Class 26: Outline
Hour 1:
Driven Harmonic Motion (RLC)
Hour 2:
Experiment 11: Driven RLC Circuit
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Last Time:
Undriven RLC Circuits
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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)
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Damped LC Oscillations
Resistor dissipates
energy and system
rings down over time
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Mass on a Spring:
Simple Harmonic Motion`
A Second Look
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Mass on a Spring
(1)
(3)
(2)
(4)
We solved this:
d 2x
F = −kx = ma = m 2
dt
2
d x
m 2 + kx = 0
dt
Simple Harmonic Motion
x(t ) = x0 cos(ω 0t + φ )
Moves at natural frequency
What if we now move the wall?
Push on the mass?
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Demonstration:
Driven Mass on a Spring
Off Resonance
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Driven Mass on a Spring
Now we get:
d 2x
F = F ( t ) − kx = ma = m 2
dt
2
F(t)
d x
m 2 + kx = F ( t )
dt
Assume harmonic force:
F (t ) = F0 cos(ω t )
Simple Harmonic Motion
x(t ) = xmax cos(ω t + φ )
Moves at driven frequency
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Resonance
x(t ) = xmax cos(ω t + φ )
Now the amplitude, xmax, depends on how
close the drive frequency is to the natural
frequency
xmax
Let’s
See…
ω0
ω
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Demonstration:
Driven Mass on a Spring
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Resonance
x(t ) = xmax cos(ω t + φ )
xmax depends on drive frequency
xmax
Many systems behave like this:
Swings
Some cars
Musical Instruments
…
ω0
ω
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Electronic Analog:
RLC Circuits
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Analog: RLC Circuit
Recall:
Inductors are like masses (have inertia)
Capacitors are like springs (store/release energy)
Batteries supply external force (EMF)
Charge on capacitor is like position,
Current is like velocity – watch them resonate
Now we move to “frequency dependent batteries:”
AC Power Supplies/AC Function Generators
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Demonstration:
RLC with Light Bulb
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Start at Beginning:
AC Circuits
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Alternating-Current Circuit
• direct current (dc) – current flows one way (battery)
• alternating current (ac) – current oscillates
• sinusoidal voltage source
V (t ) = V0 sin ω t
ω = 2π f : angular frequency
V0 : voltage amplitude
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AC Circuit: Single Element
V =V
= V0 sin ω t
I (t ) = I 0 sin(ωt − φ )
Questions:
1. What is I0?
2. What is φ ?
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AC Circuit: Resistors
VR = I R R
VR V0
=
sin ω t
IR =
R R
= I 0 sin ( ω t − 0 )
V0
I0 =
R
ϕ =0
IR and VR are in phase
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AC Circuit: Capacitors
Q
VC =
C
dQ
I 0 = ω CV0
I C (t ) =
dt
π
ϕ
=
−
= ω CV0 cos ω t
2
= I 0 sin(ω t − −π 2 )
IC leads VC by π/2
Q(t ) = CVC = CV0 sin ω t
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AC Circuit: Inductors
V
I (t ) =
sin ω t dt
∫
L
L
dI L
VL = L
dt
0
=−
V0
cos ω t
V0
I0 =
ωL
ωL
π
= I 0 sin ( ω t − π 2 ) ϕ = 2
IL lags VL by π/2
dI L VL V0
=
=
sin ω t
dt
L
L
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AC Circuits: Summary
Element
I0
Current
vs.
Voltage
Resistor
V0R
R
In Phase
R=R
ω CV0C
Leads
1
XC =
ωC
Capacitor
Resistance
Reactance
Impedance
V0L
Lags
X
=
ω
L
L
ωL
Although derived from single element circuits,
these relationships hold generally!
Inductor
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PRS Question:
Leading or Lagging?
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Phasor Diagram
Nice way of tracking magnitude & phase:
V (t ) = V0 sin ( ω t )
ω
V0
ωt
Notes: (1) As the phasor (red vector) rotates, the
projection (pink vector) oscillates
(2) Do both for the current and the voltage
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Demonstration:
Phasors
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Phasor Diagram: Resistor
V0 = I 0 R
ϕ =0
IR and VR are in phase
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Phasor Diagram: Capacitor
V0 = I 0 X C
1
= I0
ωC
ϕ =−π 2
IC leads VC by π/2
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Phasor Diagram: Inductor
V0 = I 0 X L
= I 0ω L
ϕ=
π
2
IL lags VL by π/2
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PRS Questions:
Phase
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Put it all together:
Driven RLC Circuits
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Question of Phase
We had fixed phase of voltage:
V = V0 sin ω t
I (t ) = I 0 sin(ω t − φ )
It’s the same to write:
V = V0 sin(ω t + φ ) I (t ) = I 0 sin ω t
(Just shifting zero of time)
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Driven RLC Series Circuit
I (t ) = I 0 sin(ω t )
VR = VR 0 sin ( ω t )
VL = VL 0 sin ( ω t + π2 )
VS = V0 S sin ( ω t + ϕ )
VC = VC 0 sin ( ω t + −2π
)
What is I 0 (and VR 0 = I 0 R, VL 0 = I 0 X L , VC 0 = I 0 X C ) ?
What is ϕ ? Does the current lead or lag Vs ?
Must Solve: VS = VR + VL + VC
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Driven RLC Series Circuit
V0L
I(t)
VS
V0C
V0S
I0
V0R
Now Solve: VS = VR + VL + VC
Now we just need to read the phasor diagram!
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Driven RLC Series Circuit
V0L
ϕ
I (t ) = I 0 sin(ω t − ϕ )
V0C
VS = V0 S sin ( ω t )
2
V0S
2
I0
2
V0R
2
V0 S = VR 0 + (VL 0 − VC 0 ) = I 0 R + ( X L − X C ) ≡ I 0 Z
V0 S
I0 =
Z
2
Z = R + (X L − XC )
Impedance
2
φ = tan
−1
⎛ X L − XC ⎞
⎜
⎟
R
⎝
⎠
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I
I (t ) = I 0 sin ( ω t )
0
VR (t ) = I 0 R sin ( ω t )
0
VL (t ) = I 0 X L sin ( ω t + π2 )
VL
0
VR
Plot I, V’s vs. Time
VS
VC
+π/2
VL (t ) = I 0 X C sin ( ω t − π2 )
-π/2
0
VS (t ) = VS 0 sin ( ω t + ϕ )
0
0
+φ
1
2
3
Time (Periods)
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PRS Question:
Who Dominates?
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RLC Circuits:
Resonances
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Resonance
V0
I0 = =
Z
V0
R 2 + ( X L − X C )2
;
1
X L = ω L, X C =
ωC
At very low frequencies, C dominates (XC>>XL):
it fills up and keeps the current low
At very high frequencies, L dominates (XL>>XC):
the current tries to change but it won’t let it
At intermediate frequencies we have resonance
I0 reaches maximum when
ω0 =
1
LC
X L = XC
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Resonance
V0
I0 = =
Z
V0
R 2 + ( X L − X C )2
;
1
X L = ω L, X C =
ωC
φ = tan
C-like:
φ<0
I leads
L-like:
φ>0
I lags
ω0 = 1
LC
−1
⎛ X L − XC ⎞
⎜
⎟
R
⎝
⎠
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Demonstration:
RLC with Light Bulb
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PRS Questions:
Resonance
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Experiment 11:
Driven RLC Circuit
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Experiment 11: How To
Part I
• Use exp11a.ds
• Change frequency, look at I & V. Try
to find resonance – place where I is
maximum
Part II
• Use exp11b.ds
• Run the program at each of the listed
frequencies to make a plot of I0 vs. ω
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