introduction to electrical engineering l 13: single phase...

ELL 100 - Introduction to Electrical Engineering

LECTURE 13: SINGLE PHASE AC CIRCUITS

1

Introduction

Basics of sinusoid and phasors

Average, RMS value of an AC quantity

Steady state AC response for a pure resistive circuit

Exercise/Numerical Analysis

2

OUTLINE

3

INTRODUCTION

Why alternating current (AC) power system became popular?

While direct current (DC) electricity flows in one direction through a

wire, AC electricity alternates its direction in a back-and-forth motion

AC electricity is created by an AC electric generator, which determines

the frequency

The major advantage of AC electricity is that the voltage can be readily

changed, thus making it more suitable for long-distance transmission

than DC electricity

4

AC can also be employed using capacitors and inductors in electronic

circuitry, allowing for a wide range of applications

INTRODUCTION

Various AC Waveforms

5

Single Phase Transformer

SINGLE PHASE APPLIANCES

6

Single Phase Meter


7

Single Phase Power Supply with

Battery

Single Phase Grid Connected System


8

Single Phase Fan Induction Motor


9

APPLICATIONS

Phase-Shifter Circuits

10

APPLICATIONS

BASICS OF SINUSOID

A sinusoid is a signal that has the form of the sine or cosine function

A sinusoidal current is usually referred to as alternating current (AC)

Such a current reverses at regular time intervals and has alternately

positive and negative values as shown,

11Fig. Sinusoidal Variation

A sinusoidal signal is easy to generate and transmit

It is the dominant form of signal in the communications and electric

power industries

Through Fourier analysis, any practical periodic signal can be

represented by a sum of sinusoids

Circuits driven by sinusoidal current or voltage sources are called

AC circuits

BASICS OF SINUSOID

12

We experience sinusoidal variation in,

Fig. Motion of a pendulum Fig. Vibration of a string

BASICS OF SINUSOID

13

Fig. Ripples on the ocean surface

BASICS OF SINUSOID

14

15

The terms which we most commonly use:

Waveform: The variation of a quantity such as voltage or current

shown on a graph to a base of time is a waveform.

Cycle: Each repetition of a variable quantity, recurring at equal

intervals, is termed a cycle. Period: The duration of one cycle is termed its period.

Fig. Cycles and Periods

BASICS OF SINUSOID

16

BASICS OF SINUSOID

Instantaneous value: The magnitude of a waveform at any instant in

time. They are denoted by lower-case symbols such as e, v and i.

Peak value: The maximum instantaneous value measured from its

zero value is known as its peak value.

Peak-to-peak value: The maximum variation between the maximum

positive instantaneous value and the maximum negative instantaneous

value is the peak-to-peak value.

Fig. Peak Values

17

BASICS OF SINUSOID

Peak amplitude: The maximum instantaneous value measured from the

mean value of a waveform is the peak amplitude.

Frequency: The number of cycles that occur in 1 second is termed the

frequency of that quantity. Frequency is measured in hertz (Hz)

Fig. Effect on waveforms by varying frequency

Sinusoids, play an important role in the analysis of periodic signals

The derivative and integral of a sinusoid are themselves sinusoids.

For these reasons, the sinusoid is an extremely important function in

circuit analysis

BASICS OF SINUSOID

Consider the sinusoidal voltage,

( ) sinmv t V t Where, Vm = the amplitude of the sinusoid

ω = the angular frequency in radians/s

ωt = the argument of the sinusoid18

It is evident that the sinusoid repeats itself every T seconds thus, T is

called the period of the sinusoid

From the figures, we observe that ωT = 2π2

T

Fig. Vm sin ωt as a function of ωt Fig. Vm sin ωt as a function of t

BASICS OF SINUSOID

19

The fact that v(t) repeats itself every T seconds is shown by replacing t

by t + T,( ) sin ( ) sin ( )

2m m

Tv t T V t T V t

sin( 2 ) sin ( )m mV t V t v t

Hence, ( ) ( )v t T v t

Thus, v has the same value at t + T as it does at t and v(t) is said to be

periodic.

A periodic function is one that satisfies f (t ) = f (t + nT ), for all t and

for all integers n

BASICS OF SINUSOID

20

The period T of the periodic function is the time of one complete cycle

or the number of seconds per cycle

The reciprocal of this quantity is the number of cycles per second,

known as the cyclic frequency (f) of the sinusoid. Thus,

1f

T

2 f

While ω is in radians per second (rad/s), f is in hertz (Hz)

Therefore,

BASICS OF SINUSOID

21

A more general expression for the

sinusoid is,

( ) sin( )mv t V t

where, (ωt+ϕ) is the argument and

ϕ is the phase

Let us examine the two sinusoids,

1 2( ) sin( ), ( ) sin( )m mv t V t v t V t

Fig. Two sinusoids with different phases

BASICS OF SINUSOID

Both argument and phase can be in

radians or degrees

22

23

Fig. Two sinusoids with different phases

BASICS OF SINUSOID

The starting point of v2 in Fig.

occurs first in time

Therefore, v2 leads v1 by ϕ or v1 lags

v2 by ϕ

If ϕ ≠ 0, v1 and v2 are out of phase

Moreover, if ϕ = 0 then v1 and v2 are

said to be in phase as they reach their

minima and maxima at exactly the

same time

A sinusoid can be expressed in either sine or cosine form

When comparing two sinusoids, it is convenient to express both as

either sine or cosine with positive amplitudes using,

sin( ) sin cos cos sin

cos( ) cos cos sin sin

A B A B A B

A B A B A B

With these identities, it is easy to show that,

sin( 180) sin ,cos( 180) cos

sin( 90) cos ,cos( 90) sin

t t t t

t t t t

Using these, sinusoid can be transformed from sine form to cosine form

or vice versa.

BASICS OF SINUSOID

24

A graphical approach can also be used to

relate or compare sinusoids as an alternative

to using the trigonometric identities,

The horizontal axis represents the

magnitude of cosine, while the vertical axis

(pointing down) denotes the magnitude of

sine

Angles are measured positively counter

clockwise from the horizontal, as usual in

polar coordinates

BASICS OF SINUSOID

25

Fig. A graphical means of relating

cosine and sine,

cos(ωt − 90°) = sin ωt

Similarly, by adding 180° to the argument of sin ωt gives −sin ωt, or

sin(ωt + 180°) = −sin ωt as shown below,

BASICS OF SINUSOID

Fig. A graphical means of relating cosine, sin(ωt + 180°) = −sin ωt26

+ sin ωt

The graphical technique can also be used to add two sinusoids of the

same frequency when one is in sine form and the other is in cosine

form

BASICS OF SINUSOID

27

To add A cosωt and B sinωt as shown,

Here, A is the magnitude of cosωt

while B is the magnitude of sinωt

Fig. Adding A cosωt and B sinωt

cos sin cos( )A t B t C t

where, 2 2 1, tanB

C A BA

SOLVED NUMERICALS ON SINUSOIDS

28

29

Numerical 1: Determine frequency in Hz, angular frequency in rad/s,

and the amplitude of the harmonic voltage signals as shown.

Solution: All four signals in Fig. have the same amplitude of 0.6 V,

the same frequency of f = 1 kHz, and the same angular frequency

of ω = 2πf = 6283.1 rad/s.

30

Numerical 2: Determine the frequency, amplitude, and phase of the

harmonic voltage signal as shown versus the base cosine signal.

Solution: The amplitude by inspection is observed to be Vm = 1.0V.

The frequency is determined by observing the entire interval from

0 to 3 ms containing three full periods; hence T = 1 ms, and

f = 1/T = 1000 Hz =1 kHz.

31

For the phase determination, we note that the first maximum occurs later

in time than for the base cosine, which already peaks at t = 0.

Therefore, the phase must be negative, that is, φ < 0. The absolute value

of the phase is,

2T

T

which gives |φ| = π/3 for ΔT = T/6

Alternatively, the same result can be obtained by observing that the

cosine function is equal to 0.5 Vm at t = 0, so that φ = -cos-1 (0.5) = -π/3.

Numerical 3: Find the amplitude, phase, period, and frequency of the

sinusoid v(t) = 12 cos(50t + 10°) V.

Solution:

The amplitude is Vm = 12 V

The phase is ϕ = 10°

The angular frequency is ω = 50 rad/s

The period T =2π/ω =2π/50= 0.1257 s

The frequency is f = 1/T= 7.958 Hz

32

Numerical 4: Given the sinusoid 45 cos(5πt + 36°), calculate its

amplitude, phase, angular frequency, period, and frequency.

Solution: 45, 36°, 15.708 rad/s, 400 ms, 2.5 Hz

Numerical 5: Calculate the phase angle between v1 = −10 cos(ωt + 50°)

and v2 = 12 sin(ωt − 10°). State which sinusoid is leading.

Solution: METHOD 1- In order to compare v1 and v2, we must express

them in the same form. If we express them in cosine form with positive

amplitudes, v1 = −10 cos(ωt + 50°) = 10 cos(ωt + 50° − 180°)

v1 = 10 cos(ωt − 130°) or v1 = 10 cos(ωt + 230°)

33

v2 = 12 sin(ωt − 10°) = 12 cos(ωt − 10° − 90°)

v2 = 12 cos(ωt − 100°) = 12 cos(ωt − 100° + 360°) = 12 cos(ωt + 260°)

It can be deduced that the phase difference between v1 and v2 is 30°

Thus, it can be observed v2 leads v1 by 30°

METHOD 2- Alternatively, we may express v1 in sine form:

v1 = −10 cos(ωt + 50°) = 10 sin(ωt + 50° − 90°)

= 10 sin(ωt − 40°) = 10 sin(ωt − 10° − 30°)

34

But v2 = 12 sin(ωt − 10°). Comparing the two shows that v1 lags v2 by 30°.

This is the same as saying that v2 leads v1 by 30°

METHOD 3- We may regard v1 as simply

−10 cos ωt with a phase shift of +50°.

Similarly, v2 is 12 sin ωt with a phase shift

of −10°.

Therefore, as observed v2 leads v1 by 30°.

BASICS OF PHASORS

35

BASICS OF PHASORS

A phasor is a complex number that represents the amplitude and phase

of a sinusoid.

Phasors provide a simple means of analysing linear circuits excited

by sinusoidal sources

A complex number z can be written in rectangular form as

z x jy

where j = √(−1), x is the real part of z, y is the imaginary part of z

36

The complex number z can also be written in polar or exponential

form as,jz r re

Where, r is the magnitude of z, and ϕ is

the phase of z

,z x jy z r

Rectangular form Polar form

The relationship between the

rectangular form and the polar form

is shown in Fig.,

BASICS OF PHASORS

Fig. Representation of a complex

number37

Where, x axis represents the real part and the y axis represents the

imaginary part of a complex number

BASICS OF PHASORS

Given x and y, we can get r and ϕ as

2 2 1, tany

r x yx

, cos , sinwhere x r y r

Thus, z may be written as

(cos sin )z x jy r r j

The idea of phasor representation is based on Euler’s identity

cos sinje j 38

cos Re( ),sin Im( )j je e

BASICS OF PHASORS

Given a sinusoid v(t) = Vm cos(ωt + ϕ),

( )( ) cos( ) Re( )j t

m mv t V t V e

Thus, ( ) Re( )j tv t Ve where,j

m mV V e V

By considering cos ϕ and sin ϕ as the real and imaginary parts of e jϕ,

39

BASICS OF PHASORS

The plot of Vejωt = Vmej(ωt+ϕ) on the complex plane can be observed as,

Fig. Representation of Vejωt sinor rotating counter clockwise and its projection on the real axis as

a function of time.

40

As time increases, the sinor rotates on a circle of radius Vm at an

angular velocity ω in the counter clockwise direction

BASICS OF PHASORS

Phasors V = and I = are graphically represented,

Such a graphical representation of phasors is known as a phasor diagram

41

mV mI

BASICS OF PHASORS

By suppressing the time factor, we transform the sinusoid from the

time domain to the phasor domain. This transformation is summarized

as follows,

( ) cos( )m mv t V t V V

Besides time differentiation and integration, another important use of

phasors is found in summing sinusoids of the same frequency

42

SOLVED NUMERICALS ON PHASORS

43

Numerical 1: Transform these sinusoids to phasors,

(a) i = 6 cos(50t − 40°) A

(b) v = −4 sin(30t + 50°) V

Solution:

(a) i = 6 cos(50t − 40°)A has the phasor,

(b) Since, −sin A = cos(A + 90°),

v = −4 sin(30t + 50°) = 4 cos(30t + 50° + 90°) = 4 cos(30t + 140°) V

The phasor form of v is 4 140oV V

6 40I A

44

Numerical 2: Express these sinusoids as phasors,

(a) v = −14 sin(5t − 22°) V

(b) i = −8 cos(16t + 15°) A

Solution: (a) (b)

Numerical 3: Find the sinusoids corresponding to these phasors,

(a) (b)

Solution: (a) v(t) = 25 cos(ωt − 140°) V or 25 cos(ωt + 220°) V,

(b) i(t) = 13 cos(ωt + 67.38°) A

8 165I A 14 68V V

25 40V V (12 5)I j j A

45

Numerical 4: Given i1(t) = 4 cos(ωt + 30°) A and i2(t) = 5 sin(ωt − 20°) A,

find their sum.

Solution:

Current i1(t) is in the standard form. Its phasor is

We need to express i2(t) in cosine form. The rule for converting sine to

cosine is to subtract by 90°,

i2 = 5 cos(ωt − 20° − 90°) = 5 cos(ωt − 110°)

and its phasor is

4 30oI A

2 5 110oI A

46

If we let i = i1 + i2, then

I = I1 + I2 =

= (3.464 + j2) + (-1.71 - j4.698)

= 1.754 − j2.698

=

4 30 5 110o A

3.218 56.97 A

Transforming this to the time domain,

we get i(t) = 3.218 cos(ωt − 56.97°) A

47

AVERAGE AND RMS VALUES OF

AN ALTERNATING CURRENT

48

AVERAGE AND RMS VALUES OF AN ALTERNATING

CURRENT

Let us first consider the wave shown as shown, which is typical current

waveform produced by a transformer on no load,

Fig. Average and RMS values 49

If n equidistant mid-ordinates (i1, i2, ...) are taken over either the

positive or the negative half-cycle, then average value of current over

half a cycle is,

AVERAGE VALUE OF AN ALTERNATING CURRENT

1 2 ... nav

i i iI

n

Or, alternatively, average value of current is,

50

=𝐴𝑟𝑒𝑎 𝑒𝑛𝑐𝑙𝑜𝑠𝑒𝑑 𝑜𝑣𝑒𝑟 ℎ𝑎𝑙𝑓 − 𝑐𝑦𝑐𝑙𝑒

𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑏𝑎𝑠𝑒 𝑜𝑣𝑒𝑟 ℎ𝑎𝑙𝑓 − 𝑐𝑦𝑐𝑙𝑒


CURRENT If Im is the maximum value of a current which varies sinusoidally as

shown,

Fig. Average and RMS values of a sinusoidal current51


CURRENT

The instantaneous value i is represented by,

sinmi I

where, θ is the angle in radians from instant of zero current

For a very small interval dθ radians, the area of the shaded strip is

i · dθ ampere radians

The use of the unit ‘ampere radian’ avoids converting the scale on

the horizontal axis from radians to seconds

52

Therefore, total area enclosed by the current wave over half-cycle is,

AVERAGE VALUE OF AN ALTERNATING CURRENT

0 0 0

sin cos

( 1) 1 2

m m

m m

i d I d I

I I

Therefore, average value of current over a half-cycle is Iav

2 mI

Iav = 0.637Im

53

RMS VALUE OF AN ALTERNATING CURRENT

If i is the instantaneous current through the resistance, the average

power dissipated is,

Therefore, 0.7072

mm

II I

It is normal practice to omit the RMS subscript, and just denote IRMS as I

54

< 𝑖2𝑅 > = 𝐼𝑅𝑀𝑆2 𝑅 wher𝑒 < 𝑖2 >= 𝐼𝑅𝑀𝑆

< 𝑖2 >=1

𝜋 0

𝜋

𝐼𝑚2 𝑠𝑖𝑛2 𝜃 𝑑𝜃 =

𝐼𝑚2

2

IRMS


CURRENT

RMS value of a sinusoidal current or voltage is,

0.707 mI I

Form factor (= RMS/Average) of a sine wave is,

Form factor =0.707 ∗ maximum value

0.637 ∗ minimum value1.11fk

Peak or crest factor (= Amplitude/Average) of a sine wave is

Peak factor =maximum value

0.707 ∗ maximum value1.414pk

55

NUMERICALS ON AVERAGE AND RMS VALUES

56

Numerical 1: An alternating current of sinusoidal waveform has an RMS

value of 10.0 A. What is the peak-to-peak value of this current?

Solution:

The peak-to-peak value is therefore 14.14 − (-14.14) = 28.28 A

1014.14

0.707 0.707m

II A

57

Solution: The relation is of the form v = Vm sinωt and by comparison,

( ) 141.4 2ma V V V Hence, 141.4

1002

V V

(b)Also by comparison,377

377 / 2 , 602

rad s f f Hz

(c)Finally, 141.4sin 377v t

when t = 3 × 10−3 sec,

3141.4sin(377 3 10 ) 141.4sin1.131

141.4 0.904 127.8

v

V

58

Numerical 2: An alternating voltage has the equation v = 141.4 sin 377t,

what are the values of (a) RMS voltage (b) frequency

(c) the instantaneous voltage when t = 3 ms?

Numerical 3: A current has the following steady values in amperes for

equal intervals of time changing instantaneously from one value to the

next (Fig. 32):

0, 10, 20, 30, 20, 10, 0, −10, −20, −30, −20, −10, 0, etc.

Calculate the RMS value of the current and its form factor.

Fig. Waveform of current59

Solution: 𝐼𝑎𝑣 =area under curve

𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑏𝑎𝑠𝑒

2 3 2 4 3 5 4 6 50 0 10 20 30 20 10

6 6 6 6 6 6 6 6 6 6 6

0

15A

2 2 2 2 2 2

2

2 3 2 4 3 5 4 6 50 0 10 20 30 20 10

6 6 6 6 6 6 6 6 6 6 6

0

I

60

316

316 17.8

17.81.19

15.0f

av

I A

Ik

I

STEADY STATE AC RESPONSE FOR A

PURE RESISTIVE CIRCUIT

61

STEADY STATE AC RESPONSE FOR A PURE RESISTIVE

CIRCUIT

Consider a circuit having a resistance R ohms connected across the

terminals of an AC generator G as shown,

Fig. Circuit with resistance

62

If the value of the voltage at any instant is v volts, the value of the

current at that instant is given by,


CIRCUIT

vi

R

When the voltage is zero, the current is also zero and since the current

is proportional to the voltage, the waveform of the current is exactly

the same as that of the voltage

63

The two quantities are in phase as

they pass through their zero values

at the same instant and attain their

maximum values in a given

direction at the same instant

Hence the current wave is as

shown,


CIRCUIT

Fig. Voltage and current waveforms for a

resistive circuit

64


CIRCUIT

If Vm and Im are the maximum values of the voltage and current

respectively, it follows that

mm

VI

R

But the RMS value of a sine wave is 0.707 times the maximum value,

so thatRMS value of voltage=V=0.707Vm

RMS value of current=I=0.707Im

65


CIRCUIT

Substituting for Im and Vm we have,

0.707 0.707

I V VI

R R

Hence, Ohm’s law can be applied without any modification to an AC

circuit possessing resistance only

If the instantaneous value of the applied voltage is represented by,

sinmv V t

66


CIRCUIT

Then instantaneous value of current in a resistive circuit is

sinmV ti

R

The phasors representing the voltage and current in a resistive circuit

are shown below,

Fig. Phasor diagram for a resistive circuit

The two phasors are actually coincident, but are drawn slightly apart

so that the identity of each may be clearly recognized67

UNSOLVED PROBLEMS

68

Question 1: Given the sinusoidal voltage v(t) = 50 cos (30t + 10°) V, find:

(a) the amplitude Vm, (b) the period T, (c) the frequency f, and (d) v(t) at t =

10 ms.

Question 2: A current source in a linear circuit has is = 8 cos(500πt - 25°) A

(a) What is the amplitude of the current?

(b) What is the angular frequency?

(c) Find the frequency of the current.

(d) Calculate is at t = 2 ms.

Question 3: Express the following functions in cosine form:

(a) 4 sin(ωt - 30°) (b) −2 sin(6t)

(c) −10 sin(ωt + 20°)

69

Ans.2: (a) 8 A, (b) 1570.8 rad/s,

(c) 250 Hz, (d) -7.25 A

Ans.3: (a) 4 cos(ωt-120°), (b) 2

cos(6t+90°), (c) 10 cos(ωt+110°)

Ans.1: (a) 50 V, (b) 209.4 ms, (c)

4.775 Hz, (d) 44.48 V, 0.3 rad

Question 4: Given v1 = 20 sin(ωt + 60°) V and v2 = 60 cos(ωt − 10°) V,

determine the phase angle between the two sinusoids and which one lags

the other.

Question 5: For the following pairs of sinusoids, determine which one

leads and by how much.

(a) v(t) = 10 cos(4t − 60°) and i(t) = 4 sin(4t + 50°)

(b) v1(t) = 4 cos(377t + 10°) and v2(t) = −20 cos 377t

(c) x(t) = 13 cos 2t + 5 sin 2t and y(t) = 15 cos(2t − 11.8°)

70

Ans.4: 20º and v1 lags v2

Ans.5: (a) i(t) leads v(t) by 20° (b) v2(t) leads

v1(t) by 170° (c) y(t) leads x(t) by 9.24°

Question 6: Find the phasors corresponding to the following signals:

(a) v(t) = 21 cos(4t − 15°)V (b) i(t) = −8 sin(10t + 70°) mA

(c) v(t) = 120 sin(10t − 50°) V (d) i(t) = −60 cos(30t + 10°) mA

Question 7: Transform the following sinusoids to phasors:

(a) −10 cos(4t + 75°) (b) 5 sin(20t - 10°)

(c) 4 cos (2t) + 3 sin (2t)

Question 8: Two voltages v1 and v2 appear in series so that their sum is v

= v1 + v2. If v1 = 10 cos(50t − π∕3) V and v2 = 12 cos(50t + 30°) V, find v.

71

Ans.6: (a) 21∟-15 ° (b) 8∟-160 ° (c) 120∟-140 ° (d) 60∟-190 °

Ans.7: (a) 10∟-105° (b) 5∟-100 ° (c) 5∟-36.87 °

Ans.8: 15.62 cos(50t-9.8°) V

Question 9: A linear network has a current input 10 sin(ωt + 30°) A and a

voltage output -65cos(ωt+120°) V. Determine the associated impedance.

Question 10: If the waveform of a voltage has a form factor of 1.15 and

a peak factor of 1.5, and if the peak value is 4.5 kV, calculate the average

and the r.m.s. values of the voltage.

Question11: An alternating current was measured by a DC milliammeter

in conjunction with a full-wave rectifier. The reading on the milliammeter

was 7.0 mA. Assuming the waveform of the alternating current to be

sinusoidal, calculate: (a) the r.m.s. value; and (b) the maximum value of

the alternating current.

72

Ans.11: 7.8mA, 11 mA

Ans.10: 2.61kV, 3 kV

Ans.9: 6.5 Ω

Question 12: An alternating current has a periodic time 2T. The current

for a time one-third of T is 50 A; for a time one-sixth of T, it is 20 A; and

zero for a time equal to one-half of T. Calculate the RMS and average

values of this current.

Question 13: A triangular voltage wave has a periodic time of (3/100)

sec. For the first (2/100) sec, of each cycle it increases uniformly at the

rate of 1000 V/s, while for the last (1/100) sec, it falls away uniformly to

zero. Find (a) its average value; (b) its RMS value; (c) its form factor.

73

Ans.12: 30 A RMS, 20 A average

Ans.13: 10 V, 11.55 V, 1.155

REFERENCES

[1] Edward Hughes; John Hiley, Keith Brown, Ian McKenzie Smith,

“Electrical and Electronic Technology”,10th Ed., Pearson Education

Limited, 2008.

[2] Alexander, Charles K., and Sadiku, Matthew N. O., “Fundamentals of

Electric Circuits”, 5th Ed, McGraw Hill, Indian Edition, 2013.

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introduction to electrical engineering l 13: single phase...

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