Download - V

Announcements
• 1st Midterm and Homeworks #1 and #2 will be returned this
week in workshops
• Discussion of 1st exam results Monday, Oct 13th, after Prof.
McFarland is back from Japan
• 2nd Midterm Exam is 23 October, 8:00-9:20am, Hubbell
• HW#5 due Wednesday Oct 15th 11:30
• Preflights #10 and #11 will be posted later this week
– due Monday and Wednesday, 7am, respectively
– everything with workshops also back on the normal schedule
this week
Questions
• “AWESOME job using the chalk board to do examples.
Please keep doing this! keeps people awake and makes us
follow along sicne we write it down”
– I will keep trying to do this. It’s a little uncomfortable for me
because of the awkwardness of the boards.
– “can you please not make HW sets due Tuesdays? The
ChemE students in the class all have 2 other major problem
sets due Tuesdays and this would just push us over the
edge”
– I will try to keep them on Wednesday, but I need better
participation in the Wednesday preflights
– If you think of other ways I can achieve that, please let me
know. I’m open to other ideas.
– “You have to love baseball!!! Didn't your party school U.
Chicago teach you anything about loving the sport?”
– Nope. The Brewers’ spectacular collapse doesn’t help either.
– Personal note from Prof. McFarland: thank you so much for
not mentioning football in the comments this week. Oy!
Leftover: Lecture 8, Concept
2
+Q
• Two parallel plate capacitors are identical
(same A, same d) except that C1 has half of
the space between the plates filled with a
material of dielectric constant k as shown.
– If both capacitors are given the same
amount of charge Q, what is the relation
between E1, the electric field in the air of
C1, and E2, the electric field in the air of C2
(1) E1 < E2
(2) E1 = E2
k
E1=?
C1
-Q
+Q
E2=?
C2
-Q
(3) E1 > E2
• The key here is to realize that the electric field in the air in C1 must be
equal to the electric field in the dielectric in C1!!
• WHY?
•Plates are conductors; therefore they are equipotential surfaces.
•For this to happen, the charge density on each plane must be nonuniform to create equal electric fields!!
•Another way (microscopic view): polarized dielectric attracts charge
from conductor, thus moving charge from air region of C1
• Consequently, E1 < E2.
• So V1 < V2, and thus C1 > C2. (A reasonable guess!)
Today…
• Current and How Current Works
• Devices
– Batteries
– Resistors
• Ohm’s Law
• Power in Resistors (probably next lecture)
Text Reference: Chapter 25:1-8
Examples: 25. 1,3,5,6,10
Current is charge in motion
• Charge, e.g. electrons, exists in conductors
with a density, ne (ne approx 1023 cm-3)
• “Somehow” put that charge in motion:
– effective picture – some net charge flows with an
average velocity per charge carrier, vd
– more on how this happens in a minute!
• Current density, j, is given by j = qenevd
– unit of j is C/m2sec or A/m2 (A = Ampere) and 1A = 1C/s
– current, I, is j times cross sectional area, I = j pr2
– for 10 Amp in 1mm x 1mm area, j=10+7 A/m2 , and vd is
about 10-3 m/s
Circuit Elements
• Capacitors:
Purpose is to store charge (energy).
• We have calculated the capacitance of a system
• We had to modify Gauss' Law to account for bulk
matter effects (dielectrics) … C = kC0
• We calculated effective capacitance of series or
parallel combinations of capacitors
• Batteries (“Voltage sources”):
+ -
Purpose is to provide a constant
potential difference between 2 points.
• Electrical  chemical energy conversion
• For now, we will think of them as an ideal
source of voltage
• Later: Non-ideal batteries will be dealt with
in terms of an "internal resistance".
OR
+
V
-
Circuit Elements (Cont’d)
• Resistors:
Purpose is to limit current drawn in a circuit.
Note:
I
dQ
dt
UNIT: Ampere = A = C/s
• Resistance can be calculated from knowledge of the
geometry of the resistor AND the “resistivity” of the
material out of which it is made.
• The effective resistance of series and parallel
combinations of resistors will be calculated using
Kirchhoff's Laws (which are really just statements
about potential difference and current conservation).
• “Conductors” as we have considered them before,
have resistance!
Defining Resistance
R
• Resistance
Resistance is defined to be the
ratio of the applied voltage to
the current passing through.
R
V
I
UNIT: OHM = W
I
I
V
• Is this a good definition?
i.e., does the resistance belong only to the resistor?
Recall the case of capacitance: (C=Q/V) depended on the
geometry, not on Q or V individually
Does R depend on V or I ?
It seems as though it should, at first glance...
PF09 Question #2
Consider two circuits, each with a load resistor Ri
driven by a battery with voltage Vi. V1=5 Volts, V
V2=20 Volts, R1=10 W, R2=50 W.
R
Compare the currents through the two resistors.
100%
a) I1 > I2
b) I1 = I2
c) I1 < I2
80%
95%
60%
40%
20%
0%
3%
3%
What you said …
What I would have said …
I=V/R. I1=5V/10W, I2=20V/50W
V
R
Right:
Ohm's law says that I=V/R. I1= 5/10 > I2= 20/50
For question 2 I = V/R and for question 4 P = IV. Equation roulette :)
[ok, you win this time… care to try your luck again?]
Wrong:
Current is the flow of electric charge. Resistors reduce the flow of charge, and the battery
provides the charge. Although there is less charge provided by V1, the ratio of Voltage/Resistance
for V1/R1 is larger than V2/R2. So despite the fact that more charge is flowing from V2, R2 is
reducing the charge much more. So I1>I2. [Careful not to confuse charge flow with voltage.
Voltage is just a potential difference… tells you how much higher energy of charge leaving
battery is than charge returning. #2 has smaller CURRENT at higher VOLTAGE]
The current for circuit 1 is higher because the battery is working against a smaller resistance [Right
answer, but… current is also proportional to V. More V means more force on charge carriers]
Resistance and Currents
I
R
• Let’s analyze the flow of charge
through a resistor…
– If there is a potential difference inside
the resistor, then there is also…?
V
– A constant electric field
• Constant electric field implies
what force on charge carriers?
–Constant, and therefore a constant
acceleration!
• If acceleration of charge carriers occurs, then how does
current vary with time?
–Well, it should increase linearly with time, like velocity does
under constant acceleration, right?
• This isn’t what we observe…
– We observe that current is proportional to voltage and doesn’t
change over long period of time
– Let’s understand this…
I
Resistance and Currents (cont’d)
• Charges are put in motion, but scatter in a very short
time from things that get in the way
– it’s crowded inside that metal
– defects, vibrational motion of atoms (phonons), etc
1
Typical scattering time t = 10-14 sec
charges accelerated for this time and then randomly scattered
average velocity attained in this time is v = at = eEt/m
current density, j = enev, so current is proportional to E which is
proportional to Voltage
• OHM’s LAW j = (e2net/m)E (e2net/m=s , called conductivity)
•
•
•
•
Lecture 9, Concept 1
• Which analogy would you use to explain flow of charge in a
resistor in response to an potential difference? Why?
(1) Snow falling outside your window
(2) A car rolling down a hill towards a wall
(3) Molecules of a scent traveling through the air
traveling to your nose
Lecture 9, Concept 1
• Which analogy would you use to explain flow of charge in a
resistor in response to an potential difference? Why?
(1) Snow falling outside your window
(2) A car rolling down a hill towards a wall
(3) Molecules of a scent traveling through the air
traveling to your nose
• The snow is falling at an apparently constant velocity despite the
constant acceleration due to gravity because of microscopic
collisions with air molecules in its path
• The car rolls down the hill in response to a constant acceleration,
and increases in speed until a catastrophic collision at the bottom.
• The scent molecules diffuse through the air, but they diffuse
uniformly.
Resistance
I
• How do we calculate R??
• To calculate R, must calculate current
which flows when voltage V is applied.
• Applying voltage V sets up an
electric field in the resistor. What
determines the current?
R
I
V
R
V
I
• Current is charge flowing past a point per unit time.
Need to know velocity of charges.
• Is this affected by field?
• Force on charge carriers accelerates charge between
collisions
• Empirically, we separate this into two parts
– Part of this we bury as the “resistivity” of the material
– Part depends on the geometry of the resisting material
Resistivity
• Property of bulk matter related to
resistance of a sample is the
resistivity r defined as:
E
r
j
me
r  2
s e nt
E
1
j
where E = electric field and j =
current density in conductor.
For uniform case:
j
I
,
A
A
L
V  EL
I
 ρL 
 V  EL  rjL  r L  I 

A
 A 
L
V  IR where R  r
A
So, in fact, we can compute the resistance if we know a bit about the
material, and YES, the property belongs to the material!
e.g., for a copper wire, r ~ 10-8 W-m, 1mm radius, 1 m long, then R  .01W
Makes sense?
Rr
L
A
E
j
A
L
• Increase the Length, flow of electrons impeded
• Increase the cross sectional Area, flow facilitated
• The structure of this relation is identical to heat flow
through materials
• think of window for an intuitive example
How thick?
or
How big?
What’s it made of?
2
PF09 Question #5-6
Two cylindrical resistors are made
from the same material, and they
are equal in length. The first
resistor has diameter d, and the
second resistor has diameter 2d.
100%
Compare the resistance of the two cylinders.
80%
60%
a) R1 > R2
b) R1 = R2
c) R1 < R2
82%
40%
20%
0%
If the same current flows through both resistors,
80%
compare the average velocities of the electrons 70%
60%
in the two resistors:
50%
a) v1 > v2
b) v1 = v2
c) v1 < v2
40%
30%
20%
10%
0%
5%
13%
69%
25%
7%
What you said …
Right:
R is proportional to (length)/(cross-secional area) so the same length and larger area in R2
means a lower resistance in R2. The current density is larger in R1 [for the same current] so
the average drift velocities will also be larger.
IR=V since R1 is greater the potential difference is greater therefore, the velocity is greater
Wrong:
R1=R2 because resistance is not determined by size but by Ohm's Law. .
IR=V since R1 is greater the potential difference is greater therefore, the velocity is greater [oh no,
it’s equation roulette! the “V” in Ohm’s Law is potential difference, not velocity]
The current is the same flowing through the resistors and the length of the resistors is the same so
charge will flow over the same distance with the same drift velocity. [current is flow of charges past
a given point, so it doesn’t depend on length. the density of charge carriers is a property of the
material, so for the same material what matters is the cross-sectional area… more carriers in parallel
means they can go more slowly and carry the same current.]
When A is bigger, drift velocity will be less. Intuitively, with a larger area, there will be less electron
collisions and the velocity of electrons will be smaller at equilibrium [the answer is right, but it is
because V is smaller (R smaller), so less acceleration. still as many targets to collider with.]
Lecture 9, Concept 2
• Two cylindrical resistors, R1 and R2, are made of identical material.
R2 has twice the length of R1 but half the radius of R1.
– These resistors are then connected to a battery V as shown:
I1
I2
V
– What is the relation between I1, the current flowing in R1 ,
and I2 , the current flowing in R2?
(a) I1 < I2
(b) I1 = I2
(c) I1 > I2
Lecture 9, Concept 2
• Two cylindrical resistors, R1 and R2, are made of identical material.
R2 has twice the length of R1 but half the radius of R1.
– These resistors are then connected to a battery V as shown:
I1
I2
V
– What is the relation between I1, the current flowing in R1 ,
and I2 , the current flowing in R2?
(b) I1 = I2
(a) I1 < I2
(c) I1 > I2
• The resistivity of both resistors is the same (r).
• Therefore the resistances are related as:
R2  r
L2
2 L1
L
r
 8 r 1  8R1
A2
( A1 / 4)
A1
• The resistors have the same voltage across them; therefore
I2 
V
V
1

 I1
R2 8R1 8
Ohm Example: Resistors in Series
The Voltage “drops”:
Va  Vb  IR1
I
R1
Vb  Vc  IR2
Va  Vc  I ( R1  R2 )
b
a
R2
Whenever devices are in SERIES, the
current is the same through both !
c
This reduces the circuit to:
Hence:
Reffective  ( R1  R2 )
a
Reffective
c
Resistivity Example:
Resistors in Series
R1
Consider two cylindrical resistors with
lengths L1 and L2
R1  r
L1
A
R2  r
L2
A
L1
V
Put them together, end to end to make a longer one...
L1  L2
Reffective  r
 R1  R2
A
R  R1  R2
L2
R2
What about Work and Power?
• V=IR
– V is potential difference across which a charge is moved
– How much charge? Current is the rate of flow of charge
• dQ/dt = I
• So for constant current, Q=I×time
• The work required to move a charge through a potential
difference is…?
– W=V×Q
• Where does this work go in a resistor?
– Does it go into kinetic energy of the charge?
• No, charge scatters and slows
– So where does it go?
• Heat in a toaster element, for example. Or light in a light bulb.
• Remember Power from Physics 121?
– Power is Work / Time
Power
Batteries & Resistors
Energy expended
chemical
to electrical
to heat
Rate is:
work
 power  sJ 
time
Charges per time
Therefore:
P  VI
Potential difference per charge
or you can write it as
For Resistors:
Units okay?
P  IR I  I 2 R
P  V V R   V 2 R
Joule  Coulomb J  Watt
Coulomb second s
PF09 Question #3
Consider two circuits, each with a load resistor Ri driven by a battery
with voltage Vi. V1=5 Volts, V2=20 Volts, R1=10 W, R2=50 W.
Compare the power dissipated in the two
resistors.
a) P1 > P2
b) P1 = P2
c) P1 < P2
100%
80%
83%
60%
40%
20%
0%
12%
5%
V
R
What you said …
What I would have said …
P=V2/R.
I1=(5V)2/10W, I2=(20V)2/50W
V
R
Right:
[Question 2] Using the equation V=IR, we can plug in the given values of V and R to find the current,
I, for each circuit. The current in circuit 1 is .5 A, and the current in circuit 2 is .4 A, so the I1>I2.
[Question 3] Using the equation P=IV, we can plug in the given value of V with the value of I found
in Question 2 to find the power, P, for each circuit. The power in circuit 1 is 2.5 W while the power in
circuit 2 is 8 W, So P2>P1.
Wrong:
The power in the first circuit is larger than those in the other because
current is larger [correct about current, but P=I^2R. High current but no
resistance would result in no power lost]
PF09 Question #8
Electric ovens, which must deliver a certain amount of power
to cook food, run on 240 Volts AC instead of the more
common household voltage of 120 Volts AC. Which of the
following might be a reason why? (You may choose more
than one answer).
a) It may be difficult to make a very high resistance heating
element.
b) It is dangerous to carry too much current in household
wiring.
c) A low voltage oven would have to be supplied by thin
wires which would be too fragile to run through a house.
100%
d) High voltage supply wires are safer in
the event of a short in an oven.
80%
80%
60%
40%
25%
22%
20%
0%
9%
What you said …
What I would have said …
P=V2/R. So the lower voltage (120V instead of 240V) would require ¼
the resistance for equal power. Twice the current.
(a) 120V would require lower resistance, (c) thin wires (high resistance)
would be bad with high current, (d) high voltage shorts are very bad…
Right:
Doubling the voltage requires that only half the current be drawn to achieve the same
amount of power. [true, so lower current for the 240V oven]
too much current in a wire is usually the cause of fires in homes. [high current & resistance in wires
and connections means high power dissipated there, which could create dangerous amounts of heat]
Wrong:
The heat cooking the food comes from many collisions inside the wire. An oven needs
a lot of heat, and therefore the heating element needs a high resistance. If voltage is
kept constant (at 120 V) and V=IR, resistance has to be much higher [good points, but
higher resistance will mean lower current (at fixed voltage) and therefore lower power]
Danger is augmented by the increased resistance of thin wires, since thermal energy production is
proportional to R [true: thin wires have high R. so low voltage oven would require THICK wires]
Using high voltage reduces the magnitude of the current needed to generate a specific level of power
P=IV, so high voltage wires reduce the dangers associated with a large current [true oven current is
small, but a short means a low resistance to ground, so high voltage means higher current in short]
Lecture 9, Concept 3
• Consider the circuit at the right.
– Which of the following modifications to
the circuit will result in the maximum
power dissipated in the resistor?
(a) Double R, leaving V unchanged
(b) Double V, leaving R unchanged
(c) Double R and V
I
V
R
I
Lecture 9, Concept 3
• Consider the circuit at the right.
– Which of the following modifications to
the circuit will result in the maximum
power dissipated in the resistor?
(a) Double R, leaving V unchanged
(b) Double V, leaving R unchanged
I
V
R
I
(c) Double R and V
• Power = IV = V2/R
• Therefore, want to increase V and decrease R to
maximize the power dissipated in the resistor
Household Electrical Safety
• Most dangerous aspect of everyday
household current: ventricular fibrillation
• AC current of a few mA at 60Hz
through the heart can be fatal
• this translates to a current across the chest of ~100mA
• why? current density (through the heart) vs. current
• I’ve been shocked from hand-to-hand by ~110 Volts twice in
my life (once doing a demo in Physics 122)
• so, why am I still alive?
• resistance of dry skin is ~10kΩ (wet skin is only ~1kΩ)
so V=IR says current was only ~10mA
Household Electrical Safety
• Most safety tips focus on avoiding
current through the chest
• Work on electrical panels with one
hand in your pocket, standing on insulated platforms to
avoid any return path for current
• Three-prong plugs
• Provide a separate
current path, usually
connected to a metal
case of an appliance,
in case of a short between
the “hot” wire and the case
Superconductivity…
• An amazing fact of nature…
some materials at low temperature
have a resistance so low that it cannot
be measured
• “Superconductors”
• What must the electric field be in a superconductor?
• Tremendous savings from high current applications
(like high field magnets… more on this later)
• no power lost due to resistance as current flows, as
long as you can keep your conductor very, very cold
• another amazing feature… superconductors repel
magnetic fields. Our first hint (in this course) that
magnetism is related to electricity!
Summary…
• What Current Is
• Devices
– Batteries
– Resistors
• Ohm’s Law
– Nature of conduction and resistance in materials
• Power
Next Time…
• Real Batteries
• Combining Circuit Elements
• Kirchhoff’s Rules
Text Reference: Chapter 25:9-10, 26: 1-3, 6
Examples: 26: 1, 3, 6, 8