Physics 7B-1 (A/B)
Professor Cebra
Winter 2010
Lecture 4
Review of Linear
Transport Model and
Exponential Change
Model
27-Jan-2010
Physics 7B Lecture 4
Slide 1 of 23
Linear Transport Model
Starting with Ohm’s Law (DV = -IR), which we had derived from the Energy
Density Model, we can rewrite this to solve for current (I = -(1/R)DV
Using our definition of R from the previous slide [R = r(L/A) or (1/R) = k (A/L)],
We get
I = -k (A/L) DV
If we let DV  Df, then we get
I = -k (A/L) Df
Divide through by area A,
j = -k (1/L) Df
Let L become an infinitesimal
j = -kdf/dx
27-Jan-2010
 Transport Equation
Physics 7B Lecture 4
Slide 2 of 23
Linear Transport
Which statement is not true about linear
transport systems?
a) If you double the driving potential (voltage,
temperature difference,...) , the current doubles.
b) If you double the resistance, the current is halved.
c) In linear transport systems, the driving potential
varies linearly with the spatial dimension.
d) For both fluid flow and electrical circuits, the
continuity equation requires that the current density
is independent of position.
e) All of the other statements are true.
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Physics 7B Lecture 4
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Application of Linear Transport Model
• Fluid Flow
– ϕ=Head and j=mass current density
• Electric Current (Ohm’s Law)
– ϕ=Voltage and j=charge current density
• Heat Conduction
– ϕ=Temperature and j=heat current density
• Diffusion (Fick’s Law)
– ϕ=Concentration and j=mass current density
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Physics 7B Lecture 4
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Fluid Flow and Transport
Which curve best
describes the pressure in
the above pipe as a
function of position
along the direction of
laminar flow?
Pressure
j = -kdf/dx
a
b
c
d
The correct answer is “c”
x (cm)
27-Jan-2010
Physics 7B Lecture 4
Slide 5 of 23
Heat Conduction – Fourier’s Law
Suppose you lived in a 10’ x 10’ x 10’ cube, and that the wall were made of
insulation 1’ thick. How much more insulation would you have to buy if you
decided to expand your house to 20’ x 20’ x 20’, but you did want your
heating bill to go up?
a) Twice as much
b) Four times as much
c) Eight times as much
d) Sixteen times as much
Answer:
d) The area went up
by a factor of four,
therefore you will need
four times as much
thickness => sixteen
times the total volume
of insulation
jQ = (dQ/dt)/A = -kdT/dx
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Physics 7B Lecture 4
Slide 6 of 23
Diffusion – Fick’s Laws
j = -D dC(x,t)/dx
Where j is the particle flux and C in the
concentration, and D is the diffusion constant
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Physics 7B Lecture 4
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Exponential Growth
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Physics 7B Lecture 4
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Exponential Growth
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Physics 7B Lecture 4
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Exponential Growth
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Physics 7B Lecture 4
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Physics 7B Lecture 4
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Exponential Decay
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Physics 7B Lecture 4
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Exponential decay
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Physics 7B Lecture 4
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The Parallel Plate Capacitor
e = dielectric constant
+Q = stored charge
C = ke0A/d
A = Area of plates
V = voltage across plates
-Q = stored charge
C = Q/V
d = separation of plates
A
 [F], Farads
d
12  As 
e 0  8.85 10 
Vm 
k 1 for air or vaccum
C  ke 0
Electrical Capacitance is similar to the cross sectional area
of a fluid reservoir or standpipe. Electrical charge

corresponds to amount (volume)
of the stored fluid. And
voltage corresponds to the height of the fluid column.
27-Jan-2010
Physics 7B Lecture 4
Slide 15 of 23
Non-Linear Phenomenon – Dependence on source
• Note, for a realistic tank, the vB depends on
the height of the fluid in the tank. Thus the
flow rate changes with time!
• More specifically we can say that the
current depends upon the volume of fluid in
the tank. Current  DVol / Dt
yA
I  dVol / dt
dV
  aV
dt
yB
Note V is volume
not potential and
the negative sign is
for decrease in
volume with time.
• What function is it’s own derivative?
vA  0
vB  ?
1
r(v B2  v A2 )  rg(y B  y A )  0
2
v B  2gh
27-Jan-2010
y(t )  y 0e  at
dy
 ay0e at  ay(t )
dt
Physics 7B Lecture 4
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Fluid Reservoir
h
t
h  h 0e  ,
  time const
t
What will increase the
time constant?
Cross section area of the
reservoir
Resistance of the outlet
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Physics 7B Lecture 4
Slide 17 of 23
Exponential Change in Circuits: Capacitors - Charging
V = Q/C
Q=CVC
I = dQ/dt
VR=IR
ε=20V
Voltage VC (V) = (Q/C)
ε
Current I (A) = dQ/dt
t
ε/R
e RC
I charging 
e
,
R
VC :charging
RCtime const
t

RC
 e1  e

RCtime const

27-Jan-2010
Time t (s)
Time t (s)
Physics 7B Lecture 4

Slide 18 of 23

,

Charging a Capacitor
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Physics 7B Lecture 4
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Exponential Change in Circuits: Capacitors - Discharging
•
•
•
•
Now we use the charge stored up in the battery to light
a bulb
VB=IRB
Q: As the Capacitor discharges, what is the direction of
the current? What happens after some time?
Q=CVC
Q: Initially what is the voltage in the capacitor V0? What
is the voltage in the bulb? What is the current in the
circuit?
Q: At the end, what is the voltage in the capacitor?
What is the current in the circuit? What is the voltage of
the resistor?
Voltage VC (V)
Current I (A)
t
V0
V0/RB
I discharging
V0 R

e
RB
BC
,
RB Ctime const
Time t (s)
27-Jan-2010

Physics 7B
Lecture 4
ε=20V
VC :discharging  V0 e
t
RC
RCtime const
Time t (s)
Slide 20 of 23
,
Discharging a Capacitor
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Physics 7B Lecture 4
Slide 21 of 23
Capacitors in Series and Parallel
• Circuit Diagrams: Capacitors
•
Circuit Diagrams: Resistors
A
C  ke 0
d
Rr
• Capacitors in parallel (~2xA)
•
L
A
Resistors in Series (~2xL)

Rseries  R1  R2 ...
C parallel  C1  C2 ...
• Capacitors in series (~2xd)

1
Cseries
27-Jan-2010
1
1
 
...
C1 C2
•
Resistors in parallel (~2xA)

22
1
Physics 7B Lecture 4
R parallel

1 1

 ...
R1 RSlide
2
22 of 23
Capacitors: Energy Stored in a capacitor
• Because resistors dissipate power, we wrote a an equation for the power
dissipated in a Resistor:
P  IV, using V  IR :
2
V
P  I 2 R or P 
R
Note: Since I is same for resistors in series, identical
resistors in series will have the same power loss.
Since V is the same for resistors in parallel, identical
resistors in parallel will have the same power loss
• Because capacitors are used to store charge and energy, we concentrate
on the energy stored in a capacitor.
• We imagine the first and the last electrons to make the journey to the
capacitor. What are their ΔPE’s?
ΔPEfirst=qΔV ,ΔV=20
ΔPElast =qΔV , ΔV=0
Thus on average for the whole charge:
1
PE  QV, using Q = CV
2
1
PE  CV 2
2
27-Jan-2010
VR=IR
Physics 7B Lecture 4
Q=CVC
ε=20V
Slide 23 of 23
Announcements
27-Jan-2010
Physics 7B Lecture 4
Slide 24 of 23
DL Sections
06-Jan-2010
Physics 7B Lecture 1