Physics 122B
Electricity and Magnetism
Lecture 24 (Knight: 33.9, 34.1-5)
LC and AC Circuits
Martin Savage
Lecture 24 Announcements
 Lecture HW is due tonight at 10 PM.
 Midterm Exam 3 is this coming Friday. Covers
explicitly everything not covered in the previous
exam…and assumes understanding of all previous
material.
 Lecture question and lab question are multiplechoice, tutorial is long answer.
March 7, 2007
Physics 122C - Lecture 23
2
LC Circuits
A charged capacitor bears a
certain resemblance to a stretched
spring (remember the rubber
diaphragm), storing energy even
when the charge is not moving.
An inductor similarly resembles a
moving mass (remember the
flywheel), storing energy only when
charge is in motion.
We know that a mass and spring can make an oscillator. What about a
capacitor and inductor. Consider the circuit shown in the diagram. What
happens when the switch is closed?
The capacitor discharges by creating a current in the inductor. But
where does the energy go that had been stored in the inductor? There
are no dissipative elements in the system. Therefore, when the charge
of the capacitor goes to zero, all of its previous energy must reside in
the inductor. The current in the inductor falls while charging the
capacitor in the opposite direction. And so on …
March 7, 2007
Physics 122C - Lecture 23
3
The Oscillation Cycle
March 7, 2007
Physics 122C - Lecture 23
4
LC circuits (2)
dI
1
L dt +
Q = 0
C
d2Q
dt2
Q =
+
1
Q = 0
LC
a Sin( w t + f )
w2 =
March 7, 2007
1
L C
Physics 122C - Lecture 23
5
Example:
An AM Radio Oscillator
You have a 10mH inductor. What capacitor should you use with it to
make an oscillator with a frequency of 920 kHz? (This frequency is
near the center of the AM radio band.
w  2 f  2 (9.20 105 s-1 )  5.78 106 s-1
1
1
11
C 2 

3.0

10
F  30 pF
6 -1 2
2
w L (5.78 10 s ) (1.0 10 H)
March 7, 2007
Physics 122C - Lecture 23
6
Plumber’s LC Analogy
Valve
V1
V2
P1
V3
Rubber
Diaphragm
The “plumber’s analogy” of an LC circuit is a
rubber diaphragm that has been stretched
by pressure on the top (P1) side. When the
valve starts the flow, the diaphragm forces
water past the flywheel, which begins to
spin. After the diaphragm has become flat,
the momentum of the flywheel forces the
diaphragm to be stretched in the other
direction, and the cycle repeats.
March 7, 2007
Physics 122C - Lecture 23
P2
P3
Flywheel
Valve = Switch
Rubber Diaphragm = Capacitor
Flywheel = Inductor
Pressure = Potential
Water Flow = Current
7
Chapter 33 - Summary (1)
March 7, 2007
Physics 122C - Lecture 23
8
Chapter 33 - Summary (2)
March 7, 2007
Physics 122C - Lecture 23
9
Chapter 33 - Summary (3)
March 7, 2007
Physics 122C - Lecture 23
10
AC Sources and Phasors
You can think of an AC
generator as a batterylike object with an emf
that varies sinusoidally as
E (t) = E0cos wt, where E0
is the maximum emf and w
is the angular frequency,
with w=2f, where f is the
frequency in Hz.
Alternatively, the emf and other oscillatory
quantities can be represented by a phasor
diagram. The phasor is a vector of length E0
that rotates counterclockwise around the
origin with angular frequency w, so that the
angle it makes with the horizontal axis at any
time is wt. The projection of the phasor on the
horizontal axis at any time gives the emf.
March 7, 2007
Physics 122C - Lecture 23
11
Resistor AC Circuits
Consider an AC current iR
through a resistor. Ohm’s
Law gives the potential drop
across the resistor, which we
will call the resistor voltage
vR.
vR  iR R
If the resistor is connected
in an AC circuit as shown, then
Kirschoff’s loop law tells us
that:
Vsource
soruce  VR =E  vR  0
E (t )  E0 cos wt  vR
E
v
iR  R  0 cos wt  I R cos wt
R R
In the phasor diagram, the phasors for vR and iR are parallel.
March 7, 2007
Physics 122C - Lecture 23
12
Example:
Finding Resistor Voltages
In the circuit shown, find
(a) the peak voltage across
each resistor, and
(b) the instantaneous resistor
voltages at t=20 ms.
Req  R1  R2  (5 )  (15 )  20 
E cos wt (100 V) cos 2 (60 Hz)t
v
iR  I R cos wt  R  0

 (5.0 A) cos 2 (60 Hz)t
Req
Req
(20 )
 25 V for R1 =5 
VR  I 0 R  
75 V for R 2 =15 
iR (t  20 ms)  (5.0 A) cos 2 (60 Hz)(2.0 102 s)  1.545 A
 7.7 V for R1 =5 
vR  iR R  
23.2 V for R 2 =15 
March 7, 2007
Physics 122C - Lecture 23
13
Capacitor AC Circuits (1)
Consider an AC current iC through a capacitor as shown. The capacitor
voltage vC = E = E0cos wt = VCcos wt. The charge on the capacitor will be
q = CvC = CVCcos wt.
iC 
dq d
  CVC cos wt   wCVC sin wt
dt dt
iC  wCVC cos(wt   / 2)
The AC current through a capacitor leads the capacitor voltage
by /2 rad or 900.
March 7, 2007
Physics 122C - Lecture 23
14
Capacitor AC Circuits (2)
The AC current through
a capacitor leads the
capacitor voltage by /2 rad
or 900. The phasors for vC
and iC are perpendicular,
with the iC phasor ahead of
the vC phasor.
This is analogous to the
behavior of the position
and velocity of a massand-spring harmonic
oscillator.
March 7, 2007
Physics 122C - Lecture 23
15
Capacitive Reactance
For AC circuits we can define a
resistance-like quantity, measured
in ohms, for capacitance. It is
called the capacitive reactance XC:
XC 
1
1

wC 2 f C
We can then use a form of
Ohm’s Law to relate the peak
voltage VC, the peak current IC,
and the capacitive reactance XC in
an AC circuit:
IC 
VC
XC
March 7, 2007
and VC  I C X C
Physics 122C - Lecture 23
16
Question
The instantaneous value of the emf E represented by this
phasor is:
(a) Increasing;
(b) Decreasing;
(c) Constant;
(d) It is not possible to tell without knowing t.
March 7, 2007
Physics 122C - Lecture 23
17
Example: Capacitive Reactance
What is the capacitive reactance of a 0.10 mF capacitor at a 100 Hz
audio frequency and at a 100 MHz FM radio frequency?
1
X C (100 Hz) 
 15,900 
-1
-7
2 (100 s )(1.0 10 F)
X C (100 MHz) 
March 7, 2007
1
 0.0159 
8 -1
-7
2 (1.0 10 s )(1.0 10 F)
Physics 122C - Lecture 23
18
Example: Capacitive Current
A 10 mF capacitor is connected to a 1000 Hz oscillator with a
peak emf of 5.0 V.
What is the peak current in the capacitor?
X C (1000 Hz) 
IC 
1
 15.9 
-1
-5
2 (100 s )(1.0 10 F)
VC
(5.0 V)

 0.314 A
X C (15.9 )
March 7, 2007
Physics 122C - Lecture 23
19
Voltage Dividers
r
Vout
The circuit indicates a potentiometer, a resistor with a sliding
contact. The overall resistance of the unit is R, while the resistance
from the sliding tap to the bottom is r.
What is the voltage Vout delivered between the output terminals?
I E / R
Vout  I r  E
r
R
Thus, the potentiometer divides the input voltage and delivers
some fraction of it proportional to r/R. This is a voltage divider.
March 7, 2007
Physics 122C - Lecture 23
20
Analyzing an RC Circuit
Draw the current
vector I at some
arbitrary angle.
All elements of
the circuit will
have this current.
March 7, 2007
Draw the resistor
voltage VR in phase
with the current.
Draw the capacitor
voltage VC 900
behind the current.
Make sure all phasor
lengths scale
properly.
Draw the emf E0
as the vector sum
of VR and VC. The
angle of this
phasor is wt,
where the timedependent emf is
E0 cos wt.
Physics 122C - Lecture 23
The phasors
VR and VC
form the sides
of a right
triangle, with
E0 as the
hypotenuse.
Therefore,
E02 = VR2+VC2.
21
RC Filter Circuits
Now consider a circuit that
includes both a resistor and a
capacitor. Because the
capacitor voltage VC and the
resistor voltage VR are 900
apart in the phasor diagram,
they must be added like the
sides of a right triangle:
E0 2  VC 2  VR 2  ( IR) 2  ( IX C ) 2
 ( R 2  X C 2 ) I 2   R 2   wC   I 2


E0
2
I
R 2   wC 
March 7, 2007
2
VR  IR 
Vc  IX C 
Physics 122C - Lecture 23
E0 R
R 2   wC 
2
E0 / wC
R 2   wC 
2
22
Frequency Dependence
VR  IR 
Vc  IX C 
E0 R
R 2   wC 
2
E0 / wC
R 2   wC 
2
Define the crossover frequency
where VR=VC as wC:
1
wC 
RC
At w  wC VR  VC  E0 / 2  0.707E0
March 7, 2007
Physics 122C - Lecture 23
23
Filters and Transmission
An RC filter is a circuit that passes a
signal with attenuation of some frequencies.
Define the transmission of an RC filter as
T = vout/vin with wC = 1/(RC):
TLoPass 
THiPass 
1/(wC )
R 2  (wC ) 2
R
R  (wC )
2
Low Pass
2


wC / w
1  (wC / w ) 2
1
1  (wC / w ) 2
High Pass
Cross-over Point
Note log scale
March 7, 2007
Physics 122C - Lecture 23
24
Example: Designing a Filter
For a science project you have built a radio to listen to AM
radio broadcasts at frequencies near 1 MHz. The basic circuit
is an antenna, which produces a very small oscillating voltage
when it absorbs energy from an electromagnetic wave, and an
amplifier. Unfortunately, your neighbor’s short wave broadcast
at 10 MHz interferes with your reception. You decide to place
a filter between the antenna and the amplifier. You have a 500
pF capacitor.
What frequency should you select for the filter’s cross over
frequency?
What value of resistance should be used in the filter?
fC 
R
f1 f 2  (1.0 MHz)(10.0 MHz)  3.16 MHz Tlow (w 1) = T
high
(w 2)
1
1
1


 100 
6 -1
10
wC C 2 fC C 2 (3.16 10 s )(5.0 10 F)
March 7, 2007
Physics 122C - Lecture 23
25
Question
Which of these RC filter circuits has the largest cross-over
frequency wC?
March 7, 2007
Physics 122C - Lecture 23
26
AC Inductor Circuits
Consider an AC current iR through an inductor. The changing
current produces an instantaneous inductor voltage vL.
di
vL  L L
dt
If the inductor is connected in an AC
circuit as shown, then Kirschoff’s loop
law tells us that:
vL
VL
di

dt

cos wtdt
Vsoruce  VL =E  vL  0 E (t )  E0 cos wt  vL
L
L
L
V
V
V




iL  L  cos wtdt  L sin wt  L cos  wt    I L cos  wt  
L
wL
wL
2
2


In the phasor diagram, the
inductor current iL lags the
voltage vL by 900, so that iL
peaks T/4 later than vL.
March 7, 2007
Physics 122C - Lecture 23
27
Inductive Reactance
For AC circuits we can define a
resistance-like quantity, measured
in ohms, for inductance. It is
called the inductive reactance XL:
X L  w L  2 f L
We can then use a form of
Ohm’s Law to relate the peak
voltage VL, the peak current IL, and
the inductive reactance XL in an AC
circuit:
IL 
VL
XL
March 7, 2007
and VL  I L X L
Physics 122C - Lecture 23
28
Example: Current and Voltage
of an Inductor
A 25 mH inductor is used in a circuit that
is driven at 100 kHZ. The current through
the inductor reaches a peak value of 20 mA
at t=5.0 ms.
What is the peak inductor voltage, and when, closest to t=5.0 ms,
does it occur?
X L  w L  2 (1.0 105 s -1 )(2.5 105 H)  16 
VL  I L X L  (2.0 102 A)(16 )  0.320 V
The voltage peaks ¼ cycle before the current, which peaks at 5 ms.
For f = 100 kHz, T = 10 ms, so T/4 = 2.5 ms. Therefore, the voltage
peaks at t =(5.0-2.5) ms = 2.5 ms.
March 7, 2007
Physics 122C - Lecture 23
29
Analyzing an LRC Circuit
Draw the current
vector I at some
arbitrary angle.
All elements of
the circuit will
have this current.
March 7, 2007
Draw the resistor
voltage VR in phase
with the current.
Draw the inductor
and capacitor
voltages VL and VC
900 before and
behind the current,
respectively.
Draw the emf E0
as the vector sum
of VR and VL-VC.
The angle of this
phasor is wt,
where the timedependent emf is
E0 cos wt.
Physics 122C - Lecture 23
The phasors VR
and VL-VC form
the sides of a
right triangle,
with E0 as the
hypotenuse.
Therefore, E02
= VR2+(VL-VC)2.
30
The Series RLC Circuit
The figure shows a resistor, inductor,
and capacitor connected in series. The
same current i passes through all of the
elements in the loop. From Kirchhoff’s
loop law, E = vR + vL + vC.
Because of the capacitive and inductive
elements in the circuit, the current i will
not in general be in phase with E, so we will
have i = I cos(wt-f) where f is the phase
angle between current and voltage. If
VL>VC then the current i will lag E and f>0.
E02  VR2  (VL  VC ) 2   R 2  ( X L  X C ) 2  I 2
E0
E0
I

2
2
R  (X L  XC )
R 2  (w L  1/ wC ) 2
March 7, 2007
Physics 122C - Lecture 23
31
Impedance and Phase Angle
We can define the impedance
Z of the circuit as:
Z  R 2  ( X L  X C )2
 R 2  (w L  1/ wC ) 2
Then I  E / Z
From the phasor diagram ,we
see that the phase angle f of the
current is given by:
VL  VC I  X L  X C 
tan f 

VR
IR
March 7, 2007
 X L  XC
R

f  tan 1 
Physics 122C - Lecture 23

1  w L  1/ wC 

tan



R



VR  E0 cos f
32
Resonance
I
E0
R 2  (w L  1/ wC ) 2
The current I will be a maximum when wL=1/wC.
This defines the resonant frequency of the system w0:
w0 
1
LC
March 7, 2007
I
E0
2
2 
 w0  
2
R   Lw  1    
  w  
2
Physics 122C - Lecture 23
33
Example:
Designing a Radio Receiver
An AM radio antenna picks up a 1000 kHz signal with a peak
voltage of 5.0 mV. The tuning circuit consists of a 60 mH
inductor in series with a variable capacitor. The inductor coil
has a resistance of 0.25 , and the resistance of the rest of
the circuit is negligible.
(a) To what capacitance should the capacitor be tuned to listen to
this radio station.
(b) What is the peak current through the circuit at resonance?
(c) A stronger station at 1050 kHz produces a 10 mV antenna
signal. What is the current in the radio at this frequency when
the station is tuned to 1000 kHz.
X L  X C so Z  R
w0  1/ LC  1000 kHz = 1 MHz
C
1
1

Lw02 (60 10-6 H)(6.28 106 rad/s)2
 4.23 10
-10
March 7, 2007
F  423 pF
I1  E0 / R  (5.0 103 V) /(0.25 )  0.020 A  20 mA
X L '  w ' L  396 
I2 
Physics 122C - Lecture 23
X C '  1/ w ' C  358 
E0 '
R  ( X L ' X C ')
2
2
 0.26 mA
34
Lecture 24 Announcements
 Lecture HW is due tonight at 10 PM.
 Midterm Exam 3 is this coming Friday. Covers
explicitly everything not covered in the previous
exam…and assumes understanding of all previous
material.
 Lecture question and lab question are multiplechoice, tutorial is long answer.
March 7, 2007
Physics 122C - Lecture 23
35