ELL 100 - Introduction to Electrical Engineering

LECTURE 13: SINGLE PHASE AC CIRCUITS

1 OUTLINE

 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 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

3 INTRODUCTION  AC can also be employed using capacitors and inductors in electronic circuitry, allowing for a wide range of applications

Various AC Waveforms 4 SINGLE PHASE APPLIANCES

Single Phase Transformer 5 SINGLE PHASE APPLIANCES

Single Phase Meter

6 SINGLE PHASE APPLIANCES

Single Phase Power Supply with Single Phase Grid Connected System Battery 7 SINGLE PHASE APPLIANCES

Single Phase Fan Induction Motor

8 APPLICATIONS

Phase-Shifter Circuits

9 APPLICATIONS

10 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,

Fig. Sinusoidal Variation 11 BASICS OF SINUSOID

 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

12 BASICS OF SINUSOID

 We experience sinusoidal variation in,

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

13 BASICS OF SINUSOID

Fig. Ripples on the ocean surface

14 BASICS OF SINUSOID 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 15 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 16 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 17 BASICS OF SINUSOID

 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  Consider the sinusoidal voltage, Where, V = the amplitude of the sinusoid v( tVt )sin m  m ω = the angular frequency in radians/s ωt = the argument of the sinusoid 18 BASICS OF SINUSOID

 It is evident that the sinusoid repeats itself every T seconds thus, T is called the period of the sinusoid 2  From the figures, we observe that ωT = 2π T  

Fig. Vm sin ωt as a function of ωt Fig. Vm sin ωt as a function of t 19 BASICS OF SINUSOID  The fact that v(t) repeats itself every T seconds is shown by replacing t by t + T, T v()sin()sin() tTVtTVt  mm2

Vmmsin( t  2  )  V sin  t  v ( t ) Hence, v()( tTv ) 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 20 BASICS OF SINUSOID  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, 1 f  T  Therefore,  2 f While ω is in radians per second (rad/s), f is in hertz (Hz)

21 BASICS OF SINUSOID  A more general expression for the sinusoid is,

vtVt()sin()m  where, (ωt+ϕ) is the argument and ϕ is the phase  Both argument and phase can be in radians or degrees  Let us examine the two sinusoids, Fig. Two sinusoids with different phases

v12( t ) Vmm sin( t ), v ( t )  V sin(  t   ) 22 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 Fig. Two sinusoids with different phases

23 BASICS OF SINUSOID  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(ABABAB )  sin cos  cos sin cos(ABABAB )  cos cos  sin sin  With these identities, it is easy to show that, sin(tt 180)   tt  sin ,cos(   180) cos sin(tt 90)   tt  cos  ,cos( 90) sin Using these, sinusoid can be transformed from sine form to cosine form

or vice versa. 24 BASICS OF SINUSOID

 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 Fig. A graphical means of relating cosine and sine, clockwise from the horizontal, as usual in cos(ωt − 90°) = sin ωt polar coordinates

25 BASICS OF SINUSOID  Similarly, by adding 180° to the argument of sin ωt gives −sin ωt, or sin(ωt + 180°) = −sin ωt as shown below,

+ sin ωt Fig. A graphical means of relating cosine, sin(ωt + 180°) = −sin ωt 26 BASICS OF SINUSOID  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  To add A cosωt and B sinωt as shown, Here, A is the magnitude of cosωt while B is the magnitude of sinωt

Acos t B sin  t  C cos(  t   ) B where, CA B221 ,tan  A Fig. Adding A cosωt and B sinωt

27 SOLVED NUMERICALS ON SINUSOIDS

28 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. 29 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. 30 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, T  2 T

which gives |φ| = π/3 for ΔT = T/6 Alternatively, the same result can be obtained by observing that the -1 cosine function is equal to 0.5 Vm at t = 0, so that φ = -cos (0.5) = -π/3.

31 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

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

32 ° 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 )

° ° ° 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

33 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°) ° ° 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 .

34 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 zxjy

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

36 BASICS OF PHASORS  The complex number z can also be written in polar or exponential form as, zrre j Where, r is the magnitude of z, and ϕ is the phase of z zxjy , zr Rectangular form Polar form  The relationship between the rectangular form and the polar form is shown in Fig., Fig. Representation of a complex number 37 BASICS OF PHASORS

 Where, x axis represents the real part and the y axis represents the imaginary part of a complex number  Given x and y, we can get r and ϕ as y rxy 221 ,tan  where,cos,sin xryr x  Thus, z may be written as zxjy  rrj   (cossin )  The idea of phasor representation is based on Euler’s identity ej j cos sin 38 BASICS OF PHASORS

 By considering cos ϕ and sin ϕ as the real and imaginary parts of e jϕ, cosRe(),sinIm()eejj

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

jt() v( tVtV )cos()mm e Re()

jt j  Thus, v( t ) Re( Ve ) where, V Vmm e  V 

39 BASICS OF PHASORS

jωt j(ωt+ϕ)  The plot of Ve = Vme on the complex plane can be observed as,

Fig. Representation of Ve jωt sinor rotating counter clockwise and its projection on the real axis as a function of time.

 As time increases, the sinor rotates on a circle of radius Vm at an angular velocity ω in the counter clockwise direction 40 BASICS OF PHASORS

 Phasors V = Vm and I = I m   are graphically represented, Such a graphical representation of phasors is known as a phasor diagram

41 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,

v( tVtVV )cos()mm

 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, IA640  (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 VV4 140o

44 Numerical 2: Express these sinusoids as phasors, (a) v = −14 sin(5t − 22°) V (b) i = −8 cos(16t + 15°) A

Solution: (a) VV  14  68  (b) IA8165 

Numerical 3: Find the sinusoids corresponding to these phasors, (a) VV 2540 (b) IjjA(125) Solution: (a) v(t) = 25 cos(ωt − 140°) V or 25 cos(ωt + 220°) V, (b) i(t) = 13 cos(ωt + 67.38°) A

45 ° ° Numerical 4: Given i1(t) = 4 cos(ωt + 30 ) A and i2(t) = 5 sin(ωt − 20 ) A, find their sum. Solution:

o Current i1(t) is in the standard form. Its phasor is IA430

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 ) o and its phasor is IA2 5110  

46 If we let i = i1 + i2, then o I = I1 + I2 = 4305110   A

= (3.464 + j2) + (-1.71 - j4.698) = 1.754 − j2.698 = 3.21856.97  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 AVERAGE VALUE OF AN ALTERNATING CURRENT

 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, iii ... I  12 n av n

Or, alternatively, average value of current is,

퐴푟푒푎 푒푛푐푙표푠푒푑 표푣푒푟 ℎ푎푙푓 − 푐푦푐푙푒 = 퐿푒푛푔푡ℎ 표푓 푏푎푠푒 표푣푒푟 ℎ푎푙푓 − 푐푦푐푙푒

50 AVERAGE AND RMS VALUES OF AN ALTERNATING CURRENT  If Im is the maximum value of a current which varies sinusoidally as shown,

Fig. Average and RMS values of a sinusoidal current 51 AVERAGE AND RMS VALUES OF AN ALTERNATING CURRENT  The instantaneous value i is represented by,

iI m sin 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 AVERAGE VALUE OF AN ALTERNATING CURRENT

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

  i dIdI  sincos mm  00 0

 IImm(  1)  12 2I  Therefore, average value of current over a half-cycle is I  m av 

Iav = 0.637Im

53 RMS VALUE OF AN ALTERNATING CURRENT

 If i is the instantaneous current through the resistance, the average power dissipated is, 2 2 2 < 𝑖 푅 > = 퐼푅푀푆푅 wher푒 < 𝑖 > = 퐼푅푀푆

휋 2 2 1 2 2 퐼푚 < 𝑖 > = 퐼푚 푠𝑖푛 휃 푑휃 = 휋 0 2

Im Therefore, IRMSII0.707 m 2

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

54 AVERAGE AND RMS VALUES OF AN ALTERNATING CURRENT  RMS value of a sinusoidal current or voltage is,

II 0.707 m

 Form factor (= RMS/Average) of a sine wave is, 0.707 ∗ maximum value Form factor = k f  1.11 0.637 ∗ minimum value

 Peak or crest factor (= Amplitude/Average) of a sine wave is maximum value Peak factor = k  1.414 0.707 ∗ maximum value p 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?

I 10 Solution: IA14.14 m 0.7070.707 The peak-to-peak value is therefore 14.14 − (-14.14) = 28.28 A

57 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?

Solution: The relation is of the form v = Vm sinωt and by comparison, 141.4 ()141.42a VVV Hence, VV100 m 2 (b)Also by comparison, 377 377/2,60rad sf fHz 2

(c)Finally, vt 141.4sin 377 v 141.4sin(377  3  103 )  141.4sin1.131 when t = 3 × 10−3 sec, 141.4  0.904  127.8V

58 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 current 59 area under curve Solution: 퐼 = 푎푣 푙푒푛푔푡ℎ 표푓 푏푎푠푒

232435465 00 1020302010  66666666666     0   15A

222222    23  24  35   46  5  00 1020302010      66  66  66  66   66   6    I 2           0  316  IA316 17.8 I 17.8 k f    1.19 Iav 15.0 60 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 STEADY STATE AC RESPONSE FOR A PURE RESISTIVE CIRCUIT  If the value of the voltage at any instant is v volts, the value of the current at that instant is given by, v i  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 STEADY STATE AC RESPONSE FOR A PURE RESISTIVE CIRCUIT  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 Fig. Voltage and current waveforms for a resistive circuit shown,

64 STEADY STATE AC RESPONSE FOR A PURE RESISTIVE CIRCUIT

 If Vm and Im are the maximum values of the voltage and current respectively, it follows that V I  m m R But the RMS value of a sine wave is 0.707 times the maximum value, so that RMS value of voltage=V=0.707Vm

RMS value of current=I=0.707Im

65 STEADY STATE AC RESPONSE FOR A PURE RESISTIVE CIRCUIT

 Substituting for Im and Vm we have, IVV  I 0.7070.707 RR 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,

vVt m sin

66 STEADY STATE AC RESPONSE FOR A PURE RESISTIVE CIRCUIT  Then instantaneous value of current in a resistive circuit is Vtsin i  m 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 recognized 67 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. Ans.1: (a) 50 V, (b) 209.4 ms, (c) 4.775 Hz, (d) 44.48 V, 0.3 rad ° 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. Ans.2: (a) 8 A, (b) 1570.8 rad/s, (d) Calculate is at t = 2 ms. (c) 250 Hz, (d) -7.25 A Question 3: Express the following functions in cosine form: (a) 4 sin(ωt - 30°) (b) −2 sin(6t) (c) −10 sin(ωt + 20°) Ans.3: (a) 4 cos(ωt-120°), (b) 2 cos(6t+90°), (c) 10 cos(ωt+110°) 69 ° ° 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. º Ans.4: 20 and v1 lags v2 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°)

° 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

70 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

Ans.6: (a) 21∟-15 ° (b) 8∟-160 ° (c) 120∟-140 ° (d) 60∟-190 ° 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) Ans.7: (a) 10∟-105° (b) 5∟-100 ° (c) 5∟-36.87 °

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.

Ans.8: 15.62 cos(50t-9.8°) V

71 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. Ans.9: 6.5 Ω 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. Ans.10: 2.61kV, 3 kV 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. Ans.11: 7.8mA, 11 mA

72 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. Ans.12: 30 A RMS, 20 A average 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.

Ans.13: 10 V, 11.55 V, 1.155

73 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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