Physics 212
Lecture 19
LC and RLC Circuits
- Oscillation frequency
- Energy
- Damping
Physics 212 Lecture 19, Slide 1
Music
Who is the Artist?
A)
B)
C)
D)
E)
Stephane Grappelli
Pearl Django
Mark O’Connor’s Hot Swing Trio
Miles Davis
Cassandra Wilson
Some Exam Stuff
• Exam Thu. night (Mar. 29) at 7:00
– Covers material in Lectures 9 – 18
– Bring your ID: Rooms determined by discussion section (see link)
– Conflict exam at 5:15
• Final Exam
– “Regular”: Fri. May 4, 1:30; “Conflict”: Fri. May 11, 8:00
– Sign up for the conflict in the gradebook if you wish
Physics 212 Lecture 18, Slide 3
Your Comments
“I am really confused with this stuff. The PreLecture tries to tell
us too much too fast. I have lost my intuition! :(”
“The concepts today were quite intuitive. Unfortunately, there
were a lot of very scary looking equations on the slides that are
not on our formula sheet. Are we expected to memorize these
equations and/or perform calculations with them?”
The LC solutions are easy to derive
starting from what you know, NOT RLC
“Waves (graphs) and how the signs of charge swap.”
“There are so many different combinations of circuit
components now, and I’m starting to find it increasing
difficult to keep straight how each circuit behaves. Please
go over the check points as well; I got stuck on a few of
them!!! ”
WILL DO !
“Okay I get it. But one thing...does omega equal d/dt? That’s not a term
though...it’s just notation...I don’t get that. Also, why don’t we just use
superconducting LC circuits to supply electricity???”
“If this comment makes it onto the board today, no more puppies will be killed
by horrible puns.”
Physics 212 Lecture 19, Slide 4
LC Circuit
-
I
L
dI
VL  L
dt
+
C
+
Circuit Equation:
dQ
I
dt
Q
VC 
Q
C
-
Q
dI
L 0
C
dt
d 2Q
Q

2
LC
dt
d 2Q
dt 2
  2Q
where

1
LC
Physics 212 Lecture 19, Slide 5
Checkpoint 1a
At time t = 0 the capacitor is
fully charged with Qmax and the
current through the circuit is 0.
L
C
What is the potential difference across the inductor at t = 0 ?
A) VL = 0
B) VL = Qmax/C
C) VL = Qmax/2C
“The current going through the inductor is going to be 0 so by ohm’s law there is no voltage
difference over the inductor.”
“no current, simple potential equation Q=cv ”
“V = (1/2) (Q/C)”
Physics 212 Lecture 19, Slide 6
Checkpoint 1a
At time t = 0 the capacitor is
fully charged with Qmax and the
current through the circuit is 0.
L
C
What is the potential difference across the inductor at t = 0 ?
A) VL = 0
B) VL = Qmax/C
C) VL = Qmax/2C
since VL = VC
The voltage across the capacitor is Qmax/C
Kirchhoff's Voltage Rule implies that must also
be equal to the voltage across the inductor
Physics 212 Lecture 19, Slide 7
d 2Q
dt
2
2
d x
dt 2
  Q
2
  2 x

1
LC
L
k
k

m
C
F = -kx
a
m
x
Same thing if we notice that
Pendulum…
1
k
C
and
mL
Physics 212 Lecture 19, Slide 8
Time Dependence
I
L
++
C --
Physics 212 Lecture 19, Slide 9
Checkpoint 1b
At time t = 0 the capacitor is
fully charged with Qmax and the
current through the circuit is 0.
L
C
What is the potential difference across the
inductor when the current is maximum ?
A) VL = 0
B) VL = Qmax/C
C) VL = Qmax/2C
“The change in current becomes 0 when the current is maximum.”
“This is the maximum voltage across that capacitor.”
“Since the voltage is dependent on the amount of charge that is left in the capacitor it is Q/2C”
Physics 212 Lecture 19, Slide 10
Checkpoint 1b
At time t = 0 the capacitor is
fully charged with Qmax and the
current through the circuit is 0.
L
C
What is the potential difference across the
inductor when the current is maximum ?
A) VL = 0
B) VL = Qmax/C
C) VL = Qmax/2C
dI/dt is zero when current is maximum
Physics 212 Lecture 19, Slide 11
Checkpoint 1c
At time t = 0 the capacitor is
fully charged with Qmax and the
current through the circuit is 0.
L
C
How much energy is stored in the capacitor
when the current is a maximum ?
A) U = Qmax2/(2C)
B) U = Qmax2/(4C)
C) U = 0
“All the energy is in the capacitor”
“current is maximum when energy is evenly divided between the capacitor and inductor”
“All the energy would be stored in the inductor when the current is maximum.”
Physics 212 Lecture 19, Slide 12
Checkpoint 1c
At time t = 0 the capacitor is
fully charged with Qmax and the
current through the circuit is 0.
L
C
How much energy is stored in the capacitor
when the current is a maximum ?
A) U = Qmax2/(2C)
B) U = Qmax2/(4C)
C) U = 0
Total Energy is constant !
ULmax = ½ LImax2
UCmax = Qmax2/2C
I = max when Q = 0
Physics 212 Lecture 19, Slide 13
Checkpoint 2a
The capacitor is charged such
that the top plate has a charge
+Q0 and the bottom plate -Q0.
At time t=0, the switch is closed
and the circuit oscillates with
frequency  = 500 radians/s.
L
++
C --
L = 4 x 10-3 H
 = 500 rad/s
What is the value of the capacitor C?
A) C = 1 x 10-3 F
B) C = 2 x 10-3 F
C) C = 4 x 10-3 F
“w = 1 / sqrt(LC)”
“use q=vc formula”
“All the charge is built up in the capacitor and so it is 4x10^-3”
Physics 212 Lecture 19, Slide 14
Checkpoint 2a
The capacitor is charged such
that the top plate has a charge
+Q0 and the bottom plate -Q0.
At time t=0, the switch is closed
and the circuit oscillates with
frequency  = 500 radians/s.
L
++
C --
L = 4 x 10-3 H
 = 500 rad/s
What is the value of the capacitor C?
A) C = 1 x 10-3 F
B) C = 2 x 10-3 F
C) C = 4 x 10-3 F

1
LC
C
1
 2L

1
( 25  10 4 )( 4  10  3 )
 10  3
Physics 212 Lecture 19, Slide 15
Checkpoint 2b
closed at t=0
L
C
+Q0
-Q0
Which plot best represents
the energy in the inductor as a
function of time starting just
after the switch is closed?
“Since the current is initially 0, the energy begins at zero and oscillates, allowing the
energy to be negative.”
“oscillates between positive and negative, does not start at zero”
“It starts at 0 and cannot be negative.”
Physics 212 Lecture 19, Slide 16
Checkpoint 2b
closed at t=0
L
C
+Q0
-Q0
Which plot best represents
the energy in the inductor as a
function of time starting just
after the switch is closed?
1 2
U L  LI
2
Energy proportional to I2  U cannot be negative
Current is changing  UL is not constant
Initial current is zero
Physics 212 Lecture 19, Slide 17
Checkpoint 2c
When the energy stored in
the capacitor reaches its
maximum again for the first
time after t=0, how much
charge is stored on the top
plate of the capacitor?
A)
B)
C)
D)
E)
closed at t=0
L
C
+Q0
-Q0
+Q0
+Q0 /2
0
-Q0/2
-Q0
“V is the same as before, so Q is the same”
“only 1/2 the charge is needed”
“Current direction swaps around so the signs on the capacitor plates changes as well.”
Physics 212 Lecture 19, Slide 18
Checkpoint 2c
When the energy stored in
the capacitor reaches its
maximum again for the first
time after t=0, how much
charge is stored on the top
plate of the capacitor?
A)
B)
C)
D)
E)
closed at t=0
L
C
+Q0
-Q0
+Q0
+Q0 /2
0
-Q0/2
-Q0
Q is maximum when current goes to zero
I
dQ
dt
Current goes to zero twice during one cycle
Physics 212 Lecture 19, Slide 19
Physics Truth #N:
Even though the answer sometimes looks complicated…
Q(t )  Qo cos(t   )
…the physics under the hood is still very simple !!
d 2Q
dt 2
  2Q
Physics 212 Lecture 19, Slide 20
Add R: Damping
Just like LC circuit but energy and the oscillations get smaller because of R
Concept makes sense…
…but answer looks kind of complicated
Physics 212 Lecture 19, Slide 21
The elements of a circuit are very simple:
VL  L
dI
dt
VC 
V  VL  VC  VR
I
Q
C
dQ
dt
VR  IR
This is all we need to know to solve for anything !
Physics 212 Lecture 19, Slide 22
A Different Approach
Start with some initial V, I, Q, VL
Now take a tiny time step dt
(1 ms)
VL
dI  dt
L
dQ  Idt
Q
VC 
C
Repeat…
VR  IR
VL  V  VR  VC
Physics 212 Lecture 19, Slide 23
Calculation
The switch in the circuit shown
has been closed for a long time.
At t = 0, the switch is opened.
V
What is QMAX, the maximum
charge on the capacitor?
L
C
R
• Conceptual Analysis
–
–
Once switch is opened, we have an LC circuit
Current will oscillate with natural frequency 0
• Strategic Analysis
–
–
–
Determine initial current
Determine oscillation frequency 0
Find maximum charge on capacitor
Physics 212 Lecture 19, Slide 24
Calculation
The switch in the circuit shown
has been closed for a long time.
At t = 0, the switch is opened.
IL
V
L
C
R
What is IL, the current in the inductor, immediately AFTER the
switch is opened? Take positive direction as shown.
(A) IL < 0
(B) IL = 0
(C) IL > 0
Current through inductor immediately AFTER switch is opened
IS THE SAME AS
the current through inductor immediately BEFORE switch is opened
BEFORE switch is opened:
all current goes through inductor in direction shown
Physics 212 Lecture 19, Slide 25
Calculation
The switch in the circuit shown
has been closed for a long time.
At t = 0, the switch is opened.
IL
V
L
VC=0
C
R
IL(t=0+) > 0
The energy stored in the capacitor immediately after the switch is
opened is zero.
(A) TRUE
(B) FALSE
BEFORE switch is opened:
AFTER switch is opened:
dIL/dt ~ 0  VL = 0
VC cannot change abruptly
VC = 0
BUT: VL = VC
since they are in parallel
VC = 0
UC = ½ CVC2 = 0 !!
IMPORTANT: NOTE DIFFERENT CONSTRAINTS AFTER SWITCH OPENED
CURRENT through INDUCTOR cannot change abruptly
VOLTAGE across CAPACITOR cannot change abruptly
Physics 212 Lecture 19, Slide 26
Calculation
The switch in the circuit shown
has been closed for a long time.
At t = 0, the switch is opened.
IL
V
C
L
R
IL(t=0+) > 0
VC(t=0+) = 0
What is the direction of the current immediately after the switch is
opened?
(A) clockwise
(B) counterclockwise
Current through inductor immediately AFTER switch is opened
IS THE SAME AS
the current through inductor immediately BEFORE switch is opened
BEFORE switch is opened: Current moves down through L
AFTER switch is opened: Current continues to move down through L
Physics 212 Lecture 19, Slide 27
Calculation
The switch in the circuit shown
has been closed for a long time.
At t = 0, the switch is opened.
V
C
L
R
IL(t=0+) > 0
VC(t=0+) = 0
What is the magnitude of the current right after the switch is opened?
(A) I o  V C
L
(B) I o  V
2
R
(C) I o  V
L
C
(D) I o  V
2R
R
Current through inductor immediately AFTER switch is opened
IS THE SAME AS
the current through inductor immediately BEFORE switch is opened
BEFORE switch is opened:
IL
V
IL
L
IL
R
VL = 0
VL=0
C
V = ILR
Physics 212 Lecture 19, Slide 28
Calculation
The switch in the circuit shown
has been closed for a long time.
At t = 0, the switch is opened.
IL
V
C
L
R
Hint: Energy is conserved
IL(t=0+) =V/R
VC(t=0+) = 0
What is Qmax, the maximum charge on the capacitor during the oscillations?
(A) Qmax  V LC
R
Imax
C
L
L
1
(B) Qmax  CV
2
Qmax
C
When I is max
(and Q is 0)
When Q is max
(and I is 0)
1
U  LI 2
2
2
1 Qmax
U
2 C
(C) Qmax  CV
(D) Qmax 
V
R LC
2
1 2 1 Qmax
LI 
2
2 C
Qmax  I max LC 
V
LC
R
Physics 212 Lecture 19, Slide 29
Follow-Up 1
The switch in the circuit shown has
been closed for a long time. At t = 0,
the switch is opened.
V
Is it possible for the maximum
voltage on the capacitor to be greater
than V?
(A) YES
Qmax 
V
LC
R
Vmax
C
L
R
Imax =V/R
(B) NO
V L

R C
IL
Qmax 
Vmax can be greater than V IF:
V
LC
R
L
R
C
We can rewrite this condition in terms of the resonant frequency:
0 L  R OR
1
R
0C
We will see these forms again when we study AC circuits!!
Physics 212 Lecture 19, Slide 30