Transcript PowerPoint

Physics 212
Lecture 18
Physics 212 Lecture 18, Slide 1
Music
Who is the Artist?
A)
B)
C)
D)
E)
Arturo Sandoval
Tiempo Libre
Chucho Valdes & Afro-Cuban Messengers
Freddie Omar con su banda
Los Hombres Calientes
Legendary Cuban piano player
Hadn’t been in the U.S. in many years
Stop at Krannert in Fa10
Your Comments
“glad this all makes sense before spring break...haha just kidding, im completely lost
here.”
“so if no current goes across the inductor to start, how does it have an emf? Also, how
does an emf have a direction? Flux is a dot product with the B field and and area, and a
change in a scalar (flux) over a change in another scalar (time) should give a scalar back
right? am I missing something here or just over-thinking it?”
“The final checkpoint was confusing. Please go
over how to determine current direction at
different times of the switch being open or
closed as well.”
“I get most of this stuff, but I’m confused on
the last checkpoint. Please go over it!”
Difficult checkpoints today!
“Bringing back the circuits with switches makes me nervous.”
“If you make this appear extremely intuitive in
class, I will love you forever.”
We’ll work on the qualitative
picture of inductors in circuits
“Explaining the intuition behind the solenoids and inductors would help a bunch for this
lecture!”
“I'm going to need some inDUCTance tape to keep all this info from falling out of my
brain.”
05
Physics 212 Lecture 18, Slide 3
Some Exam Stuff
• Exam THURSDAY (Mar. 29) at 7:00
– Covers material in Lectures 9 – 18 (through today’s lecture)
– Bring your ID: Rooms determined by discussion section (see link)
– Conflict exam at 5:15 – sign up in your gradebook before Fri. night
(Mar. 16, 10:00 p.m.)
– If you have conflicts with both of these, contact Prof. Park
Physics 212 Lecture 18, Slide 4
From the prelecture: Self Inductance
Wrap a wire into a coil to make an “inductor”…
e = -L
dI
dt
Physics 212 Lecture 18, Slide 6
What this really means:
emf induced across L tries to keep I constant
eL = -L
dI
dt
L
current I
Inductors prevent discontinuous current changes !
It’s like inertia!
Physics 212 Lecture 18, Slide 7
Checkpoint 1
Two solenoids are made with the same cross sectional area and
total number of turns. Inductor B is twice as long as inductor A
Compare the inductance of the two solenoids
A) LA = 4 LB
B) LA = 2 LB
C) LA = LB
D) LA = (1/2) LB
E) LA = (1/4) LB
“n halved and z doubled. Using the equation, L in case B is
half of the L in case A.”
“Both have same radius and both have 8 loops.”
“B is twice as long so is twice as large ”
Physics 212 Lecture 18, Slide 8
Checkpoint 1
Two solenoids are made with the same cross sectional area and
total number of turns. Inductor B is twice as long as inductor A
LB   0 n 2r 2 z
(1/2)2
2
LB  1 L A
2
Compare the inductance of the two solenoids
A) LA = 4 LB
B) LA = 2 LB
C) LA = LB
D) LA = (1/2) LB
E) LA = (1/4) LB
Physics 212 Lecture 18, Slide 9
WHAT ARE INDUCTORS AND CAPACITORS GOOD FOR?
” Can you have capacitors and inductors in the same circuit?
“why inductors are important as opposed to capacitors. why use one
instead of the other?”
Inside your i-clicker
Physics 212 Lecture 18, Slide 10
How to think about RL circuits Episode 1:
When no current is flowing initially:
VL
I=0
L
I=V/R
R
L
t = L/R
R
I
VBATT
At t = 0:
I=0
VL = VBATT
VR = 0
(L is like a giant resistor)
VBATT
t = L/R
At t >> L/R:
VL = 0
VR = VBATT
I = VBATT/R
(L is like a short circuit)
Physics 212 Lecture 18, Slide 11
Checkpoint 2a
In the circuit, the switch has
been open for a long time, and
the current is zero everywhere.
At time t=0 the switch is closed.
What is the current I through the
vertical resistor immediately after
the switch is closed?
(+ is in the direction of the arrow)
A) I = V/R
B) I = V/2R
C) I = 0
D) I = -V/2R
E) I = -V/R
“V_L == V, so applying KVL to the right loop, I == V/R.”
“At t=0, no current flows through the inductor. The two
resistors are in series. Hence, the overall resistance =
2R. Therefore, I = V/ 2R.”
“the current runs through the solenoid, not the resistor.”
Physics 212 Lecture 18, Slide 12
Checkpoint 2a
In the circuit, the switch has
been open for a long time, and
the current is zero everywhere.
I
At time t=0 the switch is closed.
What is the current I through the
vertical resistor immediately after
the switch is closed?
I
IL=0
(+ is in the direction of the arrow)
A) I = V/R
B) I = V/2R
C) I = 0
D) I = -V/2R
E) I = -V/R
Before: IL = 0
After: IL = 0
I = + V/2R
Physics 212 Lecture 18, Slide 13
RL Circuit (Long Time)
What is the current I through the vertical resistor after the
switch has been closed for a long time?
(+ is in the direction of the arrow)
A) I = V/R
B) I = V/2R
C) I = 0
D) I = -V/2R
E) I = -V/R
After a long time in any static circuit: VL = 0
-
+
+
-
KVR:
VL + IR = 0
Physics 212 Lecture 18, Slide 14
VBATT
How to think about RL circuits Episode 2:
When steady current is flowing initially:
VL
I=0
R
L
R
L
R
I=V/R
At t = 0:
I = VBATT/R
VR = IR
VL = VR
t = L/R
t = L/R
At t >> L/R:
I=0
VL = 0
VR = 0
Physics 212 Lecture 18, Slide 15
Checkpoint 2b
After a long time, the switch is
opened, abruptly disconnecting
the battery from the circuit.
What is the current I through the
vertical resistor immediately after
the switch is opened?
(+ is in the direction of the arrow)
A) I = V/R
B) I = V/2R
C) I = 0
D) I = -V/2R
E) I = -V/R
“After the battery is disconnected, there will be a
current of V/R because the inductor will have the initial
voltage of the battery. ”
“It dissapates from the inductor through the resistor”
“After long time V across solenoid is zero. Zero voltage
 zero current”
Physics 212 Lecture 18, Slide 16
Checkpoint 2b
After a long time, the switch is
opened, abruptly disconnecting
the battery from the circuit.
What is the current I through the
vertical resistor immediately after
the switch is opened?
(+ is in the direction of the arrow)
A) I = V/R
B) I = V/2R
C) I = 0
D) I = -V/2R
E) I = -V/R
circuit when
switch opened
L
IL=V/R
R
Current through inductor
cannot change
DISCONTINUOUSLY
Physics 212 Lecture 18, Slide 17
Why is there exponential behavior ?
I
-
V=L
dI
dt
L
+
VL
+
R
t = L/R
V = IR
t = L/R
dI
L  IR  0
dt
I (t )  I 0e -tR / L  I 0e -t / t
L
where t 
R
Physics 212 Lecture 18, Slide 18
I
L
VL
R
VBATT
t = L/R
Lecture:
Prelecture:
Did we mess up??
No: The resistance is simply twice as big in one case.
Physics 212 Lecture 18, Slide 19
Checkpoint 3a
After long time at 0, moved to 1
After switch moved, which case
has larger time constant?
A) Case 1
B) Case 2
C) The same
After long time at 0, moved to 2
“Time constant is L/R, so the lower the
resistance, the larger the time constant. ”
“more resistors to run into in case 2”
“The time it takes to get to switch one should
be the same to get to switch #2”
Physics 212 Lecture 18, Slide 20
Checkpoint 3a
After long time at 0, moved to 1
After long time at 0, moved to 2
After switch moved, which case
has larger time constant?
A) Case 1
B) Case 2
C) The same
L
t1 
2R
L
t2 
3R
Physics 212 Lecture 18, Slide 21
Checkpoint 3b
After long time at 0, moved to 1
Immediately after switch moved,
in which case is the voltage
across the inductor larger?
A) Case 1
B) Case 2
C) The same
After long time at 0, moved to 2
“It has less resistance to decrease the initial
voltage from the battery.”
“More resistance v=IR. R larger so V is larger.”
“full battery voltage appears across the
inductors”
Physics 212 Lecture 18, Slide 22
Checkpoint 3b
After long time at 0, moved to 1
After long time at 0, moved to 2
Immediately after switch moved,
in which case is the voltage
across the inductor larger?
A) Case 1
After switch moved:
B) Case 2
V
V

2R
L1
C) The same
Before switch moved: I 
V
R
VL 2
R
V
 3R
R
Physics 212 Lecture 18, Slide 23
Checkpoint 3c
After long time at 0, moved to 1
After long time at 0, moved to 2
After switch moved for finite time, “Greater time constant means that the current
in which case is the current will decay more slowly in case 1.”
through the inductor larger?
A) Case 1
“The resistance allows the curent to last for
B) Case 2
longer.”
C) The same
“The currents are 0. ”
Physics 212 Lecture 18, Slide 24
Checkpoint 3c
After long time at 0, moved to 1
After long time at 0, moved to 2
After switch moved for finite time,
in which case is the current
through the inductor larger?
A) Case 1
After awhile
B) Case 2
I1  Ie-t / t
C) The same
1
Immediately after: I1  I 2
I 2  Ie-t / t
t1  t 2
2
Physics 212 Lecture 18, Slide 25
Calculation
The switch in the circuit shown has
been open for a long time. At t = 0,
the switch is closed.
What is dIL/dt, the time rate of
change of the current through the
inductor immediately after switch is
closed
R1
V
R2
L
R3
• Conceptual Analysis
–
–
Once switch is closed, currents will flow through this 2-loop circuit.
KVR and KCR can be used to determine currents as a function of time.
• Strategic Analysis
–
–
–
Determine currents immediately after switch is closed.
Determine voltage across inductor immediately after switch is closed.
Determine dIL/dt immediately after switch is closed.
Physics 212 Lecture 18, Slide 26
Calculation
The switch in the circuit shown has
been open for a long time. At t = 0,
the switch is closed.
R1
V
R2
L
IL = 0
R3
What is IL, the current in the inductor, immediately after the switch
is closed?
(A) IL =V/R1 up
(B) IL =V/R1 down
(C) IL = 0
INDUCTORS: Current cannot change discontinuously !
Current through inductor immediately AFTER switch is closed
IS THE SAME AS
the current through inductor immediately BEFORE switch is closed
Immediately before switch is closed: IL = 0 since no battery in loop
Physics 212 Lecture 18, Slide 27
Calculation
The switch in the circuit shown has
been open for a long time. At t = 0,
the switch is closed.
R1
V
R2
L
R3
IL(t=0+) = 0
What is the magnitude of I2, the current in R2, immediately after the
switch is closed?
(A)
I2 
V
R1
(B)
I2 
V
R2  R3
(C) I 2 
V
R1  R2  R3
(D) I 2  VR2 R3
R2  R3
We know IL = 0 immediately after switch is closed
R1
Immediately after switch is closed, V
circuit looks like:
I
I
R2
V
R1  R2  R3
R3
Physics 212 Lecture 18, Slide 28
Calculation
The switch in the circuit shown has
been open for a long time. At t = 0,
the switch is closed.
R1
V
R2
L
IL(t=0+) = 0
I2
R3
I2(t=0+) = V/(R1+R2+R3)
What is the magnitude of VL, the voltage across the inductor,
immediately after the switch is closed?
(A) VL  V
R2  R3
R1
(B) VL  V
(C) VL  0
(D) VL  V
R2 R3
R2  R3
(E) VL  V
R1 ( R2  R3 )
R1  R2  R3
Kirchhoff’s Voltage Law, VL-I2 R2 -I2 R3 =0
VL = I2 (R2+R3)
VL 
V
 R2  R3 
R1  R2  R3
Physics 212 Lecture 18, Slide 29
Calculation
The switch in the circuit shown has
been open for a long time. At t = 0,
the switch is closed.
What is dIL/dt, the time rate of
change of the current through the
inductor immediately after switch is
closed
dI
V R2  R3
dI
(A) L 
(B) L  0
dt
L R1
dt
R1
R2
V
L
R3
VL(t=0+) = V(R2+R3)/(R1+R2+R3)
dI
V R2  R3
(C) L 
dt
L R1  R2  R3
dI L V
(D)

dt
L
The time rate of change of current through the inductor (dIL /dt) = VL /L
dI L V R2  R3

dt L R1  R2  R3
Physics 212 Lecture 18, Slide 30
Follow Up
The switch in the circuit shown has
been closed for a long time.
What is I2, the current through R2 ?
(Positive values indicate current flows
to the right)
(A) I 2  
V
R2  R3
(B) I 2  
V ( R2 R3 )
R1  R2  R3
R1
V
R2
L
(C) I 2  0
R3
(D) I 2  -
V
R2  R3
After a long time, dI/dt = 0
Therefore, the voltage across L = 0
Therefore the voltage across R2 + R3 = 0
Therefore the current through R2 + R3 must be zero !!
Physics 212 Lecture 18, Slide 31
Follow Up 2
The switch in the circuit shown has
been closed for a long time at which
point, the switch is opened.
What is I2, the current through R2
immediately after switch is opened ?
(Positive values indicate current flows
to the right)
(A) I 2  
V
R1  R2  R3
(B) I 2  
V
R1
R1
V
(C) I 2  0
R2
I2
IL
L
(D) I 2  -
R3
V
R1
(E) I 2  -
V
R1  R2  R3
Current through inductor immediately AFTER switch is opened
IS THE SAME AS
the current through inductor immediately BEFORE switch is opened
Immediately BEFORE switch is opened: IL = V/R1
Immediately AFTER switch is opened: IL flows in right loop
Therefore, IL = -V/R1
Physics 212 Lecture 18, Slide 32