T = – 170.5 C or 102.6 K

(pretty low temperature)

R =20 1+α(T −20 C)[ ]LA

ρ20 (Ω -m) α (C-1) L Area R (170.5oC) R (20 C)

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

6m2) 8.8 KΩ 600 KΩ

Si 640 - 0.075 1 m 1 m2 8.8 KΩ 0.640 KΩ

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Resistance at Different Temperatures

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

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

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Power dissipation resistorsI Potential energy decrease

)( Vt

Q

t

U−

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 =

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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 = εI ε 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?

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What is the relationship between Emf, resistance, current, and terminal voltage?Circuit model looks like this:

I

r

•

•

R Terminal voltage = V

V = IR (decrease in PE)

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

R)(rI

)(

+===−

+=+=

IRVIr

RrIIRIr

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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

)( +=

Rr

EI

+

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

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Combination of resistors

Resistors in series

Resistors in parallelVoltages are the same, currents add.

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2121 )(RRR

IRRIRIRV

eqiv +=+=+=

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RV

RV

RV

III

+=

+=

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21

21

111,

RRRR

R

RRRSo

equiv +=

+=

Current is the same in both the resistors

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Equivalent Resistance

=

=

R2R2

)RR()RR(R eq

=++=

R4R4R

2

eq =

RR eq =

R2R

)RR(RR eq

=+=

R3R2R

2

eq =

R32R eq =

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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

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Resistance Cube cont.

IRV eq=

CBDCAD VVVV ++=

RI65IR

3IR

6IR

3IRIR

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

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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.

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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?

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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.

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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.

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