Chapter 4 AC Networks

4.1 Introduction

 Alternating Waveforms o Sinusoidal AC Voltage o Square Wave o 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 water power o oil o gas o nuclear fusion o wind o 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

180°, p radians 90°, p /2 radians y q x 0°, 0 radians 270°, 3 p /2 radians

radians deg rees   Example p 180  180  p   deg rees  radians : 45   p 4  Example : p 3  60 

Wave Description

 Amplitude (V peak ) - 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=V peak sin q  Example 4.2: o o

Determine the period and frequency of the waveform in Figure 4.7.

(5 cycles in 4

m

s 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=20sin

q

, 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=V p sin w t w  2 p  2 p f T

Sinusoidal Voltage Relationships

    v = V peak sin q v = V peak sin w t v = V peak sin 2 p t/T v = V peak sin 2 p ft  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=V p sin(

w

t+

f

)



For initial intersections after 0



v=V p sin(

w

t-

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 f I V w

Sinusoidal Plot

Example Problems

 Example 4.6: o o o

(a) Write the sinusoidal expression for each waveform appearing in Fig. 4.12.

(Current Amplitude = 5mA,

f

i (Voltage Amplitude = 100 V,

f

v = +60°,

w

= -30°,

w

= 400 rad/sec) = 400 rad/sec)

o

(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 “equivalent DC voltage” o o “effective voltage” “rms voltage”  RMS = root mean square 

Draw the “relating effective values” plots.

V eff  1 2 V p I eff  1 2 I p

4.5 Average Value

 For a sine wave, the average value is zero.

 For other waves….

G

(

average value

) 

area

(

under waveform

)

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 o 

V p

w

= 1.414(V rms ) = 1.414(40 mV) = 56.56 mV = 2

p

f = (2

p

)(0.5 kHz) = 3.142 X 10 3 rad/s v = 56.56 mV sin(3142t + 40°)

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   V eff = V rms V peak   V ave I eff = I rms   I peak I ave  f = 60 Hz  V eff = V rms = 120 V  V peak = V rms = 170 V  V ave = 0 V  I eff = I rms = limited by breaker  I peak = I rms  I ave = 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 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

Use Effective (rms) Values

 P R = V R I R = I R 2 R = V R 2 /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: V peak =I peak R Capacitors: V peak =I peak X C Inductors: V peak =I peak X L

Inductive Reactance

 Recall the form of inductance

v

 

L



i



t

 Since I for ac is constantly changing, there is a constant opposition to the flow of current o Inductive Reactance X L = w L = 2 p fL o Measured in ohms o Energy is stored in the coil, not dissipated like in the resistor.

Example

 An inductor of 400 mH is connected a.

across a 120 V, 60 Hz ac source. Find X L and the current through the coil X L X L = 2 p fL = 2 p (60 Hz)(0.4 H) = 151 W b.

I = V/X L = 120 v/151 W I = 0.8 A

X

L

vs. f graph for ideal inductor

 X L o = 2 p fL Should give a straight line graph of slope 2 p L o Higher the L the greater the slope X L f

V and I for an Inductor

 From we see that voltage is greatest

t i

when the change of current is greatest.

Current 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 i L sin(200t +30 through a 10-mH inductor is 5



). Find the voltage v L 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 , current is greatest when 

t

voltage is greatest.

change in 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

X c

 2 p 1

f C

X c also measured in ohms What is the capacitive reactance of a 50 m f capacitor when an alternating current of 60 Hz is applied?

X c X c

 1 2 p

fC

 53 W  2 p 1 ( 60

Hz

)( 50

X

10  6

f

)

Form of Reactance (Fig. 4.28)

300 250 200 150 100 50 0 0 C=1 m F C=2 m F 1000

f (Hz)

2000 3000

Example Problems

 Example 4.14: o

The voltage across a 2-

m

F capacitor is 4mV (rms) at a phase angle of -60



. If the applied frequency is 100kHz, find the sinusoidal expression for the current i C .

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

 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 2  B 2 What is the direction of C?

q  tan  1 B A

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 X L X = X L - X C R X C

Phasors

   Form Reactance First X = X L – X c Form Impedance

Z

 Calculate phase angle

R 2



X 2

f 

Tan



1 ( X R )

X L  X  X C R R X 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

X L X L = 2 p fL = 2 p (60 Hz)(5.8 X 10 -3 = 2.19 W H) X L

Z Z



R 2



X 2



3 .

25

W 

( 2 .

4

W

) 2



( 2 .

19

W

) 2

f 

Tan



1 ( X R )



Tan



1 ( 2 .

19

W

2 .

4

W

)

f 

42 .

4

 X R Z f R

Solution (cont’d)

  I = V/Z I = 120 V/3.25 W = 36.9 A ================================    Notice that if f = 120 Hz, then X L = 2 p (120 Hz)(0.0058 H) = 4.37 Z = 4.99 W f = 61.2° W  I = 24 A

Example Series R-C Circuit

 What is the current flow through a circuit with 120 V, 60 Hz source, an 80 m f capacitor, and a 24 W resistor?

Solution

X C X C Z

 

2

p

1 f C



33

W

R 2



X 2

 

2

p

( 60 Hz ( 24

W

) 2 )( 1 80 X 10



6



( 33

W

) 2 Z tan

 f

40 .

9

W 

33

W

24



1 .

38

f 

54 .

1



f )

X C X

I



V Z



120 V 40 .

9

W 

2 .

93 A

R R f Z

Solution (cont’d)

  What happens is the frequency doubles?

X C = 16.6 W Z = 29.2 W f = 34.6°  I = 4.11 A

Example - Series RLC Circuit

X L X C R  X R  X Z R X = X L - X C Z 2 = R 2 + X 2 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 m f. Compute the circuit impedance and current.

Solution

X L



2

p

f L



2

p

( 60 Hz )( 0 .

08 H )



30 .

2

W

X C



2

p

1 f C



2

p

( 60 Hz )( 1 80 X 10



6 f )



33 .

2

W

X



X L



X C



30 .

2

W 

33 .

2

W  

3 .

0

W

Z



R 2



X 2



( 15

W

) 2



(



3

W

) 2



15 .

3

W

I



V Z



240 V 15 .

3

W 

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 At times V and I are negative and power is drawn from the circuit.

o Thus the true power must be less than IV

Power Factor

i v



I p



V p sin

w

t sin(

w

t

 f

p



iv



I p V p )

f is the phase between

i

and

v sin

w

t sin(

w

t

 f

) Power Factor



true power(P) IV P



IV ( power factor ) P



Sum of p over one cycle T P



I p V p cos

f

2



I rms V rms cos

f

Power Factor

 Power Factor = cos f

cos

f 

R



Z

X Z f R

R R 2



X 2

X 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

 a) 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… The inductive reactance b) c) d) e) f) 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



2

p

fL



2

p

(60 Hz)(1 H)



377

W

b ) Z



R 2



X L 2



( 350

W

) 2



( 377

W

) 2



514 .

4

W

c) I



V d ) PF Z



440 V 514 .

4

W 

R Z



0 .

855 A



350

W

514 .

4

W 

0 .

68 e ) P



( 100 KV



A )( PF )



68 KW

Solution (cont’d)

PF



0 .

95



R Z Z Z 2 Z 2



R 0 .

95

 

R 2



R 2



( X L



( X L 368 .

42

W 

X C



X C ) 2 ) 2 X L



Z 2



R 2



X C



377

W 

( 368 .

42

W

) 2



( 350

W

) 2



262

W

X C



2

p

1 f C C



2

p

f 1 X C



2

p

( 60 Hz 1 )( 262

W

)



10 .

1

m

F

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 X L = X C , then Z = R and the circuit is purely resistive. This will occur at a certain frequency

2

p

f L



2

p

1 4

p

2 f 2 LC



1 f C

By altering L or C, the circuit can be made to resonate at any frequency.

f res



2

p

1 LC

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 (I 2 R) meant power couldn’t be transmitted far.

 Solution is to lower the current and raise the voltage (P = IV) o We can get the same power and not suffer the I 2 R losses.

o 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 Alternating current causes an alternating magnetic field o 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

V p V s



N p N s

Strictly true only when no current flows, since counter currents are set up (remember inductors?) For N p < N s we have V p < V s a step-up transformer For N p > N s we have V p > V s a step-down transformer

Transformers (cont’d)

 If no losses occur (real transformers are 90 – 95% efficient), then o V p I p = V s I s thus,

I p I s



V V p s



N s N p

A step-down transformer gives I s > I p A step-up transformer means a decrease in current

Example

    Generator furnishes 10 A at 550 V o P = VI = 5500 W Assume the transmission line has 20 W o P loss = I 2 R = (10A) 2 (20) W = 2000 W o 2000/5500 = 36% power loss Let’s step-up the voltage to 5500 V loss o o o Then I s = (V p I p )/V s = (550V)(10A)/5500V = 1A Now the losses are P loss = I 2 R = (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  i T = i 1 + i 2

i 1



V Z 1 sin(

w

t

 f

1 ) i 2



V Z 2 sin(

w

t

 f

2 )

The possibility of different phase angles means parallel impedances do not add as simply as for dc circuits

Total Current and Definitions

i T



V

 

1 Z 1 sin(

w

t

 f

1 )



1 Z 2 sin(

w

t

 f

2 )

 

Z 1



R 1 2



X 1 2 Z 2



R 2 2



X 2 2

Define

G



R Z 2 and B



X Z 2

G is called the conductance B is called the susceptance

Rule for Adding Parallel Impedances

1



( G 1



G 2 ) 2



( B 1



B 2 ) 2 Z T tan

q

T



B 1 G 1



B 2



G 2

Note: by convention, X C always carries a negative sign in computing the impedances.

Y = 1/Z T is called the admittance

Example

Compute Branch Impedances

Z 1



Z 2



R 2 R 1 2



4

p

2 f 1 2 C 1 2 Z 3



R 3 2

  

2

p

f L 3



2

p

1 f C 3

 

2 X 1

 

2

p

1 f C G 1



R Z 1 2 1 B 1



X 1 Z 1 2 X 2



0 G 2



1 R 2 B 2



0 X 3



2

p

f L 3



1



2

p

f C 3



G 3



R Z 3 2 3 B 3



X 3 Z 3 2

Resistive Elements

 If parallel impedance contains only resistors, we have o G = 1/R B = 0

1 Z T



Y



G 2



B 2



1 R 2



1 R

And the equation of parallel resistances results.

Example

Z 1



R



5

W

Z 2



X C 2



X C X 1



0 G 1



1 R B 1



0 G 2 B 2



0 B 2



0 .

0377 S

 

X C Z 2 2

 

1 X C

 

2

p

f C Y



( G 1



G 2 ) 2



( B 1



B 2 ) 2



Z T



4 .

91

W

1 ( 5

W

) 2



( 0 .

0377 ) 2



0 .

204 S

Parallel Resonance

Add L and C in parallel

Z Z 2 3

 

X X 2 L 2 C

 

X L X C G 2



0 B 2



1 X L



2

p

1 f L G 3



0 B 3

 

X C Z 3 2

 

1 X C

 

2

p

f C

Parallel Resonance

1 Z T

  

G 2



G 3



B 2



B 3 B 2



B 3



2



B 2



B 3

 

2

p

1 f L



2

p

f C 1 Z T Z T



1 4

p

2

p

2 f f 2 LC L



1 2

p

4

p

2 f f L 2 LC

Resonance occurs when the denominator becomes zero

Parallel Resonance

f res



2

p

1 LC

 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)

Resonance Comparisons

Z Y=1/Z I V=IZ Series Min Max Max Min Parallel Max Min Min Max

Tuning a Resonant Circuit

 Both series and parallel circuits have a resonance at

f res



2

p

1 LC or,

w

res



1 LC

Resistance in the circuit is significant in determining how much current passes at frequencies different from resonance.

Z



R 2

  w

L



1

w

C



2

Factor out R, Then w L

Z



R Z



R 1



1 R 2 1

 w

L R 2

 w

L 1

w

C 2



1

w

2 LC

Quality (Q) Factor

 At resonance, w o , define  The current in the circuit is…

I



V Z



V R 1

 w

L R 2 1 Q 0 1

w

2 LC 2

 w

0 L R

At resonance (series) Z = R, so we can write the current at resonance as I M = V/R

Q-Factor

 Using the definition and a little manipulation, we can derive…

I I M



1 1



Q 0 2

  w w

0

 w w

0

 

2

Q-Factor

High Pass (RC) Filter

Z



R 2



X C 2



R 2

  

2

p

1 f C

 

2 I



V i Z



V i R 2

  

2

p

1 f C

 

2

High Pass Filter

 I is the same through both R and C

V o



RI



V o V i



RV i R 2

  

2

p

1 f C

 

2 1 1

  

1 2

p

f RC

 

2

When f is small, denominator is large and the voltage ratio is small.

When f is large, denominator is almost one

Typical Results

R = 0.1 M W C = 0.16 m F Why is it called a high pass filter?

Half Power Point

 Note the point at which

V o V i



1 2

o o This will make

(V o /V i ) 2 = ½

Recall that one form of power is V 2 /R, so the output power is half the input power at this frequency.

Low Pass (RC) Filter

I



V i Z



R 2



X C 2



R 2



1

w

C 2 Z V o



IZ C



IX C



I

w

C

Low Pass Filter

V o



I

w

C

 w

C V o



V o V i

 

R

w

V i C



2



1 1



1

 w

R C



2 V i R 2



1

w

C 2

Low frequency means denominator is small and V o / V i  1 High frequency means denominator is large and V o / V i is small

Low Pass Results

Note the half power frequency again.