Transcript PPT - LSU Physics & Astronomy

Physics 2113
Jonathan Dowling
Lecture 41: WED 02 DEC
Final Exam Review I
The Grading
•
•
•
•
Midterms Final Exam
Homework
In Class PP
100 points each
- 200 points
- 75 points
- 25 points
TOTAL: 600 points
• Numerical Grade: Total Points / 6
A+:
A:
A-:
98-100
92-97
89-91
B+:
B:
B-:
86-88
78-85
75-77
C+:
C:
C-:
72-74
63-71
60-62
D+:
D:
D-:
57-59
53-56
50-52
F:
<50
Final Exam
• 3:00PM – 5:00PM MON 07DEC
PHYS 2113-1 (Moreno) is scheduled to take their exam in Lockett 9
PHYS 2113-2 (Gaarde) is scheduled to take their exam in Lockett 10
PHYS 2113-3 (Dowling) is scheduled to take their exam in Lockett 6
PHYS 2113-4 (O'Connell) is scheduled to take their exam in Lockett 15
PHYS 2113-5 and 6 (Abdelwahab) is scheduled to take their exam in
Lockett 2
PHYS 2113-7 (Hansen) Last name starts with A - K: take their exam in
Lockett 10, last name starts with L - Z: take their exam in Lockett 2
Final Exam
• 100 PTS: CH 13, 21–30 / HW01-11
This part will be 11 multiple choice
questions one from each chapter.
• 100 PTS: CH 31–33 / HW12-14
This part will be three multiple choice
questions and three word problems, one
each from each chapter.
What do you need to make on the final to get an A, B, C, etc.?
A: 90–100
B: 75–89
C: 60–74
D: 50–59
F: <50
Solve this simple equation for x:
é mt1+ mt2 + mt 3 + (hw / 472) ´ 75
ù
ê + max[(icppc – icppx) / icpp, 0] ´ 25 + x ú / 6 = y
)
ë (
û
Where mt1=exam1, mt2=exam2, mt3=exam3, hw=total points
on your hws01–14 (out of 472), icppc=checks, icppx=X’s,
icpp=number of times you were called on, max is the binary
maximum function, and y is your desired cutoff number, y = 90,
75, 60, or 50. Then x is the score out of 200 you need on the
final to make that cutoff grade y. This assumes no curve.
Example: John D’oh wants to know what he needs to make on
the final in order to get an A- = 89 in this class.
mt1 = 81
mt2 = 70
mt 3 = 90
hw = 435
icppc = 3
icppx = 0
icpp = 3
max[(icppc - icppx) / icpp, 0] = max[1,0] = 1
{( 81+70+90 ) +[ (435/450) ´ 75 ] +(1´ 25 ) + x} = 89
Solve: x = 201.5 / 200 > 100%
It is very likely impossible for John to get an A- as he’d need better than a
perfect score on the final. How good does he need to do to avoid a C+ =
74?
{( 81+70+90 ) +[(435/450) ´ 75 ] +(1´ 25 ) + x} = 75
Solve: x = 111.5 / 200 = 56%
John is extremely unlikely to get an A, and is unlikely to get a C, so the
most probable outcome is that John will get a B in this class.
LC Circuits
PHYS2113 An Electromagnetic LC Oscillator
Capacitor initially charged. Initially, current is zero,
energy is all stored in the E-field of the capacitor.
Energy Conservation: Utot = U B +UE
A current gets going, energy gets split between the
capacitor and the inductor.
1 2
1 q2
U B = L i U E =
2
2C
Capacitor discharges completely, yet current keeps going.
Energy is all in the B-field of the inductor all fluxed up.
The magnetic field on the coil starts to deflux, which
will start to recharge the capacitor.
1 2 1 q2
Utot = L i +
2
2C
Finally, we reach the same state we started with (with
opposite polarity) and the cycle restarts.
1.5
1
0.5
Charge
0
Time
-0.5
Current
-1
-1.5
(a) T / 2
(b) T
(c) T / 2
(d) T / 4
t=0
Vc = q / C
t =T /4
t =T
t =T /2
t = 3T / 4
Electric Oscillators: the Math
q = q0 cos(w t + j 0 )
Amplitude = ?
i(t) = q¢(t) = -q0w sin(w t + j 0 )
i¢(t) = q¢¢(t) = -w q0 cos(w t + j 0 )
2
Energy as Function of Time
Voltage as Function of Time
1
1
2
2
U B = L [ i ] = L [ q0w sin(w t + j 0 )] VL = Li¢(t) = -Lw 2 q0 cos(w t + j 0 )
2
2
1 [ q]
1
2
UE =
=
q
cos(
w
t
+
j
)
[0
0 ]
2 C
2C
2
1
1
VC = [ q(t)] = [ q0 cos(w t + j 0 )]
C
C
Example
• In an LC circuit,
L = 40 mH; C = 4  F
• At t = 0, the current is a
maximum;
• When will the capacitor
be fully charged for the
first time?
1.5
1
0.5
Charge
0
Time
-0.5
Current
-1
-1.5
t=0
t =T /4
t =T /2
t = 3T / 4 t = T
w=
1
1
=
rad/s
LC
16x10 -8
•  = 2500 rad/s
• T = period of one
complete cycle
•T =  = 2.5 ms
• Capacitor will be
charged after T=1/4
cycle i.e at
• t = T/4 = 0.6 ms
Example
• In the circuit shown, the switch is in
position “a” for a long time. It is
then thrown to position “b.”
• Calculate the amplitude  q0 of the
resulting oscillating current.
1 mH
1 mF
b
E=10 V
a
i = -w q0 sin(w t + f0 )
• Switch in position “a”: q=CV = (1 mF)(10 V) = 10 mC
• Switch in position “b”: maximum charge on C = q0 = 10 mC
• So, amplitude of oscillating current =
1
w q0 =
(10mC) = 0.316 A
(1mH)(1mF)
Damped LCR Oscillator
Ideal LC circuit without resistance: oscillations
go on forever;  =
C
(LC)–1/2
L
R
Real circuit has resistance, dissipates energy:
oscillations die out, or are “damped”
Math is complicated! Important points:
– Frequency of oscillator shifts away from
1.0
 = (LC)-1/2
0.8
 QLCR=2L/R
– For small damping, peak ENERGY decays
with time constant
 ULCR= L/R
UE
– Peak CHARGE decays with time constant =
Umax
Q 2 - RtL
=
e
2C
0.6
0.4
0.2
0.0
0
4
8
12
time (s)
16
20
R2
1
≪
?
2
LC
4L
3
8
10 ≪ 10 !
w¢ =
1
R2
@
LC 4L2
1
=w
LC
Q(t)
1 2 3
•••
13
Q(0) / 2 = 50%
t(s)
IS
IP
VS
VP
=
NS NP
I S NS = I P NP
VS VP
We have that:
=
NS NP
® VS N P = VP N S
(eq. 1).
If we close switch S in the figure we have in addition to the primary current I P
a current I S in the secondary coil. We assume that the transformer is "ideal, "
i.e., it suffers no losses due to heating. Then we have: VP I P = VS I S
If we divide eq. 2 with eq. 1 we get:
IS =
(eq. 2).
VI
VP I P
= S S ® IP NP = IS NS .
VP N S
VS N P
NP
IP
NS
In a step-up transformer (N S > N P ) we have that I S < I P .
In a step-down transformer (N S < N P ) we have that I S > I P .
(31-27)
Example, Transformer:
Maxwell II: Gauss’ law for B-Fields:
field lines are closed
or, there are no magnetic monopoles
S
F = å BA = 0
d >b>c>a=0
FTa &B = 2 ´ 6 - 4 ´ 3 = 12 - 12 = 0
FTc &B = -2 ´ 6 + 2 ´ 8 = -12 +16 = +4
T &B
F Side
=
-F
=0
a
a
F Side
= -FTc &B = -4
c
F Side
=0
a
F Side
=4
c
T &B
FTb &B = -2 ´ 1- 4 ´ 2 = -2 - 8 = -10 F d = +2 ´ 3 + 3 ´ 2 = +6 + 6 = +12
F bSide = -FTb &B = 10
T &B
F Side
=
-F
= -12
d
d
F bSide = 10
F Side
= 12
d
Displacement “Current”
Maxwell proposed it based on
symmetry and math — no experiment!
B
B!
B
i
i
E
Changing E-field Gives Rise to B-Field!
32.3: Induced Magnetic Fields:
Here B is the magnetic field induced along a
closed loop by the changing electric flux FE
in the region encircled by that loop.
Fig. 32-5 (a) A circular parallel-plate capacitor, shown in side view, is being charged by
a constant current i. (b) A view from within the capacitor, looking toward the plate at
the right in (a).The electric field is uniform, is directed into the page (toward the plate),
and grows in magnitude as the charge on the capacitor increases. The magnetic field
induced by this changing electric field is shown at four points on a circle with a radius r
less than the plate radius R.
Ba > Bc > Bb > Bd = 0
dF E d
dE
= EA = A
µ slope
dt
dt
dt
32.5: Maxwell’s Equations: