Transcript + R - Purdue Physics

Macroscopic Analysis of Circuits
Microscopic treatment: insight into the fundamental physical
mechanism of circuit behavior.
Not easy to measure directly E, u, Q, v.
It is easier to measure conventional current, potential
difference  macroscopic parameters
Need a link between microscopic and macroscopic quantities.
Resistance
Many elements in a circuit act as
resistors: prevent current from
rising above a certain value.
Goal: find a simple parameter which can predict V and I in
such elements.
Need to combine the properties of material and geometry.
Conductivity
Combining the properties of a material
Geometry
Conventional current: I  q nAv  q nAuE
Different properties
of the material
Group the material properties together:
I   q nu AE
r
r
I
2
J


q
nu
E


E
(A/m
)


Current density:
A
Conductivity   q nu
 A 


 V m 
Exercise
In copper at room temperature, the mobility of electrons is about
4.5.10-3 (m/s)/(V/m) and the density of electrons is n=8.1028 m-3.
What is  ?
  q nu  1.6  1019 C8  1028 m-3 4.5  103 (m/s)/(V/m)
  5.76  107 (A/m2 )/(V/m)
What is the strength of E required to drive a current of 0.3 A
through a copper wire which has a cross-section of 1 mm2 ?
J
E
I
 E
A
E
I
A
0.3 A
3

5.2

10
V/m
7
2
6
2
5.76  10 (A/m )/(V/m) 1  10 m



Exercise
The conductivity of tungsten at RT is =1.8. 107 (A/m2)/(V/m)
and it decreases 18 times at a temperature of a glowing filament
(3000 K). The tungsten filament has a radius of 0.015 mm.
What is E required to dive 0.3A through it?
J
I
 E
A
E
I
A
0.3 A
E
 424 V/m
7
2
10
2
1.8  10 (A/m )/(V/m) / 18 7.1  10
m



Conductivity with two Kinds of Charge Carriers
ICl  q1 n1 Av1  q1 n1 Au1E
I Na  q2 n2 Av2  q2 n2 Au2 E
I  I Na  ICl  q2 n2 Au2 E  q1 n1 Au1E
J
I
  q2 n2u2  q1 n1u1 E   E
A
  q2 n2u2  q1 n1u1
Resistance
 
V    E  dl
f
i
V  EL
I
J   E
A
V
E
L
V
1
I
V  V 
L
R
R
A
I  AE
Conventional current: I  V
R
Widely known as
Ohm’s law
L
Resistance of a long wire: R 
A
George Ohm
(1789-1854)
Units: Ohm, 
Resistance combines conductivity and geometry!
Microscopic and Macroscopic View
Microscopic
Macroscopic
v  uE
J  E
i  nAv  nAuE
V
I  q nAv 
R
Can we write V=IR ?
Current flows in response to a V
Exercise: Carbon Resistor
A = 0.002 mm2
Conductivity of Carbon:
 = 3.104 (A/m2)/(V/m)
L=5 mm
What is its resistance R?
L
R
A
R
3  10
0.005 m 
4

(A/m )/(V/m) 2  10
2
9
m
2

 83 
(V/A)
What would be the current through this resistor if connected
to a 1.5 V battery?
V
I
R
1.5V
I
 0.018 A  18 mA
83
Constant and Varying Conductivity
Mobility of electrons: depends on temperature
  q nu
 A 


V

m


Conductivity and resistance depend on temperature.
Ohmic Resistors
Ohmic resistor: resistor made of ohmic material…
Ohmic materials: materials in which conductivity  is
independent of the amount of current flowing through
V
I
R
R
L
A
not a function of current
Examples of ohmic materials:
metal, carbon (at constant T!)
Is a Light Bulb an Ohmic Resistor?
Tungsten: mobility at room temperature is larger than at
‘glowing’ temperature (~3000 K)
V
I
R
V-A dependence:
3V
100 mA
1.5 V 80 mA
0.05 V 6 mA
Clearly not ohmic!
R
30 
19 
8
V
R
I
I
V
Semiconductors
Metals, mobile electrons: slightest V produces current.
If electrons were bound – we would
need to apply some field to free some of
them in order for current to flow. Metals
do not behave like this!
Semiconductors: n depends exponentially on E
  q nu
Conductivity depends exponentially on E
Conductivity rises (resistance drops)
with rising temperature
Nonohmic Circuit Elements
Semiconductors
Capacitors
V
I
R
|V|=Q/C, function of time
Batteries: double current, but |V|emf, hardly changes
V
I
R
has limited validity!
Ohmic when R is indeppendent of I!
Conventional symbols:
Series Resistance
Vbatt + V1 + V2 + V3 = 0
emf - R1I - R2I - R3I = 0
emf = R1I + R2I + R3I
emf = (R1 + R2 + R3) I
emf = Requivalent I , where Requivalent = R1 + R2 + R3
Exercise
A certain ohmic resistor has a resistance of 40 .
A second resistor is made of the same material, but is three
times longer and has a half of the cross-sectional area.
What is its resistance?
L2  3L1
L
R
A
A2  A1 / 2
Resistor 1:
Resistor 2:
L1
 40 
A1
R2 
L2
3L1
L

6 1
A2 A1 / 2
A1
R2  240 
What would be an equivalent resistance of these two resistors
in series?
R  280 
Exercise: Voltage Divider
Know R , find V1,2
R1
V1
Solution:
emf
R2
V2
V
I
R
1) Find current: I 
V  IR
emf
Requivalent
emf

R1  R2
2) Find voltage:
R1
V1  IR1  emf
R1  R2
R2
V2  IR2  emf
R1  R2
3) Check: V1  V2  emf
 R1
R2 
emf 

 emf

 R1  R2 R1  R2 
Parallel Resistance
I = I 1 + I2 + I3
emf emf emf
I


R1
R2
R3
 1
1
1
emf
I  
  emf 
Requivalent
 R1 R2 R3 
1
Requivalent
1
1
1
 

R1 R2 R3
Two Light Bulbs in Parallel
R1 = 30 
What is the equivalent resistance?
1
Requivalent
R2 = 10 
What is the total current?
V
3V

 0.4 A
I
7 .5 
R
1
1
 
R1 R2
Requivalent 
R1R2
R1  R2
300 2
Requivalent 
 7.5 
40 
V
V
3V
3V



 0.4 A
Alternative way: I  I1  I 2 
R1
R2
30  10 
Two Light Bulbs in Parallel
What would you expect if one is unscrewed?
A) The single bulb is brighter
B) No difference
C) The single bulb is dimmer
Work and Power in a Circuit
Current: charges are moving  work is done
Work = change in electric potential energy of charges
W = U e  q  V
Power = work per unit time:
P
U e q  V q


V
t
t
t
I
Power for any kind of circuit component: P  IV
Units: AV 
CJ J
 W
sC s
Example: Power of a Light Bulb
I = 0.3 A
P  IV  0.3 A3 V  0.9 W
Units: AV 
emf = 3V
CJ J
 W
sC s
Power Dissipated by a Resistor
emf
Know V, find P
R
P  IV
V
I
R
2
V 

P
R
Know I, find P
P  IV
V  IR
P  I 2R
In practice: need to know P to select right size resistor –
capable of dissipating thermal energy created by current.
Energy Stored in a Capacitor
Q  C V
V 
dU electric  dQV 
Q
Q
C
Q
dQ
C
Q
Q
Q
1
Uelectric   dUelectric   dQ   QdQ
C
C0
0
0
Alternative approach:
0E 2
Energy density:
2
E  V / s
U electric 
2
1Q
C(V )

2 C
2
 0 V 
2
2
Energy:
C
0 A
s
2s
2
 0 A V 
2
 As 
2s
C V 

2
2
Real Batteries: Internal Resistance
ideal battery
Vbattery  emf  rint I
Model of a real battery
Round trip
(energy conservation):
emf  rint I  RI  0
rint0.25 
1.5 V
R
100 
10 
1
0
R
I
emf
R  rint
Ideal
0.015 A
0.15 A
1.5 A
infinite
Real
0.01496 A
0.146 A
1.2 A
6A
VR=RI
1.496 V
1.46 V
1.2 V
0V