Download I 2 - Purdue Physics

Chapter 19
A Microscopic View
of Electric Circuits
Current at a Node
The current node rule
(Kirchhoff node or junction rule [law #1]):
i1 = i 2
i2 = i3 + i 4
In the steady state, the electron current entering a node in a
circuit is equal to the electron current leaving that node
(consequence of conservation of charge)
Gustav Robert Kirchhoff
(1824 - 1887)
Analysis of Circuits
The current node rule (Charge conservation)
Kirchhoff node or junction rule [1st law]:
In the steady state, the electron current entering a node in a
circuit is equal to the electron current leaving that node
Conventional current: I = |q|nAuE
The loop rule (Energy conservation)
Kirchhoff loop rule [2nd law]:
V1 + V2 + V3 + … = 0
along any closed path in a circuit
V= U/q  energy per unit charge
Clicker
Pick right statement:
A) i1 = i4 and i2 = i3
B) i1  i4
C) i1 = i4 and i1= i2+i3
Question
Write the node equation for this circuit.
What is the value of I2?
A)
B)
C)
D)
1A
2A
3A
4A
Exercise
Write the node equation for this circuit.
What is the value of I2?
I1 + I4 = I 2 + I 3
I2 = I1 + I4 - I3 = 3A
What is the value of I2 if I4 is 1A?
I1 + I 4 = I 2 + I 3
I2 = I1 + I4 - I3 = -2A
1A
Charge conservation:
Ii > 0 for incoming
å Ii = 0 Ii < 0 for outgoing
i
Motion of Electrons in a Wire
In a current-carrying wire there must be an electric field to drive the
sea of mobile charges.
What is the relationship between the electric field and the current?
Why is an electric field required?
Interaction between electrons and lattice of atomic cores in metal.
Electrons lose energy to the lattice. Electric field must be present to
increase the momentum of the mobile electrons.
The Drude Model
eE Dt
Average ‘drift’ speed: v =
me
Dt - average time between
collisions
For constant temperature v ~ E
v = uE ,
e
u=
Dt
me
u – mobility of an electron
Electron current: i = nAv
i = nAuE
Paul Drude
(1863 - 1906)
Typical Mobile Electron Drift Speed
Typical electron current in a circuit is ~ 1018 electrons/s.
What is the drift speed of an electron in a 1 mm thick copper
wire of circular cross section?
# electrons
= nAv
s
n » 8.4 ´ 1028 m-3
3.14 × (1 ´ 10 m )
pD
A=
»
= 8 ´ 10 -7 m 2
4
4
-3
2
2
1018 s-1
1018 s-1
-5
v=
»
=
1.5
´
10
m/s
28
-3
-7
2
nA
8.4 ´ 10 m 8 ´ 10 m
(
)(
)
Typical Mobile Electron Drift Speed
Typical electron current in a circuit is 1018 electrons/s.
What is the drift speed of an electron in a 1 mm thick copper
wire? n » 8.4 ´ 1028 m-3
v = 1.5 ´ 10-5 m/s
How much time would it take for a particular electron to move
through a piece of wire 30 cm long?
s
0.3 m
4
t= =
=
2
´
10
s » 5.5 hours!
-5
v 1.5 ´ 10 m/s
How can a lamp light up as soon as you turn it on?
Typical E in a Wire
Drift speed in a copper wire in a typical circuit is 5.10-5 m/s.
The mobility is u=4.5.10-3 (m/s)/(N/C). Calculate E.
v = uE
v
5 ´ 10 -5 m/s
-2
E= =
=
1.1
´
10
N/C
-3
u 4.5 ´ 10 (m/s)/(N/C)
Electric field in a wire in a typical circuit is very small
E and Drift Speed
In steady state current is the same everywhere in a series circuit.
Ethin
Ethick
i
i
What is the drift speed?
i = nAv
nAthin vthin = nAthick vthick
vthin
Athick
=
vthick
Athin
Note: density of electrons n cannot change if same metal
What is E?
v = uE
uEthin
Athick
=
uEthick
Athin
Ethin
Athick
=
Ethick
Athin
Clicker
A
B
Lamps are identical
Batteries are identical
In B batteries are connected in series
Lamp shines brighter in
A) A
B) B
C) Lamps in A and B have the same brightness
Two Batteries in Series
Why light bulb is brighter with two batteries?
Two batteries in series can drive more current:
Potential difference across two batteries in
series is 2emf  doubles electric field
everywhere in the circuit  doubles drift
speed  doubles current.
2emf  EL  0
emf  EL  0
emf
E
L
emf
i  nAuE  nAu
L
2
emf


P1batt  eLnAu 
 L 
2emf
E
L
Work per second:
P  (q / T )EL  ieEL
P  nAueLE 2
2emf
i  nAu
L
2
2emf


P2batt  eLnAu 
 L 
P2batt  4  P1batt
Energy in a Circuit
Vwire = EL
Vbattery = ?
Energy conservation (the Kirchhoff loop rule [2nd law]):
V1 + V2 + V3 + … = 0
along any closed path in a circuit
V= U/q  energy per unit charge
Question: Twice the Length
1
2
i1
i2
Nichrome wire (resistive)
A) i1 = i2
B) i1 = 2*i2
C) i1 = ½ i2
Question:Two Light Bulbs
In series
Parallel
Identical light bulbs are first connected in series then parallel
Will there be a difference in light brightness?
A) There will be no difference
B) In series connection will give brighter light
C) Parallel connection will give brighter light
Two Light Bulbs in Parallel
DV
i = nAuE = nAu
L
1. Path ABDFA:
-2 ( emf ) + EB L = 0
L … length of bulb filament
2 ( emf )
EB =
L
2. Path ACDFA:
-2 ( emf ) + EC L = 0
EC =
2 ( emf )
L
F
EB = EC
iB = iC
ibatt = 2iB
3. Path ABDCA:
EB L - EC L = 0
EB = EC
We can think of the two bulbs in parallel as equivalent to
increasing the cross-sectional area of one of the bulb filaments.
Two Identical Light Bulbs in Series
i = nAuE = nAu
DV
L
Two identical light bulbs in
series are the same as one light
bulb with twice as long a
filament.
Identical light bulbs
1
i2 L > iL
2
Electron mobility in metals decreases as temperature increases!
Conversely, electron mobility in metals increases as temperature
decreases. Thus, the current in the 2-bulb circuit is slightly
more than half of the one-bulb circuit.
Long and Round Bulb in Series
AR >> AL
iL = nALuEL
iR = nARuER
iR = iL
AR ER = AL EL
AR
EL >> ER
ER
AL
2emf - ER L - EL L = 0
EL =
only round bulb:
2emf - ER L = 0
2emf
ER =
L
æ
AR
AR ö
÷÷
2emf = ER L + EL L = LER + L
ER = LER çç1 +
AL
AL ø
è
2emf
2emf
ER =
<<
The round bulb doesn’t light up!
L
æ
AR ö
÷÷
Lçç1 +
AL ø
è
How Do the Currents Know How to Divide?