Transcript PowerPoint

Physics 212
Lecture 18
Physics 212 Lecture 18, Slide 1
From the prelecture: Self Inductance
Wrap a wire into a coil to make an “inductor”…
e = -L
dI
dt
Physics 212 Lecture 18, Slide 2
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 3
I(t)
e = -L
dI
dt
Suppose dI/dt > 0. Induced EMF tries to counteract
this change (Lenz’s Law).
e
+
I(t)
-
V1
V2
+
-
Voltage across inductor V1 – V2 = VL = + L dI/dt > 0
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 5
WHAT ARE INDUCTORS AND CAPACITORS GOOD FOR?
Inside your i-clicker
Physics 212 Lecture 18, Slide 6
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 7
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 8
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 9
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 10
Why is there exponential behavior ?
I
2
V1 – V2 = L
dI
dt
L
1
VL
3
R
t = L/R
V3-V4 = IR
4
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 11
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 12
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 13
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 14
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 15
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 16
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 17
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 18
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 19
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 20
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 21
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 22