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this might be a stupid question but how is it possible for an inductor to
have twice the voltage of a battery once it is disconnected from the
battery?! where did this extra voltage come from?
I have a few questions: 1) Is the induced emf from the inductor, induced on itself or
another wire going through it? 2) How is the voltage able to jump from 0 to some other
value after the circuit switch has been opened to disconnect the battery? 3) Finally, I
thought that the magnetic field could do no work, so how does it store energy, doesn't
that mean it is doing work on the charges in the inductor? I don't know if I worded that
right, but I hope you understand what I am asking.
Spring break 2013, more like one man one a lot of smart physics lectures and
studying for physics. AND the right hand rule.............
Your Comments
How are office hours going to be handled, what with
spring break and all?
Seeing as it's pi day today, I'm trying to think of a clever and witty
comment to make...but honestly all I can think about is how hungry I
am and what I would give to just be eating some pie right now.
Physics 212
Lecture 18
Today’s Concepts:
A) Induction
B) RL Circuits
Hour Exam 2 is October 31 st
Lectures 9- 18
current I
L
dI
dt
e L
= - L
emf induced across L tries to keep I constant.
Inductors prevent discontinuous current changes!
It’s like inertia!
What this really means:
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
CheckPoint 2
L n r z
B
B A
L L
2
1
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
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.
Before: IL = 0
I = + V/2R
I
I
After: IL = 0
IL = 0
CheckPoint 2a
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:
V L
RL Circuit (Long Time)
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 a long time, the switch is
opened, abruptly disconnecting the
battery from the circuit.
R
L
IL = V/R
circuit when switch
opened
Current through inductor
cannot change
DISCONTINUOUSLY
CheckPoint 2b
R
I
V = IR
dI
dt
V = L
L
Why is there Exponential Behavior?
VL
R
L =
R
L =
dt
dI
L
where R
L =
( )
tR L t I t I e I e
= =
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
R
L
2
1 =
R
L
3
2
=
CheckPoint 3a
A) Case 1
B) Case 2
C) The same
Before switch moved: R
V
I =
After switch moved:
R
R
V
V
L
1
R
R
V
V
L
2
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?
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.
R 1
L
V
R 2
R 3
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
What is IL, the current in the inductor, immediately after the switch is closed?
A) IL = V/R 1 up B) IL = V/R 1 down C) IL = 0
INDUCTORS: Current cannot change discontinuously!
Immediately before switch is closed: IL = 0 since no battery in loop
IL = 0
Calculation
The switch in the circuit shown has been
open for a long time. At t = 0 , the switch is
closed.
R 1
L
V
R 2
R 3
Current through inductor immediately after switch is closed
is the same as
the current through inductor immediately before switch is closed
IL(t = 0 +) = 0 I 2 (t^ =^0 +)^ =^ V/(R 1 +^ R 2 +^ R 3 )
Calculation
The switch in the circuit shown has been
open for a long time. At t = 0 , the switch is
closed.
R 1
L
V
R 2
R 3
I 2
A) B) C) D) E)
What is the magnitude of VL, the voltage across the inductor, immediately after the switch
is closed?
Kirchhoff’s Voltage Law,
VL - I 2 R 2 - I 2 R 3 = 0 VL = I 2 (R 2 + R 3 )
1
2 3
R
R R
VL =V VL =V VL =^0
1 (^23 )
2 3
R R R
R R VL V
= 1 2 3
2 3
R R R
R R VL V
=
1 2 3
R R
R R R
V
VL +
A) B) C) D)
VL(t = 0 +) = V(R 2 + R 3 )/(R 1 + R 2 + R 3 )
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
R 1
L
V
R 2
R 3
The time rate of change of current through the inductor (dIL /dt) = VL /L
1
2 3
R
R R
L
V
dt
d IL + = = 0
dt
d IL
1 2 3
2 3
R R R
R R
L
V
dt
d IL
L
V
dt
d IL
1 2 3
2 3
R R R
R R
L
V
dt
d IL
