Physics 1B schedule Winter 2008
Week
Mon
Friday
Wed
Lecture: the Electric field
1:Jan 7
2: Jan 14
Lecture: Intro, 15.1-15.3
Lecture: Electric Flux &
Gauss’s law
Lecture: The Coulomb law,
Lecture: Gauss
law/examples
3: Jan 21
University Holiday
Lecture: Potential
4: Jan 28
Lecture: capacitance
Lecture: capacitor
combinations
5: Feb 4
Lecture: Resistivity, Electric
power
Quiz 2: chapter 16
6: Feb 11
Lecture: Kirchhoff’s rules
Quiz 3: chapter 17&18
7: Feb 18
University Holiday
Lecture: Magnetic force
Lecture: current loop,
8: Feb 25
solenoid
Lecture: Induced EMF
Lecture: Faraday’s law,
Lecture Inductance,
9: March 3
Lenz’s law
Inductors
Lecture: Energy of the
Lecture: AC circuits, EM
10: March 10 magnetic field
wave
HW: Ch15: 1,10,11,13,15,17,20,24,27,28,30,32,36,38,43,46,48
Ch16: 1,3,5,8,12,15,19,22,23,25,29,31,33,35,43,45,47,49,60
Ch17: 1,3,8,9,11,13,16,19,20,23,31,33,39,45,52,60
Ch18: 1,3,5,7,13,17,21,26,31,33,35,
Ch19: 1,3,8,9,11,15,19,22,24,27,29,34,37,38,41,44,47,49,57,61
Ch20: 1,5,8,11,13,16,18,23,25,27,29,31,34,37,39,
Quizzes:
HW problems, problems in class, more…
4 best out of 5
No make-up quizzes for any reason
Quiz 1: chapter 15
Lecture: equi-potential
surfaces
Lecture: Electric current,
Ohm’s law
Lecture: Resistors, series
parallel
Lecture: Magnetism ch19
Lecture: torque on current
loop, Ampere’s law
Quiz 4: chapter 19
Quiz 5: chapter 20
Lecture: discussion of the
final exam
Instructor: D.N. Basov
T.A.: Eric Spencer
Instructor office hours:
Tuesdays 4:30-5:30 pm
PB: TH 7-8:50pm LEDDN AUD
Final exam: all material in ch 15-21
no make up final for any reason
Last update:
January 16, 2007
17.1 Electric Current
Chapter 15-16: electrostatics
Chapter 17-18 charges in motion
Electric current: the rate at which electric charge
passes through a surface.
∆q
I=
∆t
Moving charges: positive, negative, both
Direction of current: the direction of + charges flow!
V
x
+
+
+
+
17.1 Electric Current
A flashlight bulb carries a current of 0.1 A. (a) Find
the charge passed in 10 s. (b) How many electrons
does this correspond to?
(a)
(b)
∆q
I=
∆t
∆q = I ∆t = 0.1(10) = 1C
q = Ne
q
1
18
=
N= =
6.2
x
10
electrons
−19
e 1.6 x10
17.2 Electric Current: a microscopic view
Time between collisions
−14
τ ≈ 10
s
Collisions of electron with the lattice
(resistance) slows down the velocity.
Drift velocity vd- average velocity in the direction of the flow.
17.2 Electric Current: a microscopic view
A
∆x
Vd =
∆t
Rate of flow is
ch arg e ⎛ ch arg e ⎞⎛ volume ⎞
⎛ A∆x ⎞
=⎜
=
qn
= qnAv d
( )⎜
⎟⎜
⎟
⎟
time
⎝ volume ⎠⎝ time ⎠
⎝ ∆t ⎠
∆q
I=
= qnAv d
∆t
Vd –Drift velocity
n – no. of charge carriers/volume
q- charge per charge carrier
17.2 Electric Current: a microscopic view
17.2 Find the drift velocity of electrons in Cu. For I=10 A,
A=3x10-6 m2. Use density of Cu, ρ=8.95 g/cm3(each atom
of Cu contributes 1 carrier electron) MA=63.5g/mole
I
vd
I = qnAv d
q =e
ρ
atoms
density
N Aρ
=
=
=
n=
mass per atom M / N A
m3
M
23
NA
6 6.0 x10
= 8.95(10 )
n = ρ (10 )
63.5
MA
6
n = 8.5 x1028 e / m3
vd =
I
enA
vd =
10
1.6 x10−19 (8.5 X 1028 )(3 X 10−6 )
v d = 2.4 x10−4 m / s
Electrons in a Resistor
F= ma = qE
Lines of Electric Force
(-Electric Field)
Atoms of Resistor
Material (e.g. Carbon)
17.3 Current and voltage in circuits
When a battery is placed in a
closed loop an electric field is
established inside the
conducting wire.
V
A
The potential difference created
by the battery terminal is also
called EMF (ElectroMotive
Force). Units: Volts
17.4 Resistance and Ohm’s Law
Resistors
V
A
Ohm’s law
R: resistance,
units Volts/Ampere = Ohm (Ω)
∆V
R=
I
Resistance causes conversion
of potential energy to heat.
17.4 Resistance and Ohm’s Law
V
A
I
Slope=1/R
Ohm’s law
R: resistance,
units Volts/Ampere = Ohm (Ω)
∆V
I ∝ ∆V
1
I = ∆V
R
∆V = IR
17.4 Resistance and Ohm’s Law
non-ohmic resistance
Semiconductor
diode
I I
V
∆V
A
I
Slope=1/R
Ohm’s law
R: resistance,
units Volts/Ampere = Ohm (Ω)
∆V
I ∝ ∆V
1
I = ∆V
R
∆V = IR
17.5-6 Resistivity and its temperature dependence
Resistivity, ρ
Property of conducting material
L
A
L
R=ρ
A
ρ
Resistivity, ohms meter (Ω·m)
17.5-6 Resistivity and its temperature dependence
A block of metal has a resistance R. If each of the
dimensions of the block are doubled, what will the
resistance be?
L
h
w
Lo
Ro = ρ
w o ho
L=
2Lo
w=
2wo
h=
2ho
2Lo
Ro
=
R=ρ
2w o 2ho
2
Thermometry
A platinum resistance thermometer uses the change in
resistance to measure temperature. If a student with the
flu has a temperature rise of 4.5o C measured with a platinum
resistance thermometer and the initial R= 50.00 ohms. What is
the final resistance? α=3.92x10-3 oC-1
R∝ρ
R = Ro [1 + α (T − To )]
R = 50.00[1 + 3.92 x10 −3 (4.5)]
R = 50.00[1.018] = 50.88Ω
Example
A piece of 20-gauge wire that is one meter long has an electric resistance of
0.050 Ω. Calculate its resistivity. The cross-sectional area of 20-gauge wire is
0.5716mm2.
We need to convert to the proper units.
Lo 17.7 Superconductivity
Ro = ρ
w o ho
Hg
17.8 Electrical energy and power
Physics 1A:
power is a measure of the work done in
a given amount of time
Physics 1B:
Circuits, charge ∆q moving across
potential difference ∆V.
I
SI Units: Watt=Joule/sec
Other units: kW, HP
∆PE
R
∆V
Power: the rate at which energy is
delivered to the resistor
17.8 Electrical energy and power
Physics 1B:
Circuits, charge ∆q moving across
potential difference ∆V.
I
R
P = I ∆V
P = I (IR ) = I R
2
∆V
∆V
∆V 2
P =(
)∆V =
R
R
Three equivalent relations for the power
Power: the rate at which energy is
delivered to the resistor
17.8 Electrical energy and power
∆V
A lightbulb has an output
of 100 W when connected
to a 120V household
outlet. What is the
resistance of the filament?
∆V
P=
R
V 2 1202
=
= 144Ω
R=
100
P
2
17.8 Electrical energy and power
A heating element in an electric range is rated at 2000
W. Find the current required if the voltage is 240 V. Find
the resistance of the heating element.
P = IV
P 2000
= 8.3 A
I= =
240
V
P = I 2R
P 2000
= 29Ω
R= 2 =
2
I
8.3
17.8 Electrical energy and power
Cost of electrical power
Kilowatt hour = 1kW x1hr=1000J/s(3600s)=3.6x106J
1kW hr costs ~ $0.15
How much does it cost to keep a 100W light on for
24 hrs?
$
Cost =
kwhr = 0.15(0.10)(24) = $0.36
kwhr
17.8 Electrical energy and power
A 10 km copper power cable with a resistance of
0.24 Ω leads from a power plant to a factory. If the
factory uses 100 kW of power at a voltage of 120 V
how much power would be dissipated in the cable.
R=0.24Ω
Pf=105 W
120 V
Pf = I ∆Vf
5
Pf
10
I=
=
= 8.3 x102 A
∆Vf 120
A large current is
required to provide this
power at low voltage
Pc = I 2Rc = (8.3 x102 )2 (0.24) = 1.6 x105 W
Very lossy cable
17.8 Electrical energy and power
Why do electric companies make power
lines with very high voltages to deliver
electricity to your house?
R = 10Ω, V = 100,000V to
deliver 100kW of power.
What if V=2000 V only?
0.01 % loss only
25% loss. Huge!
17.8 Electrical energy and power
Power Transmission
High voltage
∆Vtrans=105V
Power loss=I2Rwire
Power transferred=∆Vtrans I
transformer
∆V=120 V
Low voltage
High voltage transmission- power transmitted with
lower current. Therefore lower I2R loss in the line.
18 Direct-current circuits
18.1 Sources of EMF
There are three basic types of emf:
1) Ideal Battery: a battery that lacks
internal resistance.
2) Real Battery: a battery that
contains internal resistance, r,
that hampers current flow.
3) Generator: a device that uses
mechanical energy to create
electrical energy.
Battery with internal resistance:
∆Vba = Vb – Va
∆Vba = ɛ – Ir
For this circuit:
ɛ = IR + Ir
Fig.18-1
18.2 Resistors in series
A typical circuit problem is to solve for the
power/voltage/current at a specific point
(like at a resistor).
Ohm’s Law:
∆V = IR
ɛ = IR
I = ɛ/R
More complicated circuits?
Reduce them down to the simplest one
and then apply Ohms law
18.2 Resistors in series
∆V1
∆V2
Req = R1 + R2
∆V= ∆V1 + ∆V2
Current I is the same everywhere in the circuit
∆V=IR1 + IR2
IReq = I(R1 + R2)
Req = R1 + R2
For N resistors in series:
Req = R1 + R2 +…….RN
18.2 Resistors in series
∆V1
L
R=ρ
A
∆V2
L
A
A
R~L
Rtot ~ L1 + L2
Req = R1 + R2
18.2 Resistors in series
Why do we care?
Consider Simple Circuit: Two resistors in Series
r
C
D
B
E
ε
R
A
∆V
F
-Ir
V
ε
-IR
∆V
0
A
B
C
D
E
F
Voltage can be
‘tailored’ to
produce any
voltage we
desire!
18.3 Resistors in parallel
I = I1 + I2
∆V
∆V ∆V
=I =
+
Req
R1 R2
1
1
1
=
+
Req R1 R2
For N resistors in parallel
1
1
1
1
=
+
+ .......... +
Req R1 R2
RN
Req is smaller than
any R
18.3 Resistors in parallel
1
1
1
=
+
Req R1 R2
L
R=ρ
A
R ~ 1/Area
Atot=A1 + A2
Atot=1/R1 + 1/R2
Rtot~1/Atot ~ 1/(1/R1 + 1/R2)
Example:
What is the current (I) that flows through
the battery?
=6.0 Ω
=5.0 Ω
Resistors summary
Series
Parallel
Wiring
Each resistor on the
same wire.
Each resistor on a
different wire.
Voltage
Different for each resistor.
Vtotal = V1 + V2
Same for each resistor.
Vtotal = V1 = V2
Current
Same for each resistor
Itotal = I1 = I2
Different for each resistor
Itotal = I1 + I2
Increases
Req = R1 + R2
Decreases
1/Req = 1/R1 + 1/R2
Resistance
18.4 Kirchhoff’s Rules
Kirchhoff’s Loop Rule:
Kirchhoff’s Junction Rule:
I1
I2
R1
∆V
The sum of the potential
differences across all the
elements around any
closed circuit loop must be
zero.
Conservation of energy
R3
I3
∑i
n
=0
R2
The sum of the currents
entering any junction must
equal the sum of the
currents leaving that
junction.
Conservation of
charge
E
A
How do I use Kirchhoff’s Rules?
+
R1
B
I1
R2
(a) Identify circuit loops and nodes.
(b) Assign variables and directions to each current.
(The directions you assign can be arbitrary;
the correct signs will emerge from the algebra.)
(c) Sum these assigned currents at each
node and set this sum to zero.
(d) Sum voltage gains and drops through each element
around each loop. The direction you choose around each
loop is arbitrary. It will all work out as long as you follow
these conventions:
No. equations needed=
Find I1, I2, I3
no. Junction=
I1
I2
2Ω
I3
No. loop =
4Ω
5V
10V
V=IR
10 − 2I1 − 5 = 0
10 − 5
I1 =
= 2.5 A
2
I1 = I2 + I3
I3 = I1 − I2
5 − 4I2 = 0
5
I2 = = 1.25 A
4
I3 = 2.5 − 1.25 = 1.25 A
3
1
2
How do I use Kirchhoff’s Rules?
Problem: E1 = 12 V, E2 = 3 V, E3 = 3 V,
1. Assign currents as shown.
R1 = 1 Ω, R2 = 5 Ω, R3 = 5 Ω.
Find the current through R3.
Node Law
I1 + I2 + I3 = 0
R3
I1
R1
E1
R2
I3
E1 − R1I1 + R2 I2 + E 2 = 0
I2
+
-
2. Identify loops:
Loop 1
Loop 1
E2
- Loop 2
+
+
-
E3
Loop 2
−E 2 − R2 I2 + R3 I3 − E 3 = 0
(E 3 − E1 )R2 + (E 3 + E 2 )R1 ]
[
I3 =
[R1R2 + R1R3 + R2 R3 ]
I 3 = −1.11 A
18.5 RC circuits
Loop rule:
+ + + + +
+ + +
--- --- --- --- --- --- --- ---
When the capacitor is fully
charged, the current, I, will
become zero.
18.5 RC circuits /charging/
At τ=RC:
τ: time constant
Loop rule:
When the capacitor is fully
charged, the current, I, will
become zero.
This is a differential equation, the solution
is (to find Q for a charging capacitor):
18.5 RC circuits /charging/
At τ=RC:
τ: time constant
Loop rule:
To increase the time constant, either increase
the resistance, R, or the capacitance, C, of the
circuit.
When the capacitor is fully
charged, the current, I, will
become zero.
In a circuit with a large time constant, the
capacitor charges very slowly. The capacitor
charges very quickly if there is a small time
constant.
18.5 RC circuits /discharging/
At τ=RC:
Loop rule:
18.5 RC circuits
Example
The circuit has a capacitance of C = 0.3µF, a
total resistance of 20kΩ and a battery “emf” of
12V. What is the maximum charge it can
acquire? What time does it take to reach 99%
of this value?
Loop rule:
Fully charged capacitor: