Transcript Ch 19all 20.2

Chapter 19
A Microscopic View
of Electric Circuits
Current in a Circuit
A microscopic view of electric circuits:
• Are charges used up in a circuit?
• How is it possible to create and maintain a nonzero electric
field inside a wire?
• What is the role of the battery in a circuit?
In an electric circuit the system does not reach equilibrium!
Steady state and static equilibrium
Static equilibrium:
• no charges are moving
Steady state (Dynamic Equilibrium):
• charges are moving
• their velocities at any location do not change with time
• no change in the deposits of excess charge anywhere
Current in Different Parts of a Circuit
IB = IA in a steady
state circuit
We cannot get something for nothing!
What is used up in the light bulb?
Energy is transformed from one form to another
Electric field – accelerates electron
Friction – energy is lost to heat
Battery – chemical energy is used up
Closed circuit – energy losses to heat:
not an isolated system!
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)
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, the electric field drives the mobile charges.
What is the relationship between the electric field and the current?
Why is an electric field required?
Electrons lose energy to the lattice of atomic
cores. Electric field must be present to increase
the momentum of the mobile electrons.
Collisions with Core gives frictional force.
In steady state, net force is zero.
The Drude Model
E
Dp
Momentum principle:
= Fnet
Dt
p eEDt
Speed of the electron: v =
=
me
me
eE Dt
Average ‘drift’ speed: v =
me
Dp = Fnet Dt = eEDt
Dp = p - 0 = eEDt
Dt - average time between
collisions
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)
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
Exercise
i = nAv
0.1 mm
1 mm
vthin = ?
vthick = 410-5 m/s
Ethin= 10-1 N/C
Ethick= ?
vthin = 40010-5 m/s
Ethick= 10-3 N/C
Question
i = nAv
1.5.n1 = n2
2.u1 = u2
1018
1 mm
2 mm
Every second 1018 electrons enter the thick wire.
How many electrons exit from the thin wire every second?
1018
A)
B) 1.5 x 1018
C) 2 x 1018
D) 4 x 1018
E) 12 x 1018
Question
i = nAv
v = uE
1.5.n1 = n2
2.u1 = u2
1018
1 mm
2 mm
What is the ratio of the electric field, E1/E2?
A)
B)
C)
D)
3:1
6:1
8:1
12:1
Direction of Electric Field in a Wire
E must be parallel to the wire
E is the same along the wire
Does current fill the wire?
Is E uniform across the wire?
B
C
D
A
A
B
C
D
DVABCDA = - ò E1 ×dl - ò E3 ×dl - ò E2 ×dl - ò E3 ×dl = 0
VAB
0
E1 = E2
VCD
0
A Mechanical Battery
Van de Graaff generator
Electron Current
Connecting a Circuit
When making the final connection in a circuit, feedback
forces a rapid rearrangement of the surface charges
leading to the steady state.
This period of adjustment before establishing the steady
state is called the initial transient.
The initial transient
Connecting a Circuit
1. Static equilibrium: nothing moving
(no current)
2. Initial transient: short-time process
leading to the steady state
3. Steady state: constant current
(nonzero)
Current at a Node
i1 = i 2
i2 = i3 + i 4
The current node rule
(Kirchhoff node or junction rule [law #1]):
In the steady state, the electron current entering a node in a
circuit is equal to the electron current leaving that node
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
Potential Difference Across the Battery
Coulomb force on each e
FC
non-Coulomb force on each e
FC
1. FC =eEC
EC =
e
2. FC =FNC
EC
DVbatt
FC s FNC s
= EC s =
=
e
e
Fully charged battery.
Energy input per unit charge
emf – electromotive force
The function of a battery is to produce and maintain a charge separation.
The emf is measured in Volts, but it is not a potential difference!
The emf is the energy input per unit charge.
chemical, nuclear, gravitational…
Field and Current in a Simple Circuit
Round-trip potential difference:
DVbatt + DVwire = 0
emf + ( - EL) = 0
E=
emf
L
emf
I = enAuE = enAu
L
We will neglect the battery’s internal resistance for the time being.
Field and Current in a Simple Circuit
Round-trip potential difference:
Path 1
DVbatt + DV1 + DV3 = 0
emf + ( - E1L1 ) + (- E3L3 ) = 0
Path 2
E1 L1 = E2 L2
DVbatt + DV2 + DV3 = 0
emf + ( - E2 L2 ) + (- E3L3 ) = 0
V Across Connecting Wires
The number or length of the
connecting wires has little effect on
the amount of current in the circuit.
DV
L
DV =
i
L
nAu
+ DVbattery = 0
i = nAuE = nAu
DVwires + DVfilament
DVbattery » emf
uwires >> ufilament  DVwires << DV filament
DV filament » (emf )
Work done by a battery goes mostly into energy dissipation in the
bulb (heat).
Application: Energy Conservation
V1 + V2 + V3 + Vbatt = 0
(-E1L1) + (-E2L2) + (-E3L3) + emf = 0
An electron traversing the circuit DC-B-A suffers a decrease in its
electric potential energy. Where does
this loss of energy appear?
Twice the Length
Nichrome wire (resistive)
Quantitative measurement of current with a compass
DV
i = nAuE = nAu
L
i2 L =
1
iL
2
Current is halved when increasing the length of the wire by a
factor of 2.
Doubling the Cross-Sectional Area
DV
i = nAuE = nAu
L
Loop: emf - EL = 0
Electron current in the wire increases
by a factor of two if the crosssectional area of the wire doubles.
Nichrome wire
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 that in the one-bulb circuit.
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.
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
Electron current: i = nAuE
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
Resistance
f
DV = - ò E · dl
i
DV = EL
I
J = = sE
A
DV
E=
L
DV
1
I=
DV = DV =
L
R
R
sA
I = sAE
Conventional current: I = DV
R
Widely known as
Ohm’s law
L
Resistance of a long wire: R =
sA
George Ohm
(1789-1854)
Units: Ohm, 
Resistance combines conductivity and geometry!
Microscopic and Macroscopic View
Microscopic
v = uE
i = nAv = nAuE
Macroscopic
J = sE
DV
I = q nAv =
R
Can we write V=IR ?
Current flows in response to a V
Constant and Varying Conductivity
Mobility of electrons: depends on temperature
æ A ö
÷
è V ×m ø
s = q nu ç
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
DV
I=
R
R=
L
sA
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)
DV
I=
R
V-A dependence:
3V
100 mA
1.5 V 80 mA
0.05 V 6 mA
R
30 
19 
8
DV
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
s = q nu
Conductivity depends exponentially on E
Conductivity rises (resistance drops)
with rising temperature
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: Voltage Divider
Know R , find V1,2
R1
V1
Solution:
emf
R2
V2
DV
I=
R
1) Find current: I =
DV = IR
emf
Requivalent
emf
=
R1 + R2
2) Find voltage:
R1
DV1 = IR1 = emf
R1 + R2
R2
DV2 = IR2 = emf
R1 + R2
3) Check: DV1 + DV2 = emf
é R1
R2 ù
emf ê
+
= emf
ú
ë R1 + R2 R1 + R2 û
Parallel Resistance
I = I 1 + I2 + I3
emf emf emf
I=
+
+
R1
R2
R3
æ 1
1
1ö
emf
I =ç +
+ ÷ emf =
Requivalent
è R1 R2 R3 ø
1
Requivalent
1 1
1
= + +
R1 R2 R3
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
DU e = Dq × DV
Power = work per unit time:
P=
DU e Dq × DV Dq
=
=
DV
Dt
Dt
Dt
I
Power for any kind of circuit component: P = IDV
CJ J
= =W
Units: AV =
sC s
Power Dissipated by a Resistor
emf
Know V, find P
R
P = IDV
DV
I=
R
2
DV )
(
P=
R
Know I, find P
P = IDV
DV = IR
P = I 2R
In practice: need to know P to select right size resistor –
capable of dissipating thermal energy created by current.
What is the power output of the battery?
Question
A certain capacitor with only air between its plates has capacitance
C and is connected to a battery for a long time until the potential
difference across the capacitor is equal to 3 V. The battery is then
removed from the circuit and a dielectric (K=2) is inserted between
the capacitor plates filling the entire volume. The energy stored in
the capacitor WITH dielectric compared to the energy stored
Without dielectric is:
A) The same
B) Larger by a factor of 2
C) Smaller by a factor of 2
D) Larger by a factor of 4
E) Smaller by a factor of 4
Uelectric
1 Q 2 C(DV )2
=
=
2 C
2