Transcript Lec10

Lecture 10-1
C
Capacitor Examples
C
2C
C  C1  C2 
C
C
C/2
 1 0 A / 2
d
 A    2 
 0  1

d 
2 

 2 0 A / 2
d
d/4
3d/4
1
1
1
d / 4 3d / 4




C
C1
C2
0 A
 0 A

d 1
3 1


0 A 
4 
4
Lecture 10-2
Electric Current
Current = charges in motion
q dq
Magnitude
I  lim

x 0 t
dt
rate at which net positive charges
move across a cross sectional surface
I   J d A
A
Units:
[I] = C/s = A (ampere)
Current is a scalar, signed quantity, whose
sign corresponds to the direction of motion of
net positive charges by convention
J = current density
(vector) in A/m²
Lecture 10-3
Microscopic View of Electric Current in Conductor
All charges move with some velocity ve
A
random motion with high speeds
(O(106)m/s) but with a drift in a certain
direction on average if E is present
• thermal energy
• scattering off each
other, defects, ions,
…
Drift velocity vd is orders of magnitudes less
than the actual velocity of charges,
Lecture 10-4

I  nAvd q
I
 vd 
nAq
Current and Drift Velocity in Conductor
Drift velocity vd is orders of
magnitudes less than the actual velocity
of charges.
where n =carrier density
In the following condition:
I = 1.0A,
copper: n ~1029atoms/m3
1mm radius wire.
Vd~0.01mm/s
Lecture 10-5
Ohm’s Law
Current-Potential (I-V) characteristic of a
device may or may not obey Ohm’s Law:
I V
or V  IR with R constant
V V

Resistance  R    I   A   (ohms)
tungsten wire
gas in fluorescent tube
diode
Lecture 10-6
Resistance and Resitivity for Ohmic Material
I V
I J
V E
J E
E  J
resistivity
 I
 L
 V  EL     L  I   
 A
 A
A
L
R (in )
resistance
Lecture 10-7
Resistance
Resistance
(definition)
V
R
I
R
I
V
constant R
L
R
A
Ohm’s Law
Lecture 10-8
Warm up
There are 2x1014 electrons across a resistor in 10
seconds. What is the current through the resistor?
a) 3.2mA
b) 1.6 mA
c) 3.2 A
d) 1.6A
e) 3.2 mA
Note: e = 1.6x10-19 C
R
I
V
Lecture 10-9
Temperature Dependence of Resistivity
  0 1   (T  T0 )
• Usually T0 is 293K (room temp.)
• Usually  > 0 (ρ increases as T )
Material
0 (m)
 (K-1)
Ag
1.6x10-8
3.8x10-3
Cu
1.7x10-8
3.9x10-3
Si
6.4x102
-7.5x10-2
glass
1010 ~ 1014
sulfur
1015
Copper
Lecture 10-10
Electric Current and Joule Heating
• Free electrons in a conductor gains
kinetic energy due to an externally
applied E.
• Scattering from the atomic ions of the
metal and other electrons quickly leads to
a steady state with a constant current I.
Transfers energy to the atoms of the solid
(to vibrate), i.e., Joule heating.
Lecture 10-11
Energy in Electric Circuits
• Steady current means a constant amount of charge Q flows past
any given cross section during time t, where I= Q / t.
I
a
a
b
I
∆Q
∆Q
Energy lost by Q is
U  Q  (Va  Vb )  I t V
=> heat
So, Power dissipation = rate of decrease of U =
dU
P
 IV  I 2 R  V 2 / R
dt
b
Lecture 10-12
EMF – Electromotive Force
• An EMF device is a charge pump that can maintain a potential
difference across two terminals by doing work on the charges
when necessary.
Examples: battery, fuel
cell, electric generator,
solar cell, fuel cell,
thermopile, …
• Converts energy (chemical, mechanical, solar, thermal, …)
into electrical energy.
 Within the EMF device, positive charges
are lifted from lower to higher potential.
 If work dW is required to lift charge dq,
  dW
   Volt
dq
EMF
Lecture 10-13
Internal Resistance of a Battery
load
 i r iR  0
internal
resistance
 i

Rr
terminal
voltage
, Vb  Va    ir

R
Rr
Lecture 10-14
Energy Conservation
A circuit consists of an ideal battery
(B) with emf , a resistor R, and two
connecting wires of negligible
resistance.
Energy
conservation
• Ideal battery: no internal
energy dissipation
• Real battery: internal
energy dissipation exists
Work done by battery is equal
to energy dissipated in resistor
dW  i 2 Rdt or  i dt  i 2 R dt
   iR
dW > i2Rdt or  > iR=V
Lecture 10-15
Lecture quiz A
There are1014 electrons across a resistor with
potential drop of 3.2mV in 10 seconds. What is
the resistance of the resistor?
a) 2.0 Ω
b) 1.0 Ω
c) 2.5 Ω
d) 3.0 Ω
e) 4.0 Ω
Note: e = 1.6x10-19 C
R
I
V
Lecture 10-16
Lecture quiz B
There are 1014 electrons across a resistor of
resistance 1.0Ω in 10 seconds. What is the
potential drop across the resistor?
a)3.2 mV
b)8.0 V
c)2.5 V
d)1.6 mV
e)1.6mV
Note: e = 1.6x10-19 C
R
I
V
Lecture 10-17
Lecture quiz C
The potential drop is 6.4mV across a resistor of
resistance 1.0Ω. How many electrons enter the
wire in 10 seconds?
a)3.2×1019
b)8.0×1015
c)2.5×1012
d)4.0x1014
e)1.6×1019
Note: e = 1.6x10-19 C
R
I
V