Transcript Presentation_W12D2_answers

W12D2
RC, LR, and
Undriven RLC Circuits;
Experiment 4
Today’s Reading Course Notes: Sections 11.7-11.9, 11.10,
11.13.6; Expt. 4: Undriven RLC Circuits
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Announcements
Math Review Week 13 Tuesday 9pm-11 pm in 26-152
PS 9 due Week 13 Tuesday at 9 pm in boxes outside 32-082 or 26-152
Next Reading Assignment W12D3 Course Notes: Sections 11.8-9, 11.1211.13
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Outline
Experiment 4: Part 1 RC and LR Circuits
Simple Harmonic Oscillator
Undriven RLC Circuits
Experiment 4: Part 2 Undriven RLC Circuits
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RC Circuit Charging
(
dQ
1
=Q - Ce
dt
RC
)
Solution to this equation when
switch is closed at t = 0:
Q(t) = Ce (1- e )
I(t) = I0e-t / t
t = RC : time constant
-t / t
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RC Circuit: Discharging
dQ
1
=Q
dt
RC
Solution to this equation when
switch is closed at t = 0
Q(t) = Qoe-t / RC
I(t) = (Qo / RC)e-t / t
time constant:
t = RC
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RL Circuit: Increasing Current
ö
dI
Ræ
e
=- çI - ÷ Þ
dt
Lè
Rø
I(t) =
e
(1 - e-t / ( L / R) )
R
Solution to this equation when
switch is closed at t = 0:
I(t) =
e
R
(1- e-t / t )
t = L / R : time constant
(units: seconds)
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RL Circuit: Decreasing Current
dI
- L + - IR = 0 Þ
dt
I(t) = I 0 e-t / ( L / R)
Solution to this equation when
switch is opened at t = 0:
I(t) =
e
R
(1- e-t / t )
t = L / R : time constant
(units: seconds)
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Measuring Time Constant
Pick a point 1 with y1 (t1 ) = y0 e
Find point 2 such that
By definition
y2 (t2 ) = y2 (t1 + t ) = y0e
-(t1 / t )
y2 (t2 ) = y1 (t1 )e
-1
t º t2 - t1 then
-(t1 + t )/ t
= y0e
-(t1 / t ) -1
e = y1 (t1 )e
-1
2) In the lab you will plot semi-log and fit curve (make sure
you exclude data at both ends)
y(t) = y0e-(t / t ) Þ
ln( y(t) / y0 ) = ln e-(t / t ) = -(t / t ) Þ
ln y(t) = ln y0 - (t / t ) Þ t = -1/slope
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Experiment 4:
RC and RL Circuits
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Mass on a Spring:
Simple Harmonic Motion
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Demonstration
Mass on a Spring:
Simple Harmonic Motion
Mass on a Spring (C 2)
http://scripts.mit.edu/~tsg/www/demo.php?letnum=C%202&show=0
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Mass on a Spring
(1)
(2)
What is Motion?
d x
F = -kx = ma = m 2
dt
d 2x
m 2 + kx = 0
dt
2
(3)
(4)
Simple Harmonic Motion
x(t) = x0 cos(w 0t + f )
x0: Amplitude of Motion
w0 =
f: Phase (time offset)
k
= Angular frequency
m
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Simple Harmonic Motion
1
Period =
frequency
Amplitude (x0)
1
®T =
f
2p
2p
Period =
®T =
angular frequency
w
x(t) = x0 cos(w 0t + f )
-p
Phase Shift (f ) =
2
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Concept Question: Simple
Harmonic Oscillator
Which of the following functions x(t) has a second
derivative which is proportional to the negative of the
function
d 2x
dt
1 2
1. x(t) = at
2
2. x(t) = Aet /T
3. x(t) = Ae-t /T
æ 2p ö
4. x(t) = Acos ç t ÷
è T ø
2
µ -x ?
Concept Question Answer: Simple
Harmonic Oscillator
Answer 4. By direct calculation, when
æ 2p ö
x(t) = Acos ç t ÷
è T ø
æ 2p ö
æ 2p ö
dx(t)
= - ç ÷ Asin ç t ÷
dt
è T ø
è T ø
2
2
æ 2p ö
æ 2p ö
æ 2p ö
d x(t)
= - ç ÷ Acos ç
t ÷ = - ç ÷ x(t)
2
è T ø
è T ø
è T ø
dt
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Mass on a Spring: Energy
(1) Spring
(2) Mass
x(t) = x0 cos(w 0t + f )
(3) Spring
(4) Mass
dx
= vx (t) = -w 0 x0 sin(w 0t + f )
dt
Energy has 2 parts: (Mass) Kinetic and (Spring) Potential
2
1 æ dx ö
1 2 2
K = m ç ÷ = kx0 sin (w 0t + f )
2 è dt ø
2
1 2 1 2 2
U s = kx = kx0 cos (w 0t + f )
2
2
Energy
sloshes back
and forth
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LC Circuit
1. Set up the circuit above
with capacitor, inductor,
resistor, and battery.
2. Let the capacitor become
fully charged.
1. Throw the switch from a
to b.
1. What happens?
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LC Circuit
It undergoes simple harmonic motion, just like a
mass on a spring, with trade-off between charge on
capacitor (Spring) and current in inductor (Mass).
Equivalently: trade-off between energy stored in
electric field and energy stored in magnetic field.
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Energy stored
in electric field
Energy stored in
magnetic field
Energy stored
in electric field
Energy stored in
magnetic field
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Concept Question: LC Circuit
Consider the LC circuit at
right. At the time shown the
current has its maximum
value. At this time:
1. the charge on the capacitor has its maximum value.
2. the magnetic field is zero.
3. the electric field has its maximum value.
4. the charge on the capacitor is zero.
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Concept Q. Answer: LC Circuit
Answer: 4. The current is
maximum when the charge on
the capacitor is zero
Current and charge are exactly 90 degrees out of
phase in an ideal LC circuit (no resistance), so when
the current is maximum the charge must be
identically zero.
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LC Circuit: Simple Harmonic
Oscillator
Q
dI
dQ
-L =0 ; I =C
dt
dt
d 2Q 1
+
Q=0 Þ
2
LC
dt
Simple harmonic oscillator:
Q(t) = Q0 cos(w 0t + f )
Charge:
Angular frequency:
w 0 = 1 / LC
Amplitude of charge oscillation:
Q0
f
Phase (time offset):
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LC Oscillations: Energy
Notice relative phases
2
æ
ö
Q
Q
2
0
UE =
=ç
cos
w 0t
÷
2C è 2C ø
2
æ Q02 ö 2
1 2 1 2 2
U B = LI = LI0 sin w 0t = ç
sin w 0t
÷
2
2
è 2C ø
2
Q 2 1 2 Q0
U = UE + UB =
+ LI =
2C 2
2C
Total energy is conserved !!
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LC Circuit
Oscillation
Summary
Adding Damping:
RLC Circuits
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Demonstration
Undriven RLC Circuits (Y 190)
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RLC Circuit: Energy Changes
Include finite resistance:
Q
dI
+ I R+ L =0
C
dt
dQ
Multiply by I =
Þ
dt
Q dQ
dI
2
+ I R + LI
=0Þ
C dt
dt
d é Q2 1 2 ù
2
+
L
I
=
I
R
ê
ú
dt ë 2C 2
û
Decrease in stored
energy is equal to Joule
heating in resistor
d
(U E + U B )= - I 2 R
dt
Damped LC Oscillations
Resistor dissipates energy
and system rings down over
time. Also, frequency
decreases:
w 0 = 1 / LC > R / 2L
Q(t) = Q0e-( R/ 2 L)t cos(w 't)
w ' = w 02 - (R / 2L)2
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Experiment 4: Part 2
Undriven RLC Circuits
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Appendix: Experiment 4:
Part 2
Undriven RLC Circuits
Group Problem
and Concept Questions
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Problem: LC Circuit
Consider the circuit shown in the figure. Suppose the switch
that has been connected to point a for a long time is suddenly
thrown to b at t = 0. Find the following quantities:
(a) the frequency of oscillation of the circuit.
(b) the maximum charge that appears on the capacitor.
(c) the maximum current in the inductor.
(d) the total energy the circuit possesses as a function of time t.
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Concept Question: Expt. 4
In today’s lab the battery turns on
and off. Which circuit diagram is
most representative of our circuit?
1.
2.
1.
2.
3.
4.
1
2
3
4
3.
4.
Load lab while waiting…
Concept Question Answer: Expt. 4
Answer: 1.
There is resistance in the circuit
(in our non-ideal inductor).
The battery switching off doesn’t
break the circuit but allows it to
ring down
Concept Question: LC Circuit
1.
2.
3.
4.
Current
Charge
Tlag
1.0I0
0.5Q0
0.5I0
0.0Q0
0.0I0
-0.5Q0
-0.5I0
-1.0Q0
0
40
80
Current through Capacitor
1.0Q0
Charge on Capacitor
The plot shows the charge
on a capacitor (black curve)
and the current through it
(red curve) after you turn
off the power supply. If you
put a core into the inductor
what will happen to the
time TLag?
-1.0I0
120
Time (mS)
It will increase
It will decrease
It will stay the same
I don’t know
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Concept Question Answer: LC
Circuit
Current
Charge
Tlag
1.0I0
0.5Q0
0.5I0
0.0Q0
0.0I0
-0.5Q0
-0.5I0
-1.0Q0
0
40
80
Current through Capacitor
1.0Q0
Charge on Capacitor
Answer 1.
TLag will increase.
Putting in a core increases
the inductor’s inductance and
hence decreases the natural
frequency of the circuit.
Lower frequency means
longer period. The phase will
remain at 90º (a quarter
period) so TLag will increase.
-1.0I0
120
Time (mS)
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Concept Question: LC Circuit
Current
Charge
Tlag
1.0I0
0.5Q0
0.5I0
0.0Q0
0.0I0
-0.5Q0
-0.5I0
-1.0Q0
0
40
80
Current through Capacitor
If you increase the
resistance in the circuit
what will happen to rate
of decay of the pictured
amplitudes?
Charge on Capacitor
1.0Q0
-1.0I0
120
Time (mS)
1.
2.
3.
4.
It will increase (decay more rapidly)
It will decrease (decay less rapidly)
It will stay the same
I don’t know
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Concept Question Answer: LC
Circuit
Charge on Capacitor
1.0Q0
Current
Charge
Tlag
1.0I0
0.5Q0
0.5I0
0.0Q0
0.0I0
-0.5Q0
-0.5I0
-1.0Q0
0
40
80
Current through Capacitor
Answer: 1. It will increase
(decay more rapidly)
-1.0I0
120
Time (mS)
Resistance is what dissipates power in the circuit
and causes the amplitude of oscillations to
decrease. Increasing the resistance makes the
energy (and hence amplitude) decay more rapidly.
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