Transcript Document

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
nˆ

B
d
d(BAcos )
d cos
d


 BA
 BAsin 
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
v R  Ri
  vR  vL
i  I sin(t)
  m sint
  2pf
di
 I cos(t)
dt

f  1000 Hz
di
v L  L LI cos(t)
dt

v L  LI cos( t)

I
R R
V

L = 4.22mH
I
X  L  L
V
L

VR=RI
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  R2  (L)2
I

I
m
Z
m
R  (L)
2
X L  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.
X L  L  6.28 1000  0.00422 H  26.5
Z  R2  (L)2
Z  102  (26.5)2  28.3
m
100
I 
 3.53A
Z 28.3
VR  RI  10  3.53  35.3v
VL  X L I  26.5  3.53  93.5v
Power in AC circuits
P  i 2 R  (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
Irms
2
Pavg  Irms
R
I

2
2p
1
2p
1
I 2
RI  sin d  RI  ( ) R
2
2
0
2
2
2
Averaging over a sine curve
Calculate Power lost in resistor from example
2
Pavg  Irms
R
Irms
I
3.53A


 2.50 A
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  rms Irms 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
  v R  vC  v L
VC
  m sint
i  I sin(t   )
I
R  (XL  XC )
2
m

R  (X L  XC )
2
Z  R 2  (L 
2

m
XC=1/(C)
2
VL XL=L
Z
1
L 
C
tan 
R
1 2
)
C
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