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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!
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