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PHYS 2421 - Fields and Waves 25.1 Current 25.2 Resistivity 25.3 Resistance 25.4 Electromotive force and circuits 25.5 Energy and power in circuits 25.6 Theory of metallic conduction

PHYS 2421 - Fields and Waves - Academics Portal Index > …academics.utep.edu/Portals/1788/physics 2421/class material/Ch25... · PHYS 2421 - Fields and Waves 25.1 Current 25.2 Resistivity

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PHYS 2421 - Fields and Waves

25.1 Current

25.2 Resistivity

25.3 Resistance

25.4 Electromotive force and circuits

25.5 Energy and power in circuits

25.6 Theory of metallic conduction

Batteries and power supplies are examples of

“Electromotive forces”

To have a current, a pump and a circuit are needed

Analogy Electric circuit View of flow

Current flows equally in all points of circuit

It does not get “used up”

EMF

Electromotive forces (EMF)

• are not forces (wrong name!)

• are electric potentials

• are measured in volts

• examples:• batteries

• solar cells

• electric generators

• thermocouples

• fuel cells

• Ideal EMF maintains same Vab

independent of the current flowing

abV = EIdeally, EMFs should not have any resistance:

But in practice they do: abV Ir= E

r is the internal resistance of EMF, the is reduced by Ir

0I I No current flows through A= ?

0 12 Vab abV V Ir r= ? = E E E

How to measure?Use voltmeters and ammeters

Assumptions

• Voltmeters draw zero current• Ammeters have zero resistance

Current?

Voltage between a and b??

122

2 4

V AI

r R

E

V = ?Remember: voltmeters draw zero current

and ammeters have zero resistance

Equivalent

diagram

' ' 2 4 8 A Va bV IR

12 2 8 V

or

A 2 V

abV IrE

I = ?

Homework : Problem 25.33 (11th Ed.) or 25.35 (12th Ed.)

V = ?I = ?

Homework : Problem 25.35 (11th Ed.) or 25.37 (12th Ed.)

Equivalent

diagram

0I

12

2 4

2

V

A

Ir R

E

V = ?I = ?

126

2 0

V AI

r R

E

12 6 2

0

V A

V IrE

Equivalent diagram

Homework : Problem 25.37 (11th Ed.) or 25.39 (12th Ed.)

Voltage rises and drops in a circuit

Voltage

increases

by 12 V

Voltage drops by

Ir=2 A x 2 =4 V

Voltage drops by

IR=2 A x 4 = 8 V

Summary of Section 25.4

Hmwk Sect. 25.4: Probls 25.33, 25.35 and 25.37 (11th Ed.) or

25.35, 25.37 and 25.39 (12th Ed.)

Electromotive forces (EMF)

• are not forces (wrong name!)

• are electric potentials

• are measured in volts

• Have a small internal resistance

abV Ir= E

Assumptions• Voltmeters draw zero current• Ammeters have zero resistance

Power = energy output per unit time

W Fd qEd qP V IV

t t t t

Power consumed by a resistor:

Units? Amp Volt WattC J J

Ws C s

2 2 /P IV I R V R

Power output of a source:

2P IV I Ir I I rE E

Power input to a source:

2P IV I Ir I I rE E

Rate of energy conversion in battery

122

2 4I

r R

V A

E

V = ?

I = ?

' ' 2 4 8 A Va bV IR

2 8 16P I A V WE

Rate of energy dissipation in battery22 2 2 8P I Ir I r A W

Net power output of battery2 16 8P I I r W W=8 WE

Hmwk : Problem 25.45 (11th Ed.) or 25.47 (12th Ed.)

Rate of energy dissipation in resistor22 2 4P I R A =16 W

121.2

2 8I

r R

V A

E

V = ?

I = ?

' ' 1.2 9.6a bV IR A 8 V

Rate of energy dissipation in resistor22 1.2 12P I R A 8 W

Increasing R reduces power consumption by resistor !

Question: which light bulb has a filament with a larger resistor, a 50 W or a 100 W?

Rate of energy dissipation in resistor22 2 4P I R A =16 W

Change R . . .

Hmwk : Problem 25.49 (11th Ed.) or 25.53 (12th Ed.)

Rate of energy conversion in battery

126

2

V A

I

r

EI = ?

6 12 72 A V WP IE

Rate of energy dissipation in battery22 6 2 72 A WP I Ir I r

Net power output of battery2 72 72 W W=0 WP I I rE

Summary of Section 25.5

Hmewk for Sect. 25.5: Probls 25.45, 25.49 (11th Ed.) or Probls 25.47, 25.53 (12th Ed.)

Power = energy output per unit time P IV

Power consumed by a resistor:

Units: W= WattJ

s2 2 /P IV I R V R

Power output of a source:2P IV I Ir I I rE E

Power input to a source: 2P IV I Ir I I rE E

Electrons are pushed by F=qE

They move and collide with atoms.

The less collisions the better the conduction

Volume ddQ q n qnAv dt

Before we obtained:

d

dQI nqv A

dt d

IJ nqv

A Current density

0d

qEF ma qE a v v

m

qEa

m

The drift velocity is:

Where is the average time between collisions

The longer , the better the conduction 2

d

nqJ nqv E

mThe current density becomes:

A microscopic view of resistivity

And since

Where m and q are the mass and charge of the electrons, n is

the charge density, is the average time between collisions

The longer is, the smaller the resistivity

2 2/

V EL IR I L IE J

V IR L L A A

E E m

J nq E m nq

2

m

nq

Hmwk for Section 25.5: Probl 25.52 (11th Ed.) or 25.56 (12th Ed.)

2

d

nqJ nqv E

m

Hmwk for Ch. 25:

Probs. 1, 3, 7, 11, 13, 27, 33, 35, 37, 45, 49 and 52 (11th Ed.) orProbs. 1, 5, 9, 17, 13, 27, 35, 37, 39, 47, 53 and 56 (12th Ed.)

2

m

nq

PHYS 2421 - Fields and Waves

25.1 Current

25.2 Resistivity

25.3 Resistance

25.4 Electromotive force and circuits

25.5 Energy and power in circuits

25.6 Theory of metallic conduction