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Chapter-III ACCircuits Concepts

3.1 IntroductionThe electrical quantities which vary or changes continuously with time are known as Alternating quantities

Alternating voltage and current are some of those quantities.

The magnitude of these Alternating quantities changes continuously with time and the direction also vary orchanges after definite time period. These alternating quantities may have square waveform, triangular

waveform, saw tooth waveform, sinusoidal waveform etc.

Here we are concern mainly with sinusoidal waveforms. These Alternating sinusoidal quantities are applied todifferent electrical circuits and their steady state behavior is studied.

3.2 Representation of Sinusoidal Alternating Voltage and CurrentWhen an alternator produces AC voltage, the voltage switches polarity over time, but does so in a very

particular manner. When graphed over time, the wave traced by this voltage of alternating polarity from an

alternator takes on a distinct shape, known as a sine wave (Fig 3.1).

Fig 3.1

3.2.1 Graph of AC Voltage over Time (The Sine Wave)In the voltage plot from an electromechanical alternator, the change from one polarity to the other is a smoothone, the voltage level changing most rapidly at the zero (crossover) point and most slowly at its peak. If we

were to graph the trigonometric function of sine over a horizontal range of 0 to 360 degrees, we would findthe exact same pattern as in Table 3.1.

Angle (o) sin(angle) wave Angle (o) sin(angle) wave

0 0.0000 zero 180 0.0000 zero

15 0.2588 + 195 -0.2588 -

30 0.5000 + 210 -0.5000 -

45 0.7071 + 225 -0.7071 -

60 0.8660 + 240 -0.8660 -

75 0.9659 + 255 -0.9659 -

90 1.0000 +peak 270 -1.0000 -peak

105 0.9659 + 285 -0.9659 -

120 0.8660 + 300 -0.8660 -

135 0.7071 + 315 -0.7071 -

150 0.5000 + 330 -0.5000 -

165 0.2588 + 345 -0.2588 -

180 0.0000 zero 360 0.0000 zero

Table 3.1


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Instantaneous value: The value of alternating quantity at any particular instant of time is called as

Instantaneous value.

Cycle: One complete set of positive and negative values of an alternating quantity is called as a cycle.Alternation : One half of a cycle is called as an alternation.

Time period: The time taken by the wave to complete one cycle is the time-period. It is denoted by T and unit is

in seconds.

Frequency: The no. of Oscillations by the wave in one second is called the frequency. It is denoted by f andunit is hertz.

Angular velocity: The velocity with which the coil moves or the ratio of angle turned per unit time is called

angular velocity. It is denoted by . = 2f = 2 / T

Amplitude: The maximum value of sinusoidal wave attained during one cycle is called the amplitude or peak.

It is denoted by Emax for voltage amplitude the Imax for current amplitude.

Peak Value: The maximum instantaneous value measured from its zero value is known as Peak value.

Peak to Peak Value: The value between maximum positive instantaneous value from its maximum negative

value is known as Peak to Peak value. Peak to Peak value = 2 * Peak value

3.3 Generation of Alternating Voltage and Current

According to Faradays law of electromagnetic induction when a coil rotates in a magnetic field, an e.m.f isinduced in it, which is given by

dt

dNe

= (3.03)

An alternating voltage may be generated:(a) By rotating a coil at constant angular velocity in a uniform magnetic field as shown in Fig 3.4.

Fig 3.4

(b) By rotating a magnetic field at a constant angular velocity within a stationary coil as shown in Fig 3.5. Ifa machine is constructed to rotate a magnetic field around a set of stationary wire coils with the turning

of a shaft, AC voltage will be produced across the wire coils as that shaft is rotated, in accordance with

Faraday's Law of electromagnetic induction. This is the basic operating principle of an AC generator,also known as an alternator (Fig 3.5)


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(a) (b)

(c) (d)

Fig 3.5

Let us consider the first case in which a coil is rotated in the stationary magnetic field.

As the coil is rotated in anticlockwise direction, it cut the lines of magnetic field and intern e.m.f is generated inthe coil.

When the coil (abcd) is parallel to the magnetic field minimum magnetic field lines are cut, and hence no e.m.f

is induced in the coil. As the coil moves in anti-clockwise direction, the angle between the coil and the magneticfield lines increases. Hence flux cut will be more and more e.m.f is induced in the coil. When the coil is

perpendicular to the magnetic field, maximum lines are cut and hence maximum e.m.f is generated.

Let the generating system have the following specifications:

Number of turns of the coil = NArea of the coil = A

Flux density of the Magnetic Field = B

Angular speed of the coil = The flux linked with the coil =

The flux linked with the coil can be expressed as

cos. BAAB == (3.04)

As = t, equation (3.04) will becometBA cos= (3.05)

From equation (3.05), equation (3.03) will become

dt

tBAdNe

)cos( = (3.06)

Or tsinNBAe = (3.07)The generated e.m.f will be maximum when

1sin =t (3.08)

And that maximum e.m.f is given byNBAEm = (3.09)

From equation (3.09), equation (3.07) will becometsin mEe = (3.10)


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This is an equation of alternating sinusoidal induced voltage.

Where e = Instantaneous value of Induced Voltage.

Em = Maximum value of Induced Voltage.

3.4 Average Value of Alternating QuantitiesMathematically average value is given by

Base

curveunder theAreavalueAverage = (3.11)

The average value of an alternating quantity (whether it is sinusoidal or square or triangular) if calculated over

a complete cycle will be zero because it has identical positive and negative half cycles. Hence the average valueof an alternating quantity is calculated over a half cycle.

The Sinusoidal alternating current is given bytIi m sin= (3.02)

Fig 3.6

For an alternating Sinusoidal quantity

alterationoneoflengthbase

nalternatiooneofArea)value(IAverage av = (3.12)

=

0

)(1

tdi (3.13)

Or

=

0

)()sin(1

tdtIm

(from equation (3.02)) (3.14)

m

m

av

II

I 637.02

==

(3.15)

Similarly for Sinusoidal alternating voltage the average value is given by

m

m

avV

VV 637.0

2==

(3.16)

3.5 Root Mean Square (r.m.s) or Effective ValueThe effective or r.m.s value of an alternating current is that steady current (DC) which when flowing through

a given resistance for a given time produces the same amount of heat as produced by the alternating current

(AC) when flowing through the same resistance for same time.


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For example when we say that the r.m.s or effective value of an alternating current is 5A, it means that the

alternating current will do work or produce heat at the same rate as 5A direct current under similar conditions.

The rms value of non-sinusoidal quantities can be calculated by Graphical method

n

iiiiI

n

rms

22

3

2

2

2

1 ++= (3.17)

The r.m.s value of Sinusoidal quantities over a complete cycle can be calculated by Integral method2/1

2

0

2)(

2

1

=

wtdiIrms (3.18)

From equation (3.02),tIi m sin= (3.02)

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0

2

m )(t)sinI(2

1

=

tdIrms (3.19)

Or

2/12

0

22

m )(tsinI

2

1

=

wtd (3.20)

Or

2/12

0

2

)(2

t)cos2-(1

2

=

tdIm (3.21)

Or

2/12

0

2

2

2sin

4

=

tt

Im (3.22)

Or ( )2/1

2

002

4sin2

4

=

mI

(3.23)

OR2

m

rms

II = (3.24)

Similarly for Sinusoidal alternating voltage the rms value is given by

2

mrms

VV = (3.25)

3.6 Form Factor

Form factor of an Alternating quantity is defined as the ratio of r.m.s value of an Alternating quantity to the

maximum value of that Alternating quantity.

valueAverage

valuesmrKf...

= (3.26)

For a Sinusoidal alternating current

Kf = m

m

I

I

637.0

707.0

(3.27)Or Kf = 1.11 (3.28)

From the definition of form factor it is clear that form factor enable us to find the r.m.s value of an sinusoidal

alternating quantity if average value of an sinusoidal alternating quantity is given or vice versa.Note: It is to noted that the constant value of the form factor (K f = 1.11) is for sinusoidal alternating

quantity only and not for any other alternating quantity.


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3.7 Peak Factor

Peak factor is also known as crest factor or amplitude factor. It is defined as the ratio of maximum value of an

alternating quantity to the rms value of that alternating quantity

valuer.m.s.

valueMaximumKp = (3.29)

For a Sinusoidal AC current

KP= mm

I

I

707.0 (3.30)

Or KP = 1.414 (3.31)

Note: - It is to noted that the constant value of the Peak factor (K p = 1.414) is for sinusoidal alternating

quantity only and not for any other alternating quantity. Notations of these factors may vary from book

to book.The peak factor is of much greater importance because it indicates the maximum voltage being applied to the

various parts of the apparatus. For instance when an alternating voltage is applied across a cable, the breakdown

of the insulation will depend upon the maximum voltage. The insulation must be able to withstand themaximum value of the voltage rather then the r.m.s value.

3.8 Power in AC CircuitsSuppose that the voltage and current in a circuit are out of phase and the phase difference between them be

angle .

Apparent Power

The product of the applied voltage and total current is known as apparent power. It is also known as real power.

Apparent Power = V I Volt Ampere (VA) (3.32)

Active Power

The power actually used in the circuit (or dissipated in the form of heat) is known as Active Power. It is alsoknown as true power.

Active Power = V I cos Watt (W) (3.33)

Reactive PowerPower which is returned to the source by reactive components (for e.g. inductor, capacitor etc) in the circuit is

called reactive Power.

Reactive Power = V I sin Volt Amp. Reactive (VAR) (3.34)

3.9 Complex Notation

All the phasors related to electrical quantities can be resolve in to two components namely horizontal

component and vertical component. Accordingly these phasors can be expressed in four forms:

Polar Form

Rectangular Form

Trigonometric form

Exponential formOut of these four types, two are the most important and commonly used forms to express the phasors. They are

Polar Form and Rectangular Form. These will be discussed in detail in the next articles.

3.9.1 Polar Form

The voltage and current phasor need not always be along X orY axis as indicated in Fig. 3.11.

V


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I

Fig 3.11

They may be at an inclined direction also (as in case of RL, RC and RLC series circuits discussed in the

previous sections). Let us consider four different types of voltage equations.

v1 = V1 sin t where 1 = 00

(3.35)

v2 = V2 sin (t + 45o) where 2 < 90

0 (3.36)

v3 = V3 sin (t + 90o) where 3 = 90

0 (3.37)

v4 = V4 sin (t 60o) where 4 is negative (3.38)

In Fig. 3.12 these four phasors are represented.

Fig. 3.12

These equations in polar form can be represented as

011

= Vv (3.39)

oVv 4522 = (3.40)

oVv 9033 = (3.42)

oVv 6044 = (3.43)

3.9.2 Rectangular FormAny inclined phasor can be resolved into two components, one along the horizontal or X-axis and the otheralong the vertical or Y-axis.

Phasorv1equation (3.39) is represented in polar form by its magnitude and direction as

011 = Vv (3.39)

Horizontal component ofv1

111 )0cos( VVvo

h == (3.44)

Vertical component ofv1

0)0sin(11 ==o

r Vv (3.45)

It has horizontal component alone. It is along positive X-axis, and is considered as the reference phasor.

Phasorv2equation (3.40)is in the I quadrant, leading v1 by an angle 45o and can be represented in polar form as

oVv 4522 = (3.40)


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)45cos(22o

h Vv = (3.46)


)45sin(22o

r Vv = (3.47)

Phasorv3equation (3.42)is leading v1 by an angle 90o and can be represented in polar form as

oVv 9033 = (3.42)


0)90cos(33 ==o

h Vv (3.48)


333 )90sin( VVvo

r == (3.49)

Phasorv3 is along positive Y-axis. It has only a vertical component.

Phasorv4 equation (3.43)is an inclined phasor shown in IV quadrant lagging reference phasor v1 by an angle

60o. It can be represented in polar form as

oVv 6044 = (3.43)


)60cos()60cos( 444oo

h VVv == (3.50)


)60sin()60sin( 444oo

r VVv == (3.51)Therefore, in general, any phasor is said to have:

Either a horizontal component as v1 or

Vertical component as v3or

Both horizontal and vertical components as phasors v2and v4To distinguish a vertical component from horizontal component the complex operator j is introduced.The operator j is added along with the vertical component of the phasor. Its value is

1901 =+=j (3.52)

118012 =+=j (3.53)

jj==+=

9012701

3

(3.54)10136014 ==+=j (3.55)

jj

== 9011

(3.56)

With the complex operator j, the four phasors v1, v2, v3andv4shown in Fig. 1.6 can be represented in a clearcomplex form i.e. rectangular form as

)01()0sin0(cos0 1111 jVjVVv +=+== (3.57)

)2

1

2

1()45sin45(cos45 2222 jVjVVv

o +=+== (3.58)

)0()90sin90(cos90 3333 jVjVVvo +=+== (3.59)

)2

3

2

1())60sin()60(cos(60 4444 jVjVVv

o=+== (3.60)

Conversion facility from the polar form to the complex (or rectangular) form and vice versa are available in

most of the calculators so that one need not always multiply the modulus value with sin and cosine functions.

Once the phasors are represented in the complex form, then their addition, subtraction, multiplication and

division are made as simple as with normal numbers.The only change is that the calculator mode of operation has to be changed to the complex mode and each

phasor is to be provided with two values within a set of brackets, an active or real value and another reactive or

imaginary value along with the complex operator j or sometimes represented as i.


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3.10 Resultant of Alternating QuantitiesWhen a number of similar alternating quantities are acting simultaneously in an AC circuit, then their combined

effect or the resultant can be calculated by the following methods:(i) Trigonometric method

(ii) Polar method

3.10.1 Trigonometric MethodLet the following alternating current are acting simultaneously in an AC circuit

ti sin201 = where 1 = 00

(3.61)

)45(sin150

2 += ti where 2 < 900

(3.62)

)135(sin250

3 += ti where 3 > 900

(3.63)

)60(sin100

4 = ti where 4 is negative (3.64)

)45(cos20 02 += ti where 5 is angle of cosine (3.65)

Step1

First step to find the resultant of an alternating quantity is that if any equation is given in cosine form convert it

into sine form. For example in the above equations, equation (3.65)is in cosine form.Now when we convert any Cosine equation in to Sine Equation, there is a problem with the polarity sign. Let us

draw the Cosine equation and Sine Equation:

Fig 3.13 Cosine Equation Fig 3.14 Sine Equation

Sine Equation will start from zero position or from the origin (as Sin 0 = 0) and Cosine Equation will have a

peak at zero position (as Cos 0 = 1). It means that a Cosine equation peak appear first as compare to Sine

Equations peak. That is the Cosine equation is in leading position to that of the Sine equation. Hence when weconvert a Cosine equation in to Sine equation then the Sine equation must lag by 900.

i.e. if we consider a Cosine equation as

v = 20cos (t) (3.66)Then its sine conversion will be

v = 20 sin (t - 900) (3.67)

And to remember

)90(0

= SinCos (3.68)

Now changing equation 3.65

i5 = 20cos (t + 45

o

) (3.69) = 20 sin (t + 45o - 900)

i5 = 20 sin (t - 45o) (3.70)

Step2

Draw the phasor diagram of all the above equations (Fig 3.15)


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Fig 3.15

Step3

Find X component

X = I1 cos 1 + I2 cos 2 + I3 cos 3 + I4 cos 4 + I5 cos 5 (3.71)

X =20cos (0o) +15cos (45

o) + 25cos (135o)

+10cos ( 60o) + 20cos (-45

o) (3.72)

= 20 + 10.61 17.68 + 5 + 14.14

X = 32.07 (3.73)Step4Find Y component

Y = I1 sin 1 + I2 sin 2 + I3 sin 3 + I4 sin 4 + I5 sin 5 (3.74)

Y =20sin (0o) +15sin (45

o) + 25sin (135o)

+10sin ( 60o) + 20sin (-45

o)

= 0 + 10.61 + 17.68 8.66 7.07

Y = 12.56 (3.75)

Step5

Find magnitude of the resultant vector

Im =2222 )56.12()07.32( +=+YX (3.76)

Im = 34.44 Ampere (3.77)Step6

Find angle of the resultant vector

07.32

56.12tantan

11 ==x

y (3.78)

= 21.39o (3.79)

Step7

The resultant of all the above current phasor will be

i = Im sin (t ) (3.80)

From equation (3.77) and (3.79)

i = 34.44 sin (t + 21.39o) (3.81)

[Note: For subtraction of vectors:Suppose we have to find

e1 + e2 e3 + e4 (3.82)i.e. we have to add e1, e2 and e4 and we have to subtract e3 from their sum.

In such case, after drawing the phasor diagram, draw another phasor diagram in which invert or shift the vector

(which is/are to be subtracted) by 1800 and take its magnitude negative].

3.10.2 Polar MethodConsidering the same equations of trigonometric method:

ti sin201 = where 1 = 00

(3.61)


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)45(sin15 02 += ti where 2 < 900 (3.62))135(sin25 03 += ti where 3 > 90

0 (3.63)

)60(sin100

4= ti where 4 is negative (3.64)

)45(cos20 02 += ti where 5 is angle of cosine (3.65)

Step1

First if any equation is given in cosine form convert it into sine form. For example in the above equations,

equation (3.65)is in cosine form.

i5 = 20cos (t + 45o) (3.65)= 20 sin (900 +t + 45o)

i5 = 20 sin (t - 45o) (3.83)

Step2

Now the above alternating quantities can be converted in to polar form and then in rectangular form as:

i1 = 20o

0 (3.84)

= 20 (cos 0o +jsin 0o) = 20 +j0 (3.85)

i2 = 15o

45 (3.86)

= 15 (cos 45o +jsin 45o)= 10.61 + j10.61 (3.87)

i3 = 25o

135 (3.88)

= 25 (cos 135o +jsin 135o) = -17.68 +j17.68 (3.89)

i4 = 10o60 (3.90)

= 10 (cos (-60o) +jsin (-60o)) = 5 - j8.66 (3.91)

i5 = 20o

45 (3.92)

= 20 (cos 135o +jsin (- 45o)) = -14.14 - j14.14 (3.93)

Step3

The resultant of the above alternating quantities is

i = i1 + i2 + i3 + i4 + i5 (3.94)

= (20 +j0) + (10.61 + j10.61) + (-17.68 +j17.68)

+ (5 - j8.66) + (-14.14 - j14.14)= 3.79 +j5.49 (3.95)

Or i= 6.67

55.38

o

(3.96)Or i = 6.67 sin (t + 55.38o) (3.97)

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