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Lecture 20-1
Alternating Current (AC)
= Electric current that changes direction periodically
ac generator is a device which creates an ac emf/current.
A sinusoidally oscillating EMF is induced in a loop of wire
that rotates in a uniform magnetic field.
B  NBA cos  NBA cos t   
dB
 
  NBA  sin t   
dt
where
2
  2 f 
T
ac motor =
ac generator run
in reverse
http://www.wvic.com/how-gen-works.htm
http://www.pbs.org/wgbh/amex/edison/sfeature/acdc.html
Lecture 20-2
Root-Mean-Square Values
I I
2
I rms
 rms
1 2
sin t  I peak
2
2
peak
 2 I rms  1.41 I rms

1 2 
2
 Pav   I peak  R  I rm
sR
2

1
1
  
 peak 
I peak R  I rms R
2
2
2
2
1
 I 
I peak  0.707  I peak
2
I
Similarly,
2
peak
Pav   rms I rms
Lecture 20-3
Capacitive vs Inductive Load
I(t) leads v(t) by 90
capacitive reactance
1
XC 
C
--
+
Pav  0
VC , peak  X C I peak
vL
vL(t) leads I(t) by 90
inductive reactance
X L  L
VL , peak  X L I peak
+
--
Lecture 20-4
(Ideal) LC Circuit
• From Kirchhoff’s Loop Rule
Q
dI
L 0
C
dt
dQ
I
dt
d 2Q  1 

Q  0
2
dt
 LC 
harmonic oscillator with angular
1
frequency
0 
Q  Q peak cos(0t   )
dQ
I
 0Q peak sin( 0t   )
dt
LC
Natural Frequency
Lecture 20-5
Mechanical Analogy
d 2x  k 
k

x

0



0
 
dt 2  m 
m
No friction =
No dissipation
f 0  0 / 2
U  0, K  Kmax
U  U max , K  0
1 2
1 2
dx
U  kx , K  mv , v 
2
2
dt
U  U max , K  0
E  const.  U  K
harmonic oscillator
with
k
0 
m
Lecture 20-6
LC Oscillations
Q2
1 2
dQ
UE 
, U B  LI , I 
2C
2
dt
No Resistance =
No dissipation
Lecture 20-7
More on LC Oscillations
Charge and current: Q  Q peak cos 0t
(with =0)
I
dQ
 0Q peak sin 0t
dt
x
0,
r1
.. r1
n
Energy stored in capacitor:
1
1 2
U E (t ) 
Q peak cos2 0t
2C
UE
f( x ) 0.5
Energy stored in inductor:
1 2 2
U B (t )  L0 Q peak sin 2 0t
2
1
where
0 
LC
1 2
U B (t )  Q peak sin 2 0t
2C
so
U E (t )  U B (t ) 
2
Q peak
2C
00
0
2
tx
4
6
Period is half that of Q(t)
x
0,
r1
.. r1
n
UB
1
f( x ) 0.5
00 0
2
t
x
4
6
Lecture 20-8
Non-scored Test Quiz
A LC circuit has inductance L and capacitance C,
what’s the natural frequency?
A.
C
L
B.
LC
C.
1
LC
D.
L
C
Lecture 20-9
Series RLC Circuits
The resistance R may be a separate
component in the circuit, or the
resistance inherent in the inductor
(or other parts of the circuit) may
be represented by R.
Finite R
Energy dissipation
damped oscillation
dI
Q
L  IR   0
dt
C
d 2Q
dQ 1
L 2  R
 Q0
dt
dt C
only if R is “small”
d 2x
dx
m 2  b  kx  0
dt
dt
multiply by I
d 1 2 2
d  Q2 
 LI   I R  
0
dt  2
dt  2C 

For
large R
Lecture 20-10
Driven Series RLC Circuit
Kirchhoff’s Loop Rule:
dI
Q
  L  IR   0
dt
C
  Vapp , peak sin t
d 2x
dx
m 2 b
 kx  F0 sin t
dt
dt
common current I
I  I peak sin(t   )
must be determined
Lecture 20-11
Voltage and Current in Driven Series RLC Circuit
VR  I peak R
vR (t )  I (t ) R  VR sin t
dI

vL (t )  L  VL sin(t  )
dt
2
q(t )

vC (t ) 
 VC sin( t  )
C
2
VL  I peak L  I peak X L
I peak
VC 
 I peak X C
C
 (t )  vL (t )  vR (t )  vC (t )
  peak sin(t   )
 peak  I peak R   X L  X C 
2
Z

2
Phasors
Lecture 20-12
Impedance in Driven Series RLC Circuit
 peak  I peak R   X L  X C 
2
2
Z
1 

Z  R   L  
C 

2
2
impedance, 
1
L 
VL  VC
R

C
tan  

, cos  
VR
R
Z

Lecture 20-13
Resonance
For given peak, R, L, and C, the current amplitude Ipeak will be
at the maximum when the impedance Z is at the minimum.
 peak  I peak R   X L  X C 
2
2
Z
X L  XC
res L 
1
resC
i.e., load purely resistive
This is called resonance.
Resonance angular
frequency:
1
res 
LC
, Z  R, and I peak 
 peak
R
ε and I in phase
Lecture 20-14
Resonance (continued)
angular frequency   1
res
LC
(radians/s):
frequency (Hz):
f res 
1
2 LC
Phase difference
between ε and I:
X L  XC
R
tan  
 0, cos  
R
Z
Power dissipated:
• In a steady, driven RLC
circuit, power dissipated
= power supplied by ac
source.
• This power is dissipated
only in R.
• At resonance, this power
is maximum.

Lecture 20-15
Power Delivered
1 2
1   peak   1

Pav  I peak R 
I
R
peak



2
2  Z  2

 1
 1
R

 peak  
I peak 
 2
 2
Z
  rms I rms cos 
Pav   rms I rms

R
R
2
  rms 2
Z
Z
2
 rms
R
2
1 

2

L


R

 C 

2
 rms
R 2
 2 2
2 2
L (   res
)   2 R2
Power factor
R
 cos 
Z

Lecture 20-16
Lecture 20-17
Transformer
• AC voltage can be stepped up or
down by using a transformer.
• AC current in the primary coil
creates a time-varying magnetic
flux through the secondary coil
via the iron core. This induces
EMF in the secondary circuit.
Ideal transformer (no losses and magnetic
flux per turn is the same on primary and
secondary).
(With no load)
d  B V1 V2
 turn 
 
dt
N1 N 2
N1  N 2  V1  V2
N1  N 2  V1  V2
step-up
step-down
With resistive load R in secondary, current I2 flows in secondary by the induced
EMF. This then induces opposing EMF back in the primary. The latter EMF must
somehow be exactly cancelled because  is a defined voltage source. This occurs by
another current I1 which is induced on the primary side due to I2.
Lecture 20-18
Transformer with a Load
With switch S closed:
conservation of energy
I1V1  I 2V2
Imag+I1
proportional to
average power
I1 
V2
N
I2  2 I2
V1
N1
2
N 2 V2 1  N 2 
V1

 
V

 1
2
N1 R R  N1 
 N1 / N 2  R
equivalent resistance Req
The generator “sees” a resistance of Req
Impedance Matching: Maximum energy transfer occurs when
impedance within the EMF source equals that of the load.
Transformer can vary the “effective” impedance of the load.
I2
S
Lecture 20-19
Physics 241 –Quiz 17b – March 25, 2008
An LC circuit has a natural frequency of 141 MHz.
If you want to decrease the natural frequency to
100 MHz, which of the following will accomplish
that?
a) Double L
b) Double both L and C
c) Halve L
d) Halve both L and C
e) Double L and halve C
Lecture 20-20
Physics 241 –Quiz 17c – March 25, 2008
An LC circuit has a natural frequency of 100 MHz.
If you want to decrease the natural frequency to 71
MHz, which of the following will accomplish that?
a) Double C
b) Double both L and C
c) Halve C
d) Halve both L and C
e) Double L and halve C