Analysis Macroscopic 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
I J = = ( q nu ) E = E (A/m 2 ) Current density: A
Conductivity = q nu
A V 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
I Cl = q1 n1 Av1 = q1 n1 Au1E I Na = q2 n2 Av2 = q2 n2 Au2 E
I = I Na + I Cl = 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
I = AE
V A 1 I= V = V = L R R R
Conventional current: I = V
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
v = uE
Macroscopic
J = E
i = nAv = nAuE
V I = q nAv = R
Can we write V=IR ?
A = 0.002 mm2
Exercise: Carbon Resistor
Conductivity of Carbon: = 3.104 (A/m2)/(V/m) L=5 mm
L R= A
What is its resistance R?
R=
( 3 10
( 0.005 m )
4 2
(A/m )/(V/m) 2 10
)(
-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. Conductivity may also depend on the magnitude of current.
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 R= I
V-A dependence: 3V 100 mA 1.5 V 80 mA 0.05 V 6 mA
R 30 19 8
I
Clearly not ohmic! V
Semiconductors
Metals, mobile electrons: slightest V produces current. If electrons were bound need to apply some field to free some of them and start flow current 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 For resistors made of the same material and with the same A it follows from the definition of resistance:
L R= A
Exercise: Voltage Divider
Know R , find V1,2 R1 emf R2 V2 V1 Solution:
V I= R
emf Requivalent
V = IR
emf = R1 + R2
1) Find current: I = 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
For resistors made of the same material and with the same A it follows from the definition resistance:
L R= A 1 A = R L
Aequivalent = A1 + A2 + A3
Two Light Bulbs in Parallel
R1 = 30 What is the equivalent resistance?
1 Requivalent 1 1 = + R1 R2
R1R2 R1 + R2
R2 = 10 What is the total current?
V 3V = = 0.4 A I= 7.5 R
Requivalent =
Requivalent
300 2 = = 7.5 40
V V 3V 3V + = + = 0.4 A Alternative way: I = I1 + I 2 = R1 R2 30 10
Work and Power in a Circuit
Current: charges are moving work is done Work = change in electric potential energy of charges
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
Power Dissipated by a Resistor
emf R Know V, find P
P = IV V I= R
( V )2 P=
R
Know I, find P
P = IV
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 = Q dQ C
Q
Q C
dE = dQV =
Q Q
Q 1 E = dE = dQ = QdQ C C0 0 0
Q E= 2C
2 Q
Alternative approach:
0E 2 Energy density: 2 E = V / s
0
Q2 = 2C
2
E=
C ( V ) 2
V A V Energy: 0 ( 2 ) As = 0 ( ) 2s 2s
2
2
A C= 0 s
=
C ( V ) 2
2
Real Batteries: Internal Resistance
Drift speed of ions in chemical battery:
v ~ (FNC - eEC )
In usual circuit elements:
J = E
In a battery:
force I FNC J = = - EC ~ unit charge A e
rint - internal resistance emf
EC =
FNC I F s s - , assuming uniform field: V = EC s = NC - I e A e A
Vbattery = emf - rint I
Real Batteries: Internal Resistance
Vbattery = emf - rint I
ideal battery
Model of a real battery Round trip (energy conservation):
emf - rint I - RI = 0
I= emf R + rint
R
rint0.25
1.5 V
R 100 10 1 0
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