Chapter 24: Electric Current

CurrentDefinition of current

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

• Suppose a group of charges move perpendicularto surface of area A.

• The current is the rate that charge flows throughthis area:

dt

dQdtdQI

interval time theduring

flows that charge ofamount ; ==

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

CurrentMicroscopic view of current

CurrentMicroscopic view of current (cont’d)

CurrentMicroscopic view of current (cont’d)

• In time ∆t the electrons move a distance tx d ∆=∆ υ• 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 ∆=∆ υ

• The current I is defined by:

AnqtQ

dtdQI dt

υ=∆∆

=≡→∆ 0

lim

• The current density J is defined by:

dnqAIJ υ== Current per unit area

Units: A/m2

dnqJ υrr

= Vector current density

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.

ρEJ = V/A ohmOhm’s law

mm/AV)(V/m)/(A/m Units,y resistivit : 2 Ω=⋅=ρEJ σ=vity1/resisitityconductivi =σ

Substance ρ (Ωm)Substanceρ (Ωm)

silver 81047.1 −×81072.1 −×81044.2 −×

81020 −×

5105.3 −×

1410 1010 −1310>

2300graphitesiliconcopperglassgoldteflonsteel

Resistivity

Conductors, semiconductors and insulators• Good electrical conductors such as metals are in general good heatconductors as well.In a metal the free electrons that carry charge in electrical conduction alsoprovide the principal mechanism for heat conduction.

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

• Semiconductors have resistivity intermediate between those of metalsand those of insulators.

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

Resistivity

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

)](1[)( 00 TTT −+= αρρreference temp. (often 0 oC)

temperature coefficient of resistivity

Material Material α (oC)-1α (oC)-1

aluminum iron0039.0 0050.0

00000.00043.0leadbrass 0020.0

0005.0−00393.0

manganingraphitesilver 0038.0copper

Resistivity

Resistivity vs. temperature• The resistivity of graphite decreases with the temperature, since at highertemperature 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 propertycalled superconductivity. Superconductivity is a phenomenon where theresistivity at first decreases smoothly as the temperature decreases, and thenat a certain critical temperature Tc the resistivity suddenly drops to zero.

ρ ρ

T

ρ

TTTcsuperconductorsemiconductormetal

ResistanceResistance

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

rrρ=

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

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

Er

JrI

LA

ELVJAI == ,JErr

ρ=

IALV

AI

LVE ρρ =→==

IVR ≡

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

resistance1 V/A=1Ω

Resistance

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

)](1[)( 00 TTRTR −+= α• A circuit device made to have a specific value of resistance is called aresistor.

Ι Ι

V V

semiconductordiode

resistor that obeysOhm’s law

Resistance

Example: Calculating resistance

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

b

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

flowscurrent which thefrom circle dashed aby drepresentecylinder a of area : 2 rLA π=r

Ashell thisof resistance :

2 rLdrdR

πρ

=

∫∫ ===b

a ab

Lrdr

LdRR ln

22 πρ

πρ

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 thatforms a closed loop or complete circuit.

1Er

JrI

-- --

-

++

++

+

1Er

JrI

+

1Er

JrI

+

2Er

2Er +

---

current causesconductor insideproduced field Electric 1E

r

.completely stopscurrent and 0 field total

: as magnitude same the

has short time very aAfter

1

2

=totalE

E

E

r

r

r

current reducingand field opposing

producing ends,at up build tocharge causesCurrent

2Er

Electromotive Force (emf) and Circuit

Maintaining a steady current and electromotive force• When a charge q goes around a complete circuit and returns to itsstarting 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). Units: 1 V = 1 J/C• Emf makes current flow from lower to higher potential. A device thatproduces emf is called a source of emf.

+-

current flow

EreF

r

nFr

Er

Er

source of emf

abIf a positive charge q is moved from b to a inside thesource, the non-electrostatic force Fn does a positiveamount of work Wn=qVab on the charge.This replacement is opposite to the electrostatic forceFe, so the potential energy associated with the chargeincreases by qVab. For an ideal source of emf Fe=Fnin magnitude but opposite in direction.Wn=qε = qVab , so Vab= ε =IR for an ideal source.

Electromotive Force (emf) and Circuit

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

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

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

ε –Ir = IR or I = ε / (R + r)

Electromotive Force (emf) and Circuit

Real battery

br

+

εI

a

dRc

−

Batterya b

c d

outV I rI IR R R r

ε ε∆ −= = ⇒ =

+

• Real battery has internal resistance, r. • Terminal voltage, ∆Voutput = (Va −Vb) = ε − I r.

•

Electromotive Force (emf) and Circuit

Potential in an ideal resistor circuit

ba

b

c d

dab c

Electromotive Force (emf) and Circuit

Potential in a resistor circuit in realistic situation

b

r+

ε

I

ad

R

c

Battery

ab

c d

r

+

ε-

R

a b

V

-εI r

IR

0ba

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 dQtraverses the resistor and falls in V by ∆V.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

VRIIVVdtdQP ab

abab

R

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 −= ε

rIIIVP ab2−== ε

-+

battery

(source)

a + b -II

rate of conversion ofnonelectrical energyto electrical energy inthe source

rate of electricalenergy dissipationin the internalresistance of thesource

net electricalpower outputof the source

headlight(external circuit)

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

+

b -

-

a +I

battery

small emf

+v

Fn

rIIIVPIrV abab2+==→+= εε

total electrical power inputto the battery

rate of conversion ofelectrical energy intononeletrical energy inthe battery

rate of dissipationof energy in theinternal resistancein the battery

I

Electromotive Force (emf) and CircuitExamples:

V

A

voltmeterabcd VV =am

met

er

Ω==Ω= 4V,12,2 Rr ε

V. 8 ) A)(2 (2 - V 12V. 8) A)(4 2(

.

A. 2 2 4

V 12

=Ω=−==Ω==

=

=Ω+Ω

=+

=

IrVIRVVV

rRI

ab

cd

cdab

ε

ε

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

=Ω=

===−

=Ω=

==

IR

IVrII

IrI

bc

ε

ε

ba

A V: measures pot-ential difference

: measures currentthrough it

Electric Conduction

Drude’s model

Electric Conduction

Drude’s model (cont’d)

Electric Conduction

Drude’s model (cont’d)

Exercise 1

Calculate the resistance of a coil of platinum wire with diameter 0.5 mm and length 20 m at 20° C given ρ =11×10−8 Ω m. Also determine the resistance at 1000° C, given that for platinum α =3.93×10−3 /°C.

80 3 2

20 m(11 10 m) 11[0.5(0.5 10 m)]

lRA

ρπ

−−= = × Ω⋅ = Ω

×

To find the resistance at 1000° C: ( )0 01 T Tρ ρ α⎡ ⎤= + −⎣ ⎦

lRA

ρ= ( )0 01R R T Tα⎡ ⎤= + −⎣ ⎦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 − −° = Ω + × ° ° − ° Ω

Exercise 2

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.

1000 W 8.33 A120 V

PIV

= = =∆

120V 14.48.33A

VV IR RI

∆∆ = ⇒ = = = Ω

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

2 2( ) (240V) 4000 W14.4

VPR

∆= = =

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