Unformatted text preview: Physics 212
Lecture 19
LC and RLC Circuits
 Oscillation frequency
 Energy
Energy
 Damping Physics 212 Lecture 19, Slide 1 Music
Who is the Artist?
BB A)
B)
C)
D)
E) Jefferson Airplane
Grateful Dead
Moby Grape
The Doors
Cream Why?
One reason that will be revealed on Thursday
One
(theme of the week)
(theme
San Francisco Giants are in the World Series Physics 212 Lecture 19, Slide 2 Your Comments
“All of this stuff, I wish we could go slower but I know we cant.”
TOO TRUE: Hang in there, we’ll do our best to work on the issues…. “Differential equations have taken
Differential over my life. I'm not sure I approve.”
over
“How did yawl derive the differential
equations?”
“Why do capacitors start off with a
Why charge of zero when the switch is
opened? shouldn't they start off with
charge? “
charge?
It all depends on how the circuti was
It
circuti was
started. You have to determine the initial
conditions from the problem statement
conditions Differential Equations Do Determine
Differential
Much Behavior in Physics. We will
show corresponding equations in
mechanics today
mechanics 35
30
25
20
15
10
5
0 05 Confused Confident Physics 212 Lecture 19, Slide 3 Some Exam Stuff
• Exam Tomorrow Night ( Oct 27) at 7:00
–
–
–
– Covers material in Lectures 9 – 18 (through last Thursday’s lecture)
Bring your ID: Rooms determined by discussion section (see link)
Conflict exam at 5:15 – sign up in your gradebook before Mon (Oct 25)
If you have conflicts with both of these, contact Prof Beck 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
=−
LC
dt 2 d 2Q
dt 2 = −ω 2Q where ω= 1
LC Physics 212 Lecture 19, Slide 5 d 2Q
dt 2 2 dx
dt 2 = −ω Q
2 = −ω 2 x ω= 1
LC L k k
ω=
m C F = kx
a
m x Same thing if we notice that k↔ 1
C and m↔L Physics 212 Lecture 19, Slide 6 Time Dependence
I L ++
C  Physics 212 Lecture 19, Slide 7 Preflight 2
At time t = 0 the capacitor is
At
the
fully charged with Qmax and the
and
current through the circuit is 0. L BB C What is the potential difference across the inductor at t = 0 ?
What
A) VL = 0
B) VL = Qmax/C
B)
/C
C) VL = Qmax/2C
C)
/2C since VL = VC 60
50
40 The voltage across the capacitor is Qmax/C
Kirchhoff's Voltage Rule implies that must also
be equal to the voltage across the inductor 30
20
10
0 Pendulum… Physics 212 Lecture 19, Slide 8 Preflight 4
At time t = 0 the capacitor is
At
the
fully charged with Qmax and the
and
current through the circuit is 0. BB C L What is the potential difference across the
What
inductor when the current is maximum ?
inductor
50 A) VL = 0
B) VL = Qmax/C
B)
/C
C) VL = Qmax/2C
C)
/2C 40
30
20
10 dI/dt is zero when current is maximum 0 Physics 212 Lecture 19, Slide 9 Preflight 6
At time t = 0 the capacitor is
At
the
fully charged with Qmax and the
and
current through the circuit is 0. BB C L How much energy is stored in the capacitor
How
when the current is a maximum ?
when
A) U = Qmax2/(2C)
/(2C)
B) U = Qmax2/(4C)
B)
/(4C)
C) U = 0
C)
Total Energy is constant !
ULmax = ½ LImax2
UCmax = Qmax2/2C
I = max when Q = 0 50
40
30
20
10
0 Physics 212 Lecture 19, Slide 10
10 Preflight 8
BB The capacitor is charged such
The
that the top plate has a charge
+Q0 and the bottom plate Q0.
+Q and
At time t=0, the switch is closed
t=0 the
and the circuit oscillates with
frequency ω = 500 radians/s.
frequency L ++
C  L = 4 x 103 H
ω = 500 rad/s
rad/s What is the value of the capacitor C?
50 A) C = 1 x 103 F
B) C = 2 x 103 F
B)
C) C = 4 x 103 F
C)
ω= 1
LC C= 40
30 1 ωL
2 = 1
4 (25 ×10 )(4 ×10 −3 ) = 10−3 20
10
0 Physics 212 Lecture 19, Slide 11
11 Preflight 10 closed at t=0 L C +Q0
BB Q0 Which plot best represents
Which
the energy in the inductor as a
function of time starting just
after the switch is closed? 12
U L = LI
2 35
30 Energy proportional to I ﬂ C cannot be negative
Energy
cannot
2 Current is changing ﬂ UL is not constant
Current 25
20
15
10
5 Initial current is zero 0 Physics 212 Lecture 19, Slide 12
12 Preflight 12
When the energy stored in
When
the capacitor reaches its
maximum again for the first
time after t=0, how much
time
how
charge is stored on the top
plate of the capacitor?
A)
B)
C)
D)
E) BB closed at t=0 L +Q0 C +Q0
+Q0 /2
0
Q0/2
Q0 Q0 30
25
20 Q is maximum when current goes to zero
I= dQ
dt Current goes to zero twice during one cycle 15
10
5
0 Physics 212 Lecture 19, Slide 13
13 Add R: Damping
Just like LC circuit but energy but the oscillations get smaller because of R Concept makes sense… …but answer looks kind of complicated
Physics 212 Lecture 19, Slide 14
14 Physics Truth #1:
Even though the answer sometimes looks complicated…
Q(t ) = Qo cos(ωt − φ ) …the physics under the hood is still very simple !!
d 2Q
dt 2 = −ω 2Q Physics 212 Lecture 19, Slide 15
15 The elements of a circuit are very simple:
dI
VL = L
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 16
16 A Different Approach Start with some initial V, I, Q, VL
Now take a tiny time step dt
dI = (1 ms) VL
dt
L dQ = Idt VC = Q
C Repeat… VR = IR
VL = V − VR − VC Physics 212 Lecture 19, Slide 17
17 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 ω0 • Strategic Analysis
–
–
– Determine initial current
Determine oscillation frequency ω0
Find maximum charge on capacitor Physics 212 Lecture 19, Slide 18
18 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
(B) (C) IL > 0
(C) Current through inductor immediately AFTER switch is opened
Current
AFTER switch
IS THE SAME AS
IS
the current through inductor immediately BEFORE switch is opened
the
BEFORE switch
BEFORE switch is opened:
all current goes through inductor in direction shown
Physics 212 Lecture 19, Slide 19
19 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
(B)
BEFORE switch is opened: AFTER switch is opened:
AFTER dIL/dt ~ 0 ⇒ VL = 0 VC cannot change abruptly
cannot
VC = 0 BUT: VL = VC
since they are in parallel
since
VC = 0 BB UC = ½ CVC2 = 0 !! IMPORTANT: NOTE DIFFERENT CONSTRAINTS AFTER SWITCH OPENED
CURRENT through INDUCTOR cannot change abruptly
through INDUCTOR
VOLTAGE across CAPACITOR cannot change abruptly
across CAPACITOR Physics 212 Lecture 19, Slide 20
20 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
(B) Current through inductor immediately AFTER switch is opened
Current
AFTER switch
IS THE SAME AS
IS
the current through inductor immediately BEFORE switch is opened
the
BEFORE switch BB BEFORE switch is opened: Current moves down through L
BEFORE switch
down
AFTER switch is opened: Current continues to move down through L
Physics 212 Lecture 19, Slide 21
21 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
R2 L
C (D) I o = V (C) I o = V 2R R Current through inductor immediately AFTER switch is opened
Current
AFTER switch
IS THE SAME AS
IS
the current through inductor immediately BEFORE switch is opened
the
BEFORE switch BEFORE switch is opened: IL V
IL R VL = 0 IL
L BB VL=0 C
V = ILR
Physics 212 Lecture 19, Slide 22
22 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?
V
(A) Qmax =
LC
R Imax
L 1
(B) Qmax = CV
2 Qmax
C L C (C) Qmax = CV (D) Qmax = V
R LC 2
1 2 1 Qmax
LI =
2
2C BB When I is max
When
(and Q is 0)
1
U = LI 2
2 When Q is max
When
(and I is 0) Qmax = I max LC = V
LC
R 2
1 Qmax
U=
2C
Physics 212 Lecture 19, Slide 23
23 FollowUp 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 VL
RC C L
R Imax =V/R
max =V/R (B) NO
(B) Vmax = IL Qmax = Vmax can be greater than V IF:
can
IF V
LC
R L
>R
C We can rewrite this condition in terms of the resonant frequency: ω0 L > R OR 1
>R
ω0C We will see these forms again when we study AC circuits!! BB Physics 212 Lecture 19, Slide 24
24 ...
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 Spring '11
 MESTRE
 Energy, iL, switch, Qmax

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