Chapter Two
Fundamentals of A.C
Fundamentals of A.C
Important definitions
Fundamentals of A.C
Important definitions
• Waveform.
- The path traced by quantity(such as voltage) plotted as a function time, position and so on.
• Periodic waveform.
- A waveform that repeats itself after a certain equal time intervals.
• Dc quantity.
- A quantity whose value and polarity is not changing.
• Pulsating quantity.
- A periodic quantity whose positive part is not equal to its negative part.
• Ac quantity.
- Aperiodic quantity whose positive part is equal to its negative part.
Important definitions
Fundamentals of A.C
• Period (T)
The time interval between successive repetitions of a periodic
waveform, or the time taken to complete one cycle.
• Cycle
The portion of a waveform contained in one period of time.
• Frequency (f)
The number of cycles that occur in one second. Its unit is called
Hertz(Hz=cycles/second).
• Frequency and period are inversely related.
Thus , 𝑓 =
1
𝑇
→𝑇=
1
𝑓
f in (HZ) and T in seconds(s)
Fundamentals of A.C
Important definitions
• Instantaneous value
is the value of ac or pulsating quantity at any instant of time.
• Maximum or peek value
is the maximum instantaneous value as measured from zero level.
EX: Find T, How many cycles, f, Maximum value of the wave.
T  1.2 s
T  3 ms
2.4
No of cycles 
 2 cycles
1.2
1
1
f  
Hz
T 1.2
Max value  2V
7
No of cycles   2.33 cycles
3
1
1
f  
 0.33e3 Hz  0.33kHz
T 3ms
Max value  2V
Fundamentals of A.C
a.Peak Vp  8 V
d. T  0.4 s
b. At t  0.3 s, v  -8 V; at 0.6 s, v  0 V
e. cycles 
c.Vpp  8  (8) 16 V
f. f
1.4
 3.5 cycles
0.4
1
 2.5 cps, or 2.5 Hz
T
• Average value
The average height of the of fig. a is shown in fig. b. The area under both curves is the same.
Important definitions
Fundamentals of A.C
• The average value of any current or voltage is the value
indicated on a dc meter. In other words, over a
complete cycle, the average value is equivalent to the
dc value.
𝑉𝑎𝑣𝑎
=
𝑎𝑟𝑒𝑎 𝑜𝑓 𝑎 𝑐𝑜𝑚𝑝𝑙𝑒𝑡𝑒 𝑐𝑦𝑐𝑙𝑒
𝑝𝑒𝑟𝑖𝑜𝑑
=
1 𝑇
𝑣
‫׬‬
0
𝑇
𝑡 𝑑𝑡
Since the ac quantity is symmetrical about the time axis
(+ve portion area equals the –ve portion area),so its
average value over a complete area is zero, therefore the
average value is determined over half period.
2 𝑇/2
∴ 𝑉𝑎𝑣𝑎 = න 𝑣 𝑡 𝑑𝑡
𝑇 0
Fundamentals of A.C
EX: Determine the average value of the waveforms of next Figures.
10*1e3 10*1e3
Area
under
curve
a  Averagevalue 

 0V
3
2e
Length of the curve
14*1e3  6*1e3
Area
under
curve
b  Averagevalue 

 4V
3
2e
Length of the curve
Fundamentals of A.C
EX: Find the average values of the following waveforms over one full cycle:
3
3
3*
4e
1*
4e
a  Averagevalue  Area under curve 
 1V
3
8e
Length of the curve
3
3
3
Area
under
curve
10
*
2e

4
*
2e

2
*
2e
b  Average value 

 1.6A
3
10e
Length of the curve
Fundamentals of A.C
• Effective (rms) Values
• To relate dc and ac quantities with respect to the power delivered to a load, RMS will help us determine the
amplitude of a sinusoidal ac current required to deliver the same power as a particular dc current.
• How is it possible for a sinusoidal ac quantity to deliver a net power if, over a fully cycle, the net current in any
one direction is zero, average value=0.
• Irrespective of direction, current of any magnitude through a resistor will deliver power to that resistor.
– During the positive or negative portions of a sinusoidal ac current, power is being delivered at each instant of
time to the resistor.
Snice the power is proportional to the square of current or voltage, then the average of the square of the time
varying current or voltage should be found which is equivalent dc quantity which produces the same power.
If the power dissipated by an ac current i is the power dissipated by an equivalent dc current Idceq., then
𝑃 = 𝐼2𝑅 = 𝑖2𝑅
∴ 𝐼𝑑𝑐𝑒𝑞. = 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑜𝑓 𝑖2
𝐼𝑑𝑐𝑒𝑞. is called root mean square value or effective value.
𝐼𝑟𝑚𝑠 =
1 𝑇
2𝑑𝑡
𝑖(𝑡)
‫׬‬
𝑇 0
𝑎𝑙𝑠𝑜
𝑉𝑟𝑚𝑠 =
1 𝑇
‫𝑡𝑑)𝑡(𝑣 ׬‬
𝑇 0
Fundamentals of A.C
EX: Find the effective or rms value of the waveform of the figure shown.
1
T
4
1 2
2
v (t)dt 
Vrms 
[  3 dt 

T 0
8

1
[9(4)
8
Vrms 
9* 4 1* 4
 2.236V
8
8
Fundamentals of A.C
EX: Calculate the rms value of the voltage of Figure shown.
1
2
T
4
8
1
2
2
2
v
(t)dt

Vrms 
[
(10)
dt

(4)
dt

(2)
dt]  4.899V




T 0
10 0
2
6
2
100* 2 16* 2  4* 2
 4.899V
Vrms 
10
EX: Determine the average and rms values of the square wave of Figure shown.
Solution: By inspection, the average value is zero.
Vrms 
Vrms 
1600*10e3 1600*1
1T
T
1 3 [

20e
2
v
 (t)dt
0
10e3

0
40 2 dt 
20e3
 (40)
10e3
2
dt]  40V
Fundamentals of A.C
EX: Determine the average and rms values of the square wave of Figure shown.
Vrms 
1T
T
2
v
 (t)dt
0
T  2
Vm

v   Vm
 2
0t  

  t  2 


Vrms 
2
Vm 2
1
2
[ Vm dt   (
) dt] 
2
2 0

2
1
Vm 2
2 
[[Vm t] 0 [
t] ] 
4
2
Vm2

2
2
Vm
1
[ Vm2 dt   (
)dt]
4
2 0

1
[[  0] [ (2   )]]
2
4
5
Vm2
Vm2 4  
Vm2 5

[[ ] [ ]] 
[
]
[ ]  Vm V
4
8
2
4
2 4
2
Fundamentals of A.C
• Form factor &Peak factor
RMS
Kf 
Average value
M a x value
kp 
R M S value
EX: Determine the rms value, average value, Kf, Kp of the square wave
of Figure shown.
Irms 
1
T
T /2
T
2
i
 (t)dt
0
T T
Im

i   Im
 2
0t T /2


T /2t T

T
T /2
T
2
1
Im
1
Im
[  Im2 dt   ( )2 dt] 
[  Im2 dt   (
)dt]
Irms 
T 0
2
T 0
4
T /2
T /2
2
2
T
1
T
1
Im
Im
2 T / 2 [
T
t]T / 2 ] 
[[  0] [ (T  )]]
[[Im t] 0
T
4
T
2
4
2
Im 2 [[T ] [ T ]]  Im 2 [ 4 1 ]]  Im2 [ 5 ]]  Im 5
8
8
T
2
8
8
Fundamentals of A.C
T Im
T
Im* 
*(T  ) Im
2
2
2 
Iav 
T
4
 Im
i   Im
 2
or
1 T
Iav   i(t)dt
T 0
0 t T /2


T /2 t T

T /2
T
Iav  1 [  I m dt   ( Im )dt  Im
T 0
2
4
T /2
kf 
rms

av
5
8  3.16, kp  max  Im  1.265
Im
rms
5
Im
4
8
Im
Fundamentals of A.C
The Sinusoidal Waveform
• The sinusoidal waveform is the only alternating waveform whose shape is unaffected by the response
characteristics of R, L, and C elements. In other words, if the voltage across (or current through) a resistor, coil,
or capacitor is sinusoidal in nature, the resulting current (or voltage, respectively) for each will also have
sinusoidal characteristics, as shown in next figure.
• The unit of measurement for the horizontal axis can be time, degree, or radians.
The term radian can be defined as follow: If we mark off a portion of the
circumference of a circle by a length equal to the radius of the circle, the angle
resulting is called 1 radian. The result is
1 rad  57.296   57.3
The quantity π is the ratio of the circumference of
a circle to its diameter.
D *

D
6.28 rad  360º
and 2  2*3.14  6.28 rad
2  360º
Fundamentals of A.C
  360º  2 rad  rad 

or 1º 
* rad
360
2

180

180

Radians  (
) (degrees)
Degrees  (
180º
180º

) (Radians)
Plotting a sine wave versus radians.


EX : 90º: Radians  (
) (90º)  rad
180º
Sine wave and cosine wave with the
horizontal axis in degrees.
2


30º: Radians  (
) (30º)  rad
180º
6
 rad : Degrees  (180º) (  )  60º
3

3
3
180º 3
rad : Degrees  (
) ( )  270º
2

2
• Note: α can be each of them
Fundamentals of A.C
It is of particular interest that the sinusoidal waveform can be derived from the length of the vertical
projection of a radius vector rotating in a uniform circular motion about a fixed point. We will trace a
complete sinusoidal waveform after the radius vector has completed a 360° rotation about the center.
Fundamentals of A.C
Generating a sinusoidal waveform through the vertical projection of
a rotating vector.
Fundamentals of A.C
• The velocity with which a radius vector rotates about the center, called the angular velocity () :
distance in angle(degree or red
Angular velocity 
time

(omega)  
–Since ω is typically provided in radians per second, the angle α obtained is usually in radians.
–The time to complete one revolution is equal to the period (T) of the sinusoidal waveform.
T  2 ,
2

rad / s
T
This equation states that the smaller the period of the sinusoidal waveform or the smaller the time interval before
one complete cycle is generated, the greater must be the angular velocity of the rotating radius vector.
• Since the frequency ( f ) of the generated waveform is inversely related to the period (T) of the waveform:
2

 2f rad / s
T
Fundamentals of A.C
This equation states that the higher the frequency of the generated sinusoidal waveform, the higher must be
the angular velocity.
2

 2f rad / s
T
Demonstrating the effect of ω on the
frequency and period.
Fundamentals of A.C
EX: Determine the angular velocity of a sine wave having a frequency of 60 and 50 Hz.
EX: Determine the angular velocity of a sine wave having a period of 2s.
EX: Determine the frequency and period of the sine wave of next figure.
500
  2f  f 
 79.58 Hz
2
1
1
T 
 12.57 ms
f 79.58
  2  T  2  12.57 ms
T
500
1
1
f  
 79.58Hz
T 12.57
Fundamentals of A.C
EX: Sketch e =10 sin 314t with the abscissa
a. angle (α) in degrees.
b. angle (α) in radians.
c. time (t) in seconds.
if  used in degree, the  must be in degree/s this is not usual

c : t    t 


6  1.67ms
314
0
for 0º: t 
 0ms
314
for 30º : t 
for 90º: t 

2  5ms
314
for 180º: t  10ms
for 360º: t  20ms
or : t    t 
for 360º T 
2


 20ms
314
T 20
for 180º 
 10ms
2 2
T 20
for 90º 
 5ms
4 4
T
for 30º
 1.67ms
12
Or
Fundamentals of A.C
EX: Given ω=200 rad/s, determine how long it will take the sinusoidal waveform to pass through an angle of 90°.
EX: Find the angle through which a sinusoidal waveform of 60 Hz will pass in a period of 5 ms.
  t  2ft  2 *60*5e3  1.885 rad
 º  1.885
180

 108º
EX: a. Find the angle at which the magnitude of the sinusoidal function
v =10 sin 377t is 4 V.
b. Determine the time at which the magnitude is attained.
However, the figure reveals that the magnitude of 4 V (positive) will be attained at two points
between 0° and 180°. The second intersection is determined by:
 )
23.58(
 2  180º23.578  156.422º
b) t  α  t 
t2 
α2


α


 )
156.42(
180  1.09ms
377
180  7.24ms
377
Fundamentals of A.C
General Format for the Sinusoidal Voltage or Current
The basic format for the sinusoidal waveform is:
𝑎 = 𝐴 𝑚 sin
𝜔𝑡±𝜑
𝑤ℎ𝑒𝑟𝑒: 𝑎 𝑖𝑠 𝑡ℎ𝑒 𝑖𝑛𝑠𝑡𝑎𝑛𝑡𝑎𝑛𝑒𝑢𝑠 𝑣𝑎𝑙𝑢𝑒 𝑎𝑡 𝑡𝑖𝑚𝑒 𝑡, 𝐴 𝑚 𝑖𝑠 𝑡ℎ𝑒
𝑚𝑎𝑥. 𝑣𝑎𝑙𝑢𝑒, 𝜔 𝑖𝑠 𝑡ℎ𝑒 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 & 𝜑 𝑖𝑠 𝑡ℎ𝑒 𝑝ℎ𝑎𝑠𝑒 𝑎𝑛𝑔𝑙𝑒.
Therefore the factors affecting on the shape of the waveform are:
1. The amplitude (max. value or peak) Am
E X : G i v e n i  6 * 1 0 - 3 sin 1 0 0 0 t , d e t e r m i n e i at t  2 ms.
   t  1 0 0 0 t  ( 1 0 0 0 rad/s)(2e - 3 s)  2 rad
i c a n b e f o u n d b y calculator in R A D m o d e
º 2
180

 114.59º (for calculator in D E G m o d e )
i  6 e  3 sin(114.59º )  5 . 4 6 m A
Fundamentals of A.C
2.
Angular frequency:
a  Asin 1t 
if 2  1 

b  Asin 2t
T1 T2
f 2  f 1 T1  T 2
Fundamentals of A.C
3. Phase angle
𝑎 = 𝐴 𝑚 sin 𝜔𝑡 + 𝜃
𝑎 = 𝐴 𝑚 sin 𝜔𝑡 − 𝜃
For one vector
ω
θ=+ Leading
θ=- Lagging
Fundamentals of A.C
Phase Relations for two vectors
The terms lead and lag are used to indicate the relationship between two sinusoidal waveforms of the same frequency
plotted on the same set of axes.
•The cosine curve is said to lead the sine curve by 90
•The sine curve is said to lag the cosine curve by 90
The 90° is referred to as the phase angle between the two waveforms, Sine and cosine curves are out of phase by 90
sin(t  90º )  sin(t  2 )  cos(t)
sin(t)  cos(t  90º )  cos(t  2 )
a  Asin(t)
b  B sin(t)
a & b are in phase, A  B
Fundamentals of A.C
a1  Asin(t)
a1  Asin(t)
a2  Asin(t  ),
a2  Asin(t   ),
where is  ve,a 2 leads a 1

or a1 lags a 2 by 

 A  A or diff

where is  ve,a 2



lags a1 or a1 leads a 2 by  
Fundamentals of A.C
EX: What is the phase relationship between the sinusoidal waveforms of each of the following sets?
a. v  10sin (t  30 )
i  5sin (t  70 )
a. See the Figure
b. i  5sin (t  60 )
v  10sin (t  20 )
c. i  2 cos(t 10 )
v  3sin (t 10 )
d. i  sin (t  30 )
v  2sin (t 10 )
e. i  2 cos(t  60 )
v  3sin (t 150 )
i leads v by 40°, or v lags i by 40°.
Fundamentals of A.C
c. See the Figure.
oblem
c. i  2 cos (t 10 )
pr
v  3sin (t 10 )
cos   sin(  90)

i  2 cos(t 10 )  2 sin ( t 10  90 )  2sin (t 100 )
i leads v by 110°, or v lags i by 110°.
Fundamentals of A.C
d. See the Figure.
blem
d. i   sin (t  30 )
pro
v  2sin (t 10 )
i  sin (t  30 )
 sin (t  30 180 )
 sin (t  210º )
v leads i by 160°, or i lags v by 160°.
i  sin (t  30 )
 sin (t  30 180 )
 sin (t 150º )
i leads v by 200°, or v lags i by 200°
Fundamentals of A.C
e. See the Figure.
blem problem
e. i   2 cos
pro
v
by choice
i  2 cos(t  60 )  2 cos(t  60
 2 cos(t 
v and i are in phase.
or i  2 cos(t  60 )
 2sin (t  60  90 )
 2sin (t 150 )

Fundamentals of A.C
EX : let
X1(t) = 10 sin (50 t + 60)
X2(t) = 20 cos (50 t + 30)
find the frequency,phase angle between them
𝜔t = 50t → 𝜔 = 50 rad/s
𝜔
50
𝜔 = 2𝜋f → f =
=
= 7.9577 Hz
2𝜋 2𝜋
X2(t) = 20 cos (50 t + 30) = 20sin(50 t + 30 + 90) = 20sin(50 t + 120)
X2leads X1 by 60º or X1 lags X2 by 60º
Fundamentals of A.C
𝑋(𝑡) = 𝑋𝑚 sin( 𝜔𝑡 + 𝜃)
from trigonometry:
sin(𝛼 ∓ 𝛽) = sin(𝛼)cos(𝛽) ∓ sin(𝛽)cos(𝛼)
cos(𝛼 ∓ 𝛽) = c𝑜𝑠(𝛼)cos(𝛽) ± sin(𝛼)sin(𝛽)
∴ 𝑋(𝑡) = 𝑋𝑚 sin( 𝜔𝑡 + 𝜃) = 𝑋𝑚[sin( 𝜔𝑡) cos( 𝜃) + sin( 𝜃) cos( 𝜔𝑡)]
= 𝑋𝑚 cos( 𝜃) sin( 𝜔𝑡) + 𝑋𝑚 sin( 𝜃) cos( 𝜔𝑡) = 𝐴 sin( 𝜔𝑡) + 𝐵 cos( 𝜔𝑡)
𝐴
tan( 𝜃) =
𝑋𝑚 =
𝐁
𝐵
=
𝑋𝑚 sin( 𝜃)
𝐴 𝑋𝑚 cos( 𝜃)
𝐴2 + 𝐵2
𝑋(𝑡) = 𝑋𝑚 sin( 𝜔𝑡 + 𝜃) =
=
sin( 𝜃)
cos( 𝜃)
𝐴2
→𝜃=
+ 𝐵2
tan−1
sin( 𝜔𝑡
𝐵
𝐴
+ tan −1 (
𝐵
))
𝐴
Fundamentals of A.C
The average value ,rms value, Kf, and Kp of a sinusoidal waveform.
Iav _ half cycle 
2Am
T
T


2 Im

A or
Iav  1  i(t)dt  1 [ Imsin(t)]dt  0
T 0
T 0
T /2
2
Iav _ half cycle  Imsin(t)dt
T 0
2
2
T
, Iav 
2

2
(  )/2
Imsin(t)dt 



t)]
[cos(

0

Im
Im
0

 Im

(
2
[cos(t)]0 


sin(t)dt
0
)/2
 Im
2 Im


[cos( )  cos( *0)] 
[11] 
A




 Im
The average value ,rms value, Kf, and Kp of a sinusoidal waveform.
Fundamentals of A.C
Irms 
1
T
T
 i (t)dt , T 
2
0
2

2
, Irms 
1
2


2
2
(Im
sin
(t))dt

0
2
2
 Im2  2
 Im2  1 1
[   cos(2t)]dt
sin (t))dt 


2 0
2 0 2 2
2
2


2
2
 Im2
 Im2 
1 
[ 1dt 
2 cos(2t)dt]
[ 1dt   cos(2t)dt] 


2 0
4 0
4 0
0
2
2
2
2
2
 Im2

Im
1
1

Im
2 1
2



[[t] 0  (
)]
[[t 
sin(2t))]0 
sin(2t)]0 
[[

sin(2

4
4
4
 2
2
2

 Im2 2
Im2
Im
 Im2 2 1
[[

sin(4 )] 
[[
 0] 

A

4

2
4
 2
2
Im

rms
2
Kf 


2 Im 2 2
av

max Im
Kp 

 2
Im
rms
2
Fundamentals of A.C
EX: If v=100√2 sin (100πt+30º), find
1-The maximum value. 2-The average value. 3-The rms value
4-The instantaneous value at t=0.83ms. 5-The frequency and the angular frequency. 6-The period. 7-The phase angle. 8Sketch the waveform
1Vm  100 2 V , 2 Vav 
2Vm

2*100 2 200 2


V



Vm 100 2
3
3 Vrms 

 100V , 4  v  100 2 sin(100 *0.83e  )  100V
6
2
2

1
1
5    100 rad / s,  2f  f 
 50Hz, 6  T  
 20ms
f 50
2
7  Phase angle   30º 

6
rad
wt
v
0
83.1
π/2
114.4
π
-83.1
3π/2
-114.4
2π
83.1