81

AC CircuitsUNIT 4 AC CIRCUITS

Structure 4.1 Introduction

Objectives

4.2 Introduction to Alternating Quantity

4.3 RMS and Average Value 4.3.1 RMS Value of Complex Wave 4.3.2 Average Value 4.3.3 RMS Value of Sinusoidal Wave 4.3.4 Average Value of Sinusoidal 4.3.5 Form Factor 4.3.6 Peak, Crest or Amplitude Factor

4.4 Complex Representation of Alternating Quantities

4.5 Resistive, Inductive and Capacitive Circuits 4.5.1 Series RLC Circuit 4.5.2 Parallel RLC Circuit

4.6 Concept of Power 4.6.1 Power in AC Circuit 4.6.2 Power Triangle

4.7 Power Factor

4.8 Quality Factor (Q-Factor)

4.9 Series and Parallel Resonance 4.9.1 Series Resonance 4.9.2 Parallel Resonance

4.10 Three Phase Circuit 4.10.1 Connections in Three Phase System 4.10.2 Voltage, Current and Power Relation 4.10.3 Conversion from Star Connection to Delta Connection

4.11 Maximum Power Transfer Theorem

4.12 Summary

4.13 Answers to SAQs

4.1 INTRODUCTION

This unit deals with the concept of AC circuit analysis. Unit starts with the basic definitions about the alternating quantity and then covers the concept of RMS and average value, form factor and peak factor.

At the latter stage of unit, we give the idea of resistance, inductive and capacitive circuits, calculation for their voltage, current, power and power factor. We also introduce three types of powers – active, reactive and apparent power. Detail discussions about the series and parallel RLC circuits is also given in this unit.

The unit also covers the concept of series and parallel resonance with sufficient description and examples.

Analysis of balanced three phase circuit, star-delta conversion and maximum power transfer theorem are also the part of this unit.

82

Objectives Electrical

After studying this unit, you should be able to

• explain the concept of alternating quantity,

• define the RMS and average values with form and peak factor,

• determine the voltage, current and power for pure resistive, inductive, capacitive circuits and circuit having their combinations,

• define the power factor and quality factor,

• distinguish between apparent, active and reactive powers,

• explain the concept of series and parallel resonance,

• distinguish between different three phase connection,

• determine the current, voltage and power in star and delta connected three phase balanced circuits,

• convert the star connection to delta and vice-versa, and

• apply the maximum power transfer theorem and Thevenin theorem in ac and dc circuits.

4.2 INTRODUCTION TO ALTERNATING QUANTITY

Quantity which varies periodically with time is known as the alternating quantity. It may be voltage or current. Some waveform of alternating quantities are shown in Figure 4.1.

0/2 2

ωt

(a) Sine Wave

0

f (t)

TT/2

t

(b) Triangular Wave

0

f (t)

TT/2t

(c) Square Wave

Figure 4.1 : Alternating Waveforms

83

AC Circuits4.2.1 Basic Definitions about the Alternating Quantity Here we take the example of sinusoidal form of alternating current.

0/2 2

IM

i = IM sin ωt

ωt

Figure 4.2 : Sinusoidal Waveform

Cycle It is the one complete set of positive and negative half of any alternating quantity.

Time Period It is the time required in second to complete one cycle of any alternating quantity.

Frequency Number of cycles per second is known as the frequency of alternating quantity. It is the reciprocal of the time period and its unit is cycle per second or Hertz.

f = 1/T (in Hertz) In India, the frequency of alternating supply is 50 Hertz, while in America and Japan it is 60 Hertz.

Amplitude

It is the peak or maximum value of the alternating quantity. In Figure 4.2, Im denotes the amplitude of the current wave.

Equation of Sinusoidal Alternating Quantity The sinusoidal alternating current is written as

. . . (4.1) sinmi I t= ω

ω

Similarly, we can write for voltage

sinmv V t= . . . (4.2)

where i, ν = Instantaneous value of current and voltage Im, Vm = Maximum values

ω = 2πf [f = frequency in Hz]

= 2Tπ [T = Time period in sec] . . . (4.3)

Phase Phase of any alternating quantity shows the position of wave at any time after it has passed through the zero position of reference.

Phase Angle It is the angle of alternating quantity with respect to the reference position.

Phase Difference When the maxima and minima of two sinusoidal alternating quantities (of same frequency) do not occur at the same instant of time, then these two quantities are said to have phase difference.

84

Electrical

ωt

(a) Reference Waveform

ωt

(b) Lagging Waveform by Angle φ

ωt

(a) Leading Waveform by Angle φ

Figure 4.3

Lagging Current by Angle φ

If the current waveform of Figure 4.3(a) is assumed to be reference wave, then Figure 4.4(a) shows the lagging current.

i

Im

ωt

Figure 4.4(a) : Reference Current Waveform

sin ( )mi I t= ω − φ . . . (4.4)

This equation shows that current is lagging (or running behind) the reference wave by an angle φ.

Leading Current by an Angle φ Figure 4.4(b) shows the leading current by an angle φ by the reference wave of Figure 4.4(a).

sin ( )mi I t= ω + φ . . . (4.5)

85

i

Im

AC Circuits

ωt

Figure 4.4(b) : Leading Current Waveform

4.3 RMS AND AVERAGE VALUE

These two values represent the direct value (steady values) of any alternating quantity and defined as : RMS Value (Root Mean Square Value

The root mean square current is given by that equivalent steady (dc) current which will produce the same amount of heat in a given circuit as could be produced by the alternating current when flowing through the same circuit for the same time period. Consider the circuit shown in Figure 4.5. First we connect the Direct Source (DC) to the resistor for the particular time t and find the heat i2 Rt joules. Then for the same time t, we apply the alternating current from AC source v and again calculate the heat transferred to the resistor. Now, the RMS value of current i is given by,

2

0

1 [ ( )]= ∫T

rmsI i tT

dt . . . (4.6)

T = Time period of current wave. RMS value is also known as the effective or virtual value.

Figure 4.5 : A Typical Circuit

4.3.1 RMS Value of Complex Wave Let the complex current is given by,

0 1 2 3cos cos 2 sin2m m mi I πI t i t I t⎛ ⎞= + ω + ω + ω −⎜ ⎟

⎝ ⎠

Then the RMS value is given by

2 2

2 1 2 30 2 2 2

m m mrms

I I II I ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 . . . (4.7)

86

4.3.2 Average Value Electrical

The average value of alternating current is defined by that steady value of current which transfers the same amount of charge across any given circuit as could be transferred by the alternating current for the same time.

0

1 ( )T

avI i tT

= ∫ dt . . . (4.8)

Note : Waveforms with half wave symmetry, i.e.

1( )2

f t f t T⎛ ⎞= +⎜ ⎟⎝ ⎠

where T = time period.

Then the average value of such waves for the complete time period is zero (because positive and negative halves are equal in one time period). For such waves, the average value is computed over the positive half of the period. This is sometimes called the half cycle average. (All the waveforms of Figure 4.1, have such symmetry).

4.3.3. RMS Value of Sinusoidal Wave Consider the waveform of Figure 4.2. The time period is 2π radian, i.e. T = 2π.

From Eq. (4.6) the RMS value is given by

2

0

1 [ ( )]T

rmsI i tT

= ∫ dt

Here, we integrate with respect to ωt or angle.

So, 2

2

0

1 [ ( )] ( )2rmsI i t

π= ω

π ∫ d t

2

2 2

0

1 sin ( )2 mI t d

π= ω

π ∫ tω

22

2

0

1 cos 2 1 cos 2( ) sin2 2 2mI t td t t

π − ω − ω⎡ ⎤ ⎡= ω ω = ⎤⎢ ⎥ ⎢π ⎥⎣ ⎦ ⎣∫ ∵

⎦

22

0

sin 2 0.7072 2 2m m

mI Itt

πω⎡ ⎤= ω − = =⎢ ⎥π ⎣ ⎦I . . . (4.9)

4.3.4 Average Value of Sinusoidal Wave Since the waveform has the half wave symmetry the average value for time period T = 2π will be zero. So the average value will be calculated for half time period.

From Eq. (4.8), the average value is given by

0 0

1 1( ) sin ( )T

av mI i t dt I t dT

π= = ω

π∫ ∫ tω

02

[ cos ] 0.637

2

m m mm

I I It π= − ω = = =

ππ πI . . . (4.10)

87

AC Circuits4.3.5 Form Factor The form factor is defined as

RMS valueAverage valuefK = . . . (4.11)

For sinusoidal wave,

0.707

1.10.637

mf

m

IK

I= =

4.3.6 Peak, Crest or Amplitude Factor

Maximum valueRMS valuefK = . . . (4.12)

For sinusoidal wave,

2 1.414

2

mf

m

IK I= = =

4.4 COMPLEX REPRESENTATION OF ALTERNATING QUANTITIES

Alternating quantities are represented by complex number.

Rectangular or Cartesian Representation

Any alternating quantity can be represented in complex form with real and imaginary part : Like

i a jbc jd

= + ⎫⎬ν = + ⎭

. . . (4.13)

where 1j = − , which gives the rotation through 90o in anticlockwise direction,

a = real part, and

b = imaginary part.

Representation of Alternating Quantity on Complex Plane

In complex plane, x-axis represents the real axis (reference axis) and y-axis represents the imaginary axis ( j-axis). Figure 4.6 shows the representation of complex quantity on complex plane.

Im

a + jb

0Re

a

b

Figure 4.6 : Complex Plane

88

Polar Representation Electrical

Any alternating quantity can be represented in polar form (magnitude – angle form) as under

1 1

2 2

i rr

= ∠θ ⎫⎬ν = ∠θ ⎭

. . . (4.14)

Rectangular to Polar Conversion

2 21r a b= +

and 11 tan b

a−θ = . . . (4.15)

Also 2 22r c d= +

and 12 tan d

c−θ = . . . (4.16)

In this way, rectangular quantities a + jb and c + jd can be converted into corresponding polar form, i.e.

1r 1∠θ and 2 2r ∠θ

In Figure 4.7

2 21r OA a b= = +

and θ1 is the angle made of line OA from the horizontal axis are show, A

r1

O θ1

Figure 4.7

Polar to Rectangular Conversion

1r ∠θ1 can be written as :

11 ,jr e θ

Then by Euler’s theorem :

[ ]11 1 1cos sinjr e r j 1θ = θ + θ

1 1 1cos sinr j r 1= θ + θ . . . (4.17)

So, 1 1cosa r= θ

and b r1 1sin θ =

Example 4.1 Find RMS and average value of the output of half wave rectifier.

Solution The waveform of current, output of the half wave rectifier is shown in Figure 4.8.

Figure 4.8

89

AC CircuitsTime period of wave = 2π

RMS value of current

2

2

0 0

1 ( ) 0 ( )2rmsI i d t d t

π π⎡ ⎤⎢ ⎥= ω +

π ⎢ ⎥⎣ ⎦∫ ∫ ω

2

0

1 sin ( ) ( )2 mI t d

π⎡ ⎤⎢ ⎥= ω

π ⎢ ⎥⎣ ⎦∫ tω

2

0(1 cos 2 ) ( )

4mI t d t

π⎡ ⎤⎢ ⎥= − ω

π ⎢ ⎥⎣ ⎦∫ ω

0.52m

rms mI

I I= =

Average value of current

0 0

1 1( ) sin ( )2 2

mav m

II i d t I t d t

π π= ω = ω ω =

π π∫ ∫ π

Example 4.2

Find the form factor and peak factor for the above half rectified wave.

Solution

From the above example,

0.5rms mI I=

mav

II =

π

∴ Form factor 0.51.57rms m

mav

I III

= = =

π

and peak factor 20.5

m m

rms m

I II I

= = =

Example 4.3

For the output of full wave rectifier, determine

(a) RMS value,

(b) Average value,

(c) Form factor, and

(d) Peak factor.

Solution

The output current of full wave rectifier is shown in Figure 4.9.

Figure 4.9

90

Electrical Time period = π

(a) 2 2 2

0 0

1 1( ) sin ( )rms mI i d t I t dπ π

t= ω = ω ωπ π∫ ∫

22

0(1 cos ) ( ) 0.707

2 2m m

rms mI I

I t d tπ

= − ω ω = =π ∫ I

(b) [ ]00

21 sin ( ) 1 cos 0.637m mav m m

I II I t d t t

ππ= ω ω = − ω = =

π π π∫ I

(c) Form factor 0.707

1.10.637

rms mf

av m

I IK

I I= = =

(d) Peak factor 1.4140.707

m mp

rms m

I IK

I I= = =

Example 4.4

For the saw tooth wave shown in Figure 4.10, determine the form factor.

Figure 4.10 : Saw Tooth Wave

Solution

Time period of given wave is 2 sec. Expression of function f (t) for first time period.

10( ) 5 for 0 22

f t t t t= = ≤ ≤

RMS Value

2 22 2

0 0

1 1 25( ) (5 )2 2

T

rms2

0F f t dt t dt t dt

T= = =∫ ∫ ∫

23

0

25 5.772 2rms

tF = =

Average Value

22 2

0 0 0

1 1 5 5( ) 5 (4) 52 2 2 4

T

avrF f t dt t dt

T= = = =∫ ∫ =

Form Factor

5.77 1.1545

rmsf

av

FK

F= = = .

91

AC CircuitsSAQ 1 (a) Find the rms and average value of the waveform shown in Figure 4.11,

where in the first interval . Also find the form factor. 200( ) 10 tf t e−=

Figure 4.11

(b) Find the effective (or RMS) value for the average function.

( ) 20 30 sin 60 sin 22

v t t t π⎛ ⎞= + ω + ω −⎜ ⎟⎝ ⎠

4.5 RESISTIVE, INDUCTIVE AND CAPACITIVE CIRCUITS

AC through Pure Resistive Circuit

Applied voltage, . . . (4.18) sinmv V t= ω

The resultant current, i

sin sinmm

VVi t IR R

= = ω = ωt . . . (4.19)

Current, i is in phase with applied voltage, v. The waveforms and phasor diagram are shown in Figures 4.12 and 4.13, respectively.

Figure 4.12 : Resistive Circuit

Figure 4.13 : Current and Voltage Waveforms

AC through Pure Inductive Circuit The circuit is shown in Figure 4.14. Again the applied voltage is given by

sinmv V t= ω

div Ldt

=

92

Electrical Now, sinmv V t= ω

∴ sinmdiV t Ldt

ω =

∴ sinmVdi t dt

L= ω

Figure 4.14 : Inductive Circuit

Integrating both sides, we get

sinmVi t

L= ω∫ dt

( cos ) cosm mV Vt t

L L= − ω = − ωω ω

sin2

mVt

Lπ⎛ ⎞= ω −⎜ ⎟ω ⎝ ⎠

sin2

m

L

Vt

Xπ⎛ ⎞= ω −⎜ ⎟

⎝ ⎠ . . . (4.20)

where XL = inductive reactance = ωL in ohm, if we put mm

L

VI

X= , then

sin2mi I t π⎛ ⎞= ω −⎜ ⎟

⎝ ⎠ . . . (4.21)

Here, we can see that current lags the applied voltage by angle 2π . The waveform

and phasor diagram are shown in Figure 4.15.

(a) Waveform (b) Phasor Diagram

Figure 4.15

AC through Pure Capacitor

The current passing through capacitor is given by

=dvi Cdt

. . . (4.22)

( sin )= ωmdvC Vdt

t

cos= ω ωmC V t

93

AC Circuits sin1 2

π⎛ ⎞= ω +⎜ ⎟⎝ ⎠

ω

mVt

C

sin2π⎛ ⎞= ω +⎜ ⎟

⎝ ⎠m

C

Vt

X . . . (4.23)

where 1=ωcX

C = Capacitive reactance in ohm.

Figure 4.16 : Pure Capacitive Circuit

If we put = mm

C

VI

X, then

sin2mi I t π⎛ ⎞= ω +⎜ ⎟

⎝ ⎠ . . . (4.24)

Here, we can see that current i leads the applied voltage by 2π .

The waveform and phasor diagram are shown in Figure 4.17.

(a) Waveform (b) Phasor Diagram

Figure 4.17

4.5.1 Series RLC Circuit The applied voltage is sinmv V t= ω

Inductive reactance LX L= ω

Capacitive reactance 1=ωCX

C

Figure 4.18 : Series RLC Circuit

94

Electrical

)

Here, we define the impedance of the circuit which is given by

( L CZ R j X X= + − in ohm . . . (4.25)

The magnitude of the impedance is

2 2| | ( )L CZ R X X= + −

and angle 1tan L CX XR

− −θ =

This is the angle between the applied voltage v and the current i. θ is also known as the power factor angle of the circuit.

The current i is given by

2 2 1

sin

tan−ων

= =+ ∠

mV ti XZ R X

R

where X = XL – XC

or 1sin tan| |

mV Xi tZ R

−⎛ ⎞= ω −⎜ ⎟⎝ ⎠

. . . (4.26)

If the circuit in inductive (XL > XC) then current i lags the v by angle 1tan XR

− . But if the

circuit is capacitive then i leads the v by angle 1tan XR

− .

The voltage drop across each passive elements

VR = i R, VL = XL i, VC = XC i

Also R L Cv v v v= + + (Kirchhoff’s Voltage Law) . . . (4.27)

In general, if we apply Kirchhoff’s voltage law in series RLC circuit then KVL equation is given by

1div R i L i dtdt C

= + + ∫ . . . (4.28)

4.5.2 Parallel RLC Circuit Now we connect passive elements R, L, C in parallel along with one current source.

Figure 4.19 : Parallel RLC Circuit

Since the circuit is parallel the voltage v across all the elements remain same. The source current i is distributed along the three elements by Kirchhoff’s current law.

R L Ci i i i= + +

in general,

95

AC Circuits 1v di v dt vC

R L d= + +∫ t

. . . (4.29)

Here, we define conductance, susceptance and admittance. Conductance G = Reciprocal of resistance

1 in mho ( )R

=

Inductive susceptance 1 ( )LL

BX

=

Capacitive susceptance 1 ( )CC

BX

=

Admittance Y = Reciprocal of impedance

1Z

= . . . (4.30)

4.6 CONCEPT OF POWER

We know that power is defined as the time rate of doing work and is given by P. dwPdt

= . . . (4.31)

dw dqdq dt

× v i= ×

Energy (Total work done) is given by

joulesw P dt= ∫ or Watt-sec . . . (4.32)

4.6.1 Power in AC Circuit Power in AC circuit is classified as apparent power, active power and reactive power : Apparent Power

Product of voltage and current. Its unit is volt-amp. S = VI Volt-Amp = I 2 Z . . . (4.33)

Apparent power is an important parameter in the specification of electrical equipment, as the size and cost of many electrical machines depends on their Volt-Amp rating rather than voltage rating.

Active Power or True Power It is given by P = VI cos θ Watt . . . (4.34) It is the power, which is actually consumed in any circuit. Its unit is Watt. The above formula indicates that the power received by a load is not merely the product of its rms values of its terminal voltage and current but includes an additional multiplicative factor cos θ, called power factor of the load.

Reactive Power It is the power taken by the reactance and its unit is volt-amp reactive (VAr). It is given by

Q = VI sin θ VAr . . . (4.35) Q is called 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.

96

4.6.2 Power Triangle Electrical

All the three power S, P and Q can be represented in a triangle, called the power triangle as shown in Figure 4.20.

S=VI

S=VI

P=V I cosθ

Q=V I sinθ

θ

S=VI

P=V I cosθ

Q=V I sinθ

θ

(a) For Lagging or Inductive Load (b) For Leading or Capacitive Load

Figure 4.20 : Power Triangle

Power can also be represented in complex form. S = P + jQ . . . (4.36)

Magnitude 2 2S P Q= +

4.7 POWER FACTOR In the expression for power P = VI cos θ, the term cos θ is known as power factor, its value lies between 0 to 1. It may be defined as

• Cosine of the angle of lead or lag between current phasor and voltage phasor,

• The ratio True Power Watts=Apparent Power Volt amperes

. . . (4.37)

• The ratio ResistanceImpedance

RZ= . . . (4.38)

For a constant load power P and system voltage V, current and VI requirement 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 voltage drop and power losses in supply lines due to the increased line current. High value of p.f is a desired operating condition in power supply system.

4.8 QUALITY FACTOR (Q-FACTOR) Q-factor is the reciprocal of the power factor of coil. It is the figure of merit, i.e. it should be as large as possible for any coil (combination of resistance and inductance).

Maximum energy stored2Energy dissipated per cycle

Q = π . . . (4.39)

or 1cos

ZQR

= =φ

. . . (4.40)

2 2+

= LR XL

R

If R is very less as compared to XL, then

LX LQR R

ω= = . . . . (4.41)

97

AC CircuitsExample 4.5

In the given circuit, find the active, reactive and apparent power.

220 0 ;60Hz∠ o

i

0.2H15Ω

Figure 4.21

Solution

The inductive reactance, XL = 2π fL

= 2π × 60 × 0.2

= 75.408 Ω

Impedance, Z = 15 + j 75.408 Ω

In polar form = 76.885 ∠ 78.749o Ω Current in the Circuit

o

o220 0

76.885 78.749viZ

∠= =

∠

= 2.861 ∠ − 78.749o Amp (lagging)

Power factor cos φ = cos 78.749o

= 0.195 (lag)

and sin φ = sin 78.749o

= 0.98 Active Power

P = VI cos φ

= 220 × 2.861 × 0.195 = 122.7 Watts

or P = I 2 R = 122.7 Watts

Reactive Power

Q = VI sin φ

= 220 × 2.861 × 0.98 = 616.83 volt-Amp reactive

or Q = I 2 X = 617.1 VA

Apparent Power S = VI

= 220 × 2.861 = 629.42 Volt-amp

or 2 2S P Q= +

= 629.2 Volt-amp.

Example 4.6

98

In series R-L circuit, shown in Figure 4.22, find Electrical

(a) Impedance,

(b) Resultant current,

(c) Power factor and its nature, and

(d) Quality factor.

220 0 ; 50Hzo

i

L = 0.1HR = 10

Figure 4.22

Solution

The inductive reactance XL is given by

XL = ωL

= 2π fL

= 2π × 50 × 0.1 Ω

= 31.42 Ω

The impedance in rectangular form is given by

Ω 10 31.42LZ R j X j= + = +

In polar form Z = 32.97 ∠ 72.345o Ω

The resultant current is given by

o

o220 0

32.97 72.345i

Zν ∠

= =∠

= 6.672 ∠ − 72.345o Amp.

The current is lagging (since the circuit is inductive).

The power factor angle, φ, is the angle between voltage current

φ = 72.345o (lagging)

∴ p.f. cos φ = cos (72.345o)

= 0.303 (lag)

Alternatively, the power factor can be determined by

cos RZ

φ =

1032.97

=

= 0.303 (lag)

Quality factor is given by φ

1= = 3cos

Q .3 .

Example 4.7

99

AC CircuitsAn inductive coil of resistance 32 Ω and reactance 15.7 Ω is connected in series with a capacitor of reactance 79.5 Ω. The circuit is connected across 500 V AC supply, determine (a) Current, (b) Phase difference between voltage and current, (c) Magnitude of voltage across the inductive coil, and (d) Total power absorbed.

Solution

i

R=32 x =15.72 x =79.5c

v = 500V Figure for Example 4.23

R = 32 Ω, X = XL – XC

= – 63.8 Ω (circuit in capacitive)

Impedance Z = R – jX = 32 – 63.8 j Ω

(a) Current o500 0

32 63.8viZ j

∠= =

−

= 7 63.36∠ o Amp (b) Angle between V and i

φ = 63.36 (leading) (c) Impedance of coil, ZL = R + j XL = 32 + j 15.7

| | 35.6439LZ = Ω

Voltage across coil = I ZL = 7 × 35.6439 = 249.5075 Volt

(d) Power absorbed = I2 R = 72 × 32 = 1568 watts

Example 4.8 In the capacitive circuit of Figure 4.24, find

(a) Impedance, (b) Resultant current, (c) Power factor, (d) Power absorbed by the circuit, and (e) Phasor relation between different voltages and currents.

50Hz

i

VR

C=1 FμR=120Ω

VC

100V

Figure 4.24

Solution

100

(a) The capacitive reactance is given by Electrical

1 12CX

C f= =ω π C

61

2 50 1 10−=

π × × ×

= 3182.68 Ω

So the impedance, CZ R j X= −

= 120 – j 3182.68 Ω

In polar form, Z = 3184.94 ∠ − 87.84o Ω

(b) Resultant current

o

o100 0

3184.94 87.84viZ

∠= =

∠−

= 0.03139 ∠ 87.84o Amp (leading)

(c) Power factor, cos φ = cos 87.84o

= 0.0376 (lead)

(d) Power is absorbed by resistor only and that is also known as the active power in the circuit

P = I 2 R

= (0.03139)2 × 120

= 0.1182 Watts

(e) Voltage drop across resistor :

VR = iR

= 0.03139 ∠ 87.84o × 120

= 3.7668 ∠ 87.84o Volt (in phase with current i)

Voltage drop across capacitor

( )c cV i j X= −

= 0.03139 ∠ 87.84 × 3182.68 ∠ − 90o

= 99.9 ∠ − 2.16o (lagging)

The phasor diagram is shown in Figure 4.25 and applied Voltage of 100 ∠ 0o is taken as reference.

87.84o

I

VR

VC

V=100VV V V= +R C

2.16o

Figure 4.25 : Phasor Diagram

101

AC CircuitsSAQ 2 (a) The current in a circuit is given by (4.25 + j 12) Amp, when applied voltage

is (100 + j 50) volt, determine

(i) the complete expression for the impedance,

(ii) power consumed,

(iii) apparent power, and

(iv) phase angle between the current and voltage.

(b) A sinusoidal source of e (t) = 170 sin 377 t is applied to an R-L circuit. It is found that the circuit absorbs 720 watts, when an effective current of 12 Amp flows.

(i) find the power factor of the circuit,

(ii) calculate the inductance in henry, and

(iii) compute the value of impedance.

4.9 SERIES AND PARALLEL RESONANCE

Resonance is that condition in any AC circuit at a particular frequency, when the applied voltage and resultant current are in same phase. So, at resonance the power factor of the AC circuit is unity and it behaves like a pure resistive circuit.

4..91 Series Resonance In series RLC circuit at a particular frequency when inductive reactance (XL = ωL)

becomes equal to the capacitive reactance 1CX

C⎛ =⎜ ω⎝ ⎠

⎞⎟ , then the circuit is said to be at

resonance

XL = XC . . . (4.42)

In the series RLC circuit of Figure 4.26, the circuit current I is given by

V volts, f Hz

VR

I

CLR

VL VC

Figure 4.26 : Series Resonance Circuit

VI AZ

=

where Z represents the equivalent impedance of the circuit.

1Z R j Lj C

= + ω +ω

1R j L jC

= + ω −ω

102

Electrical or ( )L CZ R j X X= + −

where 1ω andωL CX L X

C= =

R jX= +

where (XL – XC) = X (Net Reactance)

Thus Amp( )L C

V VIR j X X R jX

= =+ − +

VL

VR II

90o

VC

I90o

(across R) (across L)

(across C)

Figure 4.27 : Voltage and Current Phasor

And voltage drop across resistance, R is VR = IR (in phase with I)

Voltage drop across inductance, L is VL = I XL (leading I by 90o)

Voltage drop across capacitance, C is VC = I XC (leading I by 90o)

The phasor diagram of the variables of the entire circuit is shown in Figure 4.28. VL

VC

VR

V

(VL – Vc)

Figure 4.28 : Voltage Vector Diagram of RLC Series Circuit

Resonance condition is obtained when XL = XC in the series circuit (this is obtained by either decreasing the supply frequency when XL would decrease and XC would increase or by increasing the supply frequency making XL to increase and XC to decrease).

When XL = XC, Z = R + jO = R

Also, 0 AmpV VIZ R

= = [I0 = Current at resonance]

The p.f. (power factor) of the circuit becomes

cos 1R RZ R

φ = = = . . . (4.43)

The expression of frequency of resonance can be obtained as follows :

Let ω0 or f0 be the frequency at which XL = XC.

i.e. 00

1LC

ω =ω

103

AC Circuitsor 2

0 01 1i.e. rad/ sec

LC LCω = ω =

i.e. 01 Hz

2f

LC=

π . . . (4.44)

Voltage vector diagram at resonance is shown in Figure 4.29. VL

VC

I V

Figure 4.29 : Voltage Vector Diagram at Resonance (VL = Vc)

Impedance Curve Figure 4.30 shows variation of impedance Z, inductive reactance XL and capacitive reactance XC with frequency. At resonance XL = XC and Z = R.

R

X

0

Impe

danc

e

XL = ωL

ωω0

Xc = I/ωL

Figure 4.30 : Impedance Curve

Angle Curve Angle φ (between voltage and current) is given by

1 ( )tan L cX X

R− −

φ = . . . (4.45)

Here, φ is known as the power factor angle. Figure 4.31 shows the variation of φ with frequency.

At frequencies below ω0 the capacitive reactance is greater than the inductive reactace and the angle of the impedance is negative. If the resistance is low, the angle changes more rapidly with frequency as shown in Figure 4.31. As ω approaches zero the angle of Z approaches – 90.

0

High R

Low R+90o

-90o

φ

ω ω0

Figure 4.31 : Angle Curve

104

Admittance Curve Electrical

Admittance Y is reciprocal of impedance. Its variation with frequency is shown in Figure 4.32.

1YZ

= . . . (4.46)

Low R

Adm

ittan

ce

High R

w0 Figure 4.32 : Admittance Curve

Resonance Curve Figure 4.33 shows the variation of current with frequency. Current is maximum at

resonance frequency ω0. At rms value of current 02

I , power becomes half of its

maximum value. So the points, corresponding to 02

I on the resonance curve, are

known as the half power points. Frequency ω1 and ω2 corresponding to these two points are known as the lower and upper half power frequency.

I

I0

I / 20

ω (rad/sec)

ω1 ω0 ω2

Figure 4.33 : Resonance Curve

Band Width

The distance between upper and lower half power frequencies, measured in hertz or in rad/sec, is called the banwidth BW.

BW = ω2 – ω1 in rad/sec. . . . (4.47)

or BW = f2 – f1 in Hz . . . (4.48)

The resonant frequency ω0 is geometric mean of ω1 and ω2.

0 1 2 0 1and 2f f fω = ω ω = . . . (4.49)

4.9.2 Parallel Resonance Figure 4.34 represents a parallel resonating circuit where a coil is connected in parallel with a capacitor C and the combination is connected across an AC voltage source of variable frequency. Figure 4.35 represents the vector diagram of the given circuit.

105

R L

Coil

CIC

IL

IV

AC .f Hz

AC Circuits

Figure 4.34 : A Typical Parallel Resonating Circuit

IL

I cosL φ

I sinL φφ

IC

V

Figure 4.35 : Vector Diagram of the Parallel AC Circuit

Let IC = The current through the capacitor,

IL = The current through the coil,

I = Vector sum of IL and IC, i.e. the source current,

V = Supply voltage,

VR = Drop across R,

VL = Drop across L,

VC = Drop across C,

φ = p.f. angle of the coil (i.e. the angle of lag of IL with respect to V),

ZL = Coil impedance, and

XC = Capacitor impedance (or simple capacitor reactance).

Here CC

VIX

= . . . (4.50)

And 2 2 2 ( )

LL L

V V VIZ 2R X R L

= = =+ + ω

cos RZ

φ =

At resonance the capacitive current must be equal to the inductive part of the coil current, i.e. the imaginary components of IL and IC must cancel each other at resonance.

i.e. sinC LI I= φ

or L

C L L

XV VX Z Z

= ×

where sin ,L

L

XZ

φ =

or 2L C LZ X X=

106

Electrical Also 2

00

1 . , i.e.L LL LZ L Z

C C= × ω = =ω C

. . . (4.51)

[ω0 represents resonance frequency]

or 2 2 20

LR LC

+ ω =

or 2 2 20

LL RC

ω = −

i.e. 2

20 2

1 RLC L

ω = −

∴ 2

20 2

1 1R L RLC L CL

ω = − = − . . . (4.52)

i.e. 20

12

Lf RL Cπ

= − . . . (4.53)

If the resistance of the coil be neglected,

01 1

2 2Lf = = . . . (4.54)

L C LCπ π

Again, at resonance, since the reactive components of IL and LC balance each other, the only remaining part of the current is IL cos φ (= I)

cosLI I= φ

or, .L L

V V RZ Z ZΩ

= [ZΩ = equivalent impedance of parallel circuit.]

or, 2L

LZ LCZR R CRΩ == = L

LZC

⎡ ⎤=⎢ ⎥

⎣ ⎦∵ . . . (4.55)

Then, the equivalent impedance of the parallel resonating circuit is L/CR at resonance. This impedance is called dynamic resistance of the parallel circuit. Normally R being loss, this impedance is very high at resonance and then the current is much lower in the parallel circuit. Then, this circuit is also called rejector circuit. Different Curves at Parallel Resonance

Figure 4.36 shows admittance (Y), impedance (Z) and power factor angle (θ) curves, which show the variation of Y, Z and (θ) with frequency.

G

Y

(Y)

0

Adm

ittan

ce

High R

Impe

danc

e

Low R

0

Low R

High R+90o

-90o

(Z)

Bc = ωC

ω ω ω0ω0

BL = I/ωL ω ω0

(a) (b) (c)

Figure 4.36

BBC = ωC and BLB =1/ωL are the capacitive and inductive susceptance.

107

AC CircuitsNet susceptance B = BBC – BL B

= ωC – 1/ωL . . . (4.56)

at resonance B = 0.

Resonance Curve

Since admittance is minimum or impedance is maximum at resonance frequency ωC, so current I is also minimum at parallel resonance.

I, Z Impedance Z =L

CR

0Lagging p.f. Leading p.f.

ω ω0

Figure 4.37 : Resonance Curve for Parallel Resonance

Properties of Resonance of parallel RLC circuit

(a) Power factor is unity.

(b) Current at resonance is [V / (L / CR)] and is in phase with the applied voltage. The value of current at resonance is minimum.

(c) Net impedance at resonance of the parallel circuit is maximum and equal to (L / CR) Ω.

(d) The admittance is minimum and the net susceptance is zero at resonance.

(e) The resonance frequency of this circuit is given by

2

0 21 1

2Rf

LC L= −

π

Example 4.9

Find the resonant frequency ω0 (or f0) for the given series RLC circuit

20 Fμ2 Henry10Ω

i

v = Vm sin ωt = 50 sin ωt

Figure 4.38

Also find the expression for resonant current.

108

Solution Electrical

R = 10 Ω, L = 2 Henry, C = 20 μF

Under resonance :

XL = XC

1LC

ω =ω

∴ 01LC

ω =

36

1 0.158 102 20 10−

= =× ×

×

and 01 25.16 Hz

2f

LC= =

π

Current i is given by ViZ

=

But at resonance Z = R or 10 Ω

So sin 50 sin 5 sin Amp

10mV t ti t

Rω ω

= = = ω

Example 4.10

Find the value of R, so that the condition for resonance is fulfilled

R

5j -3j

4

Figure 4.39

Solution Admittance of first branch

1 21 5 mho ( )

5 25R jY

R j R−

= =+ +

Admittance of second branch

21 4 3 mho ( )

4 3 25jY

j+

= =−

Total admittance of the circuit :

1 2 25 4 3

2525R j jY Y Y

R− +

= + = ++

According to condition of resonance the imaginary part of the admittance must be zero

i.e. 25 3 0

2525R−

+ =+

109

AC Circuits

25 3

2525R=

+

⇒ = 4.082 ΩR

Example 4.11

A circuit shown in Figure 4.40 having a resistance of 5 Ω, an inductance of 0.4 H and a variable capacitance in series is connected across a 110 V, 50 Hz supply. Calculate :

(a) The value of capacitance to give resonance

(b) Current

(c) Voltage across the inductance

(d) Voltage across the capacitance

(e) Q-factor of the circuit.

Solution

Applied Voltage, V = 110 V

Resistance, R = 5 Ω

Inductance, L = 0.4 Henry

Resonant frequency, f0 = 50 Hz

As Resonant frequency 01

2f

LC=

π

C

110 V, 50 Hz

0.4 H5Ω

i

Figure 4.40

(a) At resonant frequency capacitance

2 2 2 20

1 14 4 (50) 0.4

Cf L

= =π π × ×

= 2.53303 × 10− 5

= 25.3303 μF

(c) At resonant condition 1105

VIR

= = .

= 22 A

(c) Voltage across inductance = I XL = 22 × 2π × 50 × 0.4

= 2764.60 V.

(d) Voltage across capacitance = Voltage across inductance

= 2764.60 V

110

(e) Q-factor 0 02L f LR R

ω π= =

Electrical

2 50 0.5

4π × ×=

= 25.1327

Example 4.12

A 20 Ω resistor is connected in series with a coil, a capacitor and an ammeter across a 25 V variable frequency supply. When the frequency is 400 Hz, the current is at its maximum value of 0.5 A and potential difference across the capacitor is 150 V. Calculate :

(a) Capacitance of capacitor, and

(b) Resistance and inductance of coil.

C

25 V, 400 Hz

LRL

VC=150VIM=0.5A

R=20

A

Figure 4.41

Solution

(a) Let capacitance = C farad

Resonant frequency, f0 = 400 Hz

Capacitive reactance, 1 1 12 400 800CX

C C C= = =ω π × × π ×

Ω

Voltage across capacity = Im XC = 150 V

Im = 0.5 A

∴ 0.5 150800 C

=π

60.5 1.3263 10 F800 150

C −= = ×× π ×

or C = 1.3263 μF

(b) Let, RL and L be the resistance and inductance of coil respectively

Current, mL

VIR R

=+

250.520 LR

=+

(∵Applied voltage V = 25 V)

RL = 30 Ω

111

AC CircuitsSAQ 3 (a) A series RLC circuit has a resonant frequency of 220.6 Hz and is fed from

125 V source. At resonance the voltage across inductance and capacitance is 4151 Volt. The resistance of circuit is 1.06 Ω. Find QLC of the circuit.

(b) A coil of inductance 0.75 H and resistance 40 Ω is a part of a series resonant circuit having a resonant frequency of 55 Hz. If the supply is 250 V, 50 Hz. Find

(i) current,

(ii) power factor, and

(iii) voltage across the coil.

(c) Determine R1 and R2 which cause the circuit shown in Figure 4.42 to be resonant at all frequencies.

Figure 4.42

4.10 THREE PHASE AC CIRCUIT

Alternating circuits so far discussed have single phase supply system and its satisfactory for domestic application like light, fan and heating, etc. But for industrial applications and large size electrical devices, etc. 1-phase system has many limitations. Hence, polyphase system (2-phase, 3-phase and 6-phase systems) are used in electrical engineering. But for generation, transmission and distribution and utilisation of electrical power a 3-phase system has been universally adopted. 2 and 6-phase are derived from the 3-phase system, if required.

A set of three single phase sinusoidal voltage currents of same frequency and magnitude

but have a progressive phase difference of 120o or 23π radians electrical constitutes

balanced 3-phase voltage/current.

Figure 4.43(a) : Phasor Diagram

112

Electrical

(b) Wave Diagram of 3-Phase Voltage

Figure 4.43 : Three-phase Voltage System

Figure 4.43 shows that 3-phase voltage and their instantaneous values are given by,

sinaa mV V′ = ω

osin ( 120 )bb mV V′ = ω −

2sin3mV t π⎛ ⎞= ω −⎜ ⎟

⎝ ⎠

osin ( 240 )cc mV V′ = ω −

4sin3mV t π⎛ ⎞= ω −⎜ ⎟

⎝ ⎠

Wave diagram is shown in Figure 4.43(b).

For a general n-phase system displacement between phases is given by o360 (electrical)

n,

so for 2-phase system it is 90o electrical.

General Notations used in 3-phase

In three phase circuits, the phases are normally identified by the colours of insulation on wires used, i.e. R for red, Y for yellow and B for blue. Often the phases are also marked as A, B, C or 1, 2, 3. These nomenclatures, i.e. R-Y-B, A-B-C or 1-2-3 are universally accepted.

The voltage in circuit VRY is positive from R to Y. the order of the subscripts denoting the direction of the voltage. So

RY YV V R=

Similarly, RY YI I R= −

This form of double subscript notation makes the calculations in 3-phase circuits simple and less ambiguous.

Phase Sequence

Phase sequence is the sequence or order in which the three current/voltages attain their maximum values one after the other.

Let the three phases, , , andaa bb ccV V V′ ′ ′ shown in Figure 4.44 refers phase A, B and C, respectively. Then for the waveform shown A-B-C is the phase sequence. Since only two directions, i.e. anticlockwise and clockwise of rotations are possible, only two types of phase sequence are possible. By convention phase sequence ABC or RYB is taken as positive and ACB or RBY as negative.

113

AC Circuits

t

= ω −

= ω −

(a) Positive Sequence (b) Negative Sequence

Figure 4.44

Correct phase sequence is required when the machines are to be operated in parallel, otherwise large circulating currents will flow.

4.10.1 Connection in Three-Phase System The three elements, i.e. three phases of 3-phase source or three impedances of 3-phase load can be inter-connected in two different ways called

(a) Star connection, and

(b) Delta connection.

In star connection similar ends the 3-phases are jointed together to form the neutral as shown in Figure 4.45(a), star connection is denoted by Y and the common or neutral point is denoted by n, it is also called star point.

(a) Star Connection (b) Delta Connection

Figure 4.45

In delta connection three-phases are connected in series, end-to-end to form a closed mesh as shown in Figure 4.45(b). Two dissimilar ends are joined together. Delta connection is denoted by Δ.

It can be shown that for 3-phase balanced system, the sum of instantaneous values of currents or voltages is always zero. Let instantaneous 3-phase voltages of a balance system are represented by :

sina mV V= ω

V V . . . (4.57) osin ( 120 )b m

V V osin ( 240 )c m

then, at any instant,

o osin sin ( 120 ) sin ( 240 )a b c m m mv v v V t V t V t+ + = ω − ω − + ω −

114

Electrical o o[sin sin ( 120 ) sin ( 240 )]mV t t t= ω + ω − + ω −

o o[sin 2 sin ( 180 ) cos 60 ]mV t t= ω + ω −

o[sin 2 sin cos 60 ]mV t t= ω − ω

= 0 . . . (4.58)

Similarly, for 3-phase currents of a balance system, we can show that

0a b ci i i+ + = . . . (4.59)

This is the important property of 3-phase balanced system which has made possible the star and delta connection of three phases, without any circulating currents.

4.10.2 Voltage, Current and Power Relations In Star Connection

Vph = Phase voltage

VL = Line voltage

Iph = Phase current

IL = Line current

Figure 4.46

In star connection

3L pV V= h . . . (4.60)

L phI I= . . . (4.61)

3 3 cos WaL LP V Iφ = θ tts

3 cosph phV I= θ . . . (4.62)

where cos θ = Power factor.

In Delta Connection

L phV V=

3L pI I= h

3 3 cos WaL LP V Iφ = θ tts

θ

. . . (4.63) 3 cosph phV I=

115

AC Circuits

Figure 4.47

Star-Delta Connection

Figure 4.48

4.10.3 Conversion from Star Connection to Delta Connection From the Star Connection

Resistance between terminals 1 and 2 when 3 is open circuited :

⇒ 12 1 2R R R= + . . . (4.64)

Figure 4.49

Similarly, resistance between terminals 2 and 3 when terminal one is open circuited.

⇒ 23 2 3R R R= + . . . (4.65)

In the same way resistance between terminals, 1 and 3 is

13 1 3R R R= + . . . (4.66)

116

Electrical

Figure 4.50

From the Delta Connection Resistance between terminals 1 and 2 when 3 is open circuited. Rc and Rb are in series. This combination is in parallel to Ra.

12( )a b c

a b c

R R RR

R R R+

=+ +

. . . (4.67)

Figure 4.51

Similarly, the resistances between terminals 2 and 3 and 1 and 3 are given by :

23( )a c

a b

b

c

R R RR

R R R+

=+ +

. . . (4.68)

13( )a b

a b c

cR R RR

R R R+

=+ +

. . . (4.69)

Now compare the results of star and delta connections :

1 2( )a b c

a b c

R R RR R

R R R+

+ =+ +

. . . (4.70)

2 3( )b a c

a b c

R R RR R

R R R+

+ =+ +

. . . (4.71)

1 3( )c a b

a b c

R R RR R

R R R+

+ =+ +

. . . (4.72)

After solving the Eqs. (4.70), (4.71) and (4.72) we can determine three unknown resistances Ra, Rb and Rc in terms of R1, R2 and R3.

1 2 2 3 3 1

3a

R R R R R RR

R+ +

= . . . (4.73)

1 2 2 3 3 1

1b

R R R R R RR

R+ +

= . . . (4.74)

1 2 2 3 3 1

2c

R R R R R RR

R+ +

= . . . (4.75)

Eqs. (4.73), (4.74) and (4.75) are the desired results.

117

AC Circuits4.10.4 Conversion from Delta Connection to Star Connection Now we have to find R1, R2 and R3 in terms of Ra, Rb and Rc.

Ra, Rb, Rc – Known resistances.

R1, R2, R3 – Unknown resistances.

Again solving the Eqs. (4.70). (4.71) and (4.72) we can determine the resistances of star connection, i.e. R1, R2, and R3.

1a c

a b

R RR

cR R R=

+ + . . . (4.76)

2a b

a b c

R RR

R R R=

+ + . . . (4.77)

3b c

a b c

R RR

R R R=

+ + . . . (4.78)

Table 4.1 : Comparison between Star and Delta Connected Systems

Sl. No. Star (Y) Connected System Delta (Δ) Connected System 1. In star connected system there is

common point known as neutral ‘n’ or star point. It can be earthed.

There is no neutral point in delta connected system

2. In star connected system we get 3-phase, three wire system and also 3-phase, 4 wire system is taken out.

Only 3-phase, 3 wire system is possible in delta connected system

3. Line voltage 3V VL ph=

or, 13ph LV V=

Line voltage = Phase voltage LV Vph=

4. Line current = Phase current I IL ph=

Line current 3LI I ph=

13ph LI I=

5. Three phase power 3 cosLV I ph= φ

3 cophV I ph= s φ

Three phase power 3 cosL LV I= φ 3 cophV I ph= φs

Example 4.13

Obtain the equivalent inductance at terminal AB in circuit shown in Figure 4.52.

Figure 4.52

Solution

Inductances also follow the same rule as the resistances for series, parallel and star-delta connections. First we convert the star connection a b c n to corresponding delta connection.

118

Electrical

Figure 4.53

For delta connection,

an bn bn cn cn anab

cn

L L L L L LL

L+ +

=

1 1 1 3H1

+ += =

Similarly, Lbc = 3 H and Lac = 3 H

Figure 4.54

So, the circuit of Figure 4.52 is redrawn as shown in Figures 4.54 and 4.55.

Figure 4.55

32eqL = H is in parallel with 3 Henry.

So, 1.5 3 1 Henry1.5 3eqL ×

= =+

Example 4.14 A balanced star connected load is supplied from a symmetrical 3-phase 440 Volt system. The current in each phase is 40 Amp and lags 30o behind the phase voltage. Find

(a) phase voltage, (b) power, (c) reactive power drawn by a the load, and (d) phase impedance.

119

AC CircuitsSolution

For three phase star connected load voltage applied VL = 440 Volt

(a) ∴ 440 254.03 Volt3phV = =

Power current o40 30 Ampph LI I= = ∠ − =

(b) Power 3 3 coL LP V Iφ s= φ

o3 440 40 cos 30= × ×

= 26400 Watt = 26.4 kW

(c) Reactive power drawn by the load :

3 3 sinL LQ V Iφ = φ

o3 440 40 sin 30= × ×

= 15242.047 Var = 15.24 kVAr

(e) Phase impedance o

o254.03 040 30

phph

ph

VZ

I∠

= =∠ −

o6.35 30= ∠ Ω

Example 4.15

A delta connected balanced 3-phase load is supplied from a 3-phase 400 V supply. The line current is 30 Amp and the power taken by the load is 10 kW. Find

(a) impedance in each branch,

(b) the phase current,

(c) power factor, and

(d) power consumed if the same load is connected in star.

Solution

Delta Connection Load

Line current IL = 30 Amp

Phase current 30 17.32 Amp3phI = =

(a) Phase impedance 400 V17.32 Amp

phph

ph

VZ

I= =

= 23.095 Ω

(b) Phase current Iph = 30 Amp (already calculated)

(c) Total 3-phase power = 10 kW

But 3-phase power 3 cosL LV I= φ

∴ 10 1000cos3 3 400 30L L

PV I

×φ = =

× ×

= 0.4811

(d) The same power will be consumed if load is delta connected, i.e. 10 kW.

120

SAQ 4 Electrical

(a) A balanced star connected load of (8 + 6j) ohms per phase is connected to a 3-phase, 230 V supply. Find the

(i) line current,

(ii) power, and

(iii) power factor.

(b) Three inductive coils each with a resistance of 15 Ω and an inductance of 0.03 Henry are connected in star to a 3-phase, 400 V, 50 Hz supply. Calculate

(i) phase current and line current, and

(ii) total power absorbed.

4.11 MAXIMUM POWER TRANSFER THEOREM

This theorem is used to determine the value of load, for which source will transfer the maximum power. Before applying the maximum power transfer theorem, we apply Thevenin theorem across load terminals. Now there are following three cases for load to receive the maximum power.

Case I : Load is Pure Resistive

Now we have to find load resistance RL so that maximum power is transferred to it.

Figure 4.56 : Pure Resistive Load

Load current iL :

( )

THL

TH L TH

ViR R j X

=+ +

2 2

| |( )

=+ +

THL

TH L TH

ViR R X

Power delivered to load RL.

2

22 2| |

( )= × = ×

+ +TH

L L LTH L TH

VP i R RR R X

L

Condition for maximum power

0L

L

dPdR

=

121

AC Circuits

2 2

22 2

( ) 2 ( )0

( )

⎡ ⎤+ + − +⎢ ⎥= =⎢ ⎥+ +⎢ ⎥⎣ ⎦

TH L TH L TH LTh

TH L TH

R R X R R RV

R R X

2 2( ) 2 (+ + = +TH L TH L TH L )R R X R R R

⇒ 2 2 2= +L TH TR R X H

or, 2 2= +L TH THR R X

. . . (4.79) | |= THZ

So load resistance RL should be equal to the magnitude of the internal impedance of circuit.

Case II : Load is Complex (Impedance) Load current

( ) (

THL

TH TH

Vi)R R j X X

=+ + +

. . . (4.80)

Figure 4.57 : Complex Load

Magnitude of load current

2 2| |( ) (

THL

TH TH

Vi)R R X X

=+ + +

Load power 2| |L LP i R= ×

2

2 2( ) ( )Th

TH TH

VR

R R X X= ×

+ + +

For getting the maximum power of PL, denominator of equation should be small. For that select X = − Xth.

Then 2

2( )Th

LTH

VP RR R

= ×+

⇒ 0LTH

dP R Rdl

= ⇒ =

So, for maximum power factor TH TH THZ R j X Z= − = (Complex conjugate of ZTH) . . . (4.81)

Maximum Power Transfer in Case of DC Network Figure 4.58, shows the DC network for maximum power transfer. The load current is given by

THL

TH L

ViR R

=+

122

Power delivered to load Electrical

2

22( )

THL L L

TH L

VP i R RR R

= × = ×+

L

Figure 4.58 : DC Network

Condition for maximum power

0L

L

dPdR

= . . . (4.82)

⇒ RL = RTH . . . (4.83)

So load resistance should be equal to the internal resistance of circuit for receiving the maximum power. The maximum power from equation

2 2

max 2 4( )Th TH

LLL L

V VP RRR R

= × =+

2

max 4TH

TH

VPR

= . . . (4.84)

Maximum power transfer theorem is useful in communication network and electronic circuits where demand of maximum power is important rather than the efficiency. The overall efficiency of a network supplying maximum power to any branch is only 50%. So use of maximum power transfer theorem is limited in electrical and distribution networks where aim is high efficiency not the maximum power. As stated earlier, before applying the maximum power transfer theorem, we assure that circuit should be in Thevenin’s equivalent form. So, here we define the Thevenin theorem and then draw the Thevenin’s equivalent circuit. Thevenin theorem states that any linear bilateral network across its load terminal can be replaced by simple network having one voltage source (VTH) and one series impedance (ZTH). Thevenin’s Equivalent Circuit

Figure 4.59 : Thevenin’s Equivalent Circuit

123

AC Circuitswhere VTH = Thevenin’s equivalent voltage

= VOC = Open circuit voltage across the load terminals.

ZTH = Thevenin’s equivalent impedance

= Drive point impedance across the load terminals when all the sources replaced by their internal resistances.

Example 4.16

Find the value of load resistance, RL for which the source will transfer the maximum power. Also find the value of maximum power.

Figure 4.60

Solution

First we apply the Thevenin theorem and draw the Thevenin’s equivalent circuit. Remove the load resistance RL.

Figure 4.61

Calculation for VTH :

VTH = Open circuit voltage across load terminals AB, and

= VAB

Figure 4.62

AB TV V H=

5 4 i= − +

124

Electrical 105 4

6⎛ ⎞= − + ⎜ ⎟⎝ ⎠

= 1.66 Volt

Calculation for RTH :

Replace all the sources by internal resistance. (Internal resistance of ideal voltage source is zero)

Figure 4.63

TH ABR R=

2 // 4 6= +

2 4 62 4×

= ++

= 7.33 Ω

Thevenin’s Equivalent Circuit

According to maximum power transfer theorem

7.33L THR R= = Ω

1.66 0.1132 Amp7.33 7.33Li = =

+

Figure 4.64

Maximum power 2max L LP i R= ×

0.09398 Watts=

93.98 m Watts=

Example 4.17

Find the load for which the source will transfer the maximum power if load is connected across the terminal A and B, under the following conditions :

(a) load is resistive, and

(b) load is complex

125

AC Circuits

Figure 4.65

Solution

Calculation for VTH

o20 0 1.692 0.461 Amp

11 3i j

j∠

= = +−

(6 3 ) 11.53 2.3 VoltTH ABV V i j j= = − = −

Figure 4.66

Calculation for ZTH

5 (6 3 ) 2.88 0.57611 3TH

jZ jj

−= = −

−Ω

Figure 4.67

Thevenin’s Equivalent Circuit Case I

Load is resistive, then value of load resistance will be equal to magnitude of ZTH.

| |L THR Z=

2 2(2.88) (0.576) 2.93= + = Ω

H

Case II Load is complex :

*=L TZ Z (Complex conjugate of ZTH)

2.88 0.576 j= + Ω

126

Electrical

Figure 4.68

SAQ 5 Determine the value of load ZL such that the maximum power is transferred for the network shown in Figure 4.69. Also find the value of maximum power in watts.

Figure 4.69

4.12 SUMMARY

This unit describes the basic concepts of single and three-phase ac system. Unit provides the detailed analysis and derivations as needed by topics. Here we determine the voltage current power and power factor for resistive, inductive and capacitive circuits. Also we apply Kirchhoff’s voltage and current laws in series and parallel RLC circuit. Complete analysis of balanced three-phase circuit along with the star-delta conversion is given in the unit. Also the proper comparison is done between star connection and delta connection.

A very detailed description of series and parallel resonance is given in the unit with all important characteristics. Different curves like impedance curve, angle curve, admittance curve, and resonance curve are drawn for both types of resonance.

At the end of the unit maximum power transfer theorem is stated for both ac and dc circuit. Since the concept of Thevenin theorem is required for applying the maximum power transfer theorem so same is also defined in the unit.

4.13 ANSWERS TO SAQs

SAQ 1

(a) RMS value 2

0

1 [ ( )]T

rmsF f t dtT

= ∫

0.05

200 2

0

1 [10 ] 5 2.240.005

te dt−= =∫ =

127

AC CircuitsAverage value

0.05200

0 0

1 1( ) 100.05

Tt

avF f t dt e dtT

−= =∫ ∫

1avF =

Form factor 2.24rms

av

FF

= =

(b) The effective or RMS value of v (t) can be determined by equation.

2 2

2 30 60(20)2 2rmsV ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

400 450 1800 51.478 Volt= + + =

SAQ 2 (a)

i

Z

ν Figure for Answer to SAQ 2(a)

Here ν = 100 + j 50 Volt

i = 4.25 + j 12 Amp

Impedance, 6.3247 6.0933vZ ji

= = − Ω

Impedance is capacitive as reactive component is negative.

Power S = VI * = (P + jQ)

= (100 + j 50) (4.25 – j 12)

= 425 + j 212.5 – j 1200 + 600

= 1025 – j 987.5

= 1423.299 ∠ – 43.9325o

Active power, P = 1025 Watts = Power consumed

Reactive power, Q = 987.5 Var (Capacitive)

Apparent power = 1423.299 VA

Angle between current and voltage = 43.93o.

(b) Applied voltage = 170 sin 377 t

= 170 0∠ o Volts

Current, I = 12 Amp,

Power, P = 720 Watts

Current resistance = R Ω

P = I2 R

2 2720 5

(12)PRI

= = = Ω

But P = VI cos φ

128

Electrical P.f. 720cos 0.35294

170 12P

VIφ = = =

×

φ = 69.33275o

Impedance 170 14.166712

VI

= = = Ω

Reactance = XL

tan LXZ

θ =

∴ XL = 37.556

= ω L

∴ 99.61 mHLXL = =ω

SAQ 3

(a)

I

R=1.06Ω L C

125V, 220.6 Hz Figure for Answer to SAQ 3(a)

At resonance, 125 125 117.92 Amp1.06

IR

= = =

given that I XL = I × 2π fL = 4151 volt

∴ 4151 4151 0.0254H2 117.92 2 220.6

LI f

= = =× π × × π ×

22 2

1 12 4 4

f f CLC LC L f

= ⇒ = ⇒ =π π π 2

1

620.5 10 FC −= ×

Quality factor, 2 33.2WL f LQR R

π= = =

(b) R=40Ω L=0.75 H C

250V, 50 Hz Figure for Answer to SAQ 3(b)

Resonance frequency, f0 = 55 Hz

129

AC CircuitsLet capacitance = C Farad

0 2 20

1 12 (2 )

f CLC f L

= ⇒ =π π

∴ C = 11.165 μF

Applied voltage = 250 V, 50 HZ

XL = 2π fL

= 2π × 50 × 0.75 = 235.6194 Ω

Reactance of capacitance, 12CX

fC=

π

61 285.096

2 50 11.165 10CX −= = Ωπ × × ×

Total Impedance 2 2(40) (235.6194 285.96) 63.6234= + − = Ω

(i) Line current, 250 3.9294 Amp63.6234

VIZ

= = =

(ii) Power factor 40 0.6287 (leading)63.6234

RZ

= = =

(iii) Voltage across coil I XL = 3.9294 × 235.6194

= 925.8429 Volt.

(c) For the given parallel circuit, the resonant frequency is given by

21

0 22

1 L CRLC L CR

⎡ ⎤−ω = ⎢ ⎥

−⎢ ⎥⎣ ⎦

Now can assume any value provided 0ω2 21 2

LR RC

= =

In present case, 3

62 10 25

80 10LC

−

−×

= =×

R1 = R2 = 5 Ω.

SAQ 4

(a) Line voltage = 230 Volt

In star connection phase voltage

230 132.79 Volt3phV = =

Per phase impedance of load

o(8 6 ) 10 36.87phZ j= + Ω = ∠ Ω

Phase current

oo

132.79 13.279 36.87 Amp10 36.87

phph

ph

VI

Z= = = ∠ −

∠

130

Electrical (i) In star connection o13.279 36.87 Ampph LI I= = ∠ −

(ii) Three-phase power 3 cosL LV I= φ

3 230 13.279 cos ( 36.87)= × × −

= 4232 Watt

(iii) Power factor = cos φ = 0.8.

(b) Pre-phase load

Resistance R = 15 Ω, inductance L = 0.03 H

Reactance of coil

2 50 0.03 9.42LX L= ω = π × × = Ω

Per phase impedance of load

o15 9.42 17.71 32.12ph LZ R j X j= + = + = ∠ Ω

Power factor = cos φ 0.847

Supply voltage = 400 V (3-phase)

(i) Phase voltage 400 230.94 Volt3phV = =

Phase current = Line current = 230.94 13.04 Amp17.71

ph

ph

VZ

= =

(ii) Total power absorbed

3 cosL LP V I= φ

3 400 13.04 0.847= × × ×

= 7652.12 Watts

SAQ 5

(a) Calculation for VTH :

Figure for Answer to SAQ 5 (a)

4TH ABV V i= =

o30 204

3 4j⎛ ⎞∠

= ⎜ ⎟⎜ ⎟+⎝ ⎠

o22.96 6.96 24 16.86 Voltj= − = ∠ −

131

AC Circuits(b) Calculation for ZTH :

Figure for Answer to SAQ 5(b)

4 (3 ) 5 1.44 3.084 3TH

jZ j jj

= − = −+

Ω

Thevenin’s Equivalent Circuit

For maximum power transfer, the load impedance ZL is complex conjugate of ZTH.

Figure for Answers to SAQ 5(c)

*L THZ Z=

1.44 3.08 j= + Ω

Power will be dissipated in resistive part only.

o24 16.86

1.44 + 1.44TH

LL TH

ViZ Z

∠ −= =

+

8.33 16.86 Amp= ∠ −

2max | | 1.44LP i= ×

= 99.92 Watts