Physics 4B Lecture Notes
331
Chapter 33  Electromagnetic Oscillations and Alternating Current
Problem Set #12  due:
Ch 33  3, 10, 17, 18, 31, 34, 39, 42, 56, 59, 61, 77, 80, 84, 85, 87
Lecture Outline
1. The LC Circuit
2. The LRC Circuit
3. Basic AC Circuits
4. Frequency Filtering Circuits
5. The RLC Tuning Circuit
6. Power in AC Circuits
7. Transformers
We will use the loop and junction theorems to analyze the behavior of circuits that exhibit oscillatory
behavior.
We will start by allowing the circuits to oscillate at their “natural” frequency, then we will
examine the effect of forcing them to oscillate at other frequencies.
1. The LC Circuit
In the circuit at the left, the capacitor is charged when the switch is connected
from a to b.
Then the switch is connected from b to c.
The capacitor begins
to lose its charge as it attempts to get current through the inductor.
The
inductor begins to build up a magnetic field.
This field peaks as the charge
on the capacitor goes to zero.
Then the energy in the inductor begins to create
current to charge the capacitor back up.
All of this is explained by the loop theorem which requires that the
potential difference across the capacitor always equal the potential difference across the inductor,
V
c
=
V
L
⇒
Q
C
= 
L
dI
dt
⇒
Q
C
= 
L
d
2
Q
dt
2
⇒
d
2
Q
dt
2
= 
1
LC
Q
This differential equation is the SHM equation,
d
2
x
dt
2
= 
k
m
x
= ϖ
2
x
⇒
x
=
Acos(
ϖ
t
+δ
) .
By analogy,
Q
=
Q
m
cos(
ϖ
t
+δ
)
The LC Circuit
where
ϖ =
1
LC
Example
1:
Show that the above expression for the charge on the capacitor satisfies the loop
theorem.
Q
=
Q
m
cos(
ϖ
t
+δ
)
⇒
dQ
dt
= 
Q
m
ϖ
sin(
ϖ
t
+δ
)
⇒
d
2
Q
dt
2
= 
Q
m
ϖ
2
cos(
ϖ
t
+δ
)
= ϖ
2
Q
Substituting into the loop theorem,
ϖ
2
/
Q
= 
1
LC
/
Q
⇒ ϖ =
1
LC
.
L
ε
C
a
b
c
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332
Example
2:
For an LC circuit find (a)the peak current in terms of the peak charge on the capacitor
and (b)the fraction of the period between the time the capacitor's charge peaks and the current
peaks.
(a)From the definition of current, I
≡
dQ
dt
= 
Q
m
ϖ
sin(
ϖ
t
+δ
)
⇒
I
m
=
Q
m
ϖ
.
(b)First let's define Q=Q
m
when t=0.
This establishes the value of
δ
,
Q
=
Q
m
cos(
ϖ
t
+δ
)
⇒
Q
m
=
Q
m
cos(
δ
)
⇒
cos(
δ
)
=
1
⇒ δ =
0
Note that I
=
I
m
when sin(
ϖ
t
+δ
)
=
1
⇒ ϖ
t
+δ
=
π
2
⇒ ϖ
t
=
π
2
⇒
t
=
π
2
ϖ
.
Recall,
ϖ =
2
π
T
⇒
t
=
π
2
2
π
T
⇒
t
=
1
4
T
.
2. The LRC Circuit
The capacitor is charged by connecting a to b.
Then the switch
is moved from b to c.
The oscillations of the LC circuit are
damped out by the heat created in the resistor.
As usual, the
loop theorem gives the details,
V
L
+
V
R
+
V
C
=
0
⇒
L
dI
dt
+
IR
+
Q
C
=
0
⇒
L
d
2
Q
dt
2
+
R
dQ
dt
+
Q
C
=
0
If we assume that R is fairly small then we would expect
damped oscillations of the form, Q
=
Q
m
e
γ
t
cos
ϖ
d
t where
Q
m
is the initial charge on the capacitor,
α
is the damping
constant, and
ϖ
d
is the frequency of oscillation.
A graph of
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 Fall '09
 Staff
 Current, Inductor, power supply

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