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R
C
e
~
L
Today...
• RLC circuits with a sinusoidal drive
• Phase difference between current & voltage for
Resistors, Capacitors, and Inductors.
• Reactance
• Phasors
• Application to frequency filters (high-, low-pass)
Driven LRC circuits
• Last time we discovered that an LC
circuit was a natural oscillator:
•
1
resonance 
in absence of resistive loss
LC
However, the resistance of any real inductor x
will cause oscillations to damp out,
unless we can supply energy at the rate the
resistor dissipates it! How? A sinusoidally
varying emf (AC generator) will sustain
sinusoidal current oscillations!
R
+ +
- 0,
r1
C
r1
.. r1
L
10
n
100
n
1
f( x ) 0
1
0
5
10
x
•
Very useful → TV, radio, computer clocks, …
Generic problem: we are given an
“ac voltage source” “driving” a circuit
eosint
~
We need to find the current that flows: I(t)=Iosin( t-f)
It is also sinusoidal at the same frequency, 
possibly with some “phase” angle relative to the voltage source
Preview
• Our goal is to understand how an AC LRC circuit works.
• Physical picture of each object:
– Source: ~ produces an oscillating voltage (supplies whatever
current the rest of the circuit “requires”)
– Resistor:
causes a voltage drop when a current flows
through. As soon as the voltage changes, so does the
current  always in phase.
– Capacitor:
resists change in charge Q  resists change in
voltage V  Q .  voltage across capacitor lags behind (90˚) the
C
current (charge leaving & entering the plates).
– Inductor:
in current.
resists change in magnetic flux  resists change
current
 always lags voltage (90˚).
AC Circuits
Series LRC
R
• Statement of problem:
Given e = emsin t, find I(t).
Everything else will follow.
C
e
L
~
• Procedure: start with loop equation
d 2Q
dQ Q
L 2 R
  e m sin  t
dt
dt C
• We could solve this equation with tons of algebra involving
sin( t) and cos( t); or with simple complex algebra.
• We will do neither, but start by considering simple circuits
with one element (R, C, or L) in addition to the driving emf.
eR Circuit
• R only: Loop eqn gives:

VR  RI R  e m si n t
x
R
IR 
em
R
s i n t
e
Voltage across R in phase with
r1
r1
x 0 , .. r1
0 , .. r1 current through R
n
enm
em 1
R
VR
0
~
1
IR
0
f( x ) 0
f( x ) 0
 em 1
00

2
t
x
4
6
IR
em
1
0
R 0
2
t
4
6
Note: this
is always,
always,
true in
LRC
circuit,
always…
x
 But voltage across R is not always in phase with source!
eC Circuit
• Now consider C only: Loop eqn gives:
0,

VC 
Q
 e m sin t
C

dQ
IC 
  Ce m c o s  t
dt
r1
.. r1
n
e
~
Ce m1
Is this
always
true?
IC
VC
0
f( x ) 0
0
f( x ) 0
1
00
IC
Voltage across C lags current through C by
r1
one-quarter cycle (90).x 0, n .. r1
e m1
 em
C
Q  Ce m sin t
2
4
x
6
t
 Ce m1
YES!
0
0
2
t
x
4
6
eL Circuit
• Now consider L: Loop eqn gives:
VL  L
r1
 d IL 
e
I L   d I L   m c o s t
L

0,
d IL
 e m s i n t
dt
.. r1
n
e m1
em
L
IL
s i n t d t
e
L
~
em

sin t   / 2
L
Voltage across L leads current through L by oner1
x 0 , .. r1
quarter cycle (90).
n
em 1
L
IL
VL
0
f( x ) 0
0
f( x ) 0
 em
1
00
2
tx
4
6
e
 m 1
L 00
2
tx
4
6
Yes, yes,
but how to
remember?
Introducing...
VL leads IL
VC lags IC
Hi kids, I’m Eli and
I’ll help you learn
physics !
…we’ll see about that
Summary thus far…
R
• LRC Circuit
– Given: e  e m sin  t
C
– Assume solution for current: I(t) = Im sin( t- )

VR  RI m sin(t   )
1
VC  
I m cos( t   )
C
e
L
~
VL   LI m cos(t   )
– Note that in all cases V  I, though there may be a phase shift:
VR  I m R
VC  I m
1
C
XC
VL  I m L
XL
reactance
What is reactance?
You can think of it as a frequency-dependent resistance.
1
XC 
C
XL  L
( " XR "  R )
For high ω, χC~0
- Capacitor looks like a wire (“short”)
For low ω, χC∞
- Capacitor looks like a break
For low ω, χL~0
- Inductor looks like a wire (“short”)
For high ω, χL∞
- Inductor looks like a break
(inductors resist change in current)
Filter Example #1
Vout
ε
~
Consider the AC circuit shown.
For very high frequencies, is Vout
big or small?
Recall: capacitor resists change in voltage.
High frequency more change  smaller reactance  smaller VC
1
  , XC  0
XC 
 Vout pulled to ground
C
1
Low frequency  capacitor has time to charge up  larger VC
  0 , XC  
 no current flows  no voltage drop across R
Vout ~ ε
Vout
What is ω0?
e
Use dimensional analysis.
  RC
0 
2
2

time RC
So, this is a circuit that only passes low frequencies: “low-pass” filter
 Bass knob on radio
0

Lecture 19, Act 1
• A driven RC circuit is connected as
shown.
– For what frequencies  of the voltage
source is the current through the
resistor largest?
(a) small (b) large (c)  
e
~

1
LC
R
C
More Filters
ω=0
a.
~
Vout
b.
~
~
ω=∞
Capacitor ~ wire
Vout ≈ ε
ω=∞
No current
Vout ≈ 0
ω=0
Inductor ~ wire
Vout ≈ ε
Vout
c.
Vout
No current
Vout ≈ 0
Vout
e
Highpass filter
0

Vout
e
Lowpass filter

0
ω=0
No current because of capacitor
ω=∞
No current because of inductor
Vout
(Conceptual
sketch only)
e
0
Bandpass
filter
AC Circuits, Quantitative
• Our goal is to calculate the voltages
across the various elements, and also
the current flowing through the circuit.
• If these were just three resistors in
series, we could calculate the net resistance
simply by adding the individual resistances.
R
C
e
L
~
• Because current and voltage are 90 out of phase for the
capacitor and the inductor, a straight sum does not work
(unless you use complex numbers, which we don’t in P212).
• Instead we use “phasors”, a geometric way to visualize an
oscillating function (avoiding nasty trig. or complex numbers).
• A phasor is a rotating “vector” whose magnitude is the
maximum value of a quantity (e.g., V or I).
–The instaneous value is the projection on the y-axis.
–The phasor rotates at the drive frequency . (Appendix)
“Beam me up Scotty –
It ate my phasor!”
Phasors
R
Problem: Given Vdrive = εm sin(ωt),
find VR, VL, VC, iR, iL, iC
C
e
~
Strategy:
We will use Kirchhoff’s voltage law that the (phasor)
sum of the voltages VR, VC, and VL must equal Vdrive.
L
Phasors, cont.
R
Problem: Given Vdrive = εm sin(ωt),
find VR, VL, VC, iR, iL, iC
C
e
L
~
1. Draw VR phasor along x-axis (this direction
is chosen for convenience). Note that since
VR = iRR, this is also the direction of the
current phasor iR. Because of Kirchhoff’s
current law, iL = iC = iR ≡ i (i.e., the same
current flows through each element).
VR, iRR
Phasors, cont.
R
Problem: Given Vdrive = εm sin(ωt),
find VR, VL, VC, iR, iL, iC
2. Next draw the phasor for VL. Since the
inductor current iL always lags VL  draw
VL 90˚ further counterclockwise. The
length of the VL phasor is iL XL = i L
C
e
L
~
VL = i XL
VR = i R
Phasors, cont.
Problem: Given Vdrive = εm sin(ωt),
R
C
find VR, VL, VC, iR, iL, iC
3. The capacitor voltage VC always lags
iC  draw VC 90˚ further clockwise.
The length of the VC phasor is
iC XC = i /C
e
L
~
VL = i XL
VR = i R
VC = i XC
The lengths of the phasors depend on R, L, C, and ω.
The relative orientation of the VR, VL, and VC phasors
is always the way we have drawn it. Memorize it!
Phasors, cont.
R
Problem: Given Vdrive = εm sin(ωt),
C
find VR, VL, VC, iR, iL, iC
•
The phasors for VR, VL, and VC are
added like vectors to give the drive
voltage VR + VL + VC = εm :
e
L
~
VL
εm
VR
VC
• From this diagram we can now easily calculate
quantities of interest, like the net current i , the
maximum voltage across any of the elements,
and the phase between the current the drive
voltage (and thus the power).
2
Lecture 19, Act 2
2A
A series RC circuit is driven by
emf e =e m sin t. Which of the
following could be an appropriate
phasor diagram?
VL
~
εm
εm
VC
VR
VR
VR
VC
(a)
2B
VC
(b)
εm
(c)
For this circuit which of the following is true?
(a) The drive voltage is in phase with the current.
(b) The drive voltage lags the current.
(c) The drive voltage leads the current.
RC Circuit, quantitative:
C
e m sin t
~
VR = iR
f
VC = iXC
 iR 
εm
2
 (iX C )2  e m
i 2 ( R2  X C2 )  e m
i
R
R X
2
C
R
VR  iR  e m
R2  X C
2
VR
em
2
2
em

R
R 2   1C 
2
2
1

1
 
0 2

RC Circuit, cont.
C
e
Vout
Vout
~
e
R

R
R 
2
1
C

2
1

1
 
0 2

1
0 
RC
Ex.: C = 1 μF, R = 1Ω
High-pass filter
"transmission"
1
0.8
0.6
High-pass filter
0.4
0.2
0
0.E+00
1.E+06
2.E+06
3.E+06
4.E+06
(Angular) frequency, om ega
5.E+06
6.E+06
Note: this is ω, f 

2
LRC Circuits, quantitative
where . . .
Im XL
R
C
e
~
L

εm
X L  L
Im R
1
XC 
C
Im XC
The phasor diagram gives us graphical solutions for f and Im:
Im XL
εm
f
Im R
Im XC
X L  XC
tan f 
R
I  I m sin( t  f )

e m2  I m2 R 2   X L  X C 2
Im 

em
R2  X L  X C 
2

Z  R2  X L  X C 
2

em
Z
Summary
• Reactances  ~ frequency-dependent resistance
– Capacitors
» look like wires for high frequencies
» look like breaks for low frequencies
» Voltage lags current by 90˚
– Inductors
» look like breaks for high frequencies
» look like wires for low frequencies
» Current lags voltage by 90˚
– Filters (low-pass, high-pass, band-pass)
• LRC Circuit
– Apply KVL using phasors
X  XC
tan f  L
R
Im 
XC 
X L  L
Im XL
I  I m sin( t  f )
em
R2  X L  X C 
2

em
Z
1
C
εm
f
Im R
Im XC
Appendix: Phasors
• A phasor is a “vector” whose magnitude is the maximum value
of a quantity (e.g., V) and which rotates counterclockwise in a 2d plane with angular velocity . Recall uniform circular motion:
The projections of r (on
the vertical y axis) execute
sinusoidal oscillation.
y
x  r cos t
y  r sin t

x
Ex. Source V  e m sin( t )
= y-component of the V-phasor
3
2
1
4
ωt=0
ωt=45˚
V=0
V
y
em
ωt=90˚
V=εm
ωt=270˚
2
V=-εm
The phasor picture corresponds to a snapshot
at some time t. The projections of the phasors
along the vertical axis are the actual values of
the voltages at the given time. One can draw
the phasors at different times, simply by rotating
the entire diagram. With this understanding,
other questions can be easily answered…
When the current through the circuit is
maximum, what is the potential difference
across the inductor?
a) VL = 0
b) VL = VLmax/2
c) VL = VLmax
When the capacitor is fully charged, what is the magnitude of the voltage
across the inductor?
a) VL = 0
b) VL = VLmax/2
c) VL = VLmax
Explanation:
Charge on C = max
Current = max
VR
VL
VC
Since the current and VL are 90 degrees out of phase, when
the current is at a maximum, VL will be at 0. When the capacitor
is fully charged, the current through the circuit will be 0, and the
magnitude of L will be at a maximum.
VL will actually be at a minimum because VC and VL are 180
degrees out of phase.