Lecture 12 Electromagnetic Oscillations and
Alternating Current Chp. 33
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Cartoon -. Opening Demo - Warm-up problem
Physlet
Topics
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LC Circuit Qualitatively
Electrical and Magnetic energy oscillations
Alternating current
\Pure R and L, circuti
Series RLC circuit
Power and Transfomers
Demos
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LR circuit
Series LRC circuit
Axis of rotation
Coil of wire

 m  B  nˆdA
 
 B  dA
 B cosdA
n̂

B
d
d(BA cos  )
d cos 
d


 BA
 BA sin 
dt
dt
dt
dt
d
 BA  sin  but   t so

dt
  BA sin t
  m sin t
2pf and f= 60 Hz
Where is the rotational angular
frequency of the generator
phase
  m sin( t  )
Instantaneous voltage

Amplitude
time
Angular frequency
m
Phasor diagram

t  
Phase constat
vR  Ri
  vR  vL
i  I sin( t)
  m sin t
  2pf
di
 I cos(t)
dt

f 1000Hz
di
v L  L LI cos(t)
dt


L = 4.22mH
vL  LIcos(t)
I
R R
V


VR=RI
I
X  L  L
V
L
VL= XLIL or VL= (L)I since I=IL

Impedance Z: New quantity for AC circuits.
This is analogous to resistance in DC circuits
Z  R  (L)
2
I
I
2
m
Z
m
R  (L)
2
XL  L
2

RL Circuit Example
Suppose m = 100 volts, f=1000 Hz, R=10 Ohms, L=4.22 mH,
Find XL, Z, I, VR, and Vl.
XL  L  6.28 1000  0.00422H  26.5
Z
R 2  (L) 2
Z  10 2  (26.5) 2  28.3
m
100
I

 3.53A
Z 28.3
VR  RI 10  3.53  35.3v
VL  XL I  26.5  3.53  93.5v
Power in AC circuits
P  i 2R  (I sin( t)) 2 R
Instantaneous power doesn’t mean anything
Need to average over time or one period of the sine wave
Pavg 
2p
1
2p
 Rd (I sin(  )) 2 
0
Note
2p
1
2p
RI 2  sin 2d  RI 2
0
I
Irms 
2
2
Pavg  Irms
R
1
I
 ( )2 R
2
2
Averaging over a sine curve
Calculate Power lost in resistor from example
2
Pavg  Irms
R
I
3.53A
Irms 

 2.50A
2 1.414
Pavg  (2.50A)210  62.5Watts
To calculate power produced by the generator you
need to take account of the phase difference between
the voltage and the current. In general you can write:
Pavg  rmsIrms cos 
For an inductor P = 0 because the phase difference between
current through the inductor and voltage across the inductor is
90 degrees
Series LRC circuit
VR
  vR  vC  vL
VC
  m sin t
i  Isin( t  )
I
R  (X L  XC )
2
m
R  (X L  XC )
2

2

m
XC=1/(C)
2
VL XL=L
Z
1
L 
C
tan  
R
1 2
Z  R  (L 
)
C
2
ELI the ICE man
Resonance
X L  XC
1
L 
C
1

LC
Series LRC demo
10 uF
4.25 mH
1
1

6.28 LC 6.28 4.25 103 H 106 F
f  2442Hz
f 
Series LCR circuit