chapter 24: electric current

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Chapter 24: Electric Current Curren t Definition of current current is any motion of charge from one region to another. • Suppose a group of charges move perpendi to surface of area A. • The current is the rate that charge flows this area: dt dQ dt dQ I interval time the during flow that charge of amount ; Units: 1 A = 1 ampere = 1 C/s

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Chapter 24: Electric Current. Definition of current. A current is any motion of charge from one region to another. Suppose a group of charges move perpendicular to surface of area A. The current is the rate that charge flows through this area:. Current. Units: 1 A = 1 ampere = 1 C/s. - PowerPoint PPT Presentation

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Page 1: Chapter 24: Electric Current

Chapter 24: Electric Current

Current Definition of current

A current is any motion of charge from one region to another.

• Suppose a group of charges move perpendicular to surface of area A.

• The current is the rate that charge flows through this area:

dt

dQdt

dQI

interval time theduring

flows that charge ofamount ;

Units: 1 A = 1 ampere = 1 C/s

Page 2: Chapter 24: Electric Current

Current Microscopic view of current

Page 3: Chapter 24: Electric Current

Current Microscopic view of current (cont’d)

Page 4: Chapter 24: Electric Current

Current Microscopic view of current (cont’d)

• In time t the electrons move a distance tx d • There are n particles per unit volume that carry charge q

• The amount of charge that passes the area A in time t is

)( tnAqQ d • The current I is defined by:

Anqt

Q

dt

dQI d

t

0

lim

• The current density J is defined by:

dnqA

IJ Current per unit area

Units: A/m2

dnqJ

Vector current density

Page 5: Chapter 24: Electric Current

Resistivity

Ohm’s law• The current density J in a conductor depends on the electric field E and on the properties of the material.• This dependence is in general complex but for some material, especially metals, J is proportional to E.

E

J Ohm’s law

mm/AV)(V/m)/(A/m Units,y resistivit : 2

V/A ohm

vity1/resisitityconductivi

Substance Substancem) m)

silver

copper

goldsteel

81047.1 81072.1 81044.2

81020

graphite

siliconglass

teflon

5105.3

1410 1010 1310

2300

EJ

Page 6: Chapter 24: Electric Current

Resistivity

Conductors, semiconductors and insulators• Good electrical conductors such as metals are in general good heat conductors as well.

• Poor electrical conductors such as plastic materials are in general poor thermal conductors as well.

In a metal the free electrons that carry charge in electrical conduction alsoprovide the principal mechanism for heat conduction.

• Semiconductors have resistivity intermediate between those of metals and those of insulators.

• A material that obeys Ohm’s law reasonably well is called an ohmic conductor or a linear conductor.

Page 7: Chapter 24: Electric Current

Resistivity

Resistivity and temperature• The resistivity of a metallic conductor nearly always increases with increasing temperature.

)](1[)( 00 TTT reference temp. (often 0 oC)

temperature coefficient of resistivity

Material oC)-1

aluminum

brass

graphitecopper

0039.0

0020.00005.000393.0

iron

leadmanganin

silver

0050.0

00000.0

0038.0

0043.0

Material oC)-1

Page 8: Chapter 24: Electric Current

Resistivity

Resistivity vs. temperature• The resistivity of graphite decreases with the temperature, since at higher temperature more electrons become loose out of the atoms and more mobile.

• This behavior of graphite above is also true for semiconductors.

• Some materials, including several metallic alloys and oxides, has a property called superconductivity. Superconductivity is a phenomenon where the resistivity at first decreases smoothly as the temperature decreases, and then at a certain critical temperature Tc the resistivity suddenly drops to zero.

T

T

TTc

metal semiconductor superconductor

Page 9: Chapter 24: Electric Current

Resistance Resistance

• For a conductor with resistivity , the current density J at a point where the electric field is E : JE

• When Ohm’s law is obeyed, is constant and independent of the magnitude of the electric field.

• Consider a wire with uniform cross-sectional area A and length L, and let V be the potential difference between the higher-potential and lower-potential ends of the conductor so that V is positive.

E

JI

AL

ELVJAI ,

JE

IA

LV

A

I

L

VE

I

VR

resistance1 V/A=1

• As the current flows through the potential difference, electric potential is lost; this energy is transferred to the ions of the conducting material during collisions.

Page 10: Chapter 24: Electric Current

Resistance

Resistance (cont’d)• As the resistivity of a material varies with temperature, the resistance of a specific conductor also varies with temperature. For temperature ranges that are not too great, this variation is approximately a linear relation:

)](1[)( 00 TTRTR • A circuit device made to have a specific value of resistance is called a resistor.

V

V

resistor that obeysOhm’s law

semiconductordiode

Page 11: Chapter 24: Electric Current

Resistance

Example: Calculating resistance

b

a

• Consider a hollow cylinder of length L and inner and outer radii a and b, made of a material with . The potential difference between the inner and outer surface is set up so that current flows radially through cylinder.

r

A flowscurrent which thefrom circle

dashed aby drepresentecylinder a of area : 2 rLA

• Now imagine a cylindrical shell of radius r, length L, and thickness dr.

shell thisof resistance : 2 rL

drdR

b

a a

b

Lr

dr

LdRR ln

22

Page 12: Chapter 24: Electric Current

Electromotive Force (emf) and Circuit

Complete circuit and steady current• For a conductor to have a steady current, it must be part of a path that forms a closed loop or complete circuit.

1E

JI

1E

JI

1E

JI

---

--

--

-

+

+

+ +

+

+

++

2E

2E

current causesconductor inside

produced field Electric 1E

current reducing

and field opposing

producing ends,at up build

tocharge causesCurrent

2E

.completely stopscurrent

and 0 field total

: as magnitude same the

has short time very aAfter

1

2

totalE

E

E

Page 13: Chapter 24: Electric Current

Electromotive Force (emf) and Circuit

Maintaining a steady current and electromotive force• When a charge q goes around a complete circuit and returns to its starting point, the potential energy must be the same as at the beginning.• But the charge loses part of its potential energy due to resistance in a conductor.

• There needs to be something in the circuit that increases the potential energy.

• This something to increase the potential energy is called electromotive force (emf).

• Emf makes current flow from lower to higher potential. A device that produces emf is called a source of emf.

Units: 1 V = 1 J/C

+-

current flow

EeF

nF

E

E

source of emf

ab

If a positive charge q is moved from b to a inside the source, the non-electrostatic force Fn does a positive amount of work Wn=qVab on the charge.This replacement is opposite to the electrostatic force Fe, so the potential energy associated with the charge increases by qVab. For an ideal source of emf Fe=Fn

in magnitude but opposite in direction.Wn=qqVab , so Vab= =IR for an ideal source.

Page 14: Chapter 24: Electric Current

Electromotive Force (emf) and Circuit

Internal resistance• Real sources in a circuit do not behave ideally; the potential difference across a real source in a circuit is not equal to the emf.

Vab= – Ir (terminal voltage, source with internal resistance r)

• So it is only true that Vab=E only when I=0. Furthermore,

–Ir = IR or I = / (R + r)

Page 15: Chapter 24: Electric Current

Electromotive Force (emf) and Circuit

Real battery

br

+

I

a

dR

c

Battery

outV I rI I

R R R r

• Real battery has internal resistance, r.

• Terminal voltage, ΔVoutput = (Va −Vb) = − I r.

a b

c d

Page 16: Chapter 24: Electric Current

Electromotive Force (emf) and Circuit

Potential in an ideal resistor circuit

ba

b

c d

dab c

Page 17: Chapter 24: Electric Current

Electromotive Force (emf) and Circuit

Potential in a resistor circuit in realistic situation

b

r

+

I

ad

R

c

-

Battery

V

ab

c d

r

+

-

R

I r

IR

0

a b

ba

Page 18: Chapter 24: Electric Current

Energy and Power in Electric Circuit

Electric power

Electrical circuit elements convert electrical energy into1) heat energy( as in a resistor) or 2) light (as in a light emitting diode) or3) work (as in an electric motor). It is useful to know the electrical power being supplied. Consider the following simple circuit.

dUe is electrical potential energy lost as dQ traverses the resistor and falls in V by ΔV.

Electric power = rate of supply from Ue.

abe dQVVdQdU

abV

abV(watts) W 1 J/s 1 C/s) J/C)(1 (1 :Units

power Electric2

2

R

VRIIVV

dt

dQP ab

abab

R

Page 19: Chapter 24: Electric Current

Energy and Power in Electric Circuit

Power output of a source• Consider a source of emf with the internal resistance r, connected by ideal conductors to an external circuit.• The rate of the energy delivered to the external circuit is given by:

IVP ab• For a source described by an emf and an internal resistance r

IrVab

rIIIVP ab2

+

a + b -

-

II

battery

(source)

headlight(external circuit)

rate of conversion ofnonelectrical energyto electrical energy inthe source

rate of electricalenergy dissipationin the internalresistance of thesource

net electricalpower outputof the source

Page 20: Chapter 24: Electric Current

Energy and Power in Electric Circuit

Power input of a source• Consider a source of emf with the internal resistance r, connected by ideal conductors to an external circuit.

alternatorlarge emf

+

a + b -

-

II

battery

small emf

+v

Fn

rIIIVPIrV abab2

rate of conversion ofelectrical energy intononeletrical energy inthe battery

rate of dissipationof energy in theinternal resistancein the battery

total electrical power inputto the battery

Page 21: Chapter 24: Electric Current

Electromotive Force (emf) and Circuit Examples:

V

A

voltmeteram

met

erabcd VV

4V,12,2 Rr

V. 8 ) A)(2 (2 - V 12

V. 8) A)(4 2(

.

A. 2 2 4

V 12

IrV

IRV

VVrR

I

ab

cd

cdab

W.16) 4(A) 2(by given also isIt

W.16A) V)(2 (8by given also isoutput power The

W.16 isoutput power electrical The

W.8) 2(A) 2( isbattery in theenergy ofn dissipatio of rate The

W.24A) V)(2 (12 isbattery in the conversionenergy of rate The

22

2

22

IR

IV

rII

Ir

I

bc

ba

A V: measures pot- ential difference

: measures current through it

Page 22: Chapter 24: Electric Current

Electric Conduction

Drude’s model

Page 23: Chapter 24: Electric Current

Electric Conduction

Drude’s model (cont’d)

Page 24: Chapter 24: Electric Current

Electric Conduction

Drude’s model (cont’d)

Page 25: Chapter 24: Electric Current

80 3 2

20 m(11 10 m) 11

[0.5(0.5 10 m)]

lR

A

Calculate the resistance of a coil of platinum wire with diameter 0.5 mm and length 20 m at 20 C given =11×10−8 m. Also determine the

resistance at 1000 C, given that for platinum =3.93×10−3 /°C.

To find the resistance at 1000 C: 0 01 T T

lR

A 0 01R R T T But so we have :

Where we have assumed l and A are independent of temperature - could cause an error of about 1% in the resistance change.

3 1(1000 C) (11 )[1 (3.93 10 C )(1000 C 20 C)] = 53R

Exercise 1

Page 26: Chapter 24: Electric Current

The hair dryer is taken to the UK where it is turned on with a 240 V source. What happens?

1000 W8.33A

120V

PI

V

120V14.4

8.33A

VV IR R

I

A 1000 W hair dryer manufactured in the USA operates on a 120 V source. Determine the resistance of the hair dryer, and the current it draws.

2 2( ) (240V)4000 W

14.4

VP

R

This is four times the hair dryer’s power rating – BANG and SMOKE!

Exercise 2