# Lecture 5

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```							Physics 212
Lecture 18

Physics 212 Lecture 18, Slide 1
Main Point 1

First, we defined the self-inductance L of a conducting loop to be the ratio of the
magnetic flux through the loop to the current that produced it. This quantity is
determined totally by the geometry of the loop. We used Faraday’s law to
determine that the emf induced in the loop was equal to minus the product of the
self-inductance with the time rate of change of the current.
Physics 212 Lecture 18, Slide 2
Main Point 2

Second, we determined that the behavior of a simple RL circuit was identical in
form to that of the corresponding RC circuit. In particular, we calculated the time
constant for the RL circuit to be equal to the inductance divided by the resistance.
Physics 212 Lecture 18, Slide 3
Main Point 3

Finally, we determined that the energy stored in an inductor is proportional to the
product of the self-inductance and the square of the current flowing through it.
This energy is stored in the magnetic field. We used a long solenoid to calculate the
energy density in this magnetic field and found it was proportional to the square of
the magnetic field, much like we found that the energy density in the electric field
was proportional to the square of the electric field.      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 5
Physics 212 Lecture 18, Slide 6
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

Physics 212 Lecture 18, Slide 7
Physics 212 Lecture 18, Slide 8
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
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

Physics 212 Lecture 18, Slide 10
Physics 212 Lecture 18, Slide 11
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

Physics 212 Lecture 18, Slide 12
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
B) Case 2
C) The same

Physics 212 Lecture 18, Slide 13
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
B) Case 2
C) The same

Physics 212 Lecture 18, Slide 15
Physics 212 Lecture 18, Slide 16
How to think about RL circuits Episode 1:
When no current is flowing initially:

VL
t = L/R
I=0                   I=V/R

L         R               L       R

I

VBATT                    VBATT                      t = L/R

At t >> L/R:
At t = 0:

Physics 212 Lecture 18, Slide 17
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                                                +                     -

KVR:
VL + IR =
After a long time in any static circuit: VL   =0             0
Physics 212 Lecture 18, Slide 18
Physics 212 Lecture 18, Slide 19
How to think about RL circuits Episode 2:
VBATT    When steady current is flowing initially:

VL        t = L/R
I=0

R           L       R             L         R

I=V/R                                               t = L/R

At t >> L/R:
At t = 0:
I = VBATT/R                 I=0
VR = IR                     VL = 0
V L = VR                    VR = 0

Physics 212 Lecture 18, Slide 20
Physics 212 Lecture 18, Slide 21
Calculation
The switch in the circuit shown has
been open for a long time. At t = 0,          R1             R2
the switch is closed.
V
What is dIL/dt, the time rate of                    L                   R3
change of the current through the
inductor immediately after switch is
closed
Conceptual Analysis

Strategic Analysis

Physics 212 Lecture 18, Slide 22
Physics 212 Lecture 18, Slide 23
Physics 212 Lecture 18, Slide 24
Physics 212 Lecture 18, Slide 25

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