Download AC Circuits - Part II

PHY-2054
J. B. Bindell
CHAPTER 22 – ALTERNATING CURRENT
PART 2
THE FUTURE
Monday – Continue with Chapter 22
 Wednesday –

 Hope
to return the exams.
 more of the same

Friday –
 Quiz
 No

10:30 AM office hours (sorry)
Next Monday – Probably a problem solving
session at 7:00AM. Confirmation later in the
week.
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
v  iR  IR cos(t )
vRmax  I R
COMPARING
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 PHASOR vL  LI cos(t  )
2
vL  LI cos(t 

2
)
cos(t 

2
)   sin(t )
direction
t
t 

2
t
REMEMBER FOR AC SERIES CIRCUITS
Compute the reactance
of a 0.450 H inductor at
frequencies of 60.0 Hz
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
=3 KHZ
R=30 W
E=60V
R=30W
XL=L=90W
=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
t
Source voltage leads
the current by the angle .
I
V  Vmax

VL  IX L
Let V  IZ
VR  IR
t
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
=3 KHZ
The drawing is obviously
NOT to scale.
I
R=30 W

VL  IX L
Let V  IZ
VR  IR
t
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
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
WHAT ABOUT THE CAPACITOR??
C:
q
vc 
c
vc 1 q 1
1

 i  I cos(t )
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(t )
C
1
Xc 
(ohms)
C
CAPACITOR
PHASOR
DIAGRAM
1
vc 
I sin(t )
C
1
Xc 
(ohms)
C
I

vC 
cos(t  )
C
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
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
AC CIRCUITS LOOK COMPLICATED
http://www.ngsir.netfirms.com/englishhtm/RLC.htm