Lecture15 -...

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Unformatted text preview: Announcement:
Test
graded
by
Thursday
 
 
 
Average
for
mul7ple
choice
7.5/15:
exam
will
be
graded
by
Thursday
 Office
Hours:
TR
10:30‐11:30
am
 
 
 
Physical
Science
Building
 
 
 
Enter
through
the
garage
door
behind
the
Harris
engineering
building
 Current:
charge
per
unit
7me
 dQ I= dt 1
Amp
=
1
C/sec
 What
is
the
speed
of
electrons?
 1
volt
 0
volt
 Poten7al
energy:
1
electron
volt


 12 U = mv 2 1.6 × 10 −19 J = v= 1 × 9.109 × 10 31 kg × v 2 2 3.2 × 10 −19 m / s = 0.35 × 10 6 m / s 9.109 × 10 −31 But
in
reality,
charge
carriers
diffuse
at
mm/sec,
why?
 Carrier
velocity
small
because
of
scaRering

 vd :
“DriU
velocity”
 Current
 L
 A
 
n:
volume
density
of
electrons
 Q = neAL dQ dL I= = neA = neAvd dt dt Current
Density
 L
 A
 I J= A Ohm’s
Law
 J = σE σ:
conduc7vity
 Ohm’s
Law
 L
 A
 Ohm’s
Law
 J = σ E 1 J=E σ L J = LE σ LI = LE = V σA L I =V σA L =R σA RI = V Conduc7vity,
Resistance
and
Resis7vity
 L R= σA 1 ρ= σ ρL R= A Resis7vity
 Silver:
1.59×10‐8
Ωm
 Copper:
1.7×10‐8
Ωm
 Gold:
2.44×10‐8
Ωm
 Glass:
1010
~1014Ωm
 Resistance
:
Ω
 Resis7vity
:
Ω
m
 Example
 L
 A
 A=
1
cm2,
L=
4
m,
Resis7vity:
Copper:
1.7×10‐8
Ωm
 ρL R= A 1.7 × 10 −8 Ωm × 4 m R= = 6.8 × 10 −4 Ω 0.0001m 2 Example
2
 L
 A
 A=
1
cm2,
L=
4
m,
Resis7vity:
Copper:
1.7×10‐8
Ωm
 1.7 × 10 −8 Ωm × 4 m R= = 6.8 × 10 −4 Ω 0.0001m 2 If
you
have
1
µV
along
the
wire,
how
much
current
will
pass
through
the
wire?
 V = IR 1 × 10 −6 V = I × 6.8 × 10 −4 Ω I = 0.147 × 10 −2 A Resis7vity
 Resis7vity
 Temperature
dependence
of
resis7vity
 Temperature
 Metals
 Temperature
 Semiconductors
 Resis7vity
 Superconduc7vity
 Temperature
 Transi7on
Temperature
 • 
zero
resis7vity
 • 
perfect
resistance
to
any
magne7c
field
 Tc
=138
K
HgBa2Ca2Cu3O8
 Onnes,
1911:
in
Mercury
 RC
circuit
 Ini7ally,
voltage
across
the
capacitor
is
10
volts
and
it
discharges
 Ini7al
energy
stored
is

 1 CV 2 2 Board
calcula7on
 Energy
stored
in
capacitor
 Time
 Power:
rate
of
energy
dissipated
in
circuit
 Poten7al
drop
across
the
resistor
 Poten7al
energy
drop
across
the
resistor
 V
 qV
 0
 0
 If
voltage
is
held
constant,
rate
of
energy
dissipa7on
is

 d (qV ) dq =V = IV = P dt dt P:
power
[J/s],
WaR
 Direc7on
of
current
 Conven7on
assumes
posi7ve
charges
 Resistors
in
series
 Resistors
in
parallel
 ...
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This note was uploaded on 07/16/2011 for the course PHY 2049 taught by Professor Saha during the Fall '08 term at University of Central Florida.

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