Download Set 1 - UCF Physics

PHY-2054
ALTERNATING CURRENT
PART 1
THIS WEEK
We will do the lab session on the inductor’s
time constant.
 We will discuss what happens when the applied
voltage is a sinusoid.
 We will do some experiments on AC circuits
 We will have a quiz on Friday.


NOTE: Start
Circuits
reading chapter 23! – AC
MONDAY
MONDAY
WEDNESDAY
FRIDAY
29
AC
AC
AC
5
WAVES
EXAM III
OPTICS
12
OPTICS
OPTICS
OPTICS
19
OPTICS
OPTICS
OPTICS
26
EXAM IV
Date and Time of Final is being investigated.
AC GENERATOR – SIMILAR TO THE MOTOR BUT
REALLY DIFFERENT … WHATEVER THAT MEANS!
“OUTPUT” FROM THE PREVIOUS DIAGRAM
 2
DC /AC
BUT NOT ALWAYS! (CAPACITOR)
SO, LET’S TALK ABOUT PHASE
Y=F(X)=X2
30
25
20
15
10
5
0
0
1
2
3
4
5
6
Y=F(X-2)=(X-2)2
y
30
x2
25
20
15
10
(x-2)2
2
5
0
0
1
2
3
4
5
6
x
THE “RULE”
f(x-b): shift a distance b in the POSITIVE
direction
 f(x+b): shift a distance n in the NEGATIVE
direction.


The signs switch!
THE SINE

2
LET’S TALK ABOUT PHASE
f(t)=A sin(wt)
A=Amplitude (=1 here)
f(t)=A sin(wt-[/2])
A=Amplitude (=1 here)

sin( wt  )   cos(wt )
2
A SINE WAVE SHIFTED BY P RADIANS IS
A. cosine
B. - sine
C. -cosine
D. sine
E. tangent
FOR THE FUTURE

sin( wt  )   cos(wt )
2

cos(wt  )  sin( wt )
2
w  2f
AC APPLIED VOLTAGES
This graph corresponds
to an applied voltage
of V cos(wt).
Because the current
and the voltage are
together (in-phase) this
must apply to a Resistor
for which Ohmmmm said
that I~V.
OOPS – THE AC PHASER
i  I cos(wt )
THE RESISTOR
v  iR  IR cos(wt )
PHASOR DIAGRAM
Pretty Simple, Huh??
VR  IR
HERE COMES TROUBLE ….
We need the relationship between I (the current through)
and vL (the voltage across) the inductor.
FROM THE LAST CHAPTER:
i
vL  L
t
* unless you have taken calculus.
CHECK IT OUT---
 means change or difference .
(thing)  thing final  thing initial
(t)  t final  t initial
SOvL  L
i
t
L
 ( I cos wt )
L
I (cos(wt  t )  cos(wt ))
t
t

cos(wt ) cos(wt )  sin(wt ) sin(wt )  cos(wt )
 LI
cancel
vL
t
When t gets very small,
cos (wt) goes to 1.
Let's look at what's left :
sin(wt ) sin(wt )
vL  wLI
w

t
?
?
r  
lim 0
sin( )

1
THIS LEAVES
vL  w sin(wt )
The resistor voltage looked like a cosine so we would like the
inductor voltage to look as similar to this as possible. So let’s
look at the following graph again (~10 slides back):
f(t)=A sin(wt)
A=Amplitude (=1 here)
f(t)=A sin(wt-[/2])
A=Amplitude (=1 here)

sin( wt  )   cos(wt )
2
BREAK
Where bee we?
 Equipment didn’t work on Monday but it should be working
today. You finished all of the calculations in the previous
hand-out.
 Today we will begin with a look at LR circuits:
 LR with a square wave input so you can determine the time
constant.
 LR with AC so you can look at phase relationships as well as
inductive reactance
 Add a capacitor and look at an RLC circuit to determine
resonance conditions as well as phase relationships.
 The previous will take at least one additional session. Maybe
two!
Keepeth in Mind
 Note the appearance of a new WebethAssignment.
 Quiz on Friday
 Exam Next Wednesday – Magnetism  AC circuits
 Monday – we will try to begin optics. Some of the
AC may spill over into that session
 Starting Monday, Dr. Dubey will take over the class
as lead instructor. She is a better teacher than I am.
 START STUDYING FOR EXAM III!!!
RESULT - INDUCTOR
vL  wLI sin(wt )

sin( wt  )   cos(wt )
2
I is the MAXIMUM
current in the
circuit.
sin(wt )   cos(wt 
vL  wLI cos(wt 

2

2
)
)
RESISTOR
v  iR  IR cos(wt )
vRmax  I R
COMPARING
INDUCTOR
vL  wLI cos(wt 

2
vLMax  I wL
(wL) looks like a
resistance
XL=wL
Reactance - OHMS
)
FOR THE INDUCTOR
vLMax  I wL  IX L  VL
FOR THE RESISTOR
vRMax  IR  VR
SLIGHTLY CONFUSING POINT
We will always use the CURRENT as the basis for calculations
and express voltages with respect to the current.
What that means?
We describe thecurrent as varyingas :
i  I cos(wt)
and the voltageas
v  Vcos(wt   )
where  is thephaseshift between the
currentand the voltage.

THE PHASOR vL  wLI cos(wt  )
2
vL  wLI cos(wt 

2
)
cos(wt 

2
)   sin(wt )
direction
wt
wt 

2
wt
REMEMBER FOR AC SERIES CIRCUITS
In the circuit below, R=30 W and L= 30 mH. If the angular frequency
of the 60 volt AC source is is 3 K-Hz
WHAT WE WANT TO DO:
(a) calculate the maximum current in the circuit
(b) calculate the voltage across the inductor
(c) Does Kirchoff’s Law Work?
R=30 W
E=60V
L= 30 mH
w=3 KHZ
R=30 W
E=60V
R=30W
XL=wL=90W
w=3 KHZ
L= 30 mH
The instantaneous voltage across
each element is the PROJECTION
of the MAXIIMUM voltage onto
the horizontal axis!
This is the SAME as the sum of the
maximum vectors projected onto
the horizontal axis.
I
VL  IX L
VR  IR
wt
Source voltage leads
the current by the angle .
I
V  Vmax

VL  IX L
Let V  IZ
VR  IR
wt
Z  Impedance
I Z  I R  I X L or
2
2
2
2
Z  R 2  X L2
VL
  tan ( )
VR
2
2
R=30 W
E=60V
1
L= 30 mH
w=3 KHZ
The drawing is obviously
NOT to scale.
I
R=30 W

VL  IX L
Let V  IZ
VR  IR
wt
L= 30 mH
Z  Impedance
I Z  I R  I X L or R  30W
2
2
2
2
Z R X
2
2
2
L
w=3 KHZ
E=60V
2
V
60
I 
 0.632A
Z 94.9
X L  wL  90W
Z  (30) 2  (90) 2  94.9W
VL
  tan ( )
VR
1
 90 
0
  tan    71
 30 
1
R  30W
X L  wL  90W
Z  (30) 2  (90) 2  94.9W
 90 
0
  tan    71  1.25 rad
 30 
1
V
60
I 
 0.632A
Z 94.9
vL  IwL sin(wt )  0.63x30000x0.03sin(wt )  567sin(wt )
vR  0.63x30x cos(wt )  18.9 cos(wt )
Let's look at thesum of these two voltagesthat
should look like theAC source.
vL   IwL sin(wt )  0.63x3000x0.03sin(wt )  56.7 sin(wt )
vR  0.63x30x cos(wt )  18.9 cos(wt )
Let's look at thesum of these two voltagesthat
should look like theAC source.
wt
56.7Sin(wt)
0
18.9 cos(wt)
0
18.9
SUM
18.9
0.2 11.26455106 18.52325832 7.258707
0.4 22.08002001 17.40805279 -4.67197
0.6 32.01522824 15.59884312 -16.4164
0.8 40.67409035 13.16775681 -27.5063
1 47.71140484 10.21171358 -37.4997
1.2 52.84661617
6.84856156 -45.9981
1.4 55.87499969 3.212379001 -52.6626
IT DOES!
80
2
60
40
20
56.7Sin(wt)
18.9 cos(wt)
0
0
-20
-40
-60
-80
1
2
3
4
5
6
7
8
SUM
WHAT ABOUT THE CAPACITOR??
C:
q
vc 
c
vc 1 q 1
1

 i  I cos(wt )
t
c t c
c
Without repeating what we did, the question is what function will have
a f/t = cosine? Obviously, the sine! So, using the same process
that we used for the inductor,
1
vc 
I sin(wt )
wC
1
Xc 
(ohms)
wC
CAPACITOR
PHASOR
DIAGRAM
1
vc 
I sin(wt )
wC
1
Xc 
(ohms)
wC
I

vC 
cos(wt  )
wC
2
NOTICE THAT
The voltage lags
the current by 90
deg
 I and V are
represented on
the same graph
but are different
quantities.

SUMMARY
FOR AC CIRCUITS
RESONANCE
When X L  X C
Z  R Current is Maximized
 0
1
wL 
wC
1
2
w 
LC
w  resonance
1
LC
AC CIRCUITS LOOK COMPLICATED
http://www.ngsir.netfirms.com/englishhtm/RLC.htm
When you get a chance – check this site out.
stoppeth here
An AC source with ΔVmax = 125 V and f = 25.0 Hz
is connected between points a and d in the
figure.
Calculate the maximum voltages between the following
points:
(a) a and b 62.8 V
(b) b and c 45.6 V
(c) c and d 154 V
(d) b and d 108 V