Transcript AC Circuits - Part I

PHY-2054
J. B. Bindell
CHAPTER 22 – ALTERNATING CURRENT
PART 1
UPCOMING ITEMS FOR YOUR CONSIDERATION
No problem session on Monday (Not much to
do!)
 Watch for a Mastering Physics Assignment
 Quiz next Friday on AC … will it ever end?
 Today we start AC Circuits


We will review the exam when they are
returned.
EXAM STRUCTURE
Question Number
1
Section 003
Section 004
Multiple Choice (5)
Multiple Choice (5)
2-Wire Problem – Different
currents in each class and
different from the quiz
2-Wire Problem – Different
currents in each class and
different from the quiz
3 (similar or exactly from
past.)
Inverted coil problem.
Coil problem from class.
4 (New)
Charged particle orbit.
Electron & proton in same
orbit.
2 (seen before)
We will retain this schedule.
PHY2054 Problem Solving/Office Hours Schedule
Room MAP-318
Bindell
Monday
8:30-9:15AM
Tuesday
Bindell
Dubey
11:00-12:00PM
12:00-1:00PM
Wednesday
8:30-9:15AM
10:30 - 11:15
AM*
Thursday
Friday
8:30-9:15AM
10:30-11:30AM
1:30-2:45PM
These sessions will be used both for office hours and problem solving. Students from any
section of 2054 are invited to stop by for assistance in course materials (problems, etc.)
Note: There will be times when the room may not be available. In that case we will use our individual offices.
* In Office
Dr. Dubey's hours are for problem
solving only.
If something else is going on in the room, come to my office!
If I am not there …. come to my office.
OK … HOW WAS THE TEST?
A.
B.
C.
D.
Easy
OK
Difficult
Impossible
HOW DID YOU DO?
A.
B.
C.
D.
E.
great
less than stellar
ok
poor
bombed
THIS TIME I DID
A.
B.
C.
better than last time
about the same as last
time
worse than last time
AC GENERATOR
“OUTPUT” FROM THE PREVIOUS DIAGRAM

2
NUCLEAR
DC /AC
HOME GENERATORS
FUEL
WHAT WORKS ON AC?
BUT NOT ALWAYS! (CAPACITOR)
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
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.
PHASOR
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
RESULT
vL  wLI sin(wt )

sin( wt  )   cos(wt )
2
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
)
SLIGHTLY CONFUSING POINT
We will 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
current and the voltage.
BACK TO THE PHASOR THING
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