Physics 2113
Jonathan Dowling
Lecture 33: WED 12 NOV
Electrical Oscillations, LC Circuits, Alternating Current I
Nikolai Tesla
What are we going to learn?
A road map
• Electric charge
 Electric force on other electric charges
 Electric field, and electric potential
• Moving electric charges : current
• Electronic circuit components: batteries, resistors, capacitors
• Electric currents  Magnetic field
 Magnetic force on moving charges
• Time-varying magnetic field  Electric Field
• More circuit components: inductors.
• Electromagnetic waves  light waves
• Geometrical Optics (light rays).
• Physical optics (light waves)
Energy Density in E and B Fields
uE =
e0E
2
2
B
uB =
2 m0
2
Oscillators in Physics
Oscillators are very useful in practical
applications, for instance, to keep time, or
to focus energy in a system.
All oscillators can store energy in
more than one way and exchange it
back and forth between the different
storage possibilities. For instance, in
pendulums (and swings) one
exchanges energy between
kinetic and potential form.
We have studied that inductors and capacitors are devices
that can store electromagnetic energy. In the inductor it is
stored in a B field, in the capacitor in an E field.
PHYS2110: A Mechanical Oscillator
E = K +U = ENERGY
1
1 2
2
E = mv + k x
2
2
dE
1 æ dv ö 1 æ dx ö
= 0 = m ç 2v ÷ + k ç 2x ÷
dt
2 è dt ø 2 è dt ø
dv
®m +kx=0
dt
Solution :
a = v¢(t) = x ¢¢(t)
Newton’s law
d 2x
m 2 +k x =0
F=ma!
dt
k
x(t ) = x0 cos(w t + f0 ) w =
m
x0 :
amplitude
w :
f0 :
frequency
phase
v = x ¢(t)
PHYS2113 An Electromagnetic LC Oscillator
Capacitor initially charged. Initially, current is zero,
energy is all stored in the E-field of the capacitor.
Energy Conservation: Utot = U B +UE
A current gets going, energy gets split between the
capacitor and the inductor.
1 2
1 q2
U B = L i U E =
2
2C
Capacitor discharges completely, yet current keeps going.
Energy is all in the B-field of the inductor all fluxed up.
The magnetic field on the coil starts to deflux, which
will start to recharge the capacitor.
1 2 1 q2
Utot = L i +
2
2C
Finally, we reach the same state we started with (with
opposite polarity) and the cycle restarts.
Electric Oscillators: the Math
Utot = U B + UE
1 2 1 q2
Utot = L i +
2
2C
dUtot
1 æ di ö 1 æ dq ö
= 0 = L ç 2i ÷ +
çè 2q ÷ø
è
ø
dt
2
dt
2C
dt
Energy Cons.
æ di ö 1
VL + VC = 0 = L ç ÷ + ( q )
è dt ø C
Or loop rule!
Both give Diffy-Q:
Solution to Diffy-Q:
d 2q q
0=L 2 +
dt
C
q = q0 cos(w t + j 0 )
wº
1
LC
i = q¢(t)
i ¢(t) = q¢¢(t)
LC Frequency
In Radians/Sec
i = q¢(t) = -q0w sin(w t + j 0 )
i¢(t) = q¢¢(t) = -w 2 q0 cos(w t + j 0 )
T a µ LC
T b µ L(2C)
T c µ L(C / 2)
T >T >T
b
a
wº
c
T=
1
LCeq
2p
w
µ LCeq
Ceqa = C
Ceqb = 2C
Ceqc = C / 2
Electric Oscillators: the Math
q = q0 cos(w t + j 0 )
i(t) = q¢(t) = -q0w sin(w t + j 0 )
i¢(t) = q¢¢(t) = -w 2 q0 cos(w t + j 0 )
Energy as Function of Time
Voltage as Function of Time
1
1
2
2
U B = L [ i ] = L [ q0w sin(w t + j 0 )] VL = Li¢(t) = -Lw 2 q0 cos(w t + j 0 )
2
2
1 [ q]
1
2
UE =
=
q
cos(
w
t
+
j
)
[0
0 ]
2 C
2C
2
1
1
VC = [ q(t)] = [ q0 cos(w t + j 0 )]
C
C
LC Circuit: At t=0 1/3 Of Energy Utotal is on Capacitor C and Two
Thirds On Inductor L. Find Everything! (Phase φ0=?)
1
2
U B (t) = L [ q0w sin(w t + j 0 )]
2
1
2
UE (t ) =
q0 cos(w t + j 0 )]
[
2C
2
1
é
ù
L
q
w
sin(
j
)
0 û
U /3
U B (0) 2 ë 0
=
= total
2
1
2U total / 3
UE 0
éë q0 cos(j 0 ) ùû
2C
()
tan(j 0 ) =
(
2
LC éë q0w sin(j 0 ) ùû
1 w = 1 / LC
=
2
2 q0 = VC
éë q0 cos(j 0 ) ùû
q = q0 cos(w t + j 0 )
i(t) = -q0w sin(w t + j 0 )
1
2
1
2
U B (0) = L [ q0w sin(j 0 )] = U total / 3
2
1
2
UE ( 0) =
q
cos(
j
)
[
0
0 ] = 2U total / 3
2C
)
j 0 = arctan 1 / 2 = 35.3°
i ¢(t) = -w 2 q0 cos(w t + j 0 )
q
VL (t) = - 0 cos(w t + j 0 )
C
q
VC (t) = 0 cos(w t + j 0 )
C
Analogy Between Electrical
And Mechanical Oscillations
d 2q q
0=L 2 +
dt
C
1
w=
LC
d 2x
m 2 +k x =0
dt
k
w=
m
q = q0 cos(w t + j 0 )
x(t ) = x0 cos(w t + f0 )
i = q¢(t) = -q0w sin(w t + j 0 )
v = x¢(t) = -w x0 sin(w t + j 0 )
i¢(t) = q¢¢(t) = -w q0 cos(w t + j 0 )
a = x¢¢(t) = -w 2 x0 cos(w t + j 0 )
2
q®x
1/C ® k
i®v
L®m
Charqe q -> Position x
Current i=q’ -> Velocity v=x’
Dt-Current i’=q’’-> Acceleration a=v’=x’’
LC Circuit: Conservation of Energy
q = q0 cos(w t + f0 )
1.5
1
0.5
0
Time
-0.5
Charge
Current
dq
i=
= -w q0 sin(w t + f0 )
dt
UB =
-1
1 q2
1 2
UE =
=
q0 cos2 (w t + j 0 )
2C
2C
-1.5
1.2
And remembering that,
1
0.8
0.6
0.4
0.2
0
Time
1 2 1
Li = Lw 2 q02 sin 2 (w t + j 0 )
2
2
Energy in capacitor
Energy in coil
1
cos x + sin x = 1, and w =
LC
2
2
1 2
Utot = U B + U E =
q0
2C
The energy is constant and equal to what we started with.
LC Circuit: Phase Relations
q = q0 cos(w t + f0 )
1.5
1
0.5
Charge
0
Time
-0.5
Current
dq
i=
= -w q0 sin(w t + f0 )
dt
Take j 0 = 0 as origin of time.
-1
-1.5
q µ cos(w t)
i µ -sin(w t)
Trigamarole: - sin(w t - p / 2) = cos(w t)
The current runs 90° out of phase with respect to the charge.
1.5
wº
1
1
LC
0.5
Charge
0
Time
-0.5
Current
T=
-1
-1.5
t = 0´T
t =T /4
t =T /2
t = 3T / 4
t =T
2p
w
1.5
1
0.5
Charge
0
Time
-0.5
Current
-1
-1.5
(a) T / 2
(b) T
(c) T / 2
(d) T / 4
t=0
Vc = q / C
t =T /4
t =T
t =T /2
t = 3T / 4
Example 1 : Tuning a Radio Receiver
The inductor and capacitor in my
car radio have one program at L =
1 mH & C = 3.18 pF. Which is the
FM station?
(a) KLSU 91.1
(b) WRKF 89.3
(c) Eagle 98.1 WDGL
FM radio stations: frequency
is in MHz.
w=
1
LC
=
1
-6
1 ´ 10 ´ 3.18 ´ 10
= 5.61 ´ 10 8 rad/s
w
f =
2p
= 8.93 ´ 10 7 Hz
= 89.3 MHz
-12
rad/s
Example 2
• In an LC circuit,
L = 40 mH; C = 4  F
• At t = 0, the current is a
maximum;
• When will the capacitor
be fully charged for the
first time?
1.5
1
0.5
Charge
0
Time
-0.5
Current
-1
-1.5
t=0
t =T /4
t =T /2
t = 3T / 4 t = T
w=
1
1
=
rad/s
LC
16x10 -8
•  = 2500 rad/s
• T = period of one
complete cycle
•T =  = 2.5 ms
• Capacitor will be
charged after T=1/4
cycle i.e at
• t = T/4 = 0.6 ms
Example 3
• In the circuit shown, the switch is in
position “a” for a long time. It is
then thrown to position “b.”
• Calculate the amplitude  q0 of the
resulting oscillating current.
1 mH
1 mF
b
E=10 V
a
i = -w q0 sin(w t + f0 )
• Switch in position “a”: q=CV = (1 mF)(10 V) = 10 mC
• Switch in position “b”: maximum charge on C = q0 = 10 mC
• So, amplitude of oscillating current =
1
w q0 =
(10mC) = 0.316 A
(1mH)(1mF)
Example 4
In an LC circuit, the maximum current is 1.0 A.
If L = 1mH, C = 10 mF what is the maximum charge q0 on
the capacitor during a cycle of oscillation?
q = q0 cos(w t + f0 )
dq
i=
= -w q0 sin(w t + f0 )
dt
Maximum current is
i0= q0 Maximum charge: q0=i0/
Angular frequency w=1/√LC=(1mH 10 mF)–1/2 = (10-8)–1/2 = 104 rad/s
Maximum charge is q0=i0/ = 1A/104 rad/s = 10–4 C