Download 2 - Stephen F. Austin State University

Chapter 4
AC Networks
4.1 Introduction

Alternating Waveforms
o
o
o
Sinusoidal AC Voltage
Square Wave
Triangular Waveform
Function Generators are devices that can
create all of these waveforms.
 These devices will be used in lab.
 Sketch the waveforms and circuit symbols
for the waveforms above.

4.2 Sinusoidal AC Waveform

AC Generators (a.k.a. alternators) are fueled by:
o
o
o
o
o
o
water power
oil
gas
nuclear fusion
wind
solar power

Power to a shaft turns a rotor inside of a stator.

Rotor - rotating magnetic

Stator - stationary coil of wire
Simple Harmonic Motion (SHM)

Position versus time for simple harmonic
motion is shown as sine or a cosine
Uniform Circular Motion
Video
90°, p/2 radians
y
180°, p radians
q
x
270°, 3p/2 radians
0°, 0 radians
 p 
radians  
deg rees 
 180 
 180 
deg rees  
radians 
 p 
p
Example : 45 
4
p
Example :  60
3
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

Example Problems
Sketch the following waveform: v=Vpeaksinq
 Example 4.2:

o
o

Determine the period and frequency of the
waveform in Figure 4.7.
(5 cycles in 4ms for v versus t)
Example 4.3:
o
How long will it take a sine wave with a frequency
of 0.2 kHz to complete 10 cycles?
Example Problems

Example 4.4:
o
For v=20sinq, at what angle will the voltage
rise to 5 Volts?
Angular Velocity (w) - rotation rate of a
vector (or generator shaft)
 w is measure in radians per second.
 New Waveform Equation: v=Vpsinwt

2p
w
 2pf
T
Sinusoidal Voltage Relationships
v = Vpeaksin q
 v = Vpeaksin wt
 v = Vpeaksin 2pt/T
 v = Vpeaksin 2pft


Which form you use depends on what you know
Example Problems

Example 4.5:
o
Given i=2010-3 sin 400t, find the time at
which i rises to 10 mA.
Phase Relationship



f = phase angle
For initial intersections before 0
v=Vpsin(wt+f)
For initial intersections after 0
v=Vpsin(wt-f)
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
Current and Voltage out of Phase

View as two vectors rotating around the
circle
o
Must mean they have the same w
V
f
I
w
Sinusoidal Plot
Example Problems

Example 4.6:
o
o
o
o
(a) Write the sinusoidal expression for each
waveform appearing in Fig. 4.12.
(Current Amplitude = 5mA, f i = +60°, w = 400 rad/sec)
(Voltage Amplitude = 100 V, f v = -30°, w = 400 rad/sec)
(b) What is the phase relationship between the
two waveforms?
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°)
Graphs of Ex. 4.7
4.4 Effective (RMS) Value

These mean the same thing for AC circuits:
o
o
o
“equivalent DC voltage”
“effective voltage”
“rms voltage”
RMS = root mean square
 Draw the “relating effective values” plots.

1
Veff 
Vp
2
I eff
1

Ip
2
4.5 Average Value
For a sine wave, the average value is zero.
 For other waves….

area (under waveform)
G(average value) 
T ( period )
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 = 2pf = (2p)(0.5 kHz) = 3.142 X 103 rad/s
v = 56.56 mV sin(3142t + 40°)

o
Example Problems

Example 4.10:
Calculate the average value of the waveform
below over one full cycle.
12
10
8
6
4
2
0
-2 0
-4
-6
2
4
6
8
10
12
What are the following for a wall outlet?







f
Veff = Vrms
Vpeak
Vave
Ieff = Irms
Ipeak
Iave







f = 60 Hz
Veff = Vrms = 120 V
Vpeak = 2 Vrms = 170 V
Vave = 0 V
Ieff = Irms = limited by breaker
Ipeak = 2 Irms
Iave = 0 A
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
o

Yet the resistor gets hot – power is dissipated
I and V are not always zero – you can get
shocked from ac just like dc
We need a new definition for power
Use Effective (rms) Values
PR = VRIR = IR2R = VR2/R
 Look just like the dc forms of power

o
But we use rms values instead
4.7 The R, L, and C Elements

Ohm’s Law for Peak Values
Resistors: Vpeak=IpeakR
Capacitors: Vpeak=IpeakXC
Inductors: Vpeak=IpeakXL
Inductive Reactance

Recall the form of inductance
i
v  L
t

Since I for ac is constantly changing, there is a
constant opposition to the flow of current
o
o
o
Inductive Reactance XL = wL = 2pfL
Measured in ohms
Energy is stored in the coil, not dissipated like in the
resistor.
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.
b.
XL = 2pfL = 2p(60 Hz)(0.4 H)
XL = 151 W
I = V/XL = 120 v/151 W
I = 0.8 A
XL vs. f graph for ideal inductor

XL = 2pfL
o
o
Should give a straight line graph of slope 2pL
Higher the L the greater the slope
XL
f
V and I for an Inductor

From
when
i
v  L
t
Current
we see that voltage is greatest
the change of current is greatest.
Highest
positive v
Zero here
Most negative
here
Voltage and Current (Coil)
Current
Voltage
Voltage leads current by 90°
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.
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.
Capacitive Reactance

V is changing in ac circuits
o
o
But Q = VC, so the charge changes also
Since I  Q , current is greatest when change in
t
voltage is greatest.
Current
Voltage
I leads V by
90°
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.

Capacitive Reactance
1
Xc 
2p f C
Xc also measured in ohms
What is the capacitive reactance of a 50 mf capacitor
when an alternating current of 60 Hz is applied?
1
1
Xc 

2pfC 2p (60 Hz )(50 X 10 6 f )
X c  53 W
Form of Reactance (Fig. 4.28)
300
250
XC (W )
200
150
100
C=1mF
C=2mF
50
0
0
1000
2000
f (Hz)
3000
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.
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
o
Scalar: 50mph
Vector: 50mph North

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
j  1
Imaginary
C=A+Bj
q
Real
What is the magnitude of C?
C A B
2
2
What is the direction of C?
B
q  tan
A
1
Phasor Diagrams
We make the real axis the resistance
 The imaginary axis is reactance

o
o

Inductive reactance is plotted on the +y axis
Capacitive reactance is plotted on the –y axis
The vector addition of R and X is Z
XL
XC
X = XL - XC
R
Phasors



Form Reactance First X = XL – Xc
Form Impedance Z  R 2  X 2
Calculate phase angle f  Tan 1( X
XL
XC


X
R
R
X
R
Z
R
)
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
Solution
XL = 2pfL = 2p(60 Hz)(5.8 X 10-3 H)
XL
XL = 2.19 W
R
Z  R 2  X 2  ( 2.4W )2  ( 2.19W )2
Z  3.25W
f  Tan 1 ( X R )  Tan 1 ( 2.19W 2.4W )
f  42.4
X
Z
R
f
Solution (cont’d)
I = V/Z
 I = 120 V/3.25 W = 36.9 A
================================
 Notice that if f = 120 Hz, then
 XL = 2p(120 Hz)(0.0058 H) = 4.37 W
 Z = 4.99 W f = 61.2°
 I = 24 A

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?
Solution
1
1
XC 

2p f C 2p ( 60 Hz )( 80 X 10 6 f )
X C  33W
XC
Z  R 2  X 2  ( 24W )2  ( 33W )2
Z  40.9W
tanf  33W
R
R
24
 1.38
f  54.1
V
120V
I 
 2.93 A
Z 40.9W
f
X
Z
Solution (cont’d)
What happens is the frequency doubles?
 XC = 16.6 W
Z = 29.2 W f = 34.6°
 I = 4.11 A

Example - Series RLC Circuit
XL
XC

R

X
R
X = XL - XC
X
Z
R
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.
Solution
X L  2p f L  2p ( 60 Hz )( 0.08 H )  30.2W
1
1
XC 

 33.2W
6
2p f C 2p ( 60 Hz )( 80 X 10 f )
X  X L  X C  30.2W  33.2W  3.0W
Z  R 2  X 2  ( 15W )2  ( 3W )2  15.3W
I V
Z
 240V
15.3W
 15.7 A
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
o
At times V and I are negative and power is
drawn from the circuit.
Thus the true power must be less than IV
Power Factor
i  I p sin wt
v  V p sin( wt  f )
f is the phase between i and v
p  iv  I pV p sin wt sin( wt  f )
true power(P)
Power Factor 
IV
P  IV ( power factor )
Sum of p over one cycle
P
T
I pV p
P
cos f  I rmsVrms cos f
2
Power Factor

Power Factor = cos f
X
Z
f
X
R
R
cos f  
Z
R
R2  X 2
For purely inductive or
purely capacitive circuits,
f = 90° and cos f = 0 no power is dissipated
Control of the Power Factor
Most household appliances are highly
resistive and require high PF.
 Industry tends to be rather inductive.
 Some substations have capacitor banks
timed to break in at certain times when
inductive load changes, adjusting the PF to
the desired value

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)
b)
c)
d)
e)
f)
The inductive reactance
The impedance
The current
The power factor
The power used by the motor
The size capacitor required to change PF to 95%.
Solution
a) X L  2pfL  2p (60 Hz)(1 H)  377 W
b ) Z  R 2  X L2  ( 350W )2  ( 377 W )2  514.4W
c) I  V
Z
d ) PF  R
 440V
Z
514.4W
 350W
 0.855 A
514.4W
 0.68
e ) P  ( 100 KV  A )( PF )  68 KW
Solution (cont’d)
PF  0.95  R
Z
 368.42W
0.95
Z 2  R 2  ( X L  X C )2
ZR
Z 2  R 2  ( X L  X C )2
X L  Z 2  R 2  X C  377 W  ( 368.42W )2  ( 350W )2  262W
1
XC 
2p f C
1
1
C

 10.1mF
2p f X C 2p ( 60 Hz )( 262W )
Lightning Arrestors

A high voltage line is likely to transmit a
high voltage, high frequency electric
discharge if struck by lightning.
o
The high frequencies may travel to a substation,
causing arcing in generators.
Solution
Choke Coil is an
inductor of very few
turns (low L). It offers
low reactance to 60 Hz.
but stops high
frequencies
Alternate path to ground offered (spark gap and series
resistor). This path has high R to low frequencies and low R
to high frequencies.
Series Resonance
Z  R 2  ( X L  X C )2
If XL = XC, then Z = R and the circuit is purely resistive.
This will occur at a certain frequency
1
2p f L 
2p f C
4p 2 f 2 LC  1
1
f res 
2p LC
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
Variable Capacitors (varicaps)
One set of plates moves with
respect to the other. Since C
depends on plate area, C varies as
plates are shifted. Used in tuning a
radio.
All frequencies are excited on the
primary (left) coil and induced onto
the secondary coil. We tune the
capacitor to reach resonance, which
maximizes the current flow.
Power Transmission

In the early days power was transmitting as dc.
o
o

Applications used low voltage (10 – 110 v)
Power losses (I2R) meant power couldn’t be transmitted
far.
Solution is to lower the current and raise the
voltage (P = IV)
o
o
We can get the same power and not suffer the I2R
losses.
Problem: applications cannot use the high voltages
Transformers

Used to change the voltage either down
(step-down) or up (step-up)
Principle of Operation

Primary is connected to the ac source.
o
o

Alternating current causes an alternating
magnetic field
Voltage is induced on the secondary
The same number of lines of magnetic force
pass through each coil
o
Induced voltage in each turn of wire should be
the same
Law of Transformers
Vp
Vs

Np
Ns
Strictly true only when no current flows, since
counter currents are set up (remember inductors?)
For Np < Ns we have Vp < Vs a step-up transformer
For Np > Ns we have Vp > Vs a step-down transformer
Transformers (cont’d)

If no losses occur (real transformers are 90 –
95% efficient), then
o
VpIp = VsIs
thus,
Ip
Vs N s


I s Vp N p
A step-down transformer gives Is > Ip
A step-up transformer means a decrease in current
Example

Generator furnishes 10 A at 550 V
o

Assume the transmission line has 20 W loss
o
o

Ploss = I2R = (10A)2(20) W = 2000 W
2000/5500 = 36% power loss
Let’s step-up the voltage to 5500 V
o
o
o

P = VI = 5500 W
Then Is = (VpIp)/Vs = (550V)(10A)/5500V = 1A
Now the losses are Ploss = I2R = (1A)2(20) W = 20 W
Or 0.36% loss
When the power gets to the user, we put in a step-down
transformer
Parallel RC Circuit

General Scheme
o
Form impedance for each branch
o
From Kirchhoff’s Current Law
 iT
= i1 + i2
i1 
V
sin( w t  f1 )
Z1
i2 
V
sin( w t  f2 )
Z2
The possibility of different
phase angles means parallel
impedances do not add as
simply as for dc circuits
Total Current and Definitions
1

1
iT  V  sin( w t  f1 ) 
sin( w t  f2 )
Z2
 Z1

Z 1  R12  X 12
Define
GR
Z 2  R22  X 22
Z
2
and
BX
G is called the conductance
B is called the susceptance
Z2
Rule for Adding Parallel Impedances
1
 ( G1  G2 )2  ( B1  B2 )2
ZT
tanqT 
B1  B2
G1  G2
Note: by convention, XC always carries a negative sign in
computing the impedances.
Y = 1/ZT is called the admittance
Example
Compute Branch Impedances
1
Z1  R  2 2 2
4p f C1
1
X1  
2p f C
R1
X1
G1  2 B1  2
Z1
Z1
Z 2  R2
X2  0
2
1
G2 


1


Z 3  R   2p f L3 
2p f C3 

2
3
2
1
R2
B2  0
X 3  2p f L3  1
G3 
R3
Z 32
B3 
2p
X3
Z 32
f C3 
Resistive Elements

If parallel impedance contains only resistors, we
have
o
G = 1/R
B=0
1
 Y  G2  B2 
ZT
1
1
2 
R
R
And the equation of parallel resistances results.
Example
Z1  R  5W
X1  0
1
G1 
R
Z2  X  X C
G2  0
B2 
2
C
B1  0
 XC
1


 2p f C
2
Z2
XC
B2  0.0377 S
Y  ( G1  G2 )2  ( B1  B2 )2 
ZT  4.91W
1
( 5W )
2
 ( 0.0377 )2  0.204 S
Parallel Resonance
Add L and C in parallel
Z2  X  X L
G2  0
B2 
1
1

X L 2p f L
Z 3  X C2  X C
G3  0
B3 
 XC
1


 2p f C
2
Z3
XC
2
L
Parallel Resonance
1

ZT
G2  G3 2  B2  B3 2

B2  B3 2  B2  B3 
1

 2p f C
2p f L
1 1 - 4p 2 f 2 LC

ZT
2p f L
ZT 
2p f L
1 - 4p 2 f 2 LC
Resonance occurs when the
denominator becomes zero
Parallel Resonance
f res

1

2p LC
Same relationship as for series resonance
o
o
In series resonance the impedance is a
minimum (Z=R)
In parallel resonance the impedance is a
maximum (Y=0)
Resonance Comparisons
Series
Parallel
Z
Min
Max
Y=1/Z
Max
Min
I
Max
Min
V=IZ
Min
Max
Tuning a Resonant Circuit

Both series and parallel circuits have a resonance at
f res 
1
or, wres 
2p LC
1
LC
Resistance in the circuit is significant in determining how
much current passes at frequencies different from resonance.

Z  R  w L 1
2
Factor out R,
Then wL
Z  R 1
Z  R 1


2
wC
1
w L - 1w C
2
R
wL 

2
1 
1



R 2  w 2 LC 
Quality (Q) Factor


At resonance, wo, define
Q0 
The current in the circuit is…
I V
V

Z R
w0 L
R
1
w L  
2
1 
1 
 1  2

 R   w LC 
2
At resonance (series) Z = R, so we can write the current at
resonance as IM = V/R
Q-Factor

Using the definition and a little manipulation, we
can derive…
I

IM
1
 w w0 
1  Q   
 w0 w 
2
0
2
Q-Factor
High Pass (RC) Filter
 1 

Z  R 2  X C2  R 2  
 2p f C 
I
Vi

Z
Vi
 1 

R  
 2p f C 
2
2
2
High Pass Filter

I is the same through both R and C
Vo  RI 
Vo

Vi
RVi
 1 

R 2  
2
p
f
C


1


1

1  
 2p f RC 
2
When f is small,
denominator is large and the
voltage ratio is small.
2
When f is large, denominator
is almost one
Typical Results
R = 0.1 MW C = 0.16 mF
Why is it called a high pass filter?
Half Power Point

Note the point at which
o
o
Vo
Vi
1
2
This will make (Vo/Vi )2 = ½
Recall that one form of power is V2/R, so the output
power is half the input power at this frequency.
Low Pass (RC) Filter
 1 
2
2
2
Z  R  XC  R  

w C 
I
Vi
Z
2
I
Vo  IZ C  IX C 
wC
Low Pass Filter
I
Vo 

wC
Vo 
Vi
 1 
w C R 

w C 
Vi
2
2
Rw C 2  1
Vo
1

2
Vi
1  w RC 
Low frequency means denominator is small and Vo / Vi  1
High frequency means denominator is large and Vo / Vi is small
Low Pass Results
Note the half power frequency again.