bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb nnnnnnnnnnnnnnn jjjjjjjjjjjjjjjjjjjjjjbllllllllllll ttttttttttttttttt

1. AC Networks
Chapter 2

2. Wave Description
 Amplitude (Vpeak) - maximum displacement
from the zero line
 Period (T) - the time required for one cycle
 Frequency (f) - the number of cycles that
appear in a time span
 1 cycle per second = 1 Hertz
 T=1/f

3.  Angular Velocity (w) - rotation rate of a
vector (or generator shaft)
 w is measure in radians per second.
 New Waveform Equation: v=Vpsinwt
f
2
T
2




w

4. Sinusoidal Voltage Relationships
 v = Vpeaksin q
 v = Vpeaksin wt
 v = Vpeaksin 2t/T
 v = Vpeaksin 2ft
 Which form you use depends on what you know

5. Phase Relationship
 f = phase angle
 For initial intersections before 0
v=Vpsin(wt+f)
 For initial intersections after 0
v=Vpsin(wt-f)

6. Resistive Circuit
 Only for a resistive circuit can current and
voltage be in phase.
 If the current and voltage go through zero
at different times, they are said to be out
of phase

7. Sinusoidal Plot

8. Example Problems
 Example 4.7:
o Determine the phase relationship between the
following two waveforms:
v=8.6 sin(300t+80°)
i=0.12 sin(300t+10°)

9. 4.4 Effective (RMS) Value
 These mean the same thing for AC circuits:
o “equivalent DC voltage”
o “effective voltage”
o “rms voltage”
 RMS = root mean square
 Draw the “relating effective values” plots.
p
eff V
2
1
V  p
eff I
2
1
I 

10. 4.5 Average Value
 For a sine wave, the average value is zero.
 For other waves….
)
(
)
(
)
(
period
T
waveform
under
area
value
average
G 

11. Example Problems
 Example 4.9:
o Write the sinusoidal expression for a voltage having
an rms value of 40 mV, a frequency of 500 Hz and and
initial phase shift of +40.
o Vp = 1.414(Vrms) = 1.414(40 mV) = 56.56 mV
 w = 2f = (2)(0.5 kHz) = 3.142 X 103 rad/s
o v = 56.56 mV sin(3142t + 40°)

12. Example Problems
 Example 4.10:
Calculate the average value of the waveform
below over one full cycle.
-6
-4
-2
0
2
4
6
8
10
12
0 2 4 6 8 10 12

13. What are the following for a wall outlet?
 f
 Veff = Vrms
 Vpeak
 Vave
 Ieff = Irms
 Ipeak
 Iave
2
 f = 60 Hz
 Veff = Vrms = 120 V
 Vpeak = Vrms = 170 V
 Vave = 0 V
 Ieff = Irms = limited by breaker
 Ipeak = Irms
 Iave = 0 A
2

14. Effective Values
 Average values for I or V are zero
 If we use IV to measure power with a dc
meter, we will get zero
o Yet the resistor gets hot – power is dissipated
o I and V are not always zero – you can get
shocked from ac just like dc
 We need a new definition for power

15. Use Effective (rms) Values
 PR = VRIR = IR
2R = VR
2/R
 Look just like the dc forms of power
o But we use rms values instead

16. 4.7 The R, L, and C Elements
 Ohm’s Law for Peak Values
Resistors: Vpeak=IpeakR
Capacitors: Vpeak=IpeakXC
Inductors: Vpeak=IpeakXL

17. Inductive Reactance
 Recall the form of inductance
 Since I for ac is constantly changing, there is a
constant opposition to the flow of current
o Inductive Reactance XL = wL = 2fL
o Measured in ohms
o Energy is stored in the coil, not dissipated like in the
resistor.
t
i
L
v





18. Example
 An inductor of 400 mH is connected
across a 120 V, 60 Hz ac source. Find XL
and the current through the coil
a. XL = 2fL = 2(60 Hz)(0.4 H)
XL = 151 W
b. I = V/XL = 120 v/151 W
I = 0.8 A

19. XL vs. f graph for ideal inductor
 XL = 2fL
o Should give a straight line graph of slope 2L
o Higher the L the greater the slope
XL
f

20. Current
V and I for an Inductor
 From we see that voltage is greatest
when the change of current is greatest.
t
i
L
v




Highest
positive v
Zero here
Most negative
here

21. Voltage and Current (Coil)
Current
Voltage
Voltage leads current by 90°

22. Example Problems
 Example 4.13:
o The current iL through a 10-mH inductor is
5 sin(200t +30). Find the voltage vL across
the inductor.

23. Capacitive Reactance
 The effect of a capacitor is to prevent
changes in voltage
o Keep the potential difference from building up
in the circuit
 Since the build up of potential is minimized,
the flow of current is reduced.
o Capacitors act like a resistance in an ac circuit.

24. Capacitive Reactance
 V is changing in ac circuits
o But Q = VC, so the charge changes also
o Since , current is greatest when change in
voltage is greatest.
t
Q
I



Voltage
Current
I leads V by
90°

25. CIVIL
 CIVIL is a memory aid.
 For a capacitor C, current I leads the voltage
V by 90 degrees.
 For an inductor L, voltage V leads the
current I by 90 degrees.

26. Capacitive Reactance
C
f
Xc
2
1

 Xc also measured in ohms
What is the capacitive reactance of a 50 mf capacitor
when an alternating current of 60 Hz is applied?
W


 
53
)
10
50
)(
60
(
2
1
2
1
6
c
c
X
f
X
Hz
fC
X



27. Form of Reactance (Fig. 4.28)
0
50
100
150
200
250
300
0 1000 2000 3000
f (Hz)
X
C
(
W
)
C=1mF
C=2mF

28. Example Problems
 Example 4.14:
o The voltage across a 2-mF capacitor is 4mV
(rms) at a phase angle of -60. If the applied
frequency is 100kHz, find the sinusoidal
expression for the current iC.

29. 3.6 Phasors and Complex Numbers
 A vector is a quantity has both magnitude and
direction.
 A scalar quantity has only a magnitude.
 Examples:
o Scalar: 50mph
o Vector: 50mph North

30.  A phasor is a complex number used to
represent a sine wave’s amplitude and
phase.
 A complex number can be written as
C = A + Bj
where C is a complex number,
A and B are real numbers and
1
j 


31. Real
Imaginary
C=A+Bj
q
What is the magnitude of C?
What is the direction of C?
2
2
B
A
C 

A
B
tan 1


q

32. Phasor Diagrams
 We make the real axis the resistance
 The imaginary axis is reactance
o Inductive reactance is plotted on the +y axis
o Capacitive reactance is plotted on the –y axis
 The vector addition of R and X is Z
R
XL
XC
X = XL - XC

33. Phasors
 Form Reactance First X = XL – Xc
 Form Impedance
 Calculate phase angle
2
2
X
R
Z 

)
R
X
(
Tan 1


f
XL
XC R

R
X 
R
X Z

34. Example Series R-L Circuit
 A coil has a resistance of 2.4 W and a
inductance of 5.8 mH. If it is connected to
a 120-V, 60 Hz ac source, find the reactance
and the current

35. Solution
XL = 2fL = 2(60 Hz)(5.8 X 10-3 H)
XL = 2.19 W
W
W
W
25
.
3
Z
)
19
.
2
(
)
4
.
2
(
X
R
Z 2
2
2
2








 

4
.
42
)
4
.
2
19
.
2
(
Tan
)
R
X
(
Tan 1
1
f
W
W
f
XL
R
X
R
Z
f

36. Solution (cont’d)
 I = V/Z
 I = 120 V/3.25 W = 36.9 A
================================
 Notice that if f = 120 Hz, then
 XL = 2(120 Hz)(0.0058 H) = 4.37 W
 Z = 4.99 W f = 61.2°
 I = 24 A

37. Example Series R-C Circuit
 What is the current flow through a circuit
with 120 V, 60 Hz source, an 80 mf
capacitor, and a 24 W resistor?

38. Solution
W


33
X
)
f
10
X
80
)(
Hz
60
(
2
1
C
f
2
1
X
C
6
C


 
W
W
W
9
.
40
Z
)
33
(
)
24
(
X
R
Z 2
2
2
2





A
93
.
2
9
.
40
V
120
Z
V
I 


W
R
XC
R
X
Z
f




1
.
54
38
.
1
24
33
tan
f
W
f

39. Solution (cont’d)
 What happens is the frequency doubles?
 XC = 16.6 W Z = 29.2 W f = 34.6°
 I = 4.11 A

40. Example - Series RLC Circuit
XL
XC R

R
X 
R
X Z
X = XL - XC Z2 = R2 + X2
A 240 V, 60 Hz line is connected to a circuit containing a
resistor of 15 W, an inductor of 0.08 H, and a capacitor of 80
mf. Compute the circuit impedance and current.

41. Solution
W

 2
.
30
)
H
08
.
0
)(
Hz
60
(
2
L
f
2
XL 


W


2
.
33
)
f
10
X
80
)(
Hz
60
(
2
1
C
f
2
1
X 6
C 

 
W
W
W 0
.
3
2
.
33
2
.
30
X
X
X C
L 





W
W
W 3
.
15
)
3
(
)
15
(
X
R
Z 2
2
2
2






A
7
.
15
3
.
15
V
240
Z
V
I 


W

42. Power Factor
 In dc circuits we had P = IV and this still
holds in purely resistive circuits with ac
 In general, I and V are out of phase in ac
circuits.
o At times V and I are negative and power is
drawn from the circuit.
o Thus the true power must be less than IV

43. Power Factor
)
t
sin(
t
sin
V
I
iv
p
)
t
sin(
V
v
t
sin
I
i
p
p
p
p
f
w
w
f
w
w






)
factor
power
(
IV
P
IV
power(P)
true
Factor
Power


f is the phase between i and v
f
f cos
V
I
cos
2
V
I
P
T
cycle
one
over
p
of
Sum
P
rms
rms
p
p




44. Power Factor
 Power Factor = cos f
X
R
Z
f X
2
2
X
R
R
Z
R
cos



f
For purely inductive or
purely capacitive circuits,
f = 90° and cos f = 0 -
no power is dissipated

45. Example
 An industrial plant service line has an impressed
power of 100 KV-A on a 60 Hz line at 440 V.
An installed motor offers a total ohmic resistance
of 350 W and an inductance of 1 H. Find…
a) The inductive reactance
b) The impedance
c) The current
d) The power factor
e) The power used by the motor
f) The size capacitor required to change PF to 95%.

46. Solution
W
W
W 4
.
514
)
377
(
)
350
(
X
R
Z
)
b 2
2
2
L
2





W

 377
H)
Hz)(1
(60
2
fL
2
X
a) L 


A
855
.
0
4
.
514
V
440
Z
V
I
c) 


W
68
.
0
4
.
514
350
Z
R
PF
)
d 


W
W
KW
68
)
PF
)(
A
KV
100
(
P
)
e 



47. Solution (cont’d)
W
42
.
368
95
.
0
R
Z
Z
R
95
.
0
PF




W
W
W
W 262
)
350
(
)
42
.
368
(
377
X
R
Z
X
)
X
X
(
R
Z
)
X
X
(
R
Z
2
2
C
2
2
L
2
C
L
2
2
2
C
L
2
2













F
1
.
10
)
262
)(
Hz
60
(
2
1
X
f
2
1
C
C
f
2
1
X
C
C
m
W








48. Series Resonance
2
C
L
2
)
X
X
(
R
Z 


If XL = XC, then Z = R and the circuit is purely resistive.
This will occur at a certain frequency
LC
2
1
f
1
LC
f
4
C
f
2
1
L
f
2
res
2
2







By altering L or C, the circuit can
be made to resonate at any
frequency.
This condition makes X = 0
V and I will be in phase
minimum Z and maximum I

49. Parallel RC Circuit
 General Scheme
o Form impedance for each branch
o From Kirchhoff’s Current Law
 iT = i1 + i2
)
t
sin(
Z
V
i
)
t
sin(
Z
V
i
2
2
2
1
1
1
f
w
f
w




The possibility of different
phase angles means parallel
impedances do not add as
simply as for dc circuits

50. Total Current and Definitions









 )
t
sin(
Z
1
)
t
sin(
Z
1
V
i 2
2
1
1
T f
w
f
w
2
2
2
2
2
2
1
2
1
1 X
R
Z
X
R
Z 



Define 2
2
Z
X
B
and
Z
R
G 

G is called the conductance
B is called the susceptance

51. Rule for Adding Parallel Impedances
2
1
2
1
T
2
2
1
2
2
1
T
G
G
B
B
tan
)
B
B
(
)
G
G
(
Z
1







q
Note: by convention, XC always carries a negative sign in
computing the impedances.
Y = 1/ZT is called the admittance

52. Example

53. Resistive Elements
 If parallel impedance contains only resistors, we
have
o G = 1/R B = 0
R
1
R
1
B
G
Y
Z
1
2
2
2
T





And the equation of parallel resistances results.

54. Example
W
5
R
Z1 
 0
B
R
1
G
0
X 1
1
1 


C
2
C
2 X
X
Z 

S
0377
.
0
B
C
f
2
X
1
Z
X
B
0
G
2
C
2
2
C
2
2







 
W
W
91
.
4
Z
S
204
.
0
)
0377
.
0
(
)
5
(
1
)
B
B
(
)
G
G
(
Y
T
2
2
2
2
1
2
2
1









55. Parallel Resonance
C
2
C
3 X
X
Z 
 C
f
2
X
1
Z
X
B
0
G
C
2
3
C
3
3 







L
2
L
2 X
X
Z 

Add L and C in parallel
L
f
2
1
X
1
B
0
G
L
2
2





56. Parallel Resonance
   
   
3
2
2
3
2
2
3
2
2
3
2
T
B
B
B
B
B
B
G
G
Z
1








C
f
2
L
f
2
1




LC
f
4
-
1
L
f
2
Z
L
f
2
LC
f
4
-
1
Z
1
2
2
T
2
2
T






Resonance occurs when the
denominator becomes zero

57. Parallel Resonance
 Same relationship as for series resonance
o In series resonance the impedance is a
minimum (Z=R)
o In parallel resonance the impedance is a
maximum (Y=0)
LC
2
1
fres



58. Resonance Comparisons
Series Parallel
Z Min Max
Y=1/Z Max Min
I Max Min
V=IZ Min Max

59. Quality (Q) Factor
 At resonance, wo, define
 The current in the circuit is…
R
L
Q 0
0
w

2
2
2
LC
1
1
R
L
1
1
R
V
Z
V
I
















w
w
At resonance (series) Z = R, so we can write the current at
resonance as IM = V/R

60. Q-Factor
 Using the definition and a little manipulation, we
can derive…
2
0
0
2
0
M
Q
1
1
I
I











w
w
w
w

61. Q-Factor