Download Text sections 27.4, 27.6, 28.1 • Drude Model • Electrical work and

Text sections 27.4, 27.6, 28.1
• Drude Model
• Electrical work and power
• “emf”
Practice problems:
Chapter 27, problems 17, 24, 27, 47, 49
Chapter 28, problems 1, 3
From before: I =nqAvd
• We assumed that the drift velocity is proportional
to the electric field, thus leading to Ohm’s Law,
•  vd is the average velocity of an electron in a metal
(net displacement per second) and is small
•  Typical speeds are in km/s for electrons in metals
1
•  Random velocities of electrons are large (several
km/s)
With no field, the
electrons scatter
randomly (due to
impurities and ions in the
metal), so on average
they go nowhere.
An electric field will
cause the electron’s
path to curve slightly
between collisions, but
always in the same
direction, leading to a
slow but steady net
“drift”.
end
start
E
-
end
start
end
net displacement
2
Assumptions:
1) Electrons have large random velocities
(no field).
2) Electrons collide with atoms (time
);
is not affected by the field.
3) After a collision, velocity is random
again.
q
Averages:
3
So…
drift velocity:
and
(if |q| = e)
So…
Therefore…
I
+
V1
Resistance R
V2
I
Current I flows through a potential difference ΔV
Follow a charge Q : at positive end, U1 = QV1
at negative end, U2 = QV2
P.E. Decreases:
  Electrical energy is converted to other forms
of energy.
4
Quiz
The derivation assumed positive charges, and
found that the potential energy decreases with
time. If the current is actually carried by negtive
charges, then:
A) the potential energy will increase with time
B) the potential energy will still decrease
C) the potential energy will not change
Quiz
If potential energy is steadily lost, what happens
with the kinetic energy of the moving electrons ?
A) the average kinetic energy increases steadily
B) the average kinetic energy decreases steadily
C) none of the above
5
Electrical resistance converts electrical potential
energy to thermal energy (heat), just as friction in
mechanical systems converts mechanical energy to
heat.
The average kinetic energy of the electrons
doesn’t increase once the current reaches a
steady state; the electrons lose energy in
collisions as fast as it is supplied by the field.
Negative charge carriers would move in the
opposite direction (from low V to higher V); but
this still means they lose potential energy.
P = work per unit time
Power = current x (potential difference)
Units: 1 volt x 1 amp = 1 watt (= 1 J/s)
Eg. Resistor: ΔV = IR
 P = I2 R (rate of “Joule heating” in resistor)
6
1)  What is the resistance of a “60 watt” bulb? (for
a 120-V supply)
2)  Find R for a 60-W headlamp (12-V battery).
3)  What power do you get from a “60-W” household
bulb if you connect it to a 12-V car battery?
4)  How many 200-A, 750 kV lines are needed to
carry the power from a 1,500,000 kW hydro
dam?
E ≡ external work per unit charge
Units: J/C = volts (not actually a force)
e.g.: Battery
(chemical energy  electrical energy)
Generator
(mechanical energy  electrical energy)
Solar cell, etc.
7
-
Example: battery.
I=2A
Eg.
What does the energy balance
look like in this circuit?
R
12 V
What is the resistance R?
-
Example: battery
Eg.
12 V
+
+
I=2A
R
Battery: supplies power,
P = IE = 24W
(Chemical energy  electrical energy)
Resistor: Electrical energy  heat,
P = 24W=I2R ⇒ R = 6Ω
8
I
A
r
C • 
I
I
r = “internal resistance”
RL (external “load”)
B
VA – VB = “terminal voltage”  measured
VC – VB = E (but “C” is not accessible for measurement)
And VA – VC = - I r
“Terminal voltage”
At terminals
Find:
Automobile battery:
12.8 V (with 20 A current)
9.2 V (with 200 A current)
E and rinternal of battery
What is the maximum current the battery can provide?
9
Notes:
i)  As I 0, Vload E
ii)  Imax = E /r (when RL = 0, a short circuit)
Question: How would you select a load (e.g., lightbulb)
to get the maximum power output, for a given battery?
A famous theorem: Maximum power transfer is
achieved when the load resistance matches the
source resistance.
 Maximum load power when RL= r
Proof:
Set
and solve for RL (exercise)
10
VL = E – rIL
Max. power point
VLoad
ILoad
Pmax
PLoad
r
RLoad
11