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Transition to AC
W11-3
What’s Happening??
Quiz Today
Labs look pretty crappy
– There is an education issue here!
Exam #3 Next Wednesday
Study Session Monday Morning if there is
interest!
Next topic is AC Circuits
First, let’s do a bit of a clicker review …




Inductors and how they work
How they work in circuits … purely DC
circuits but not necessarily constant.
There are a few topics that we have missed
and I want to go over them today.
Then, if time, you can start the AC unit.
 0 NI
B
2r
B=0 outside
4
Magnetism
 B 2  The magnetic flux in
coil #2 due to the current in
coil #1.
 B 2 ~ i1
 B 2  ki1
i2
N 2  B 2  M 21 i1
 21
i1
emf2  N 2
 M 21
t
t
mutual Inductanc
Induction
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5
M
N something
UNIT: henry
i
Induction
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6
Minimum Change@t
V1
Maximum Change@t
Induction
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7
The Same

emf1  V1  N1
t
FLUX is the same through
both coils (windings).

emf 2  V2  N 2
t
V1 V2

N1 N 2
Induction
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8
Induction
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9
Induction
5/23/2017
10
V2 N 2

V1 N1
V2
V1 
(N 2 / N1 )
Powerin  Powerout (Lossless)
V2
V22 V12 (N 2 / N1 ) 2
I1V1  I 2V2  V2 

R
R
R
V1 (N 2 / N1 ) 2
I1 
R
So
V1
R

I1 ( N 2 / N1 ) 2
 looks like an input 


resistance!


Induction
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11
U=Area=(1/2)LI2
Power  Vi
i
V L
t
i
P  Li
t
U  Lii
U
Li
LI
i
Li

DU
 Lii
interval
i
Induction
i
I
Induction
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12
Energy Stored in a capacitor
The energy stored in a capacitor with capacitance C and a voltage
V is
U=(1/2)cV2
U=(1/2)LI2
Induction
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13
Consider a solenoid with N turns that is very long. We assume that
the field is uniform throughout its length, ignoring any “end
effects”. For a long enough solenoid, we can get away with it for
the following argument. Maybe.
N
B   0 ni   0 i
l
Induction
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14
N B
l
1 2 1 N B 2
U  Li 
i
2
2 l
 B  BA
L
1
NiBA
2
N
B  0 i
l
Bl
Ni 
U
0
1 Bl
1 B 2lA 1 B 2V
U
BA 

2 0
2 0
2 0
U 1 B2
Energy Density  
V 2 0
1
Capacitor ED   0 E 2
2
SUPER IMPORTANT !
Induction
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15
Induction
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16
Up and Down and Up and Down and …..
5/23/2017
Induction
17
Frequency
1

LC
5/23/2017
Induction
18
Alternating Currents &
Voltages
Some Math First!
r  
ac generator
“Output” from the previous diagram
 2
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(t)
A=Amplitude (=1 here)

sin( t  )   cos(t )
2
f(t)=A sin(t-[/2])
A=Amplitude (=1 here
For the future

sin( t  )   cos(t )
2

cos(t  )  sin( t )
2
  2f
AC Applied voltages
This graph corresponds
to an applied voltage
of V cos(t).
Because the current
and the voltage are
together (in-phase) this
must apply to a Resisto
for which Ohmmmm sa
that I~V.
phasor
oops – the ac phaser
i  I cos(t )
the resistor
v  iR  IR cos(t )
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 t )
L
I (cos(t  t )  cos(t ))
t
t

cos(t ) cos(t )  sin(t ) sin(t )  cos(t )
 LI
cancel
vL
t
When t gets very small, Let's look at what's left :
cos (t) goes to 1.
sin(t ) sin(t )
vL  LI
?
?
t
r  
lim 0
sin( )

1
this leaves v L
  sin(t )
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(t)
A=Amplitude (=1 here)
f(t)=A sin(t-[/2])
A=Amplitude (=1 here)

sin( t  )   cos(t )
2
result - inductor
vL  LI sin(t )

sin( t  )   cos(t )
2
I is the MAXIMUM
current in the
circuit.
sin(t )   cos(t 
vL  LI cos(t 

2

2
)
)
Resistor
comparing
v  iR  IR cos(t )
vRmax  I R
inductor
vL  LI cos(t 

2
)
vLMax  I L
(L) looks like a resistance
XL=L
Reactance - OHMS
For the inductor
vLMax  I L  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(t)
and the voltageas
v  Vcos(t   )
where  is thephaseshift between the
currentand the voltage.

the phasorvL  LI cos(t  2 )
vL  LI cos(t 

2
)
cos(t 

2
)   sin(t )
direction
t
t 

2
t
remember for ac series circuits
In the circuit below, R=30 W and L= 30 mH. If the angular fr
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
=3 KHZ
E=60V
L= 30 mH
R=30 W
=3 KHZ
E=60V
L= 30 mH
R=30W
XL=L=90W
The instantaneous voltage
across
each element is the
PROJECTION
This is the SAME as the
of
theofMAXIIMUM
voltage
sum
the
onto
maximum vectors
the
horizontal
projected
ontoaxis!
I
the horizontal axis.
VL  IX L
VR  IR
t
Source voltage leads
the current by the angle .
I
V  Vmax

VL  IX L
Let V  IZ
t
VR  IR
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
=3 KHZ
E=60V
1
L= 30 mH
The drawing is obviously
NOT to scale.

VL  IX L
Let V  IZ
I
t
VR  IR
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
=3 KHZ
R=30 W E=60V
2
V
60
I 
 0.632A
Z 94.9
X L  L  90W
Z  (30) 2  (90) 2  94.9W
VL
  tan ( )
VR
1
 90 
0
  tan    71
 30 
1
R  30W
X L  L  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  IL sin(t )  0.63x30000x0.03sin(t )  567sin(t )
vR  0.63x30x cos(t )  18.9 cos(t )
Let's look at thesum of these two voltagesthat
should look like theAC source.
vL   IL sin(t )  0.63x3000x0.03sin(t )  56.7 sin(t )
vR  0.63x30x cos(t )  18.9 cos(t )
Let's look at thesum of these two voltagesthat
should look like theAC source.
wt
56.7Sin(wt)
18.9 cos(wt)
SUM
0
0
18.9
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
Enough!