a c circuits

UNIT 3 A.C. CIRCUITS

Structure 3.1 Introduction

Objectives '

3.2 Sinusoidal Signals 3.2.1 Importance of Sinusoidal Signals 3.2.2 Effective Value and Form Factor 3.2.3 Phasor Representation

3.3 Phasors and their Treatment 3.3.1 Phasor as a Complex Number 3.3.2 Algebra of Complex Numbers 3.3.3 Kirchhoff's Laws in Phasor Domain

3.4 Impedance Concept 3.4.1 Response of Single Elements to Sinusoidal Excitation 3.4.2 Concept of Impedance and Admittance 3.4.3 Mutual Inductance

3.5 A.C. Circuit Analysis 3.5.1 Simple Series and Parallel Circuits 3.5.2 Series-parallel Reduction Techniques 3.5.3 Solution by Loop and Node Equations

3.6 Concepts Relating to Power 3.6.1 Power. Apparent Power and Power Factor 3.6.2 Reactive Power 3.6.3 Power Factor Correction

3.7 Summary

3.8 Answers to SAQs

3.1 INTRODUCTION

In Unit 2, we considered various important methods of analysis of d.c. circuits. We need to deal with such circuits wherever d.c. soprces qre present, whether in the form of d.c. generators mechanically driven by prime movers, or rechargeable or non-rechargeable d.c. cells or d.c. sources used in modelling various electrical and electronic devices. Incidentally, the only practically useful mode of storage of electrical energy at the present time is through d.c. storage batteries (rechargeable d.c. cells) and hence the latter are widely used in transport vehicles, electronic instruments and equipment, portable tools and the like.

However the bulk of electrical energy utilisation, whether in domestic installations, or in industry or in public service organisations is through alternating current (a.c.) systems involving sinusoidal voltages and currents. Strictly speaking, the adjective alternating indicates any signal whose direction alternates with time but in practice it is invariably used to refer to sinusoidal signals.

In this Unit, you will first learn about thc characteristics and representation of sinusoidal voltages and currents. You will then get to know the types of response of standard circuit elements to sinusoidal excitation. This will be a prelude to thc study of methods of analysis of circuits formed by various combinations of simple circuit elements and operating under the influence of a.c. sources. Thephasor concept which simplifies the expression of steady-state response-excitation relations in these circuits would be adopted as the framework for the development of the various techniqucs of analysis. You will also be introduced to the various ramifications of power in an a.c. circuit.

Objectives After a study of this unit, you should be able to

explain the reasons for the widespread use of sinusoidal signals in clcctrical engineering,

Intfoduction to Circuits deduce the different parameten of a sinusoidal waveform like peak value, phase, effective value, form factor etc.,

work with complex numbers as needed for the manipulation of phasors in a.c. circuit anaiysis,

express the steady state response both in time domain and in phasor domain of R, L, C elements to sinusoidal inputs,

calculate the impedance 2 and adniittance ?of simple elements and combinations thereof in 2-terminal networks,

analyse siinple a.c. circuits through series-parallel reduction techniques or through the method of loop or node equations,

describe the phenomenon of resonance in siinple series and pawllel circuits and deduce the associated characteristics,

explain the factors on which power in an a.c. circuit depends and distinguish between active power, apparent power and reactive power, and

explain the need for power factor improvement and calculate the capacitance values needed for the purpose in typical circuits.

3.2 SINUSOIDAL SIGNALS

You were already introduced in Unit 1 to periodical signals having sinusoidal waveform. Sinusoidal signals have a vital role both in electrical power engineering and in communication engineering. In power supply systems, the voltages and currents are invariably of this waveform with a frequency of 50 Hz in most countries including India. In the field of communication engineering, we have to deal with sinusoidal signals having a wide frequency range extending from a few Hz to a few GHz.

Formally, a sinusoidal function of time is defined as one having the general form A sin (wt + 8) and is characterised by three parainelers viz amplitude A, angular frequency w and phase angle 8. The argument (angle) of the sine function viz., (wt + 8 ) is measured in radians and increases at the rate of w radians per second. Since a sine function repeats itself at intervals of 2n radians of its angle, we immediately see that the period T of the sinusoid is related to w by

We also know (vide Eq.1.20) that the frequency f of the signal is the reciprocal of the period T. Therefore we have

The parameter w is thus a measure of the frequency of the signal and is called angular frequency. The phase angle 8 is the value of the argument of the sine function at the origin of time and goverlls the instantaneous value at t = 0 of the sinusoidal signal of a given amplitude. Strictly speaking, 8 should be expressed in radians as wt is expressed in radians. However since many of us have a better feel for angles measured in degrees, we

Fig. 3.1: Waveform of a general sinusoidal voltage v ( t ) - V,,, sin (wt + 0)

often indulge in the somewhat irregular practice of mixing up units by writing expressions like 325 sin ( 3 1 4 + 60") instead of the more proper 325 sin (314t + J55). However the actual evaluation of the sine function is done only after expressing the two terms in its argument in the same units (radians or degrees) before their addition.

Figure 3.1 shows the waveform of a general sinusoidal voltage of frequency5 As shown therein, the x-axis can be graduated either in terms of time t in the conventional manner or alternatively in terms of the angle wt. The latter has the effect of normalizing the period of the waveform to ~.JC radians irrespective of the actual value of frequency 8 Example 3.1

Express the following signals in the standard form, A sin (wt + 0).

(a) v1 = 20 cos wt

(b) vz = 100 sin wt + 75 cos wt

(c) v3 = 10 cos (wt - d 2 )

Solution

(a) vl = 20 cos wt = 20 sin (wt + JL / 2)

(b) 100 sin wt + 75 cos wt = (1001125 sin wt + 751125 cos wt)

= 125 (sin wt cos 0 + cos wt sin 0)

= 125 sin (wt + 0), where 0 = tan- (0.75) = 37" = 0.64 radians

Hence v, = 125 sin (wt + 37") or 125 sin (wt + 0.64)

(c) v3 = 10 cos (wt - d 2 ) = 10 cos ( d 2 - wt)

= 10 sin wt.

SAQ 1. Express the following sinusoidal signals in the standard form, A sin (wt + 0).

(a) il = 2 cos (wt + 30")

(b) vz = 4 sin 314t + 4 cos (314t + 150")

I 3.2.1 Importance of Sinusoidal Signals What is so special about sinusoidalfunctions to give them a centralplace in circuit and signal analysis? As an answer to this question, we present in the following paragraphs some useful and interesting characteristics of sinusoidal signals.

A sinusoidal function of time represents the simplest periodic process that occurs in the physical world like the vibration of a tuning fork, the small amplitude oscillation of a pendulum, and the current in a simple LC (inductor-capacitor) circuit supplied with some initial energy. A superposition of such functions is the characteristic or natural behaviour of any non-dissipative linear system of arbitrary complexity, a network comprised of

! pure inductors and capacitors being an example of such a system. 1 A sinusoidal function has certain unique characteristics which no other periodic function ! can claim. If two sinusoids of the same frequency are added to or subtracted from each ! other, the result is another sinusoid of the same frequency. If a sinusoid is integrated or

differentiated, once again a sinusoid of the same frequency arises. Therefore when a sinusoidal voltage is applied to a resistor, an inductor or a capacitor, the resulting current waveform is sinusoidal and vice versa. That the waveform is retained under the above four linear operations is a very significant and unique property. What is true of a single element, R, L or C, is also true of a complex interconnection of such elements. Thus it turns out that if any arbitrary linear electric network is excited by a sinusoidal source A sin (wt + 0), the resulting steady state current I voltage response in any element of the network is another sinusoid of the same frequency. The response can differ from the

A.C. Circuits

Introduction to Circuits excitation only in its amplitude and phase but in no other characteristic i.e., it should have the form B sin (wt + a). This property leads to particularly simple techniques of a.c. circuit analysis using concepts like impedance and sinusoidal transfer function, which we shall get to know later on.

The voltage developed by a rotating electrical generator in a power station is necessarily periodic. Leaving aside the case where this voltage is converted to d.c. form through a commutator, the particular periodic form employed in a commercial electrical generator is the simplest form of alternating quantity, namely one which varies sinusoidally with respect to time. The preference for this form is obvious from what is stated in the previous paragraph. An electrical power system may be enormously complex, containing scores of generators, hundreds of kilometres of transmission lines and different kinds of motors and other user equipment. But as long as all the generators develop sinusoidal voltages of the

lihl same frequency, as indeed they do, and the system is linear, then the voltage available at every power outlet is a sinusoid of the same frequency. Tf the generator voltages are of any other waveform, the voltages at different locations could be a maddening medley of complex waveforms with no easily recognizable relation between one and the other.

There do occur situations, especially in electronic circuits and power electronic applications, where the source functions are periodic b ~ ~ t non-sinusoidal and thcrcfore the response functions are not only non-sinusoidz! bur u a v b waveforms different from those of the source functions. Here one takes advantage of the simplicity of system analysis . under sinusoidal steady state by treating a given source function as a linear combination of sinusoidal functions, using the well known Fourier series approach and superposing the responses to the individual sinusoidal excitation functions.

Any repetitive signal has little information content and therefore the signals of importance in communication engineering are aperiodic (non periodic). Even so, the Fourier integral method enables one to express any arbitrary waveform as the superposition of sinusoids occupying a continuous band of frequencies. Here again, therefore, the response of a system to a sinusoidal input, expressed as a function of frequency becomes important and is widely used to characterise the system. You will learn about this particular characterisation in Unit 5.

The foregoing observations clearly highlight the importance of developing efficient techniques for analysis of circuits under sinusoidal excitation. Not only can these techniques be directly used in many practically useful circuits but also they can be extended to find effective solutions for the behaviour of circuits and systems working under non-sinusoidal periodic and aperiodic excitations. I

3.2.2 Effective Value and Form Factor You have already learnt in Section 1.6 that it is the effective value of a periodic signal which counts as far as power calculations are concerned and that for the particular case of a sinusoidal signal, the effective value is (l/\/Z) times the peak value. It is conventional to indicate the strength of a sinusoidal voltage or current in terms of its effective value (RMS value) and to simply use the capital letter V or I as the symbol for this quantity, discarding the subscripts in the symbols Veffand Ief A 230 V a.c. voltage would mean a voltage having an effective value of 230 V. This point has to be clearly understood. Unless otherwise stated, any numerical value assigned to an a.c. signal implies that it is the rms value and not either its peak value or its absolute average value. A 5 A, 50 Hz a.c. current would have a time variation fi x 5 sin ( 1 0 W + 8 ) since the peak value I,,, = \/Z x 5 and w = 2 ( 5 0 ) = 100n. We hereafter indicate a general sinusoidal voltage and current as fl V sin(wt + 8) and I sin (wt + 8 ) respectively, Vand I being the corresponding effective values.

In Section 1.6, we also observed that the only meaningful average value that can be associated with a sinusoid is the absolute average vallue (also equal to the average over the positive half cycle and hence called half-cycle auerage) and that the latter is equal to ( 2 1 ~ ) times the peak value.

We now define the form factor of a symmetric periodic waveform as follows:

Form factor = RMS Value

Absolute average value

From the name of the term, it is clear that form factor gives an indication of the shape of the wave. The peakier the waveform is, the larger is the form factor. For a flat waveform

like the symmetrical square wave of Figure 1.34(a), the form factor has a value equal to 1 since the rms and absotute average values of this wave are equal. For a sinusoid we have,

Form factor of a sinusoid = Peak Value /a n =--

(Peak Value) (2/n) fi - 1.11.

i The form factor is a parameter to be considered in certain applications like the determination of the effective value of voltage induced in a coil due to changing

! magnetic flux of a given amplitude.

Eiample 3.2

Find the fonn factor of a syininetric triangular voltage uravefonn shown in Figure 3.2.

4 I A

I Fig. 3.2 : Symmetric triangular waveform

Solution

To find v ; ~ , we need to find the average of 3. As the area under the curve of 3 vs t for the first quarter period is equal to the areas for the subsequent 3 quarter periods, we can find the required average of by considering only the interval 0 < t < T/4. Thus

A v = - r m v3

Likewise A

Hence form factor of the waveform = - = - - -1.154 A/2

3.2.3 Phasor Representation If a point moves around a circle with uniform angular velocity, its projection on a straight line varies sinusoidally. By virtue of this property, a sinusoid permits an extremely simple graphical representation viz a directed line segment indicating the position of the revolving point at t = 0.

1 Let the point P move around a circle of radius A in the anticlockwise direction with an angular velocity of w radlsec and pass through the angular position 0 at t = 0, as shown in Figure 3.3(a). Its position at t = 0 and a general t are marked in the figure.

A Y t OM

I

A.C. Circuits

Fig. 3.3 : Revolving point and its vertical projection. (a) Revolving point P and its projection M on Y-axis (b)Variation of m w i t h time

Introduction to Circuits The displacement from origin 0 of the projection of P on the vertical axis (viz OM) clearly varies as A sin (wt + 0) as shown in Figure 3.3(b), taking upward displacement as positive.

On this basis, we evolve a simple representation of a sinusoidA sin (wt + 0) in the form of a directed line segment at an angle 0 with respect to x-axis (i.e., directed towards the position of the revolving point at t = 0)and having a length equal t o A 1 n (i.e, rms value of the sinusoid). The choice of the length of the line as A m is dictated by convenience as we invariably deal with the nns values of sinitsoids rather than peak Values. Such a representation is called the phasor of the particular sinusoid. For the sinusoidal signal A sin (wr + 0), the phasor would thus be a directed line of length ( A m ) at an angle 0 relative to the horizontal as shown in Figure 3.4. If we imagine the phasor to mtate anticlockwise about the origin with an angular velocity w, starting from the postition

/L shown at t = 0, then the projection of the phasor on the vertical axis at any time r has a length equal to (llfi) times the instantaneous value of the related sinusoid at that time.

Any si~iusoid is uniquely determined by three quantities viz rms value, frequerrcy and phase. In a.c. circuit arnalysis, we normally deal with situations in which all currents and voltages have the same frequency and the value of this frequency is known. In this

Fig3.4 : Phasor o ia sinusaid situation, one sinusoid differs from another only in respect of rms value and phase, both A sin (wt + 0) of which are prominently displayed by a phasor. There is therefore a one-to-one

correspondence between a sinusoid of a given frequency and its phasor and we can deduce one from the other. Different voltages and currents occurring in an a.c. circuit (operating with a single frequency excitation) can therefore be represented by an assembly of directed line segments having a common origin, a representation which is far more concise and clearer than a display of all the pertinent waveforms on a common time base. For example, tlne phase difference between two voltages is clearly visu:~lised as the angle between the respective phasors . What is more, several mathematical operations on sinusoidal functions can be interpreted and more conveniently handled in t e r m of phasors either graphically or analytically as we shall see in the subsequent sections.

As an example, consider two voltages vA = 2m sin (wr + 30") and VB = 4m sin (wt + 90") whose waveforms are shown in Figure 3.5(a).

Fig. 3.5 : Two sinusoids with a phase difference of 60' (a) Waveforms (b) Phasors

The phase difference between vA and vB is 60". vB is said to lead vA by 60" as successive similar events (e.g. upward zero crossings, positive peaks, negative peaks, downward zero crossings) occur with vB at an angular interval of 60" earlier than with vA (vide Figure 3.5(a)). By the same token vA is said to lag vB by 60".

Note : In this example, we could as well have stated that VA leads v~ by 300" (vA could as well be expressed as fi x 20 sin (wt + 390")). However to avoid ambiguity, it is customary to limit the magnitude ofphase difference to 180". Thus a leading angle of 300" (= 180" + 120") is taken to be a lagging angle of 60" (= 180" - 120")).

The phasors of the two voltages are shown as 07 and OS in Figure 3.5(b). The phase difference of 60" (smaller of the two angles between OX and 07) with vB leading vA is clearly seen from the phasor diagram.

Example 3.3

Sketch the phasors for the following voltages on the same diagram and state for each pair the leading voltage and the angle of lead.

vA = fi x 100 sin (wt + 30")

VB = - a x 120 cos wt

vc = 200 cos (wt + 60")

Solution We first express VB and vc in the standard form of a sinusoid.

vB = - fi x 120 cos wt = fi x 120 cos (wt + 180")

= fi x 120sin (wt + 270")

vc = 200 cos (wt + 60") = (200 / a ) sin (wt + 150") --

The three voltages are represented by the phasors OA, OB and OY in Figure 3.6. From the phasor diagram we observe that vA leads vB by 120°, vB leads vc by 120" and vc leads vA by 120".

Fig. 3.6 : Phasors of voltages of Example 3.3

Example 3.4

Identify the sinusoidal functions of time which correspond to the phasors shown in Figure 3.7. The frequency of the signals is 50 Hz.

Solution VA = fi x 5 sin ( 1 0 W + 225")

vB = x 10 sin (100n.C - 30")

SAQ 2

i Sketch the phasors of the following sinusoidal signals: i

A.C. Circuits

Fig. 3.7 for Example 3.4

(a) - 100 sin (wt + 30") (b) 4 a cos (wt + 30")

Introduction to Circuits SAQ 3

Identify the sinusoids corresponding to the phasors shown.

Fig. 3.8 : For SAQ 3

3.3 PHASORS AND THEIR TREATMENT

In the previous Section, you saw how a sinusoidal signal can be compactly represented by a directed line segment, designated as its phasor. This is called phasor transformation. We now proceed to present an algebraic representation of the phasor in the form of a complex number. This would greatly facilitate carrying out various linear operations on sinusoids, as would be needed in a.c. circuit analysis. Let us first familiarise ourselves with complex numbers and their algebra.

For those who are already familiar with complex numbers and their algebra, Sections 3.3.1 and 3.3.2 may be used for a rapid revision.

3.3.1 Phasor as a Complex Number You are aware that real numbers can be represented by points on the x-axis (also called horizontal axis or the real line), positive numbers to the right of the origin and negative numbers to its left. Thus P in Figure 3.9 or equivalently the directed line OF represents a

Fig 3.9 : Introduction to the imaginary number j

positive real number a. The directed line OZ, which is obtained by rotating 07 in the positive (anticlockwise) direction by JC radians represents (-1) a. In fact you can readily see that multiplication of a by (- 1)" is equivalent to rotation of 07 by nn; radians for all integral values of n. We can extend this idea to fractional values of n as well. To do this, it is adequate if we define that multiplication by (- l)lR is equivalent to positive rotation by 90". Thus the directed line 07 = (- 1)'12 O? = j OY, where the symbol j stands for \/-1. Multiplication by - j then implies a counterclockwise rotation of 180" + 90" = 270". Since the square root of a negative number does not exist in the real number system, j and its product with any real number (eg. j4) are called imaginary numbers. We now see that any imaginary number has the formjy, where y is real. It call be represented by a point on the y-axis (also called vertical axis or the imaginary axis) bl units above the origin i fy is positive and below the origin if y is negative.

Note : Mathematicians use the symbol i for fi. In electrical engineering, however, the symbol i is reserved for current for historical reasons arising from the French word intensite du courant for current strength. Hence we use the next letter in the alphabet, viz. j for \/-1.

The symbol "j" can be interpreted either as an operator rotating the operand by 90" or merely as a number. Using either interpretation, we have

j 2 = - 1 ; j 3 = - j ; P Z l e t c . (3.5)

In the foregoing paragraph, we saw that real and imaginary numbers are represented respectively by points on the real axis (x-axis) and the imaginary axis (y-axis). Now the entire x - y plane is called the complex plane. An arbitrary point P on the plane with coordinates (x, y) defines a complex number ?= x + jy, where x and jy are the real and imaginary parts of 2. We indicate a complex number by a line on top eg, t to differentiate it from the real positive number z which indicates its magnitude also called modulus.

Note : Other ways of indicating a complex number are the use of boldface type (z), the use of a cap on top ( 2 ) and underlining ( t ).

We can interpret the complex numberzeither as the point P itself or the directed line segment z. The complex number O? in Figure 3.10 has a magnitude z and a direction at an angle 8 to the real line. This leads to two alternative forms of representation of a complex number.

Rectangular coordinate representation - - - - ~ = O P = O M + O N = X + ~ Y

= z ~ 0 ~ 8 + j z s i n 8 = z ( c o s 8 + j s i n 9 )

Polar coordinate representation

Recalling the Euler identity eie = cos 8 + j sine, we can p u t t in the following fonn: - z = z e j e = Z L ~

Fig. 3.10 : Representation of a complex number:- x + jy

The last expression, read as z at an angle 8 is a short fonn for the mathematically more rigorous expression ze$. To convert a complex number from one form to the other, we use the following formulas :

Polar to rectangular form :

Rectangular to polar form :

z = (x2 +y2)U2 ; 8 = tan- ' (y/x) (3.7)

Two angles differing by JC radians have the same tangent. Therefore, while applying the second expression in Eq. (3.7), you should independently deduce from the signs of y and x the quadrant into which the particular 8 falls.

The three expressions, z cos 0 + j z sin 8, z eie and z L 8 being alternative descriptions of a directed line segment, can be used to algebraically denote phasors introduced in the last section. We therefore take thepllasor for a sinusoidal voltage v = d?Vsin(wt + 8) as the complex number,

You should carefully distinguish between the phasor Vand its magnitude V. The former is a complex number, and provides information of both the'rms value and the phase of the sinusoid. The latter indicates merely the rms value. We sometimes use the notation (VI as an alternative to indicate the magnitude of V. The angle 8 of a complex number 2 is often indicated as A ~ ~ ( Z ) .

A.C. Circuits

readily established through thestandardseries expansions of the exponential function 2 and the trigonometric functions ws 0 and sin 0. Note also that - (as0 + j sine)"' - cosme+ jsinme

introduction to Circuits Example 3.5

Express the phasors of the three sinusoidal voltages of Example 3.3 as complex numbers in both rectangular and polar coordinate fonns.

Solution - VA = 100 L 30" = 100 eid6 = 100 (COS 30' + j sin 30")

(Note that ejd2 = j and e-jd2 = - j)

200 200 vc = - L 150" = - (cos 150" + j sin 150") a dz

Example 3.6

Identify the sinusoidal signals of 50 Hz frequency which have the following phasors :

Solution

(a) 7 = L tan- '4/3 = 50 L 53"

(b) C= V 3 d + 4d L tan- ' (- 413) = 50 L - 530

* v = 50 fi sin (314t - 53")

Of the two solutions of tan- ' ( -413) , the angle in the fourth quadrant is chosen because the imaginary component of ?is negative and the real component is positive

- (c) I = d(- 8)2 + (- 6)' L tan- '(- 81- 6) = 10 L 217"

i = l o a s i n (314t + 217") -

((4 I = a L taf1(2/- 2 )= f i L 135"

=. i = 4 sin (314 t + 135")

SAQ 4 Fill up the missing quantities in the table. The entries in each row should correspond to one another.

Algebraic representation of phasor Graphical

Function of time representation

Rectangular of phasor Polar form

form a. v(t) = 20 cos (wt + 200")

-. e. v(t) = 5 (sin wt + cos wt)

3.3.2 Algebra of Complex Numbers The advantage of the phasor approach in a.c. circuit analysis is that it converts many complicated mathematical operations involving trigonometric functions of time into equivalent simple algebraic manipulations on complex constants. Let us therefore first look at the important rules relating to the algebra of co~nplex numbers. In what follows, let us take

- z = x + j y = z L B

- Z, = x2 + jy, =z2 L e2

(a) Equality

Two complex numbers < and 2, are equal if and only if

x1 = X, and yl = y2

or equivalently z, = z2 and B1 = e2 Example 3.7

I f 5 = 10 L el is equal to y2 = 6 + jy,, find 8, and yZ.

Solution

l~~l=m=l~~l= 1 0 * y 2 = ~ = ~ . 3

Arg( F2) = tan- '(& 8/6) = Arg( zl) = 81 * 81 = * 53"

(b) Negation

The negative of a complex number has both its real and imaginary parts reversed in sign or equivalently it5 angle shifted by x radians.

- - z = - x - j y = z ~ 8 + n (3.9)

The validity of the last term is clear if you consider reversing the orientation of the directed line segment representing the complex number.

Example 3.8

If (4 + jb) and (a + j5) are negatives of each other, find a and b.

Solution

(c) Addition and suhtraction

These operations are conveniently effected in t e r n of rectangular components and involve addition or subtraction of the real and imaginary parts separately.

You may be fa~niliar with addition of vectors representing forces by the parallelogran~ law of forces. The addition of complex numbers follows an identical pattenl. Figure 3.1 1(a) illustrates this. Figure 3.11(b) shows an alternative way of adding the two directed line segments. Note that%, 7, and their sum fonn a closed triangle in this case.

From the geollletry of Figures 3.11(a) and (b) the following can be deduced :

z2 sin B2 + zl sin 8, tan e3 =

2, cos e2 + Z1 cos el

You can casily show that the relations in Eq. (3.11) agree with Eq. (3.10a). Subtraction of :i conlplex nun~hcr can always be considered to be the addition of its negative, and the saine parallclogram law call he applied.

A.C. Circuits

Introduclion to Circuits

(a> Fig. 3.11 : lllusttating the addition of complex numbers

Example 3.8 If zl = 10 L 30" and z2 = 4 - j6, - - 2, - 2, = 10 L 30°- (4 - j6)

= (10 cos 30" + j10 sin 30') - 4 + j6 ,

= (8.67 - 4) + j(5 + 6) = 4.67 + jll = 11.95 L 67" -

Note that the magnitude of gl - z2) is larger than the magnitudes of & and This can occur with complex numbers when their angles differ considerably.

(d) Multiplication

Multiplication is more conveniently effected in polar fonn and involves the product of magnitudes and the addition of angles, a result easily seen when the numbers are expressed in exponential form.

- - z1 . z2 = z1 ejel z2 ei ez = zlzzej(el + '2) = z, z2 L 8, + 8,

In rectangular fonn, the result can be obtained through direct multiplication.

As a special form of the above results, multiplication of a complex number? by a real number k call be shown to yield

k?=ky+jky=kzL8.

Exirmple 3.9 - -

~ i v e l 1 A = 4 - j 6 a l l d B = 6 + j 8 , f i n d ~ . B .

A.B=(4-j6)(6+j8)=24+j32-j36-j248=72-j4.

Alternatively,

A. = [w L tan- ' (- 6/4)] [m L tan- ' (8/6)]

= [7.21 L - 56 - 3"][10 L 53 . lo] = 72 -1 L -3 .2"=72 - j4

(e) Conjugation

The conjugate ?* of a complex number i has the sign of the imaginary part reversed in the rectangular representation or equivalently the sign of the angle reversed in the polar form

- z * = x - j y = z L - 0 (3.14)

The product of a complex number and its conjugate is a real number equal to the square of the magnitude of either.

-- ~ Z ' = [ ~ L ~ ] [ Z L - ~ ] = ~ L O = ~ ? ( ~ (3.15)

(f) Division

In polar form, division involves division of magnitudes and subtraction of angles.

I If division is to be effected directly in rectangular coordinate form, we first convert the denominator to a real number by multiplying both numerator and denominator by the conjugate of the latter. This process is called rationalization of the denominator.

I Example 3.10

! ~ o r z and idef ined in Example 3.9, I

= 0.721 L - 109.4"

Alternatively,

SAQ 5 State if the following are true or false.

(a) - [ lo L 30'1 = 10 L - 30"

(b) 10 L 30" + 20 L 60" = j30

(c) [ lo L 30'1 [20 L lo0] = 200 L 300"

(d) [40 L 60'1 / [lo L - 30'1 = 30 L 90"

SAQ 6 Evaluate the following

A.C. Cirmits

3.3.3 Kirchhoff's Laws in Phasor Domain As Kirchhoff's laws involve addition of voltages or cu-,ents, let us first examine the methods of addition of two sinusoidal voltages, taking the specific example of vA and vB of Figure 3.5, reproduced in Figure 3.12(a).

(a) Addition of waveforms :

Introduction to Circuits

Fig. 3.12 : Addition of sinusoids (a) Waveform addition (b) Addition of phasors

The two waveforms can be added point by point as shown in Figure 3.12(a) yielding the waveform of vc = vA + vB . This is not only laborious but also would not yield an accurate result.

(b) Manipulation o f trigonometric functions :

vc = VA + vB = 2 M sin (wt + 30") + 4 M sin (wt + 90")

= 2 M [sin wt cos 30". + cos wt sin 30" ]+

40&? [sin wt cos 90" + cos wt sin 90'1

= 1 0 6 sin wt + (1047 + 4 M ) cos wt

= 1 0 6 sin wt + 5 M cos wt

= d600 + 5000 sin (wt + tan- ' ( 5 M / 1 0 6 ) )

= 74.8 sin (wt + 70.9") = 52.- sin (wt + 70.9")

It is clear that this is also an involved process.

(c) From the geometry o f phasor diagram :

If FA and FB are the phasors of vA(t) and vB(t) as shown in Figure 3.12(b) then = + would be the phasor of vA(t) + vB(t). This can be argued from the fact that the

projection of Vc on the vertical line equals the sum of the projections of and not only for the position shown (at t = 0) but also for a general t when all the three phasors as a body rotate by an angle wt. From the geometry of the phasor diagram,

V sin 30" + VB 110+40 ~ r g ( % = tan- ' AvA cis 300 = tan- ----- - 1 m

- 70.9"

Thus = 52.9 L 70.9" and vc(t) = 52.- sin (wt + 70.9")

(d ) Using phasor algebra : -

= 20 L 30" = 20 (cos 30" + j sin 30") = 17.32 + j10

= 40 L 90" = j40

E = E +E = 1 7 . 3 2 + j 5 0 = ~ ~ t a n - l ( 5 0 / 1 7 . ) 2 )

= 52.9 L 70.9"

Hence vc(t) = 52.- sin (wt + 70.9").

From the foregoing, the si~nplicity of methods (c) and (d) using phasors in co~nparisoii with either waveform addition or ~nailipulation of trigonometric functions is evident. Nor~nally o ~ i c prefers the algebraic manipulation of pllasors as it givcs thc required result in a compact and direct manner. Even so, it is a good practice to supplement this work with a sketch of the phasor diagram, as the latter gives a better physical fcel for the sizc and relative phases of the quantities involved and also can be used to obtaiu a quick rough estimate of the final result. You are strongly urged to cultivate the practice of making rough sketches of phasor diagrams evcn when you obtain the solution algebraically.

I11 the above example, you would have ~loticcd that twc~ si~lusoidal voltages of 20 V and 40 V rlns values add up to a sinusoid of 52.9 V rms value. This is 1101 a co~~lradiction, but merely a reflection of the fact that there is a phase differcncc bctwcen the two sinusoids. Looking at Figure 3.12(a), you would note that the peaks of vA arid v,] do uot occur at thc same time and hence the peak value of vc is not equal lo the sum of the pcak values of vA and vg. When considering the addition or subtractio~l of two a.c. quantities you nlust alert yourself to take into account thcir phasc differc~~cc also and not inercly thcir rnis values.

Our prinlary collcern in this Unit is the a~ialysis of a.c. circuits, in wliich all voltages and currents are sinusoids of the sanle frequcncy. To illustrate fonuulation of Kirchhoff's laws in phasor domain for such circuits let us consider ihe portion of an a.c. circuit shown in Figure 3.1 3(a). Incidentally, the sy~nbol for a si~lusoidal voltage source is what

Fig. 3.13: C'i~cuil to illustrate Kirclll~ofCs laws in a.c. circuits (a) time domain quantities and (b) pbasor quan~ities marked

is marked between terminals A and D therein. You would recall from the discussion in Unit I that Kirchhoff's voltage law for the loop ABCD can be formulated as

vm+vg(-+v,-v,=0,

where the equation holds at every instant of tinlc. Noting that all these voltages are sinusoidal in nature, we can fonn the equivale~~t coostraint in terms of the respective phasors as follows

i&j + G / B C + F D - E = O (3.17)

The four phasors should fonn a closed polygo~l as shown in Figure 3.14(a) to satisfy the above equation. This lnalnler of drawing a phasor diagranl as an allernative to the conventional way illustrated in Figure 3.14(b) has the merit of bringing out clearly that ii=VAB+GC+VCA

Fig. 3.14 : Phasor diagrams pertaining to Eq. (3.17)

95

Turning to the Kirchhoff's current law, we have at node B,

which could be transformed into the phasor form

The complex numbers standing for the phasors -I,, 5 andT3 should therefore add up to zero . To summarise the foregoing discussion, we may state that KirchhofPs laws for a.c. circuits can be formulated in terms of the respective phasors in the same manner as done for d.c. circuits in terns of the d.c. quantities. The only difference is that each term is a complex number instead of being a real number. Hence both the real parts and the imaginary parts of the terms on the left hand side of the Eqs. (3.17) and (3.18) should separately add up to zero. When marking the various voltages and currents along with their references on the circuit diagram, we can choose to indicate either the time domain quantities as in Figure 3.13(a) or their phasors as in Figure 3.13(b). Kirchhoff's laws are satisfied in either form; however for the solution of a network, it is the phasor form that we would make use of for the most part.

Example 3.11

Refer to circuit of Figure 3.15.

Fig. 3.15 : for Example 3.11

Given that = 10 + ja, with a > 0 - - -

V B = ~ ~ L ~ , V C = ~ O + ~ L ~ , VD=c+j4 , Z A = 6 + j 8 , z B = 4 + j d , $ = e + j 2 ,

evaluate 9, a, b, c, d, and e.

Solution

From KVL applied to the two loops, we obtain

i = G = G + G Wehave l O + j a = 2 6 L 8 * 1 0 ~ + a ~ = 2 6 * , t a n 8 = ( a / l O ) * a = 2 4 , 9 ~ 6 7 . 4 "

Now 1 0 + j 2 4 = 2 0 + j b + c + j 4 ~ 2 0 + c = = l O a n d b + 4 = 2 4 ;

From KCL, we have

e = 2 , d - 6

Example 3.12

If in the circuit of Figure 3.15, ZB = 6A and Zc = 4A, what can you say about the value 0fJA ?

A.C. Chcuits


(a)iB andic in phase (b)iB andic out of phase (c)iB and& with a phase difference of 0

Solution

r We h o w l A =TB +Ic. However, the data gives only the nns values and has no phase information. Hence we can only fix the limits within which the value ofIA should lie. The maximum possible value of IA would occur whenjB and& are in phase (i.e.,

I have no phase difference) as shown in Figure 3.16(a). Now 151 = IB +IC = 10 A.

The minimum value of IA occurs when& andTc are in phase opposition (i.e., have 180" phase difference) as shown in Figure 3.16(6). Now IA = IB - IC = 2 A. For any other general phase difference 8 (see Figure 3.16(c)), the value ofIA would be in between the extreme values of 2 A and 10 A.

Hence 2 A s z A s 10A.

SAQ 7 Express the phasors of the following sinusoidal signals in both rectangular and polar coordinate forms.

(a) v(t) = 100 sin (wt - 45")

(b) ~ ( t ) = 5 ~ c o s ( w t + 1 3 5 ~ )

(c) i(t) = - 1 M sin (wt +, 1 20")

(d) i(t) = - 20 cos (wt + 30")

r SAQ 8 GiventhatA=5+j5,B=-6+j10, C = 8 + j 4 , f ind(a )A+B-C (b )A/B (c) AB'C (d) A + (BE)

i

SAQ 9 I Two sinusoidal voltages vl and v2 each of 10 V effective value add up to a sinusoid I of 15 V effective value. Find the phase difference between vl and v2.

Introduction to Circuits SAQ 10 If in the circuit of Figure 3.13, we have il = 5 sin (wt + 60') and I2 = 4 + j l , find i3(t).

3.4 IMPEDANCE CONCEPT

In the previous Section we saw how the phasor coilcept provides an effective means of applying Kirchhoff's voltage and current constraints in a.c. circuits. In this Sectioil we extend the applicatioii of this concept to characterisation of terminal relatioils of single elements and two-terminal networks. To this end, let us first examine the respoilse of RLC elements to sinusoidal excitation.

3.4.1 Response of single Elements to Sinusoidal Excitation Resistance

Coilsider a resistor of R ohms connected to a sinusoidal voltage source of v = fi Vsin (wt + 0) volts as show11 in Figure 3.17(a). From the fundamental terminal relatioilship v = Ri of a resistor it follows that i = (vlR) = ~ ( v I R ) sin (wt + 0). We note the following:

(b) Fig. 3.17 : Response of a resistor to sinusoidal excita~ion

(a) The current and voltage are in phase (is., they have zero phase difference). Bdth vary in step. The posilive peaks, negative peaks, zero crossings etc occur in both at the sallle time as shown in Figure 3.17(b).

(b) The nns values of voltage and curreut are related by

V = RI, (3.19)

a fornlula idelltical to Ohm's law in d.c. domain.

(c) The voltage and current phasors being V = V L 0 and 7 = (VIR) L 0 , as shown in Figure 3.17(c), they are related by

V = R T (3.20)

Note that the above relation is independent of w and 0.

Inductance %

Let now an inductor of L henrys have a voltage v = fi Vsin (wt + 0) applied across it as shown in Figure 3.18(a). From the fundame~~tal terminal relationship of an inductor we have

1 1 i = - 5 v dt = -In Vsin (wt + 0)dt

L L

d z v = -- cos (wt + 0) = - WL WL

We take the constant of integration in the above to be zero as for'the a.c. circuits of concern to us, there cannot be a d.c. current in any element.

(b) (cl Figure 3.18: Response or an inductor to sinusoidal excitation

The chief characteristics of this response are: 1

(a) The current has a phase difference of 90" with respect to the voltage and lags behind it. Similar events (positive peaks;~~egative peaks, upward zero crossings, downward zero crossi~lgs etc) occur in the voltage wave a quarter-period (equivalent to 90" of the angle wt) earlier than in the current wave, as seen from Figure 3.18(b).

(b) The rms values of voltage and current are related by

v = (wL) I

4c) The phasors of v and i being V = V L 0 and j= I L 8 - n/2 as shown in Figure 3.18(c). they are related by

v= (~wL)? (3.22)

Capacitance

Figure 3.19(a) shows a capacitor of C farads applied with a voltage v = fi Vsin (wt + 0). The current through the capacitor would therefore be

The chief characteristics of this response are: P

(a) The current has a phase difference of 90" with respect to the voltage and leads the latter. Similar events occur in the current wave a quarter-period (equivalent to 90")

1 earlier as shown in Figure 3.19@).

Fig. 3.19 : Response of a capacitor to sinusoidal excitation

A.C. Circuits

99

Introduction to Circuits (b) The rms values of voltage and current are related by

(c) The phasors Vand?being V L 0 and wCV L 0+ n / 2, as shown in Figure 3.13(c), they are related by

The properties derived earlier are characteristic of the three elements in a.c. circuits and hold irrespective of where they are connected in a circuit. A little reflection will show that item (c) in each case is a complete statement of the relevant properties and ilicorporates in itself the properties specifically stated under (a) and (b).

To suminarise, the three elelnents R, L and C respond differently to sinusoidal excitations (current or voltage). The current and voltage in a resistor a r e in phase, while they are in quadrature (i.e., have a phase difference of 90") in a n inductor o r a capacitor. In a n inductor, the voltage leads the current while in a capacitor, the c r~ r r en t leads the voltage. The rms values of the voltage across and the current through the element satisfy the proportio~lality relationship give11 in each case by Eq. (3.19), (3.21) and (3.23) respectively. For a given value of current to be driven through it, an inductor requires more voltage as the frequency increases. The capacitor on the other hand, needs only a smaller voltage as the frequency increases.

Example 3.13

A capacitor draws a current of 5 mA from 200 V, 50 Hz a.c. supply. What current does i t draw from 40 V, 400 Hz supply?

Solution

As mentioned earlier, all values relating to voltages and currents in a.c. circuits are Ic be taken as rms values uliless specifically stipulated.otherwise. From Eq. (3.23) we have

For a given C,

Hence the current with 40 V, 400 Hz supply = S 11~4.

SAQ 11 Fill up the blanks:

In a capacitor the current 1 the voltage by degress, while in a 3

the current and voltage are in phase.

For a given applied voltage an inductor permits a 4

current as the frequency is raised.

SAQ 12 Find the inductance of an inductor which draws a current of 1.1 A when connected to 230 V, 50 Hz voltage. What current will it draw if the supply voltage is changed to 150 V, 25 Hz?

3.4.2 Concept of Impedance and Admittance We note from Eq. (3.20), (3.22) and (3.24) that the ratio of V t o l i n each of the three cases considered is a constant which is a function of only the element value and frequency and is independent of the value of the applied voltage or current. These

relations are reminiscent of Ohm's law, except that now the quantities involved are complex constants. This happy. turn of events has come about because differentiation in time domain is equivalent to multiplication by jw in the phasor domain and integration in time domain is equivalent to division by jw in the phasor domain. Thus if i(t) transforms

A.C. Circuits

into 7, then dildt transforms into jGandJdt transforms into 71 jw. We shall now develop

this theme for a more general situation. To this end, consider a 2-terminal network N comprising linear circuit elements and let v and i be its terminal voltage and current when connected in an a.c. circuit, as shown in Figure 3.20(a).

q-q Network

Fig. 3.20 : A general linear 2-terminal network in an a.c. circuit.

The relation between v and i would in general be an involved differential equation with time as the independent variable. But under sinusoidal regime with an angular frequency w, the phasors VandTwould have a proportionality relationship independent of time. This proportionality constant is termed the impedance Zof the network N. Thus

Z= ~ / T = z L ~ (3.25)

The following are the characteristics of 2. - 2 is a function only of w and the values of elements in N. It is independent

of time and the value of Vorl . You will learn later how to compute Zfor any given N.

- 2 has the same dimensions as resistance and is measured in ohms. I

- From the definition of 2, we have

v-57, (3.26) which plays the same role in a.c. circuits as Ohm's law in d.c. circuits.

- The magnitude Z of the impedance is the ratio of the effective values of voltage and current. Sometimes, Z itself is referred to as the impedance instead of 2 when there is no room for confusion.

- a is called the angle of the impedance Zand denotes the phase angle by which V leads i.

- In rectangular coordinate form, Z can be expressed as

where the real and imaginary components R and X are respectively called the effective resistance and reactance of N. The latter may in general be positive or negative, but for a network N comprising no active elements (e.g., dependent sources), R is always non-negative.

- 5 is to be viewed purely as a complex number and cannot be associated with any sinusoidal signal as its phasor.

Note : Strictly speaking, we should distinguish between a physical element and its circuit parameter. But expressions like "a resistor having a resistance of 20 51 is connected inparallel with a capacitor of 1 pF capacitance" are not only inconvenient but may even sound pedantic. Hence you often find in literature an element itself being referred to by its circuit parameter - . resistance, inductance or capacitance. Thus the previous expression would usually be rewritten "a 20 52 resistance is connected in parallel with a 1 pF

Introduction to Circuits capacitance". In the same vein, where it is conveent , we may call the two-terminal network N itself as the impedance 2.

As a dual concept to impedance, we define the admittance 7 of the network N as - - - Y = I I V - Y L p (3.27)

Obviously y i s reciprocal to Z, is measured in siemens and enables determination of current for a given applied voltage by

Also, ~ = z - ' a n d $ = - a

In rectangular coordinate form, 7 may be expressed as

Y = G + ~ B ,

where G and B are referred to as the effective conductance and susceptance of N.

Let us now return to the behaviour of single elements considered in subsection 3.4.1. From Eq. (3.20), (3.22) and (3.24) and the definition of Z. we deduce the following expressions for their impedance.

Thus the impedance of a resistor is purely real, equal to its resistance R and has no imaginary component. It is in anticipation of this that we have used the symbol R for the real part of Z. On the other hand, inductors and capacitors have purely imaginary impedances, inductive reactance XL being positive and capacitive reactance (- X,) being negative.

How does one calculate the impedance of a nenvork with several elements? To obtain a clue, let us consider a series combination of several subnetworks with impedances 2, , & ... Z,, as shown in Figure 3.21.

b

Fig. 3.21: Series combination of impedances

When an a.c. current havingj for its phasor passes through the series combination, we have

Using KVL in phasor form, the terminal voltage of the combination is

Thus the equivalent impedance of the series combination is

In a similar fashion we can show using KCL that the equivalent impedance and admittance of the parallel combination shown in Figure 3.22 are given by

Fig. 3.22 : Parallel combination of admittances.

A.C. Circuits

or &=Yl+&+ ... +c (3.30b)

Eqs. (3.29) and (3.30) are similar to rules for combining resistors in series and in parallel. We will encounter a similar pattern in future. All the techniques that we have learnt for d.c. circuit analysis can be applied to a.c. circuits as well with certain modifications. The techniques include finding series-parallel equivalents, current arid voltage division rules, star-delta conversions, loop current and nodc voltage arialyses etc. The departures from d.c. methods are:

- KVL and KCL relations are expressed in phasor form.

Impedance Z (or admittance 7) is used in place of resistance (or conductance).

- Terminal relations V = 2 ?and?= ??are used instead of V = RI and I = GV in d.c. domain.

This is the power and beauty of the phasor concept. Essentially the same methods and formulations applicable to circuits with constant currents and voltages are made applicable to circuits with currents and voltages varying in time sinusoidally. The price we have to pay for this simplification, namely dealing with complex numbers, is indeed less taxing than the alternative of working in time domain with trigono~netric functions and carrying out differentiation and integration operations thereof. We shall study in Section 3.5 the actual application of the phasor technique and the impedance concept to a.c. circuit analysis.

SAQ 13 'If a sinusoidal voltage v(t) applied to a circuit element delivers a current i(t), then the impedance of the element is v(t) / i(t)'. Comment on the above statement.

i SAQ 14 Fill up the following table

Example 3.14

Find the impedances of the element combinations shown in Figure 3.23, taking the frequency to be 400 Hz.


introduction to Circuits Solution

(a) The impedance of the series combination is the sum of the two impedances.

(b) ~ , ~ ~ ~ + ~ ~ = 1 0 0 + j ~ x 4 0 0 x 1 0 0 x 1 0 ~ ~

= 100 + j2510

(c) . Alternative I

The two impedances being in parallel,

= 35.7 L - 26.7" - 31.9 - j16.0 D

, Alternative I1

The two admittances being in parallel,

?p = + F2 = 0.025 + j0.0126 = 0.028 L 26.70

Then & = (Y+)- = (0.028 L 26.7")- = 35.7 L - 26.7" = 31.9 - j16.0 O

Example 3.15

When a current i(t) = 1.2 cos (wt + 60") passes through a network, the terminal voltage of the latter is v(t) = 100 sin (wt + 120"). Find the impedance of the network.

Solution

i(t) = 1.2 cos (wt + 60") = 1.2 sin (wt + 150")

v(t) = 100 sin (wt + 120") * V = (100h/Z) L 120"

Example 3.16

When the element combination in Figure 3.23(a) is connected to 200 V, 400 Hz supply, what would be the current drawn? What would be the voltage across the resistance and capacitance?

Solution

As the phase of the supply voltage is not specified, we need to compute only the rms value of the current.

SAQ 15 In Example 3.16, VR and Vc do not add up to the magnitude of the supply voltage. Is this not a violation of KVL?

SAQ 16 If a current i(t) = fi sin (400t + 30") rnA passes through the element combination in Figure 3.23(b), find an expression for the voltage across the conzbination.

Example 3.17

If a voltage of 200 V, 400 Hz is applied across the element combination in Figure 3.23(c), find the total current taken by the combination.

Solution

Since the phase of the supply voltage has not beell specified, let us take it as 0". That is, we are taking the applied voltage phasor as the so-called reference phasor.

v = 2 0 0 e 0"

NOW'^^ = 200 L 0" /R = 200140 = 5

The phasor diagram showing the relative positions of the different phasors is given in Figure 3.24.

+ v Fig. 3.24 for answer to Example 3.17

If 7 has any other phase angle say a" it only means that the entire figure will rotate by a". There will be no change in either the magnitudes of the voltages and currents or in their phase differences. This is an important point to note. Wlrat no contrary information is specified, we are at liberty to arbitrarily assume one co~ivenient quantity as the reference plzasor.

3.4.3 Mutual Inductance You would recall the property of an inductance coil to set up a magnetic field when carrying a current. Some of the magnetic flux lines so set up may also link with another coil (inductor) in its proximity. In such an event, the two coils are said to be magnetically coupled. When the current in the first coil changes, not only do its self flux linkages change but also the flux linkages (called mutual flux linkages) produced by it in the second coil. Consequently there is a voltage induced in the second coil due to a change of current in the first coil. The action described above is reciprocal in that a change of

A.C. Circuits

Introduction to Circuits current in the second coil would also induce a proportionate voltage in the first coil. The induced voltage in each coil produced per unit rate of change of current in the other is defined to be the mutual inductance M between the coils. M is another passive circuit parameter like R, L and C that we discussed in Unit 1 and arises whenever two inductors are coupled. It is measured in the same units as L viz., henrys. The principle of mutually induced voltage forms the basis of transformer action which you will study in Block 2.

Figure 3.25 shows two coils of N, and N, turns having inductances L, and L2 '(called self inductances) coupled through a mutual inductance M. The self and mutually induced voltages when either coil carries a current are illustrated therein.

Fig. 3.25: Self and mutually induced voltages (a) when coil 1 only is carrying current (b) when w i l 2 only is carrying current.

In the event when the two coils carry currents i, and i2 simultaneously as in Figure 3.26 each coil has both self and mutually induced voltage components. The terminal voltages of the coils are then given by

di, di, v2=M-+L dt - dt

Two terminals called corresponding terminals, one from each coil, are specially marked with dots (also called polarity markings). If the reference directions,of the two currents i, and i2 both enter the coils (or both leave the coils) through the dotted terminals, then the signs of both the self and mutually induced voltages in each terminal voltage equation would be the same; otherwise they would be different. In Figure 3.26, a and c are corresponding te'rminals and since the references for both i, and i2 enter the respective coils through these terminals, Eq. (3.31) has the same signs for both self and mutually induced voltages.

Fig. 3.26 : Coupled wils with dot markings

What we have discussed so far are terminal equations of a pair of coupled coils in a general framework. In d.c. circuits, the terminal voltages are obviously zero as currents are steady. In a.c. circuits the tenninal equations, in terms of the respective phasors, reduce to

& + 0'c(:M>T2 - (3.32) v, = O'wM) 7, + O'WL,&

Recall that if i(t) has 1 as its phasor, then the phasor of dildt is jw 1. This principle is used in the phasor transformation of Eq. (3.31) into Eq. (3.32). From the latter, we can say that two coupied coils have a mutual reactance of wM ohms.

Example 3.18

In the circuit of Figure 3.27, L1 = 4 H, M = 2 H, L2 = 3 H. If 200 V at 50 Hz is applied to coil 1, find Il and V2.


I Solution From the figure, we have Vl = 200, I2 = 0

Also

Example 3.19

The impedance of a network is R + jX ohms. Find its admittance in terms of R and X. What will be the relation between the angles of the impedance and the admittance?

Solution

Z = R +jX

~ r g ( Y ) = - ~ r g ( Z )

I Example 3.20 In the a.c. circuit shown in Figure 3.28, IR = 8.4 mA and I. = 9.1 mA. What would be Ic? If the frequency of the source is increased to 288 Hz, what would be the new value of lo?

b Fig. 3.28: Circuit for Example 3.20

Solution

Take E b as reference phasor i.e., v$, = 200 L 0' - - Ic will lead E b by 90". i.e., lc = jIc

But ITo] = 9.1 = (8.42 +I:)"~

Thus Ic = (9.12 - 8.42)1R = 3.5 mA

At 288 Hz, IR will remain at 8.4 mA, as the same voltage is applied across R. Let the new capacitor current br I,-'

kc. Circuits

introduction to Circuits

(a ) (b)

Fig. 3.29: Phasor diagram for Example 3.20 at (a) f = 35 Hz (b) f = 288 Hz.

I& - w'C V w' 288 I- P-P-

288 * I=' = 3.5 x - = 28.8 mA zc WCV w 35 35

SAQ 17 Fill up the blanks:

In an inductor the leads the ' by ' radians.

The reactance of a given inductor increases with 4

Two inductors in proximity of each other may have j

between them. The property of mutual inductance is to induce a 6

in one coil when the in the other coil varies.

SAQ 18 Fill up the table below :

Symbol with v - i relationship Circuit reference element directions for 11.11 D.C. steady A.C. (in terms

v and i General of phasors)

Resistor R v = Ri V = R I v= ~7

Inductor L \ I I I

Capacitor C I I I I

Coupled coils with mutual inductance M

SAQ 19 The voltage and current at the terminals of a 2-terminal network are given by

v(t) = 20 sin (1000d + 45") ; i ( t ) - 0.5 sin (1000~r.t + 75")

Calculate the impedance of the network and identify an element combination which provides this impedance.

SAQ 20 A resistance of 20 P and an impedance 40 + j60 P are connected across an a.c. supply source as shown in Figure 3.30. If the voltage across the resistor is 50 V, find the source voltage. Draw a phasor diagram.

20 9 40 + j60 Q

Fig. 3.30 for SAQ 20

SAQ 21 A fluorescent lamp may be considered to be a pure resistance. A 40 W lamp is designed to operate at a voltage of 130 V at 50 Hz. This lamp is connected in series with a choke coil (which may be considered a pure inductor) across 220 V, 50 Hz supply. Calculate the required value of inductance of the choke coil.

3.5 A.C. CIRCUIT ANALYSIS

We are now in a position to take up the analysis of general a.c. circuits. The systematic application of KCL and KVL in phasor form and use of element impedanceladmittance relations constitute the main base of the related techniques. As already remarked in Section 3.4 the phasor transformation and the impedance concept serve to replace differential equations by algebraic equations and it is in the latter form that we formulate the pertinent network equations. The starting point in the solution is usually a known or assumed reference phasor.

3.5.1 Simple Series and Parallel Circuits A series RLC-circuit is shown in Figure 3.31(a). Let Z be the equivalent impedance seen

- "L

- v,=v,~ mi "-, 7 "4- - "s

- VR -

I

( a ) ( b )

Fig. 3.31 : Series a s . circuit. (a) Circuit diagram (b) Phasor diagram.

by the source. By the rule for series combination of impedances, we have,

2- R + jwL + (l/jwC) = R + j[wL - (IltvC)] = Z L a ,

k c . Circuits

Introduction to Circuits where Z = [R2 + (wL - I / w c ) ~ ] ~ / ~

and a = tan- '[(wL - 11wC) I R]

Now thecurrent?= C/z= V.L O / Z L a = (Vs/Z) L - a

I = (V, l Z) = Vs / [R2 + (wL - l / w ~ ) ~ ] ~ / ~

The phasor diagram is shown in Figure 3.31(b), taking wL > (IlwC). The current lags the applied voltage by a, where a = tan- '[(wL - l/wC)/R]. If wL c (l/wC), the net reactance is capacitive in character and the current leads the voltage by tan- '[(llwc - wL)/R].

Example 3.21

A practical lossy induclor coil tiavingR = 10Q and L = 200 mH is connected in series with a 10 pF capacitor across a 200 V, 50 I-Iz source. Find the current in the circuit and voltages across the two elements. Draw a phasor diagram.

Solution

The circuit is shown in Figure 3.32. Here the practical illductor is lilodelled ;IS a series co~nb i~ la t i o~~ of R and L. Since no phasc information is given, let us take Fs = 200 L 0"


The phasor diagram is shown. Note that thc capacitor voltage is largcr than even the source voltage in this example.

Resonance

Reverting to the circuit of Figure 3.31, let us exanline the variation of currcnt as the frequency of the source is varied, keeping Vs coilstant. From Eq. (3.33) i t is clear that Z is purely real and hence 7 is in phase F, when the source angular frequency is w,, where w J . - (l/w,C). This is said to be the resonant coriditiort of the circuit and w, is called the resonancefrequency. By definition the current is in phase with the supply voltage a t resonance. The following characteristics of a series resonant circuit are of interest.

- Source frequency at resonance wo = ( 1 I r n ) (3.35)

- Impedance of the circuit at resonance 2, = R - As ,w varies, Z varies as shown in Figure 3.33(a) and becomes minimum at

w = wo

(a) (b) Fig. 3.33 : Characteristics of series resonance

(a) Variation of Z & I with frequency (b) Phasor diagram at resonance.

- As w varies, I varies as shown in Figure 3.33(a) and attains a maximum at w = w ,

- At resonance VL = Vc since the magnitudes of reactances of the capacitor and inductor are equal.

Vs = since & + = 0 (Vide phasor diagram in Figure 3.33(b)).

Resonant circuits are widely employed in communication engineering to obtain selective response at a desired frequency at the expense of ather frequencies. For example, in a radio receiver, an antenna captures signals broadcast by several stations but an LC circuit is tuned through a variable capacitance so that its resonant frequency coincides with the transmitting frequency of the desired radio station. Thus the desired signal is made to yield maximum response and the response of other signals minimised. In a practical resonant circuit R is the unavoidable resistance of a practical inductor (like in Example 3.21) and this should be minimised to enhance the selectivity of a resonant circuit. Note that the effective impedance of a series LC circuit (R = 0) is zero at resonance and hence ideally the current will be infinite.

Example 3.22

A practical inductor withR = 10 52 and L = 0.16 mH is used in series with a variable capacitor in a resonant circuit. To what value should C be adjusted to obtain resonance at 2 MHz? With this value of C, what would be the voltage across it for an input signal of 10 mV at a frequency of (a) 2 MHz (b) 1.5 MHz ?

Solution

At 2 MHz :

I= 10 mV110 S2 = 1 mA ;

At 1.5 MHz :

= 10 + j(l508 - 2679) = 10 - j1171= 1171 L - 89.51"

Therefore

Vc = IlwC = (VIZ) (lIwC) = ( l o x 10-3/1171) (2679) = 0.0229 V

Note the reduction in the level of response to 1.1 9% when the excitation frequency shifts by 25% from the resonance frequency.

A.C. Circuits

Introduction to Circuits

Fig. 3.35: A practical form of parallel resonant circuit

The parallel RLC circuit excited by a current source has characteristics dual to those of a series RLC circuit. Referring to Figure 3.34, we have

- I ,= ~ l j w ~ = ( I , l w ~ Y ) ~ - ~ - x 1 2

i c = ~ ~ ~ ) = ( ~ S ~ / ~ ~ n / 2 - p

The phasor diagram for the case wC > ( l lwL) is given in Figure 3.34.

Fig. 3.34 : Illustrating a parallel RLC circuit

As in the series circuit, we define the resonant condition for the circuit of Figure 3.34 as one when the source current and voltage are in phase. This will occur when the imaginary part of 7 is zero i.e., when JJC = 1, a condition identical to the one obtained for the series case. Also Y is mii~imuln and V is maximum at w = wo. The curves of Y v s wand

V vs w will be similar to the curves Z vs w and I vs w of Figure 3.33. Note that the effective impedance Zo = ( l /Yo) at resonance of the parallel circuit becomes infinitely large when the resistance branch does not exist (R = oc ). In other words, an ideal parallel LC circuit is equivalent to an open-circuit at resonance.

Example 3.23

The cigcuit of Figure 3.35 is a more practical form of a parallel resonant circuit and consists of a coil of resistance R and inductance L with a capacitor C in parallel. Find the resonant frequency of the circuit and its impedance at resonance.

Solution

- Y is purely real at the resonant frequency wo. Hence

Yo, value of admittance at resonance = Rl(R2 + w g 2 ) = RCIL

Zo , the impedance at resonance = LICR

In practical circuits, w&IR >> 1 , wo - and

SAQ 22 Fill up the blanks :

In a series RLC circuit fed by a sinusoidal source, the current leads the source voltage when I. This circuit is said to be in resonance when the impedance of theRLC combination is

2

A series LC circuit is equivalent to a 3

at resonance, while a parallel LC circuit is equivalent to 4 at resonance.

SAQ 23 A 100 S2 resistor, a 100 mH inductor and a 100 pF capacitor are all connected in parallel across a 200 V, 50 Hz a.c. source. Find the current supplied by the source and its relative phase with respect to the voltage.

t 3.5.2 Series-Parallel Reduction Techniques 1

You would recall from your study of Section 2.2 how the use of simpler equivalents for series or parallel connected branches facilitated the analysis in d.c. circuits. The same technique can be employed for a.c. circuits as well. The only difference -and we have to be constantly attentive to this -is that the impedances and admittances to be processed are complex numbers and both their magnitudes and angles are to be taken into account. A couple of examples are now worked out to illustrate these techniques.

Example 3.24

Find the current supplied by the source in the circuit of Figure 3.36.

l a


Solution

R, +jwL = 10 +j400(0.05) = 10 +j20 SZ

Za, impedance of the circuit to the right of line aa

Fa and R1 are in series. Hence Zb = Za + R1 = 18 + j9 S2 - Zb and the capacitance C are in parallel.

= (llz*) + jwC = [1/(18 + j9)] + j400 x 20 x

= 0.0444 - j0.0142 = 0.0466 L - 17.7"

Source current = Yc V = (0.0466) 10 = 0.466 A

Example 3.25

The circuit shown in Figure 3.37 is e~nployed in some fornls of electronic oscillators (generators of sinusoidal voltage signals). It is required that the output voltage

A.C. Circuits

Introduction to Circuits should have a phase difference of 180" with respect to the input voltage 7; at a frequency of w rls. Find the required conditions on C and R and the value of Vi/Vo when this condition is fulfilled.


Solution

We shall work out this problem by the alternative method suggested in Section 2.2.4. Starting from E, we move towards the source expressing all intermediate variables in terms of E. Let jX = (- jlwC) be the impedance of each capacitor (at the frequency w) and let G be the admittance (conductance) of each resistor R.

We have, - I~ - GV,

V, = il(ix, = O V o - - go = V, + VO = c (1 +jGX)

- r2 = G V ~ = Vo (G + jG2X) - I3 =% +T2 = Vo (2G + jG2X)

6 =i3w= E ( - G 2 x 2 + j 2 ~ X )

k = E q +% = E ( 1 - G 2 x 2 + j 3 ~ X ) - I4 = G% = Vo (G - G3x2 + j3G2X) - - - I, = I, +z4 = VO (3G - G3x2 + j4G2X)

Er = G,) OX) = % [- 4G2x2 + j(3GX - G32)]

= ir + to = [(1 - 5G2x2) + jGX(6 - G2x2)]

Now (GIG) should be a negative real number say .-A, for a phase difference of 180" to exist between the two voltages.

Hence G2x2 = 6 3 ~ w ~ c ~ R ~ = 1 is the required condition.

(Vi / VoI = A = 5G2x2 - 1 = 29 at this condition.

SAQ 24 Find the values of R and L in the circuit of Figure 3.'38 if the source supplies a current of 0.5 A withjlagging 7 by cos- ' 0.8. Take w = 400 rad / s.


SAQ 25 The circuit in Figure 3.39 is used to produce an output voltage having 90' phase difference with the input v; at a specified frequency w. Given w = 1000 r/s and C = 10 pF, find the value of R required and the ratio Vi/Vo under these conditions.

Pig. 3.39 : Figure for SAQ 25

3.5.3 Solution by Loop and Node Equations 111 the previous Section, we saw that the series-parallel reduction methods used for a.c. networks are essentially no different from those used for resistive networks under d.c. operation. This is indeed true for the whole range of methods of a.c. network analysis iilcluding star-delta conversion methods, loop current and node voltage analyses and the use of Thevenin and Norton equivalents. All of them are straightforward adaptations of d.c. methods, with phasors replacing the pertinent d.c. values and generalised Ohm's law relatioils (3.26) and (3.28) replacing the d.c. Ohm's law. Since the equations now involve cotnplex numbers, the computations are that much more involved. As no additional theory is involved in the use of these techniques, we limit our treatment here to the illustration of loop current and node voltage methods by an example each.

Example 3.26

Consider the circuit of Figure 3.40, where the phasors of the two voltage sources and the impedances of elements in ohins are marked. We wish to find the currents in the two sources.

A

Fig. 3.40: Circuit for Examples 3.26 and 3.27

Solution

Let us first solve this circuit by the loop current method using the two loop currents as marked on the figure. The loop equations (KVL equations in phasor form) are now formulated using the concepts of self-loop impedance and mutual loop impedance. We then have

A.C. Circuits

Introduction to Circuits ~ l i m i n a t i n ~ ? ~ , we have,

[(loo + j 50 )~ - 10dl = (100 + j50) 100 - 100 (50 - j86.6)

Similarly,

Thus Il = 1.41 A ; I2 = 0.60 A.

Example 3.27 Let us now solve the same problem applying the node voltage method with node numbers as marked on the figure. We have Vlo = 100 L 0' and V20 = 100 L - 60' = 50 - j86.6. Rr~nember that we should use admittances in forming the node equations. Node equation (KCL in phasor form) at node 3 is now formed.

Multiplying throughout by j50,

( 2 + j0.5) v30 = 150 - j86.6

The source currents are then obtained.

SAQ 26 In what manner would a loop equation (KVL in phasor form) written for an a.c. circuit differ from one written for a d.c. circuit ?

SAQ 27 "Two impedances in series have an effective impedance whose magnitude is always larger than that of either impedance." Is this statement true? If not, why not?

SAQ 28 Fill up the blanks in the following statements.

In a parallel RLC circuit (Figure 3.34) fed from a sinusoidal current source, the source voltage leads the source current when 1

exceeds 2. The largest magnitude of im edance for this cifcuit occurs at a frequency w= ' and the smallest magnitude at w= and

5

Example 3.28

Evaluate & in the circuit of Figure 3.40 using Thevenin's theorem.

Solution

After disconnecting the 100 Q resistor, the Thevenin equivalent for the remaining network will be as shown within the box in Figure 3.41(a). The voltage across the 100 Q resistor will be V30. the open-circuit voltage is found from the circuit of Figure 3.41(b). Using the node voltage method,

Fig. 3.41: Circuits for Example 3.28 (a) Thevenin equivalent (b) Circuit for determining 90 (c) Circuit for determining 20

The Thevenin impedance is calculated from the circuit of Figure 3.41(c). 2, = j50 / 2 = j25.

Now using the circuit of 3.41(a),

SAQ 30 Find the currents supplied by the sources in the circuit of Figure 3.43.


-.

Introduction to Circuits SAQ 29 Calculate the currents in all the elements of the circuit shown in Figure 3.42.

230V 50Hz T-gvi YI o.2H


118

Since a.c. circuits have periodically varying voltages and currents, the power delivered to an element or a section of a circuit is also a periodically varying quantityp(t). You would recall from our discussion in Unit 1, that a meaningful measure of power in such situations is the average power over a cycle. The tenn power when used in the context of an a.c. circuit without any additional qualification means this average value and is denoted by the symbol P. In a d.c. circuit, the power delivered to a 2-terminal network is equal to VI, the product of the terminal voltage and current. In what follows, we develop the corresponding formula applicable to a.c. circuits.

3.6.1 Power, Apparent Power and Power Factor Consider the circuit of Figure 3.44, in which a sinusoidal voltage v = fi V sin (wt + 0) supplies power to a 2-terminal network N having an equivalent impedance Z= Z L a = R + jX. In this context, N is also referred to as a load and Z as the load impedance.

I - n

+ @v-fiVsin(wt+El) Z L a

N

Fig. 3.44: Circuit used for discussion oEpower in an a.c. circuit

The phasor 1 of the resultant current i in this circuit is - I= [ V L 0 / Z L a ] =(V/Z) L ( 0 - a )

Thus i = \ / Z ( ~ l z ) s i n ( w t + 0 - a ) = \ / Z ~ s i n ( w t + 0 - a )

The instantaneous power supplied by the source to the load is

.- p ( t ) = v i = \ / Z ~ s i n ( w t + 0 ) . \ / Z Z s i n ( w t + 0 - a )

= 2V7 sin (wt + 0) sin (wt + 0 - a )

= Vqcos a - cos (2wt + 20 - a) ]

The variation of p(t) is shown in Figure 3.45

/P

A.C. Circuits

Fig. 3.45 : Instan~aneous power p(t) delivered to N of Fig. 3.44

It is seen from Eq. (3.36) and Figure 3.45 (hatched portion) thatp(t) consists of a constant component VI cos a and a second component of peak value VZvarying sinusoidally at a frequency 2w. As the average of the second component over a period of the input voltage or current is zero, the average ofp(t) is equal to the first component itself. Thus,

P = VZ cos a watts (3.37)

The above formula is of great significance. It indicates that the power received by a load is not merely the product of the rms values of its terminal voltage and current but includes an additional multiplicative factor cos a, called the power factor of the load. Power factor is the cosine of the impedance angle a and hence is a property of the concerned load. Formally, power factor (p.f.) may be defined as

Power delivered to load p.f. = - - - (3.38)

Product of effective values of terminal voltage and current of the load k7

The power factor is said to be of the leading type if 7 leads v(i.e., X < 0) and of the logging type if 1 lags (i.e., X > 0).

In a contrast to power P, the product VZ is termed apparent power S and is indicated in units of volt amperes (abbreviated VA). Though dimensionally 1 VA equals 1 watt, two different names of the unit are adopted to emphasize the distinction between apparent power S and power P. Apparent power is an important parameter in the specifications of electrical equipment, as the size and cost of many electrical machines depend on their VA rating rather than wattage rating. For instance, a 500 kVA distribution transformer is rated in terms of its ability to handle S upto 500 kVA level rather than power P it can deliver to a load.

Table 3.1 gives the particular fonns of the relations discussed above for special categories of loads. Note that a pure inductor and capacitor have zero p.f. since they are only energy storage elements and not energy dissipating elements.

I Table 3.1 I

9: a Load Power P Apparent Power S Power Factor (W,

Resistor R 0" V Z = Z ~ R = V ~ I R 1.0 (unity) VI = Z'R = (v2 1 R)

Inductor jwL 90" VZ = z2wL = (v2 / wL) zero (lagging) zero

Capacitor 1ljwC -90" VZ = z2 /wC = v2 wC zero (leading) zero

Introdaction to Circuits Figure 3.46 illustrates the dependence of p.f. on the nature of Z. For passive loads, the value of p.f is clearly non-negative.

pf - cos a 0.0 lagging

0.0 leading

Lagging p.f. Leading p.f.

Inductive Impedance Capacitive Impedance C I-! 2 jXL (lnductor) R (Resistor) -jXc (Capacitor)

Fig. 3.46 : Dependenke of p.f. on nature of Z

The expression for P in Eq.(3.37) could be considered either as the power delivered by the voltage source or the power received by N of Figure 3.44. In the former case the reference direction for voltage is a rise in the reference direction for current and hence vi represents instantaneous powerp delivered by the source. The average power P delivered by the source equals the product of the effective values of voltage and current and the cosine of the phase difference between v and i. Viewed from the point of N, the reference direction for voltgge is a drop in the direction of current and hence vi represents instantaneous power p received by Nand having an average value VZ cos a


The power in an a.c. circuit is given by tke product of 1 2

'and J

The power factor is the cosine of 4. P.F. for a resistive network equals '. For a purely capacitive network it is b. Inductive loads have a ' p.f.

SAQ 32 A voltage 20@ sin (314t + 60") is applied across a load comprising R = 60S2 and L = 200 mH in series. Find the power drawn by the load and its p.f.

Example 3.29

Find the power supplied by each source of Example 3.26. Show that the total power supplied by the sources equals the sum of powers received by the passive elements in the circuit.

Source 1 :

Vl = 100 V, ZI = 1.41 A, Arg (vl) - ~ r ~ ( i ~ ) = 0" + 34" = 34"

P1 = 100 x 1.41 x cos 34" = 117 W

Source 2 :

V2 = 100 V, I2 = 0.60 A,

Source 2 is actually receiving power. The two inductive reactances do not consume any power as their p.f. is zero. The current in the resistor has a phasor - zl +Tz = 1.41 L - 34" + 0.60 L 160" = 0.84 L - 44".

Power PR absorbed by resistor = (0.84)' x 100 = 71 W . P1 + P Z = P R

Example 3.30

, Two impedances zl = 7 -j24 & and Z2 = 6 + j 8 & are connected in parallel across a 200 V source. Find the power and apparent power supplied by the source.

Solution

yl = (7 - j24)- ' = (7 + j24)/625 = 0.0112 + j0.0384 S - y2 = (6 + j8)- = (6 - j8)/100 = 0.06 - j 0.08 S

y = 0.0712 - j0.0416 = 0.0825 L - 30.3"

1f 7 = 200 L 00, 7 = Y 7 = 16.5 L - 30.3"

Therefore S = 200 x 16.5 = 3300 VA , P = 200 x 16.5 x cos 30.3" = 2848 W.

Note that P is also equal to v'R~( fl where Re( Y) is the real part of y ( = 0.0712 in this example). You are asked to show this result in SAQ 33.

SAQ 33 A load having an admittance Y= G + jB is connected across a supply voltage of V volts. Show that the power taken by the load is 9 ~ .

3.6.2 Reactive Power Let us examine again the expression in Eq. (3.36) for the instantaneous powerp taken by the load in the circuit of Figure 3.44. This expression could be put in the alternative form,

p(t) = VZ cos a [I - cos (2wt + 2 8)] - VIsin a sin (2wt + 28) =p,(t) +px(t),

where ~ , ( t ) = P [ l - cos(2wt + 28)] C

px(t) = Q sin (2wt + 28 + n),

and Q = VZ sin a

In this decomposition o fp , we note thatp, is always non negative. It represents the effect of power dissipated in the resistive elements of N. The average value of thisp, is P. On the other handpx(t) varies sinusoidally with a peak value of Q. It represents the'energy exchange between the source on one hand and the energy storage elements (LC elements) in N on the other. Q is called the reactive power as it is associated with the reactive elements (LC) and does not contribute to the net transfer of energy from the source to the load. In contrast P represents the average rate of transfer of net energy and is therefore also termed active power. Q is measured in volt amperes reactive (abbreviated VAR). We therefore have the following three power related concepts.

Power P (also called active power) = M cos a' watts (W) (3.3 9a)

Reactive power Q = M sin a volt amperes reactive (VAR) (3.39b)

Apparelit power S = VZ volt amperes (VA) (3.39~)

kc. Circuits

Introduction to Circuits It is clear that

s 2 = $ + e 2

and Q is positive or negative depending on whether a is positive (inductive loads) or negative (capacitive loads).

There is a simple way by which we can calculate the above quantities directly from Vand f without finding a. Let

~ = V L O ; 7 = 1 ~ ( 8 - a )

Consider the product of Pwith the conjugate of 1 - - V I * = ( V L ~ ) ( I L ~ - 8 ) = W L a = W c o s a + j ~ s i n a

We recognise the real and imaginary components on the right side of the above equation as P and Q respectively and refer to P + jQ as complex power 3. The magnitude of 3 is

which is the apparent power S. We then have

?= ~ T * = P + ~ Q (3.41a)

Where Re ( ) means the real component of ( ) and Im ( ) means the imaginary component of ( ). When Vand 7 are given in rectangular co-ordinates, P and Q can be readily 'calculated using Equations (3.41b) and (3.41~) without the need to convert Vand 7 into polar form.

The expressions for reactive power taken by the three elements R, L, C reduce to the simple forms given in Table 3.2.

Load a Reactive Power Q FAR)

Resistance R 0" zero

Inductance ~ W L = jXL 90" V ~ / X , = I ~ X ,

Capacitance ( j ~ c ) - ' = -jx, -90° -V2 / X, = -I2xC

As a resistor is not an energy storage element, its ability to exchange stored energy with the source is zero and hence its Q is zero.


Reactive power supplied to a circuit is the ' value of the alternating power component, representing the energy interchange between the

elements in the circuit and the source. Its value is positive for an load and negative for a load.

SAQ 35 Find the reactive power taken by an impedance Z = 100 - j100, when connected to a 200 V a.c. source.

Example 3.31

Find the values of P and Q taken by a load having V = 100 + j40 and?= 20 - j10 for its terminal voltage and current phasors.

Solution

s= ~ ~ * = ( l O O + j 4 0 ) ( 2 0 + j l O ) = 1600+j1800

Therefore,

Example 3.32

From the solution obtained in Example 3.26, verify that the sum of the reactive powers supplied by the sources equals the reactive powers taken by the other elements in the circuit.

Solution

Source 1:

V1 = 100 V, Il = 1.41A, Arg ( pi) - ArgGl) = 0" + 34" = 34"

Qsl = 100 x 1.41 x sin 34" = 78.85 VAR

Source 2 :

V2 = 100 V, I2 = 0.60 A, Arg ( V2) - k g ( & = - 60" + 200" = 140"

Qsz = 100 x 0.60 x sin(140°) = 38.57 VAR

Inductive reactances

QL1 = I : xL1 = 1.41' x 50 = 99.41 VAR

QL2 = 12 xL2 = 0.60' x 50'= 18 VAR

The resistance of 100 & does iiot take any reactive power. It is seen that Qsl + Qa = QL1 + QL2 within the limits of computational error, due to rounding off.

SAQ 36 A series RLC-circuit takes 120 W of active power and 90 VAR of reactive power from a 200 V, 50 Hz source. The voltage across the capacitor is found to be 30 V. Calculate the values of R, L and C.

3.6.3 Power Factor Correction Electrical power distribution systems operate at a nearly constant voltage level and a variable current governed by the demands of the connected load. The size and cost of the various iteins of equipment installed by the power supply organisation - like the a.c. generators, transformers and distribution lines - depend essentially on the maximum VA demand to be catered to. Now, for a given load powerP and system voltage V, the current and VA requirements vary inversely as the load power factor. A lower value of p.f. leads not only to enhanced cost of power supply equipment but also to increased power losses and voltage drop in the supply lines due to the increased line current. A high value of p.f., preferably close to unity, is therefore a desired operating condition in power supply systems.

The normal industrial electrical loads largely comprise electromagnetic machines and equipment operating at power factors of around 0.8 lagging. To encourage drawal of power at a higher p.f., electricity supply organisations bill an industrial coilsumer on the basis of a rate structure (called tariff) which penalises low p.f, conditions. Therefore it is often expedient for an industrial consumer to improve the overall p.f. of his load, even if it implies expenditure on installing necessary additional equipment. The key to the

kc. Circuits

Introduction to Circuits process of power factor improvement, usually termed power factor correction, lies in the fact that normal industrial loads with their lagging p.f. draw positive reactive power, which can be compensated by connecting in parallel with the loads, capacitors which , draw negative reactive power. It isthereby possible to effect a sizeable reduction or even elimination of overall Q taken from the supply. Remembering that p.f. equals P I-, it is clear that reduction in Q leads to an increase in p.f.

The process of p.f. correctiol~ is illustrated in Figure 3.47. Let the current taken by the load impedance 2, be 7, lagging the supply voltage 7 by an angle a.

Fig. 3.47: Illustrating power factor correction. Condition (a) with load alone (b) with capacitor connected

Let a capacitor C, connected in parallel with the load, draw a current & This current

leads f b y 90". Let - 1, = jZL sin a

The total current taken from the supply is therefore

which is in phase with t. Thus the source sees a unity p.f. combined load. The required value of capacitance can be calculated from the relation

VwC = I , = I, sin a

C = ZL sin a/wV = sin a/wZL (3.42)

It is illustrative to also view the process as cancellation of reactive power supplied by the source

Reactive power Q, taken by Z, = (v2/ZL) sin a VAR

Reactive power Qc taken by C = - V%C VAR

To make QL + Q, = 0, we require

C = (sin a / wz,),

which is the same relation given in Eq. (3.42).

In individual cases, one may desire to improve the p.f. to a suitably high value but not 1.0, so as to reduce the cost of capacitors. The decision would be based on what condition leads to the best overall economy.

SAQ 37 Explain briefly the necessity for p. f. correction.

Example 3.33 The electrical load of a consumer comprises 1 kW of lighting load and 10 kVA of motor load at 0.75 p.f. lagging. The lighting load may be taken to operate at a p.f. of 1.0. The power supply is at 240 V, 50 Hz. Evaluate the capacitance needed to be connected in parallel with these loads to bring the overall p.f. of the installation to 0.95 lagging.

Solution The circuit representation of the loads is given in Figure 3.48 where M represents motor load.


Alternative I

Let V=240L0° - I, = (10001240) L 0" =4.17+jO; Ic= +jIc

IM = (100001240) = 41.67A , cos- 0.75 = - 41.4"

Therefore, - IM = 41.67 L-41.4" = 31.26 - j27.56; - I, =TI +IM +Tc = 35.43- j(27.56 -

- Is should lag behind f by an angle cos- '(0.95) = 18.2". So

(27.56 - Ic)/35.43 = tan 18.2" - 0.33 - Ic = 15.87 A

Ic = VwC * C = 15.87/(314 x 240) F = 211 pF.

Alternative 2

We consider P and Q of each load

P(W Q (VAR) Lighting load^ 1000 0

Motor 10000 x 0.75 = 7500 10000 dl - (0.75)' = 6614

Capacitor 0 Q, Overall 8500 6614 + Q,

Overall p.f. =PI[$ + a2lU2 = 0.95

Overall Q = I d 1 - 0.95~10.95 = 2794 VAR

6614-PwC-2794 - V%C=3820 * C=211pF

Introduction to Circuits SAQ 38

An inductive load draws 1000 W from a 200 V, 50 Hz source. A capacitor of 25.3 p F connected in parallel with the impedance raises the overall p.f. of the combination to unity. What is the p.f. of the inductive load?

Example 3.34

A load impedanceZL = RL +jXL is fed from an active network whose Thevenin equivalent is as shown in Figure 3.49. Find the value of ZL which draws maximum


power from the network.

Solution

~ e t j b e the phasor of current in the circuit

- z o - R ~ + j X o

Power PL delivered to load = 1' RL = vi RL

(Ro + RL)' + (XL + ~ 0 ) '

r\

- zL = RL + jXL

r\

If XL is varied, PL becomes maximum when XL = - X,. With this value ofXL,

d PL is maximum when - [(R: / RL) + 2Ro + RL] = 0 - RL = RO

~ R L

For maximum power transfer to ZL, a = R, - JX, = Z,*.

This is an extension of the maximum power transfer theorem in d.c. circuits (vide Section 2.6.4) to the a.c. situation.

SAQ 39 What are the power factom of the following loads/circuits?

(a) A pure capacitor

(b) A series RLC-circuit at resonance

(c) A series combination of R and L with equal rms voltages across the two elements.

(d) A parallel combination of R and C with the rms values of the currents in the two branches (R & C) being in the ratio 2 : 1.

SAQ 40 Two loads are connected in parallel across a 2200 V supply. The first load is capacitive, taking a power of 20 kW at p.f. of 0.8. The second load is inductive and takes a power of 30 kW and a reactive power of 30 kVAR. Calculate the current taken from the supply. If the current is to be minimised by connecting a capacitor in parallel with the two loads, what should be its capacitanceXAssume f = 50Hz.

SAQ 41 When the terminal voltage and current of a load have the following phasors, the power and reactive power taken by the load are 40 Wand 180 VAR respectively. Evaluate a and b.

- I = b - j l

3.7 SUMMARY

Sinusoidal voltages and currents - also called a.c. voltages and currents - play a key role in the theory and practice of electrical engineering. In this Unit you were first introduced to the various useful parameters of a sinusoidal signal like period, frequency, angular frequency, peak value, effective value and form factor.

When a linear circuit is driven by sinusoidal sources of a given frequency, all voltages and currents in the circuit under steady state are also sinusoids of the same frequency. One signal differs from another only in its effective value and phase. Each sinusoidal signal can then be represented in a compact fashion by a directed line segment called a

r '?asor. A phasor can be considered a complex number. You were therefore introduced to the nature of complex numbers aud to the related algebra.

You have seen how linear operations like addition, subtraction, differentiation and integration of sinusoidal functions of time are much more conveniently handled in the phasor domain by equivalent simpler algebraic operations on complex constants. This property leads to simpler formulation of Kirchhoff's laws and element terminal relations. The latter relate the phasors of the terminal voltages and currents by proportionate complex constants called impedances/admittances. You learnt how to calculate impedance and admittance parameters of R, L, C elements and saw how the reactance parameter relates the rms values of terminal voltage and current in an energy storage element L or C. In this Unit, you were also introduced to an additional circuit element, viz., magnetically coupled coil pair and the related parameters L,, L2 and M. You have learnt how to formulate the terminal equations of this 2-terminal pair element in phasor form.

In terms of voltage and current phasors and element impedances and admittances, the a.c. circuit equations are formulated in a completely analogous matlner to the d.c. circuits. The solution of these a:gebraic equations is however more complicated since they involve complex numbers. The particular methods of a.c. circuit analysis you had studied in this Unit are the series-parallel reduction techniques and those based on loop currents and node voltages as variables. The phenomenon of resonance in series and parallel circuits and related properties formed another topic of study in this unit.

A.C. Circuits

Introduction to Circuits In the last section of the unit, you learnt how to calculate power P in an a.c. circuit and the significance of power factor. You werepalso introduced to the concepts of apparent power S, which essentially fixes the size of the power supply equipment, and the reactive power Q which represents the peak value of the idle exchange of power between the energy storage elements on one hand and the source on the other. You also gained an appreciation of the need for p.f. improvement of electrical installations and learnt how to calculate the capacitance value needed for the purpose.

3.8 ANSWERS TO SAQs

SAQ 1 :

(a) il = 2 cos (wt + 30") = 2 sin (wt + 120")

(b) vz = 4 sin 314t + 4 (cos 314t cos 150" - sin 314t sin 150")

= 2 sin 314t - fi cos 314t

= 4 [ ( I 1 2) sin 3 1 4 - ( f i12) cos 314tl - 4 sin ( 3 1 4 - 60")

SAQ 2 :

Figure for Answer to SAQ 2

SAQ 3 :

v, = 3 f i sin (wt + 90") ; vg = 4 f i sin (wt + 180")

v, = 5 f i sin (wt - 37")

SAQ 4 :

(a) V = 1 m L 290" = 4.84 - j 13.29, since cos (wt + 200") = sin (wt + 290")

(b) v(t) = 1 0 m sin (wt + 120") ; V = - 50 + j86.7

(c) i = l .h/Zsin(wt + 112.62") ;1= 1.3 L 112.62"

(d) i = 2 f i sin (wt + 75") ; 1= 2 L 75" = 0.52 + j1.93

(e) V = 5 L 45" = 3.54 + j3.54

(4 ( 4 Figure for Answer to SAQ 4

SAQ 5 :

All are false.

SAQ 6 :

10 L 30 = 10 (cos 30" + j sin 30") = 8.67 + j5

4 - '6 - - (5 +j2) = (- 0.24 - j0.68)(5 + j2) (see Ex. 3.10) (6 +I@ = - 1.20 - j3.4 - j0.48 + 1.36 = 0.16 - j3.88 - z = (0 + j1) + (8.67 + j5) + (0.16 - j 3.88) = 8.83 + j2.12

SAQ 7 :

(a) ?=(100/fi) L - 45" = 50 - j50

(b) v(t) = 5 m cos (wt + 135") = 5 m sin (wt + 225")

c= 50 L 225" = - 35.35 - j 35.35

(c) i (t) = - 1 m sin (wt + 120") = 1 m sin (wt - 60") - I= 10L-60°=5-j8.66

(d) i(t) = - 20 cos (wt + 30") - 20 sin (wt - 60") - I = (20fl) L - 60" = 7.07 - j12.25

SAQ 8 :

(a) A + B - C = ~ + j 5 - 6 + j 1 0 - ( 8 + j 4 ) = - 9 + j l l

= 5 + j 5 - 0.09 + j 1.30 = 4.91 + j 6.30

SAQ 9 :

Refer to (c) of Section 3.3.3.

15 = (lo2 + lo2 + 2 x 10 x 10 x cos 0)112

SAQ 10 : - 11 = (5/Q L 60" = 1.77 + j 3.06

Using K C L , ~ ~ =TI -I , = 1.77 + j3.06 - (4 + jl)

= - 2.23 + j2.06 = 3.04 L 137"

Thus ig(t) = 3.04fl sin (wt + 137") = 4.3 sin (wt + 137")

SAQ 11 : (1) leads (2) 90 (3) resistor (4) smaller

SAQ 12 :

230 = (h x 50) L x 1.1 + L = 0.666 H

150 50 Changed current = 1.1 x - x - = 1.435 A 230 25

kc. Circuits

Introduction to Circuits SAQ 13 : The statement is false. It is the ratio ofphasors of v(t) and i(t) which is equal to the impedance.

SAQ 14 :

R lljwL, jwC

SAQ 15 :

No. V R and vc have a phase difference of 90" since one is in phase with the current and the other lags the current by 90". The magnitude of their phasor sum is (v: + v:)" = (89.8' + 178.7~)"~, which is indeed the rms value of the supply voltage vs. KVL implies that vs = vR + vc or equivalently Vs = VR + %and not Vs = VR + VC.

SAQ 16 : - I = L 30" ; Z= 100 + j251- 270 L 68.3"

V = 27 = [270 L 68.3'1 [lo- L 30'1 = 0.27 L 98.3"

Hence v(t) = 0.27fi sin (400t + 98.3")

SAQ 17 :

(1) voltage (2) current (3) rc/2 (4) frequency (5) mutual inductancelmagnetic coupling (6) voltage (7) current

SAQ 18 :

SAQ 19 : - v = l m L 4 5 " ; 1 - 0 . 2 5 f l L 7 5 "

2 = 10 f l L 45'10.25fl L 75" = 40 L - 30" = 34.6 - j20 B

A resistance R and a capacitance C in series would provide this 3, where R = 34.6 B and C = (1110001~ x 20)F = 15.92 pF.

SAQ 20 : - - W e have Vs = VR + & Let us take VR as reference. - vR = 50 L 0" - I = (50120) L 0" = 2.5 L 0" = 2.5 - - Vz= Z I = (40 +j60) 2.5 = 100 +j150

Figure for Answer to SAQ 20 vs=-=212v.,

SAQ 21 :

From the discussion in Unit 1, we luiow power in a resistor equals v%/ R. Here V, = 130V.

Impedance of the combination = 22010.308 = 714.3 S2

We have R ; ~ + d ~ ' = 714.3'

SAQ 22 :

(1) wL < (llwC) (2) purely real (3) short-circuit (4) an opencircuit

SAQ 23 : P

The same voltage appears across each of these elements. Taking its phasor as 200 L 0°, - zR = 200 L 0°/100= 2 L 0°= 2+jO

- Zc = (200 L 0") j314 x 100 x = j6.28 - zs =IR +IL +Ic = 2 - j0.09 = 2 L - 2.6".

The source current Is is 2 A and lags the source voltage by 2.6"

SAQ 24 : 1 cos- 0.8=37"=Arg($-kg(?)

Impedance 2 seen by the source = (150 / 0.5) L 37" = 240 + j180 S2

Impedance of capacitor = - j I400 x 15 x 10- = - j167 S2

Impedance of parallel RC branch = (125) (- j167) = 80 - j60 125 - j167

Z = R +jwL+80-j60 => (R+jwL)=(240+ j180) - (80 - j60 )

R - 160S2

L =240/400= 0.6H

SAQ 25 :

Let jX be the impedance of each capacitor.

I V,=%jxl(R+jX) => Vb,=K(R+jX)/ jx c= Go +R[(&,lIX)+ ~ / I X ) ] = ( ~ ( R + j X ) / f l + ( ~ ~ / j X )

L = VO [(R / jX) + (R + jX)'~(jx)~] = [(x' - R2 - 3 j W / x'] Vo For 90" phase difference (Vi / V,) is to.be purely imaginary. Therefore,

R' = X' => dc2R2 - 1 => R - I/& - (10~/1000 x 10) - 100 S2

With this value of R, (VjV,) = 3RIX = 3.

SAQ 26 : The variables are current phasors instead of real quantities. The coefficients of the variables are complex numbers representing impedances instead of real numbers representing resistances. The right hand side expression is a complex number representing the phasor of net source voltage in the loop instead of a real number representing net d.c. voltage source strength.

SAQ 27 :

False. 5 = zl + 2' does not mean Zs = Z1 + Z,. The angles of 2, and ZZ may make the resultant have a smaller magnitude than either. For example, if zl = 10 + j20 and - ~ ~ = - j 2 0 , Z ~ + ~ = 1 0 .

kc. Circuits

SAQ 28 :

(1) llwc (2) WL (3) 1m (4) 0 (5) 00

SAQ 29 :

Fp of the parallel combination - (11100) + j314 x 10 x 10- ti + 14314 x 0.2

= 0.01 - j0.0128 = 0.0162 L - 52"

Zp of the combination = ($)- = 61.73 L 52" = 38.00 + j48.64

Impedance Zs seen by source - 20 + j314 x 0.05 + & = 58.00 + j64.34 = 86.62 L 47.97"

Current Is in source = 2301 86.62 = 2.66 A

Voltage across the parallel combination = IsZp = 2.66 x 61.73 = 164 V

Current in 10052 resistor = 1641100 = 1.64A

Current in 10 pF capacitor = 164 x 0.00314 = 0.51 A

Current in 0.2 H inductor = 164 I 62.8 = 2.61 A

SAQ 30 : - Vsl = 100 L 0°, V , = 100 L - 30"

zl = 10 + j314 x 0.159 = 10 + j50 = z2 - Z3 = 100 + j314 x 1.59 = 100 -t j500

Let us assume reference directions for loop currentsT1 and& similar to what are marked in Figure 3.40. We then have :

(110 + j550)T1 + (100 + j500)T2 = 100

(loo+ j500)T1 + (110+j550)T2- 100L - 30°,

yielding?l = 0.52 L - 13.8" andT2 = 0.52 L 186.4"

SAQ 31 :

(1) V (2) I (3) p.f. (4) angle of load impedance (5) 1.0 (6) zero leading (7) lagging.

SAQ 32 :

z= 60 + j314 x 0.2 = 60 + j62.8 = 86.86 L 46.3"

I = 200186.86 = 2.30 A , cos a = 0.69

P = W cos a = 200 x 2.30 x 0.69 = 317.4 W,

P.F. = 0.69 lagging

SAQ 33 : Let the supply voltage have the phasor V L 0" - I - (G + jB)V- v(G2 + B2)ln L tan- '(BIG)

E'.I;= = m s [ - tan- '(BIG)] = [G/(G2 + B2)ln]

P = W (pn = (V) v(G2 + B2)'" [G/(G2 + B ~ ) ~ ] = v2G watts

SAQ 34 : (1) peak (2) reactive or energy storage (3) inductive (4) capacitive

SAQ 35 :

I = 200/(100~ + 1 0 0 ~ ) ~ = fi A

a=tan- ' L-15-45"

~ = W s i n a = 2 0 0 x f i x ( - 1 ~ = - 2 0 0 V A R

SAQ 36 :

S ISOVA OVA W- 150 + I=150/200=0.75A

I ~ R = 120 + R = 120/(0.75)~ = 213.3 52

kc. Circuits

Net reactance of the circuit = XL - XC

SAQ 37 : The current for a given load power and system voltage is reduced with increased p.f. The size of the supply equipment, the line losses and the line voltage drop can thereby be reduced. The power supply authorities therefore adopt a tariff which penalises low p.f. operation of consumer loads. It is therefore economical for the consumer to save on electricity bills by installing p.f. correcting capacitors. This also helps him to get a better regulated supply voltage (one without large voltage dips).

SAQ 38 :

Let supply voltage be V = 200 + j O and the current taken by the inductive load be - r zL = Zl - jZ2. Current taken by capacitorjc = jwCV = j1.59.

~ u t 7 ~ + jc should not have an imaginary part. Therefore, Z2 = 1.59 A

P.F. of load = cos [tan- ' (ZdZl)] = 0.95 lag.

SAQ 39 :

(a) Zero leading (b) unity (c) cos(tan- ' 1) = 0.707 lagging (d) cos[tan- '(112)l = 0.89 leading.

SAQ 40 :

Fl = PI + jQl = 20000 - j20000 tan (cos- '0.8) = 20000 - j15000

(negative sign taken for Q as the load is capacitive) - s2 = P2 + jQ2 = 30000 + j30000

s = + r2 = 50000 + j15000

S = (50000~ + 1 5 0 0 0 ~ ) ~ ~ = 52,202 = 2200 1 * 1 = 23.73 A The additional capacitor takes a negative reactive power say Qc and reduces S. To minimise the line current and hence S, we require

Qc = - 15000 VAR 3 V(VwC) = 15000

Therefore C = [15000/(2200)~(314)] F = 9.87pF.

SAQ 41 :

r F= PI* = (20+ja)(b + j l )

= (20b - a) + j(ab + 20) = 40 + j180

20b - a = 40, ab = 160 I

(20b - 40)b = 160 3 b2 - 26 - 8 = 0 * (b - 112 = 9 b = 4 o r - 2 ; a - 4 0 or-80

Hence either a = 4 0 , b = 4 o r a = - 8 O , b = - 2