current

31

Current, Electricityand Its EffectUNIT 2 CURRENT, ELECTRICITY AND ITS

EFFECT

Structure 2.1 Introduction

Objectives

2.2 Magnetic Effect of Electric Current

2.3 Concept of Self and Mutual Inductance

2.4 Capacitance

2.5 DC Generator

2.6 AC Supply

2.7 AC Generator

2.8 Transformer

2.9 Inductive and Capacitive Reactance

2.10 Impedance

2.11 Resistive, Inductive and Capacitive Circuits

2.12 Series and Parallel Resonance

2.13 Power, Power Factor and Quality Factor

2.14 Summary

2.15 Answers to SAQs

2.1 INTRODUCTION

In the year 1820, Oersted found by his experiment that electric current shows magnetic effect. This experiment gave rise to the concept of electro magnetism. In 1831, Faraday discovered that time varying magnetic field can produce the electric field. Then he gave very popular Faraday’s law of electromagnetic induction. He also gave the concept of self and mutual inductance which is related to the concept of flux. In 1835, he made the capacitor and found the capacitance of that. The unit of capacitance was on his name that is Farad.

Faraday’s law of electromagnetic induction became basis of electrical machines. Transformers and generators are based on this principle.

At the latter stage of the unit, we give the idea of resistive, inductive and capacitive circuit, calculation for their voltage, current and power. We also introduce three types of powers – active power, reactive power and apparent power. Concept of series and parallel resonances is also given in the unit with sufficient description and examples.

Objectives After studying this unit, you should be able to

• explain the magnetic effect of current,

• define the basic laws of magnetism,

• determine the relation for magnetic field for different magnetic structure like straight conductor, solenoid, coil etc.,

• define the self and mutual inductance,

• establish the relation for different series and parallel arrangement of inductors,

32

Electricity • define the concept of capacitance,

• explain the principle, working and construction of DC generator and transformer,

• define the single phase AC and 3 phase AC supply along with important terms related to them,

• explain the principle, working and construction of AC generator,

• distinguish between inductive and capacitive reactance,

• describe the effect of AC in resistive, inductive and capacitive circuits,

• explain the concept of series and parallel resonance,

• distinguish between apparent, active and reactive powers, and

• define the power factor and quality factor.

2.2 MAGNETIC EFFECT OF ELECTRIC CURRENT

In the year 1820, Oersted identified the magnetic field with electric current. Then he gave the statement that “electric current produces the magnetic field”.

First we shall study some basic Laws related to magnetic effect of electric current.

2.2.1 Biot-Savart Law (Laplace Law) This law gives the relation between magnetic flux density (B in wb/m2) and the electric current I (in Amp). Consider the current carrying conductor shown in Figure 2.1. Let us find the flux density B at any point P at a distance r from the small element dl.

dl

r

I

P

Figure 2.1 : Biot-Savart Law

Now flux density dB due to small element dl is given as

dB I∝

dB dl∝

sindB ∝ θ

21dBr

∝

So flux density dB is given by dB ∝ 2

sinI dlr

θ

33

Current, Electricityand Its EffectThe proportionality constant in SI unit is

4μπ

.

∴ 22sin. w

4I dldB

rμ θ

=π

b/m . . . (2.1)

where μ = permeability of medium in Henry/metre

= 0

r

μμ

μ0 = absolute permeability = 4π × 10− 7 h/m

μr = relative permeability. Flux density is vector quantity and its direction is perpendicular to both I dl and r.

22sin ˆ. w

4 nI dldB a

rμ θ

=π

uurb/m

Magnetizing force or magnetic field intensity is defined as BH =μ

.

∴ 21 sin ˆ. A

4 nI dldH a

rθ

=π

uuurmp/m . . . (2.2)

2.2.2 Ampere’s Circuital Law for Static Magnetic Field Ampere’s Circuital law is also known as the work law. According to this law “Total work done for moving an unit north pole in a closed path is equal to current enclosed”. Let us consider Figure 2.2.

I

Current CarryingCoductor

Closed Path

Figure 2.2 : Ampere’s Circuital Law

So by Ampere’s circuital law : Work done per unit pole = I

⇒ .H dl I=∫ . . . (2.3)

But BH =μ

then .c

B dl I= μ∫ . . . (2.4)

If there are N number of conductors carrying current I in same direction and placed in a closed path then by Ampere’s circuital law :

.H dl NI=∫

This is also known as Magnetomotive force or mmf. So, mmf = NI (Amp-turn) . . . (2.5)

34

2.2.3 Magnetic Field due to Long Straight Conductor Electricity

P

I

r

Figure 2.3 : Ampere’s Circuital Law

Consider a point P at which H and B have to be found. P is at a distance ‘r’ from long straight conductor. For applying Ampere’s Circuital law draw a circle of radius ‘r’ which encloses the current carrying conductor. Then :

.H dl I=∫uur uur

(by Ampere Circuital Law)

.H dl I=∫

2H r I× π =

2IH

r=

π . . . (2.6)

So, flux density B, B = μH

or 2

IBr

μ=

π . . . (2.7)

2.2.4 Magnetic Field due to Parallel Conductors Let the two conductors placed in parallel carry currents I1 and I2 amperes. Depending on the direction of flow of current, two cases evolve.

(a) Current in same direction (b) Current in opposite direction

Current in Same Direction Let the currents be flowing in upward direction. The field strength in between the two conductors P and Q is decreased due to the two fields there being in opposite direction. Hence, the two conductors are attracted towards each other.

PP QI1 I2

d

Figure 2.4 : Current in Same Direction

Current in Opposite Direction When currents flow in opposite direction the field strength is increased in the space between the two conductors due to two fields being in the same direction. Because of the lateral repulsion of the lines of force, the two conductors experience a mutual force of repulsion.

35

PP Q

I1 I2

d

Current, Electricityand Its Effect

Figure 2.5 : Current in Opposite Direction

2.2.5 Magnetic Field due to a Long Solenoid Assumptions

(a) Magnetic field is constant throughout the length ‘l’ of the solenoid.

(b) Magnetic field outside the solenoid is negligible compared to inside the solenoid.

Let No. of turns = N, and Length of solenoid = l

Figure 2.6 : Solenoid

By Ampere Circuital Law,

H × l = NI

AT/mNIHl

=

Flux Density, 2wb/mNIBl

μ= . . . (2.8)

2.2.6 Magnetic Effect Due to Circular Coil Let there be a circular coil having only one turn, and current flowing be I. Small element dl of circular coil produces magnetic field dH at point P. The direction of dH is at right angles to the line AP joining point P to the element ‘dl’. Now, dH can be resolved into two components :

(a) The axial component dH ′ = dH sin θ

(b) The vertical component dH ″ = dH cos θ

B

A

O

dH

dH’

dl

DH’’

Px

r

Figure 2.7 : Magnetic Effect Due to Circular Coil

36

Electricity Now, the vertical component dH cos θ will be cancelled by an equal and opposite vertical component of dH due to element dl at point B. Hence, the resultant magnetizing force at P will be equal to the sum of all the axial components. ∴ H = ∑ dH ′ = ∑ dH sin θ ∫ dl

2 2 3/ 2 2 2

. . sin4 ( )

I dl r rdlr x r x

⎛ ⎞⎜ ⎟= θ⎜ ⎟π + +⎝ ⎠

∑ ∫ Q =

2

2 2 3/ 20

.4 ( )

rI r dlr x

π

=π + ∫

2

2 2 3/ 2 2 2 3/. . 2

4 ( ) 2 ( )I r r I rr x r x

π= =

π + + 2

3

2 2 3/.2 ( )I rr r x

=+ 2

∴ 3sin AT/m

2IH

rθ

=

or 3sin AT/m2NIH

r= θ (where N = Number of turns of coil) . . . (2.9)

2.3 CONCEPT OF SELF AND MUTUAL INDUCTANCE

Concept of inductance was introduced by Faraday in 1831. Mutual inductance was the important result of Faraday’s well known law of electromagnetic induction. Before stating the mutual inductance, we shall develop the concept of self induction which has already been discussed. According to Faraday’s law of electromagnetic inductance, whenever the magnetic flux linked with the coil (due to current in the same coil) changes with time then emf is induced in that coil. The magnitude of this induced emf is given by

LdNdtφ

ν = . . . (2.10)

where N = number of turns in the coil, and φ = flux per turn (in weber)

From Eq. (2.10) ( )

Ld N d

dt dtφ ψ

ν = = . . . (2.11)

where Ψ = total flux linked with the coil (in weber) But total flux ψ is directly proportional to the current i in the coil ∴ iψ ∝

or, Liψ = . . . (2.12)

where L is the proportionality constant and known as the self inductance of coil. Unit of inductance is Henry. Put the value of ψ from Eq. (2.12) to Eq. (2.11)

LdiLdt

ν = . . . (2.13)

In a similar manner, we can define the mutual inductance. Consider two isolated coils shown in Figure 2.8, in which coil 1 is carrying the current i1.

37

1 2

N2N1

i1

φ12

φ11


Figure 2.8 : Two Isolated Coils with Current i1

Let current i1 is alternating current and producing flux φ1 (per turn). Then 12111 φ+φ=φ . . . (2.14)

where φ 11 = flux linked with coil 1 only (leakage flux), and

φ 12 = flux reaches to coil 2 (mutual or useful flux).

Now the emf is induced in both the coils. In coil 1, emf e1 is induced due to its self inductance L1 and in coil 2 emf e2 is induced due to mutual inductance M between coils 1 and 2.

11 1

die Ldt

= (by Eq. (2.13))

Similarly, we can write the expression for e2

12

die Mdt

= . . . (2.15)

where M = mutual inductance between coils 1 and 2 (in henry) e2 can also be written as :

122 2

de Ndtφ

= . . . (2.16)

On equating Eqs. (2.15) and (2.16)

1 12

di dM Ndt dt

φ= 2

⇒ 122

1

dM Ndiφ

=

Assuming the zero initial conditions

122

1M N

iφ

= . . . (2.17)

Now consider the reverse case, in which we give the supply to the secondary coil and determining the induced emf in primary coil (Figure (2.9)).

N2N1

i2

22

e1

21

Figure 2.9 : Two Isolated Coils with Current i2

38

Electricity 22212 φ+φ=φ . . . (2.18)

where φ 21 = mutual flux,

φ 22 = leakage flux, and

φ 2 = total flux produced by i2.

So the induced emf 22 2

die Ldt

= . . . (2.19)

(Due to self inductance L2 of the coil). Induced emf in coil 1 (due to mutual inductance M between the two coils)

21

die Mdt

=

also 211 1

de Ndtφ

=

or 2 21

di dM Ndt dt

1φ=

211

2

dM Ndiφ

= . . . (2.20)

Assuming the zero initial conditions

211

2M N

iφ

= . . . (2.21)

Multiply Eq. (2.17) and (2.21)

2 12 212 1

2M N N

i i⎛ ⎞φ φ⎛ ⎞= ⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠ . . . (2.22)

Now we define the coefficient of coupling

12 21

1 2K φ φ= =

φ φ . . . (2.23)

or Useful fluxTotal flux

K =

1≤ Under ideal conditions, when there is no leakage, then K = 1 (i.e. 100% coupling between two coils). In any practical case, K is always less than 1. From Eq. (2.23)

12 1 21 2andK Kφ = φ φ = φ

Put these values of φ 21 and φ 12 in Eq. (2.22)

2 2 1 21

1 2

N K KM Ni i

⎛ ⎞ ⎛ ⎞φ φ= ⎜ ⎟ ⎜⎝ ⎠ ⎝

⎟⎠

2 1 1 2 2

1 2

N NKi i

⎛ ⎞ ⎛ ⎞φ φ= ⎜ ⎟ ⎜

⎝ ⎠ ⎝⎟⎠

= K2 L1 L2

or 1 2M K L L= ± . . . (2.24)

So M may be positive or negative depending upon the winding sense.

39


2.3.1 Inductances in Terms of Reluctance Self Inductance

l

I

N

Figure 2.10 : Solenoid

Figure 2.10 shows the solenoid of length ‘l’ meter, cross-section of ‘A’ m2, have turns ‘N’ over it. If this solenoid carries a current I, then magnetizing force H is given by :

AT/mNIHl

= . . . (2.25)

Flux linked per turn : BA Hφ = = μ A . . . (2.26)

[ ]B H= μQ

where μ = permeability. From Eq. (2.12)

LIψ =

or NLI Iψ φ

= =

Put value of from Eq. (2.26) φ

( )N HALIμ

=

Put value of H from Eq. (2.25)

N N IALl Iμ

=

2N Alμ

=

2N

lA

=⎛ ⎞⎜ ⎟μ⎝ ⎠

2N

S= . . . (2.27)

where S = reluctance of magnetic circuit in AT/wb

lA

=μ

. . . (2.28)

In practice, self inductance of short solenoid in given by 2NL K

S= . . . (2.29)

where K = Nagoka’s constant which depends on the ratio of length to diameter of solenoid.

40

By analogy, formula for M is, Electricity

1 2 HenryN NMS

= . . . (2.30)

where N1 = no. of turns of coil 1

N2 = no. of turns of coil 2

S = reluctance 0 1

lA

=μ μ

2.3.2 Inductances in Series and Parallel Inductances in Series

(a) Let the two coils be connected in series such that their fluxes are additive,

A Be1 e2

e1’ e2’

Figure 2.11 : Series Inductors

Self induced emf in 1 1diA e Ldt

= = −

Mutually induced emf in A due to change in 1diB e Mdt

′= = −

Self induced emf in 2 2diB e Ldt

= = −

Mutually induced emf in B due to change in 2diA e Mdt

′= = −

Total induced emf 1 2( 2di L L Mdt

= − + + ) . . . (2.31)

If L is the equivalent inductance then total induced emf in that single coil is given as,

diLdt

= − . . . (2.32)

From Eqs. (2.31) and (2.32) we get

1 2( 2L L L M )= + + . . . (2.33)

(b) When coils are so connected that their fluxes are in opposite direction,

A Be1 e2

e1’ e2’i

Figure 2.12 : Inductor in Series Opposition

41

Current, Electricityand Its EffectNow, 1 1 1anddi die L e M

dt dt′= − =

2 2 2anddi die L e Mdt dt

′= − =

Total induced emf 1 2( 2di L L Mdt

= − + − )

∴ Equivalent inductance . . . (2.34) 1 2 2L L L M= + −

Inductors in Parallel

A B

L1

L2

i1

i2

Mi

Figure 2.13 : Parallel Inductors

Let the two inductors connected in parallel be L1 and L2, and current flowing in them be i1 and i2 respectively. Let the mutual inductance in between the two be M.

Now, 1 2i i i= +

Case 1

When the two fields assist each other, then

21 2

1 2 2L L ML

L L M−

=+ −

. . . (2.35)

Case 2

When the two fields oppose each other, then

21 2

1 2 2L L ML

L L M−

=+ +

. . . (2.36)

2.3.3 Energy Stored in an Inductor Consider current through the coil of inductance L henry increases by di amperes in dt seconds then

Induced emf = VoltsdiLdt

Energy absorbed by magnetic field during time dt second is

. . . joulediiL dt Li didt

= s

Hence, total energy absorbed when current increases from O to I amperes is

2 20

0

1 1.2 2

I IL i dt L i LI⎡ ⎤= × =⎣ ⎦∫

21=2

W LI . . . (2.37)

42

SAQ 1 Electricity

(a) A coil having 150 turns and it is linked with the flux of 0.01 wb when carrying the current of 10 Amp, calculate the self inductance of coil. Now if the current in the coil is reversed in 0.01 second, calculate the induced emf in the coil.

(b) A coil of 100 turns is wound on a toroidal magnetic core having reluctance of 104 AT/wb. When the coil current is 5 amp and is increasing at the rate of 200 Amp/sec. Determine

(i) Self inductance of coil,

(ii) Energy stored, and

(iii) Self induced emf in coil.

(c) A coil of 50 turns having a mean diameter of 3 cm is placed co-axially at the centre of solenoid 60 cm long wound with 2500 turns and carry a current of 2 amp. Determine mutual inductance of the arrangement.

(d) Find Leq for the arrangements shown in Figure 2.14.

2Hi

4H

1H

5H

i

8H

M=2H

(b)

(a)

Figure 2.14

2.4 CAPACITANCE

Capacitor

A capacitor essentially consists of two conducting surfaces separated by a layer of an insulating medium called dielectric. The conducting surface may be in the form of either circular plates or be of spherical or cylindrical shape. The purpose of capacitor is to store electrical energy by electrostatic stress in the dielectric.

A B

Figure 2.15 : Capacitor

43


Capacitance

It is defined as, “ the amount of charge required to create a unit potential difference between its plates”.

If we give Q coulomb of charge to one of the two plates of capacitor and if a p.d. of V volts is established between the two, then its capacitance is,

ChargePotential Difference

QCV

= = . . . (2.38)

2.4.1 Capacitance Between Two Parallel Plates A parallel plate capacitor consisting of two plates M and N each of area A m2 separated by thickness d metres of medium of relative permittivity εr. If charge + Q coulomb is given to plate M, then flux passing through medium is Qψ = coulomb.

Flux density in medium, QDA Aψ

= = .

M

Q

N

d

ε

Figure 2.16 : Capacitor with Dielectric

Electric intensity andVE D Ed

= = ε

or Q VA d= ε

∴ Q AV d

ε=

∴ 0 r AC

dε ε

= farad (in a medium of permittivity, ε = ε0 εr) and

0 ACd

ε= farad (in air as medium) . . . (2.39)

2.4.2 Capacitance in Series and Parallel Capacitors in Series

Let, C1, C2, C3 = capacitance of each capacitor,

V1, V2, V3 = voltage gradient across each capacitor, and

C = equivalent capacitance.

In this case, charge on each capacitor is same but potential difference is different.

C2

Q+

C3

Q+

C1

Q+

V3V2V1

V Figure 2.17 : Capacitors in Series

44

V = V1 + V2 + V3 (as capacitors are in series) Electricity

or 1 2 3 1 2

1 1 1 1Q Q Q QC C C C C C C C= + + ⇒ = + +

3

⇒ 1 2 3

1 2 2 3 3 1

C C CC

C C C C C C=

+ + . . . (2.40)

Capacitors in Parallel

In this case, potential difference is same but charge is different on each capacitor

∴ 1 2Q Q Q Q3= + +

or 1 2 3CV C V C V C V= + +

or 1 2C C C C3= + + . . . (2.41)

V

C3+Q3

C2+Q2

C1+Q1

Figure 2.18 : Parallel Capacitors

Energy Stored in a Capacitor Suppose the potential difference across a capacitor to be increased from V to (V + dv) volts in dt seconds.

We know . dvi Cdt

=

Instantaneous value of power to capacitor is

. wattsdviv vCdt

=

Energy supplied to capacitor during interval dt is

. . JoudvvC dt Cv dvdt

= = les

Hence, total energy when potential difference is increased from O to V volts is

2 21 1. Joules2 2

V V

OO

Cv dv C V CV⎡ ⎤= =⎣ ⎦∫

∴ = 212

W CV . . . (2.42)

also 21

2QWC

= .

Example 2.1 A long single layer solenoid has an effective diameter of 10 cm and is wound with 1000 turns/m. There is a small concentrated coil having its plane lying in the centre cross-sectional plane of the solenoid. Calculate the mutual inductance between two coils if the concentrated coil has 100 turns and an effective diameter of 4 cm.

45


Solution

Figure 2.19 is shown as follows :

0.1 m 0.04

Figure 2.19

Let current I1 is flowing through the solenoid. Then the flux density B is given by

1 10 0

1

N IB Hl

= μ = μ

= 0 1 2wb1000m

Iμ .

So the flux linked with concentrated coil: φ2 = BA2

The cross-sectional area of concentrated coil :

22

22

4 102

A r−⎛ ⎞×

= π = π ⎜ ⎟⎜ ⎟⎝ ⎠

= 12.568 × 10− 4 m2

∴ Flux φ2 = μ0 1.256 I1 wb Mutual inductance between coil and solenoid

0 12 20

1 1

100 1.256125.6

INMI I

× μ ×φ= = = μ Henry

Example 2.2 If a coil of 150 turns is linked with a flux of 0.01 wb when carrying a current of 1.0 Amp, calculate the inductance of coil. Now, if current is uniformly reversed in 0.1 sec, calculate the induced emf.

Solution The self inductance of coil is

150 0.0110

NLIφ ×

= =

0.15 Henry=

The induced emf die Ldt

= −

2 1( )i iLdt−

= −

[ 10 10]Ldt

− −= −

= 30 volt. Example 2.3

The equivalent inductance of two series connected coil (with mutual inductance between them) is 0.8 or 0.2 Henry depending on the relative directions of currents

46

in the coil. If self inductance of one coil is 0.4 Henry then determine (i) Mutual inductance; and (ii) coefficient of coupling.

Electricity

Solution For M-Positive :

Leq = L1 + L2 + 2M

0.8 = L1 + L2 + 2M . . . (i)

L1

M

L2

L eq.Leq.

Figure 2.20

For M negative Leq = L1 + L2 – 2M

0.2 = L1 + L2 – 2M . . . (ii) From Eqs. (i) and (ii)

0.6 = 4M ⇒ M = 46.0 = 0.15 Henry

But L1 = 0.4 Henry ∴From Eq. (i)

0.8 = 0.4 + L2 + 2 ⎟⎠⎞

⎜⎝⎛

46.0

∴ L2 = 0.1 Henry

The coefficient of coupling 1 2

MkL L

= = 0.75

Example 2.4 Three capacitors C1, C2 and C3 have capacitance 20 μf, 15μf, 30μf respectively. Calculate :

(a) Charge on each capacitor when connected in parallel with 220 V supply.

(b) Total capacitance with parallel connection. (c) Voltage across each capacitor when connected in series and 220 V

supply is given to this series connection. Solution

(a) Capacitors are in parallel :

C2

C3

220V

C1

Figure 2.21

47


Charge 6 6

1 1 20 10 220 4400 10 CoulombQ C V − −= = × × = ×

6 62 2 15 10 220 3300 10 CQ C V − −= = × × = ×

6 63 3 30 10 220 6600 10 CQ C V − −= = × × = ×

(b) Total capacitance :

C = C1 + C2 + C3

= (20 + 15 + 30) × 10− 6

= 65 μF

(c) Since capacitors are in series so charge Q remains same for all capacitors.

220V

C2 C3C1

Figure 2.22

Equivalent capacitance :

eq 1 2 3

1 1 1 1C C C C

= + +

eq

1 1 1 1 0.05 0.0666 0.033320 15 30C

= + + = + +

Ceq = 6.667 μF

Total charge Q = Ceq × V

= 1466.74 μC

Now, Voltage across C1 is 11

73.337 voltQVC

= =

Voltage across C2 is 22

97.78 voltQVC

= =

Voltage across C3 is 33

48.89 voltQVC

= =

Example 2.5 An air capacitor has two parallel plates 10 cm2 in area and 0.5 cm apart. When a dielectric slab of area 10 cm2 and thickness 0.4 cm was inserted between the plates, one of the plate has to be shifted by 0.4 cm to achieve the same value of capacitance. What is the dielectric constant of slab?

Solution Capacitance without dielectric slab :

12 40

1 21

8.85 10 10 100.5 10

AC

d

− −

−ε × × ×

= =×

128.85 10

5

−×=

48

Capacitance with dielectric slab : Electricity

02

2

r AC

dε ε

=∑

Air

0.5 cm Figure 2.23

40

22

10 100.5 0.4 10

r

C−

−

ε × ×=⎛ ⎞

+ ×⎜ ⎟ε⎝ ⎠

120 8.85 10

5 54 4r r

−ε ×= =

+ +ε ε

But 1 2C C=

∴ 12 128.85 10 8.85 10

55 4r

− −× ×=

+ε

∴ εr = 5

Air

0.4 cm0.5 cm

εr

Figure 2.24

Example 2.6 Two capacitors C1 = 50 μF and C2 = 100 μF are connected in parallel across 250 V supply. Find the total energy loss.

250V

C2

C1

Figure 2.25

Total capacitance C = C1 + C2

= 50 + 100 = 150 μF

Energy loss 212

Cv=

6 21 (150 10 ) (250)2

−= × 4.687= joules

49


SAQ 2 (a) A capacitor is made of two plates with an area of 11 cm2 which are

separated by a mica sheet of 2 mm thick. If for mica εr = 6, find its capacitance.

(b) Find the equivalent capacitance and charge across each capacitor for the following arrangement.

200V

C1 = 1 μF

C2 = 2 μF

C3 = 4 μF

Figure 2.26

(c) A parallel plate capacitor having plates 100 cm2 area has three dielectric 1 mm each and of permittivities 3, 4, and 6. If peak voltage 2000 V is applied to the plates, calculate :

(i) Voltage gradient across each dielectric, and

(ii) Energy stored in each electric.

2.5 DC GENERATOR

DC generator takes mechanical input (through prime mover or through another motor) and gives electrical output to the external circuit (load). This mechanical input is given to the shaft of DC generator, field circuit is excited by DC and then emf is induced in the armature circuit. This emf is of alternating nature which is converted to DC by using commutator and brush arrangement.

2.5.1 Principle When a conductor is moving in the uniform magnetic field then emf is induced in the conductor because it cuts the lines of force of magnetic field, as given by Faraday’s law of electromagnetic induction. The polarity of this emf is decided by the Fleming’s right hand rule. This is known as generator principle. The required field flux (magnetic field) is produced either by permanent magnet (in small machines) or by field winding.

N S

CB

A D

M IMica

a

Segment

Brushes

b

(a) (b)

Figure 2.27 : DC Generator Principle

50

Generated emf Electricity

Figure 2.27 shows the simple principle of generator action when coil ABCD rotating in the uniform magnetic field created by permanent magnet. By Faraday’s law of electromagnetic induction alternating emf is induced in the coil

de Ndtφ⎛ =⎜

⎝ ⎠⎞⎟ . By using slip rings and collecting brushes this emf is converted into

direct voltage and fed to the external circuit (load).

Expression for Generated emf

Let Eg is the generated emf in the armature, according to Faraday’s law of electromagnetic induction

where No. of turnsgdE N Ndtφ

= =

N = 1 for single conductor, then

gdEdtφ

=

Change in flux in one revolution = dφ

= φ P wb

where φ = flux per pole, P = No. of poles

If N = speed in revolution per minute.

The number of revolution/sec = N/60

∴ Time for one revolution dt = 60/N

So volts/conductor60g

d PNEdtφ φ

= =

If z = total no. of conductors then

60gP NzE φ

=

if A is the no. of parallel paths then

volts60g

P NzEA

φ= . . . (2.43)

For wave winding A = 2.

For lap winding A = P.

2.5.2 Voltage Current Expressions and Circuit for Different Types of Generators

Separately Excited Generator

rf

NPrimeMover

LoadVtEg

If

Ia

rd S

Vf

Te

Figure 2.28 : Separately Excited Generator

51

Current, Electricityand Its EffectaTerminal voltage t g aV E I r= −

where Ia = armature current in Amp,

ra = armature circuit resistance in Ω,

Eg = generated emf in volts,

If = field current in Amp,

Vf = DC field supply voltage, and

rf = field resistance in Ω.

In field circuit ff

f

VI

r=

If brush contact resistance is also included, which is approximately 1 to 2 volt, then,

t g a a bV E I r V= − − r . . . (2.44)

where Vbr = drop in brushes.

If the above equation is multiplied by current Ia, then

2t a g a a a br aV I E I I r V I= − − . . . (2.45)

Electrical power output = Mechanical power converted into electrical in armature circuit – Ohmic losses in armature – Brush contact loss.

Series Generator rf

NPrimeMover Series

Field

LoadVtEg

rd

Te

Figure 2.29 : Series Generator

Here a fI I I= = L

For operation of these generator poles must have the residual magnetism. These generator must start under loaded condition because at no load 0a f LI I I= = = and machine will not start generation.

Shunt Generator

rf VtEg

IL

rLoad

L

Ishra

Ia

Figure 2.30 : Shunt Generator

52

Electricity La shI I I= +

, ,t tL sh t g a a

L f

V VI I V E I

I r= = = − r

Voltage Regulation If is defined as change in terminal voltage when load is reduced from rated load to no load, expressed as percentage of rated load voltage.

0. . 100%V V

V RV−

= × . . . (2.46)

2.5.3 Characteristics Generator characteristics are volt ampere characteristics. Main characteristics are given below. Magnetization Curve (No Load Curve)

This is the curve between generated emf, Eg, and the field current, If, with constant speed. For Separately Excited Generator

The experiment set-up is shown in Figure 2.31. Switch is kept open for this no load test. Field is excited by separated DC source and rotor is rotated by another motor. Voltmeter reads the generated emf Ef. Now the field current is varied by using rheostat and If is recorded by ammeter in field circuit.

Field

AIf

Vf

A

V LoadVt

Ef

Ia S

Figure 2.31 : Separately Excited Generator

0aa

Ef

If

Figure 2.32 : Magnetizing Curve for Separately Excited Generator

For large value of If, magnetic core becomes saturated, due to which the curve is not straight line but it is bent as shown in Figure 2.32. This curve also known as saturation curve. Voltage oa is due to residual magnetism of the poles.

For Shunt Generator

No extra DC source is used for field excitation but the residual magnetism produces the initial small field which induces the small emf in armature which in turn produces more field because of the Ia (Ish = Ia in No load condition). In Figure 2.33, switch S is kept open for magnetization curve and by varying the field current (using the rheostat) the variation in generated

53


emf Ef is recorded and the characteristic is drawn between Ef and Ish. The shape is same as that of separately excited generator (Figure 2.34).

A

A

LoadVt

Ish S

VEfField

Figure 2.33

0

Ef

Ish

Figure 2.34 : Magnetizing Curve for Shunt Generator

For Series Generator

At no load condition in series generator, there will be no field current, because Ia = If = IL = 0. So for getting the no load characteristic in series generator, it is necessary to excite field separately as in case of separately excited generator and by keeping Ia = 0. The shape of No load characteristic is same as that of for separately excited and shunt generator.

iL

If

ia

Figure 2.35 : Series Generator

0

Ef

If

Figure 2.36 : Magnetizing Curve for Series Generator

Compound Generator

54

Compound generator characteristic is same as that of shunt generator. Electricity

Load Characteristic This is the curve between load terminal voltage Vt and the field current with speed and armature current kept constant. For Separately Excited Generator

For obtaining this characteristic switch S is closed and load is connected to the generator as shown in Figure 2.31. Then by varying the field current If, corresponding terminal voltage Vt is recorded and then graph is plotted between them. For the complete experiment, armature current and speed is maintained constant. In each step, the load is varied and field current is adjusted for rated If. This characteristic is parallel to the no load characteristic oa = If corresponding to zero terminal voltage. This condition is obtained when the terminals are short circuit so oa is the field current required to circulate the rated armature current under short circuit condition.

0

Vt

Ifa Figure 2.37 : Load Characteristic

For Shunt Generator Here the shape of the Load Curve is similar to that of separately excited machine. Method for obtaining this is also same as that of for separately excited machine. Since armature current in shunt generator is given by Ia = IL = If while that in separately excited generator is given by Ia = IL so the armature reaction is different for two machines. Due to which there is slight difference in the characteristic.

For Series Generator In this case also, the characteristic is similar to that of separately excited generator.

External Characteristic This is the curve between terminal voltage Vt and the load current IL for constant speed and constant field current. For Separately Excited Generator

The switch S is closed in Figure 2.31 and the load is connected to the generator. Now the load resistance is varied in steps and the terminal voltage Vt and the load current IL are recorded.

Curve I = internal characteristic Curve II = external characteristic

Drop in characteristics is due to (I) Armature reaction (II) Armature resistance drop

55


0

Vt

IL

I

II

Voltage Drop due toArmature Reaction

Ia ra Drop

Figure 2.38 : External, Internal Characteristics for Separately Excited Generator

For Shunt Generator

The procedure for obtaining the external characteristic in shunt generator is same as that of for separately excited generator. Here the drop in characteristic (in terminal voltage) is due to :

(a) Armature reaction

(b) Armature resistance drop

(c) Due to I and II there is further reduction in field current If which again reduces the terminal voltage due to this commutative process characteristic turns back and has the shape as shown in Figure 2.39.

0

Vt

IL

Figure 2.39 : External Characteristic for Shunt Generator

For Series Generator

In Figure 2.35, when load is connected to the series generator, current IL = Ia = If starts flowing in the series circuit which causes the drop Vt across the Load. As the load increases, load current increases so the drop Vt also increases. That is why the series generators have rising external characteristic as shown in Figure 2.40, while the separately excited and shunt excited generators have the dropping characteristic.

0

Vt

IL

ExternalCharacteristic

Figure 2.40 : External Characteristic for Series Generator

56

Armature Characteristic (Regulation Curve) Electricity

This is the curve between field current If and load current IL with constant Vt and speed.

Shape of this characteristic is same for all types of generators. For obtaining this characteristic, field current is adjusted to give the rated terminal voltage for all values of load (including the no load condition when IL = 0). At each step the load current is recorded.

0

If

IL

Figure 2.41 : Regulation Curve

This characteristic is normally used for determining the series turns in compound generator.

Example 2.7 A 4-pole, DC Shunt generator has lap connected armature winding. It has 500 armature conductors and is rotating with 700 rpm. Determine the generated emf if the field flux per pole is 3 mwb.

Solution

The generated emf is given by 60g

P NZEA

φ= .

where P = 4 φ = 3 × 10− 3 wb N = 700 rpm

Z = 500 A = P = 4 (lap connection)

∴ 34 3 10 700 500

60 4gE−× × × ×

=×

17.5 volt=

Example 2.8 A series generator has armature and field resistances 0.4 and 0.8 Ω respectively. It generates 400 volts with voltage across the load as 350 volts. Determine the armature current flowing in the circuit. Also, find the load and field current.

Solution The series generator is shown in Figure 2.42.

Field

R LoadLVt

r =0.8fr =0.4a

Ia

Eg

Figure 2.42

57


( )g t a a fE V I r r= + +

400 = 350 + Ia (0.4 + 0.8) = 350 + Ia (1.2)

∴ 400 350 41.667 Amp1.2aI −

= =

For Series Generator Armature current = Field current = Load current

= 41.667 Amp.

SAQ 3 (a) A 4- pole, shunt generator with lap-connected armature having field and

armature resistance of 50 Ω and 0.1Ω respectively supplies 60 numbers of 40 watts lamps at 100 volt. Calculate the total armature current, current per parallel path and generated emf. Assume a constant drop of 1 volt per brush.

(b) A 4-pole, long shunt, lap would generator supplies 25 kW at a terminal voltage of 500 V. The armature resistance is 0.03, series field resistance is 0.04 Ω, and shunt field resistance is 200 Ω. The brush drop may be taken as 1 volt. Determine the emf generated.

2.6 AC SUPPLY

Alternating Quantity Quantity which varies periodically with time is known as the alternating quantity. It may be voltage or current. Some waveforms of alternating current are shown in Figure 2.43.

0/2 2

sin ωt

ωt

(a) Sine Wave

0

f (t)

TT/2

t

(b) Triangular Wave

0

f (t)

TT/2t

(c) Square Wave

Figure 2.43 : Alternating Waveforms

58

2.6.1 Basic Definitions of the Alternating Quantity Electricity

Here we take the example of sinusoidal form of alternating current.

i = IM sin ωt

0/2 2

IM

ωt

Figure 2.44 : 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 2.44, Im denotes the amplitude of the current wave.

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

sinmi I t= ω . . . (2.47)

Similarly, we can write for voltage

sinmv V t= ω . . . (2.48)

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] . . . (2.49)

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.

59


ωt

(a) Reference Waveform

ωt

(b) Lagging Waveform by Angle φ

ωt

(c) Leading Waveform by Angle φ

Figure 2.45

Lagging Current by Angle φ

If the current waveform of Figure 2.45(a) is assumed to be reference wave, then Figure 2.46(a) shows the lagging current.

i

Im

ωt

Figure 2.46(a) : Reference Current Waveform

sin ( )mi I t= ω − φ

φ

. . . (2.50)

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

Leading Current by an Angle φ Figure 2.46(b) shows the leading current by an angle φ by the reference wave of Figure 2.45(a).

sin ( )mi I t= ω + . . . (2.51)

60

i

Im

Electricity

ωt

Figure 2.46(b) : Leading Current Waveform

2.6.2 Phasor Diagram Phasor diagram or vector diagram shows the graphical representation of alternating quantities in space and shows the phase relation between these quantities. In phasor diagram, RMS values (Root Mean Square values) are represented with their angles. All phasors represent the sine wave of same frequency.

Figures 2.47(a), (b) and (c) show the phasor diagrams of voltage and current phasors. Horizontal line is taken as reference. Leading phasors are drawn with angle measured in anticlockwise direction with the reference line while the lagging phasors are drawn with angle measured in clockwise direction with reference line.

Voltage in reference and current is leading by an angle φ. o0V V= ∠

I I= ∠φ

I

Vφ

Figure 2.47(a)

Voltage in reference and current is lagging by an angle φ. o0V V= ∠

I I= ∠− φ

Vφ

I

Figure 2.47(b)

Voltage is not reference and current is leading by an angle φ as shown in Figure 2.47(c).

V

Reference Line

I

θ

Figure 2.47(c)

; | | (V V I I )= ∠θ = ∠ θ + φ

61


2.6.3 Complex Representation of Voltage and Current 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 = + ⎭

. . . (2.52)

where 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 2.48 shows the representation of complex quantity on complex plane.

Im

a + jb

0Re

a

b

Figure 2.48 : Complex Plane

Polar Representation

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

1 1

2 2

i rr

= ∠θ ⎫⎬ν = ∠θ ⎭

. . . (2.53)

Rectangular to Polar Conversion

2 21

11and tan

r a bba

−

⎫= + ⎪⎬

θ = ⎪⎭

. . . (2.54)

2 22

12

Also

and tan

r c ddc

−

⎫= + ⎪⎬

θ = ⎪⎭

. . . (2.55)

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

1r ∠θ1 and 2 2r ∠θ

In Figure 2.49

2 21r OA a b= = +

and θ1 is the angle made of line OA from the horizontal axis.

62

Ar1

O

Electricity

θ1

Figure 2.49

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 . . . (2.56)

so 1 1cosa r= θ

and b r1 1sin θ =

The j Operator j operator is used to indicate the anticlockwise rotation of a phasor through 90o

1j = − . . . (2.57)

Properties of Operator j

j = 90o anticlockwise rotation of a phasor 1= −

j2 = 180o anticlockwise rotation of a phasor = − 1

j3 = 270o anticlockwise rotation of a phasor = − j j4 = 360o anticlockwise rotation of a phasor = 1

and 1 jj= − .

2.6.4 3-Phase AC Supply A three-phase AC supply can be produced by a three-phase AC generator having identical windings (or phases) displaced 120o electrical apart. When these windings are rotated in a stationary magnetic field or when these windings are kept stationary and the magnetic field is rotated, an emf is induced in each windings or phase. These emfs are of same magnitude and frequency but are displaced from one another by 120o electrical.

a1

N

a2

b1

b2

c2

c1

S

ωs

(a) Three-phase AC Generator

63


(b) 3φ emf Waveform

120o

E (c c )m 1 2

E (a a )m 1 2

E (b b )m 1 2

120o

120o

(c) Phasor Diagram

Figure 2.50 : Three Phase AC Supply

When the coils mounted on same axis are rotated in anticlockwise direction at ω rad/sec, their magnitude and direction, at this instant, are given below :

(a) The emf induced in coil a1 a2 is zero, and is increasing in the positive direction as shown by wave .

1 2a ae

(b) The coil b1 b2 is 120o (electrical) behind coil a1 a2. The emf induced in this coil is negative and is becoming maximum negative as shown by wave

. 1 2b be

(c) The coil c1 c2 is 120o (electrical) behind coil b1 b2 and it is 240o (electrical) behind coil a1 a2. The emf induced in this coil is positive and is decreasing as shown by wave .

1 2c ce

Equations

1 2 sina a me E= ωt

1 2o2sin sin ( 120 )

3b b m me E t E tπ⎛ ⎞= ω − = ω −⎜ ⎟⎝ ⎠

. . . (2.58)

1 2o4sin sin ( 240 )

3c c m me E t E tπ⎛ ⎞= ω − = ω −⎜ ⎟⎝ ⎠

2.7 AC GENERATOR

2.7.1 Principle The AC generator or synchronous generator is the most commonly used machine for the generation of electrical power all over the world. Such a synchronous generator is also referred to as an ‘alternator’ since it generates voltage. A synchronous generator like any other electrical rotating machine has two main

64

parts, i.e. the stator and the rotor. The part of machine in which voltage is induced is called armature. In a synchronous generator the armature winding is placed on the stator slots. The rotor carries the field poles which produce the required magnetic lines of force.

Electricity

A generator is built utilizing the basic principle that emf is induced in a conductor when it cuts magnetic line of forces. The synchronous machine is a doubly excited machine, i.e. its field winding is excited by DC supply and armature winding deals with AC. The synchronous machine runs only on synchronous speed corresponding to frequency of operation and is given by

2 120rps or rpms sfN N

P p= =

f . . . (2.59)

where f = frequency in Hz, P = number of poles, rps = rotation per second, and rpm = rotation per minute.

2.7.2 Construction In synchronous machine field, winding is normally placed on rotor and armature winding is placed on stator. This construction is more economical and provides better efficiency. The advantages of having low power DC field winding on rotor and high power AC armature winding on stator are

(a) Less number of slip rings and insulation requirement If we use low power DC field winding on rotor then only two slip rings (with less amount of insulation) are required. While in case of armature placed on rotor at least three slip rings with heavy insulation are required. So, providing field winding on rotor is more economical.

(b) It is easier to insulate stationary armature winding, since it has high voltage rating.

(c) Stationary armature winding can be cooled easily in comparison to rotating one.

(d) Since less number of slip rings are required, so there will be low slip ring losses and, therefore, high efficiency can be achieved. Also the maintenance required by slip ring will be less.

(e) Field winding is lighter in comparison to armature winding which reduces the weight of rotating part so high speed can be achieved.

Because of the above mentioned advantages, field winding is placed on rotor and armature winding on stator in synchronous machine. Stator Construction

Stator of synchronous machine is made of laminated cast iron frame. Stator is cylindrical in shape and has slots in its inner periphery (Figure 2.51). Following types of laminations are used for making the stator : Hot Rolled Steel

It reduces the core loss, but its laminations had many imperfection, like variation in thickness between individual sheets. Hot rolled sheets leave a thin insulating scale on the surface during their processing. So, now-a-days they are not in much use.

Cold Rolled Steel It is ideal for transformer core but its use is not so efficient in case of synchronous machine because of high surface loss in pole face. But cold

65


rolled process produces sheets with a bright finish which has no insulating property.

Isotropic Non-directional Cold Rolled Steel Modern synchronous machines use this type of laminations. Electrical characteristics of isotropic cold rolled steel are similar to hot rolled steel but it has much improved mechanical characteristics like uniform thickness of laminations, smoothness of surface, higher fatigue strength and lower clamping pressure.

Cube Oriented Sheet Steel Use of this type of steel is still limited because its use is very uneconomical. It has minimum losses and magnetization along the direction of rolling and at right angles to it, thus making it an ideal material for core.

Stator Frame

Stator Slots for 3-phase ArmatureWindings

Stator Core (Laminated)

Foundation

Figure 2.51 : Stator of Synchronous Machine

Stator Winding (Armature Winding)

3-phase star connected armature winding with neutral earthed is used. Star connection has the advantage that it eliminates all triple frequency harmonics from the line voltage.

Types of Armature Windings

Armature windings of synchronous machine may be classified as under :

(a) Concentrated and distributed windings.

(b) Full pitched coil winding and short pitched coil winding.

(c) Single layer and double layer winding.

(d) Integral slot and fractional slot winding.

Winding Factor

w pK K K= × d . . . (2.60)

where Kp = Pitch factor emf with distributed windingemf with concentrated winding

= ,

Kd = Distribution factor emf induced in a short-pitched coilemf induced in a full-pitched coil

= , and

1wK < .

Comparison of Cylindrical Rotor Alternator (Non-salient Pole Type or Turbo Alternator) and the Salient Pole Alternator (Hydro or Water Wheel Generator)

66

Stator frameStator core

Uniform air gap

N

S

Concentratedfield winding

N

a2

c1N

SS

a1

a’2 a’1

b’1c’1

b2

b1c2

c’2b’2

Distributedarmaturewinding

Electricity

Cylindrical Rotor Type Salient-Pole Rotor Type

Figure 2.52 : Turbo Alternator Figure 2.53 : Hydro Alternator

Turbo Alternator (Cylindrical Rotor Type)

Hydro Alternator (Salient Polar Rotor Type)

Air gap is uniform because of the cylindrical rotor.

Air gap is non-uniform.

Reluctance and hence reactance is constant, i.e. direct axis (polar axis) reluctance and reactance = quadrature axis (inter polar axis) reluctance and reactance. i.e. Xd = Xq (theoretically) (Practically Xd is about 10% more than the Xq)

Here direct axis reactance > quadrature axis reactance, i.e. Xd > Xq. Xd is about 60% more than the Xq and direct axis reluctance is less than quadratic axis reluctance.

These are used for high speed operations. (1500 to 3000 rqm)

These are used for low and medium speed operations (150 to 600 rpm).

These machines are more stable Less stable as compared to turbo alternators. Rotor normally has large axial length and small diameter.

Its rotor has larger diameter and smaller axial length.

Steam and gas turbines are used as the prime mover, because they have their best characteristics at high speeds.

It uses the hydraulics turbines like pelton wheel, Francis and Kaplan as the prime mover.

The radial air gap flux density wave is more nearer to the sine wave.

Wave has larger harmonic as compared to the cylindrical rotor machine.

Since air gap is constant and Xd = Xq, analysis of these machine is simple.

Analysis is complicated and two reaction theory is used.

They are made in large sizes for larger ratings.

They are smaller in size and lower in ratings.

2.7.3 Length of Air Gap Length of air gap between stator and rotor is very important design parameters, since it affects the performance and stability of synchronous machine. Advantages of Large Air Gap

Stability A higher value of stability limit.

Regulation Smaller value of inherent regulation.

Synchronization Power

Higher synchronization power that makes the machine less sensitive to load vibrations.

Cooling

Better cooling at the gap surface.

67


Noise

Reduction in the noise level.

Magnetic Pull

Smaller unbalanced magnetic pull.

Disadvantages of Large Air Gap

Field mmf

Larger value of field mmf is required.

Size

Bigger size of the machine with a larger diameter of the stator.

Magnetic Leakage

Increased magnetic leakage.

Weight of Copper

Greater weight of copper in the field winding.

Cost

Increased over all cost.

Example 2.9

Calculate the highest speed at which the alternator (AC generator) can be operated for (a) 50 Hz, and (b) 60 Hz frequency.

Solution

Since it is not possible to have poles fewer than 2, so minimum value of P = 2

120

sPNf =

Synchronous speed, 120s

fNP×

=

(a) Frequency is 50 Hz

120 50 3000 rpm2sN ×

= =

(b) Frequency is 60 Hz

120 60 3600 rpm2sN ×

= =

Example 2.10

A 3-phase 6-pole star connected alternator revolves at 1000 rpm. The stator has 90 slots and 8 conductors per slot. The flux per pole is 0.05 wb (sinusoidally distributed). Calculate the voltage generated by the machine if winding factor is 0.96.

Solution

120sPNf =

6 1000 50 Hz120×

= =

Total number of stator conductors

= Conductors per slot × Number of slots

68

Electricity = 8 × 90 = 720 Stator conductor per phase

720 2403pZ = =

Winding factor kw = 0.96 Generated voltage per phase

EP = 2.22 kw f φ ZP

= 2.22 × 0.96 × 50 × 0.05 × 240 = 1278.7 volt

Generated line voltage

3 3 1278.7pE= = ×

= 2214.7 volt

SAQ 4 (a) A 6-pole AC generator rotates at 1000 rpm. Calculate the frequency of

generated emf.

(b) The stator of 3-phase, 8 pole, 750 rpm alternator has 72 slots, each of which contains 10 conductors. Calculate effective value of emf per phase if the flux per pole is 0.1 wb [Kw = 0.96].

2.8 TRANSFORMER

Transformer is the static device which transfers electrical energy from one electric circuit (primary circuit) to another electric circuit (secondary circuit) which are not electrically connected, but magnetically coupled.

Transformer is the main part of ac transmission and distribution system. Almost the entire world production of electrical energy is transformed twice, thrice or even more before being utilized, by means of transformer.

Transformer does not change the frequency of electrical supply. It merely changes the amplitude of the voltage from one electric circuit to another.

Transformer is a static device, so has no rotational losses and the efficiency as high as 99% can be obtained.

Transformer is not an energy conversion device, but it is an important component in most of the energy conversion systems. It is the main reason behind the widespread use of AC power systems. It makes possible generation, transmission and distribution of electrical energy at their most economical and suitable voltages.

2.8.1 Principle Transformer is based on the principle of electromagnetic induction. The need for the transformer action is only the existence of time varying mutual flux linking the two electric circuits. The induced emf in secondary due to alternating current in primary or establishment of an alternating flux in the magnetic circuit is given by;

1dedtψ

= ψ = N1 φ = Total flux linked with the primary

69

Current, Electricityand Its Effect 1

dNdtφ

=

where N1 = Number of turns in primary, and φ = Flux per turn.

Let sinm tφ = φ ω

then 1 1 cosme N= ω φ ωt

1max 1 12m mE N f N= ω φ = π φ

where f = supply frequency in Hz.

1max1(rms) 12 v

2 mE

E f N= = π φ olts . . . (2.61)

14.44 Voltsmf N= φ

Similarly 2 22 vmE f N= π φ olts

∴ Turn ratio 2 2

1 1

E NE N

= . . . (2.62)

I1

ZL

S

V

Figure 2.54 : Air Cored Transformer

ZL

S

φ

V

i

Figure 2.55 : Iron Cored Transformer

For getting the minimum leakage of flux, to get the maximum efficiency, core is made of soft iron rather than air (shown in Figures 2.54 and 2.55). Advantages of Iron Core Transformer over Air Core Transformer

(a) Permeance of the magnetic circuit is increased so it increases the total flux linked with the transformer.

(b) Magnetizing current (no load current) is reduced. (c) Power factor of transformer is improved. (d) Minimum leakage flux.

Classification of Transformers

Transformers can be classified on different basis like :

On the Basis of Construction

(a) Core type

(b) Shell type

On the Basis of Working

(a) Step down

70

Electricity (b) Step up

On the Basis of Application

(a) Power transformer

(b) Distribution transformer

On the Basis of Number of Phases

(a) Single phase

(b) Three phase

On the Basis of Number of Windings

(a) Single winding (auto transformer)

(b) Two winding (normal transformer)

(c) Three winding (with delta connected tertiary winding)

On the Basis of Cooling Method Used

(a) Air or natural cooled

(b) Oil immersed

Special Units

(a) Instrument Transformer (current and potential transformer)

(b) Bell transformer

(c) Welding transformer

(d) Pulse transformer

2.8.2 Construction Following materials are used for transformer construction :

Conducting Material

For the windings of transformer Aluminum, Copper and Silver can be used.

Magnetic Material

Following magnetic materials are used for core of the transformer :

(a) Silicon steel

(b) Nickle iron

(c) Hot rolled grain oriented (HRGO)

(d) Cold rolled grain oriented (CRGO)

Insulating Material

Insulating material is used for insulating the windings with core. Also it is used between the turns of windings, between low voltage and high voltage windings. Following insulating materials are most commonly used :

(a) Press board

(b) Mica

(c) Double covered copper

(d) Double covered silk

(e) Asbestos

(f) Insulating oil (transformer oil)

On the basis of construction, transformer is classified as :

(a) Core type

(b) Shell type

71


Comparison between Core and Shell Type Transformers

Core Type

HV LV LV HV

Shell Type

HVLV

Core

HVLV

LV

Limb

Figure 2.56 : Cross-section View of Figure 2.57 : Cross-section View of Core Type Transformer Shell Type Transformer

Core Type Shell Type

In core type, windings surround a considerable part of steel core.

In shell type, steel core surrounds a major part of windings.

They require less iron and more copper. It requires more iron but less conductor material.

The flux has a single path around the core. Flux is distributed because it has three limbs.

Core type transformer has concentric winding.

It has sandwiched winding.

Cooling and repairing of windings are easy.

Cooling and repairing are difficult.

It has larger value of leakage reactance. Leakage reactance is smaller because of better linkage between LV and HV windings.

Various Forms of Core Cross-section

Generally circular coils are used for transformer windings. Hence ideally cores of circular cross-section are desirable but it is very complicated to manufacture circular cone. In practice for small transformers square cone and for large transformers stepped core is used.

These cores are shown is Figure 2.58.

(a) Square Core (b) Three Shaped Core (c) Two Stepped Core

Figure 2.58 : Various Forms of Core

Example 2.11

A 10 kVA, 500/250 V, 50 Hz, single phase transformer has a net area of cross-section 90 cm2 and maximum flux density is 1.2 wb/m2. Calculate the number of turns on both primary and secondary.

Solution

Primary induced emf, E1 = V1 = 500 V

Secondary induced emf, E2 = V2 = 250 V

Supply frequency, f = 50 Hz

Maximum flux density, BBmax = 1.2 wb/m2

Net cross section area of the core, Ai = 90 cm2

72

= 0.009 m2Electricity

Maximum flux, φm = BBm × Ai

= (1.2) (0.009) = 0.0108 wb

Number of primary turns, 11 4.44 m

ENf

=φ

5004.44 50 0.0108

=× ×

= 208.54

Number of secondary turns, 22 4.44 m

ENf

=φ

2504.44 50 0.0108

=× ×

= 104.27 Example 2.12

A single-phase transformer has 400 primary and 1000 secondary turns. The net cross-sectional area of the core is 60 cm2. If the primary winding be connected at 50 Hz supply at 500 V, calculate :

(a) The peak value of the flux density in the core. (b) The voltage induced in the secondary winding.

Solution

Primary induced emf, E1 = 500 V Supply frequency = 50 Hz

Primary turns, N1 = 400

Net cross section area, Ai = 60 cm2

= 0.006 m2

Maximum value of flux, 1

14.44mE

f Nφ =

5004.44 50 400

=× ×

= 0.00563 wb

(a) Peak value of flux density in the core, maxmax

iB

Aφ

=

= 0.3384 wb/m2 (b) Voltage induced in the secondary winding

22 1

1

1000500400

NE EN

⎛ ⎞= = ×⎜ ⎟

⎝ ⎠

= 1250 volt.

SAQ 5 (a) A 25 kVA transformer has 500 turns on the primary and 40 turns on

secondary winding. Primary is connected to 3000 volt, 50 Hz mains. Calculate :

73


(i) Primary and secondary current at full load,

(ii) Secondary emf, and

(iii) Maximum flux in core.

Neglect magnetic leakage, resistance of winding and primary no load current.

(b) A 200 kVA, 3300 / 240 volts, 50 Hz single phase transformer has 80 turns on secondary winding. Assuming an ideal transformer calculate :

(i) Primary and secondary currents at full load,

(ii) Maximum value of flux, and

(iii) Number of primary turns.

2.9 INDUCTIVE AND CAPACITIVE REACTANCE

Reactance is the opposition offered to the flow of current due to the inductance and capacitance in the circuit. This hindrance is represented by ‘X’ and has unit ohm (Ω). Inductive Reactance

It is the opposition offered to flow of current due to inductance in the circuit. It is represented by ‘XL’.

XL = ω L Ω = 2π f L Ω . . . (2.63)

where f = frequency in Hz, and ω= frequency in rad/sec.

As the frequency increases, inductive reactance increases. Capacitive Reactance

It is the opposition offered to flow of current due to capacitance in the circuit. It is represented by ‘XL’.

1LX

C= Ωω

12 f C

= Ωπ

. . . (2.64)

As frequency increases, capacitive reactance decreases.

2.10 IMPEDANCE

It is the complex summation of resistance and reactance. Its unit is ohm (Ω) and it is denoted by letter Z.

Z R j X= +

( )L CR j X X= + −

1R j LC

⎛ ⎞= + ω −⎜ ⎟ω⎝ ⎠

2 2

1

Magnitude

Angle tan

Z R XXZR

−

⎫= + ⎪⎬⎪∠ = =⎭

φ . . . (2.65)

74

This is also the angle between voltage and current in the circuit (power factor angle). Electricity

Impedance Triangle

It is the right angle triangle in which φ is the angle between voltage and current in the circuit and is given by the equation.

2 2Z R X= +

1tan XR

−φ =

R

XZ

φ

Figure 2.59 : Impedance Triangle

2.11 RESISTIVE, INDUCTIVE AND CAPACITIVE CIRCUITS

2.11.1 AC Through Pure Resistance Let single phase AC voltage sinmV tν = ω is applied to the resistive circuit shown in Figure 2.60.

R

i

v = Vm sin ωt

Figure 2.60 : Resistive Circuit

Then by Ohm’s law

sinmVv vi tz R R

= = = ω . . . (2.66)

= Im sin ωt So voltage v and current i are in same phase the phasor diagram and waveforms are shown in Figure 2.61.

ωt

(a) Phasor Diagram (b) Waveform Figure 2.61

Angle between voltage and current φ = 0o, so power factor cos φ = 1 (unity). Power Consumed

p = νi (instantaneous power) = (Vm sin ωt) (Im sin ωt) p = Vm Im sin2 ωt

75

Current, Electricityand Its Effect = (1 cos 2 )

2m mtV I − ω

Average power dissipated watts2

m mV I= .

2.11.2 AC Through Pure Inductance Now consider the circuit shown in Figure 2.62.

L

i

v = Vm sin ωt

Figure 2.62

Voltage across inductor is given by

div Ldt

=

∴ 1 1 sinmi dt VL L

= ν = ω∫ ∫ t dt

cosmVt

L= − ω

ω

sin2

mVt

Lπ⎛ ⎞= ω −⎜ ⎟ω ⎝ ⎠

. . . (2.67)

Here current i is lagging behind the voltage ν by 2π .

Inductive reactance, XL = ωL Ω

then sin2

m

L

Vi t

Xπ⎛ ⎞= ω −⎜ ⎟

⎝ ⎠

sin2mI t π⎛ ⎞= ω −⎜ ⎟

⎝ ⎠ . . . (2.68)

Angle between voltage and current φ = 90o (lagging) p.f. = cos φ = cos 90o = 0.

The voltage, current phasor diagram and waveform are shown in Figure 2.63.

Average power (Pav = VI cos φ) consumed in pure inductive circuit is zero.

V

i

I

90o

ωt

(a) (b)

Figure 2.63

2.11.3 AC Through Pure Capacitance Circuit of Figure 2.64 shows the pure capacitive circuit with applied voltage.

ν = Vm sin ωt.

76

C

i

Electricity

v = Vm sin ωt

Figure 2.64

Current di Cdtν

= . . . (2.69)

( sinmdC V tdt

)= ω

cosmC V t= ω ω

sin1 2mV

t

C

π⎛ ⎞= ω +⎜ ⎟⎝ ⎠

ω

Here current I is leading the voltage v by 2π .

Capacitive reactance, 1CX

C= Ωω

.

∴ sin2

m

C

Vi t

Xπ⎛ ⎞= ω +⎜ ⎟

⎝ ⎠

sin2mI t π⎛ ⎞= ω +⎜ ⎟

⎝ ⎠ . . . (2.70)

where mm

C

VI

X=

Angle between voltage and current φ = 90o (leading).

∴ p.f. = cos φ = cos 90o = 0

Average power consumed by pure capacitive circuit is zero. The phasor diagram and waveform voltage and current are shown in Figure 2.65.

/2

i

V

I

90o

ωt

(a) (b)

Figure 2.65

2.11.4 AC Through R-L-C Circuit The series RLC circuit is shown in Figure 2.66.

C L R

77

C

i

LR


v = Vm sin ωt

Figure 2.66

KVL equation for the given circuit is

1diRi L i ddt C

ν = + + ∫ t

Total impedance ( )L CZ R j X X= + −

where XL = Inductive reactance = ωL

XC = Capacitive reactance 1C

=ω

By ohm’s law, current I is given by

iZ R jXν ν

= =+

where L CX X X= −

2 2 1

sin

tan

mV ti XR X

R−

ω=

+ ∠

12 2

sin tanmV Xi tRR X

−⎛ ⎞= ω −⎜ ⎟⎝ ⎠+

. . . (2.71)

Angle between voltage and current 1tan XR

−φ =

This angle φ will be angle of lead if XC > XL (circuit is capacitive)

Angle φ will be angle of lag if XL > XC (circuit is inductive)

In this AC circuit there are three types of powers :

(a) Apparent power

(b) Active power

(c) Reactive power

These powers are explained in detail later.

Table 2.1 shows the summary of results obtained.

Table 2.1 : Supply Voltage ν = Vm sin ωt

Circuit Impedance Current i (Amp)

Power Factor Angle, Phasor Diagram

78

Electricity Z (Ω) (φ) R

V = V sinwtm

V

Z = R Ωs sin

Vmi tR

= ω φ = 0o

VI

L

ν

Z = j XL = ωL Ω osin ( 90 )Vmi tX L

= ω − φ = 90o (lagging) V

I

90o

C

ν

Z = − j XC Ω

= − j / ωC osin ( 90 )

Vmi tXc

= ω + φ = 90o (leading)

V

I

90o

L

ν

R

Z = R + j XLsin ( )

Vmi tZ

= ω − φ 1tanX LR

−φ = (lagging)

0 < φ < 90o (lag)

Vφ

I

C

ν

R

Z = R – j XCsin ( )

Vmi tZ

= ω + φ 1tanXCR

−φ = (leading)

0o < φ < 90o (lead)

I

Vφ

CL

ν

R

Z = R + j (XL − XC) sin ( )

Vmi tZ

= ω ± φ 1tanX XL C

R

−−φ =

Solved Examples on Single Phase R-L-C Circuit

Example 2.13

In series R-L circuit, shown in Figure 2.67, find

(a) Impedance,

(b) Resultant current,

(c) Power factor and its nature, and

(d) Quality factor.

220 0 ; 50Hzo

i

L = 0.1HR = 10

Figure 2.67

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

79


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 2.14

An 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 2.14

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

80

Electricity | | 35.6439LZ = Ω

Voltage across coil = I ZL = 7 × 35.6439

= 249.5075 Volt

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

Example 2.15

In the capacitive circuit of Figure 2.69, 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 2.69

Solution

(a) The capacitive reactance is given by

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

81


= 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 2.70 and applied Voltage of 100 ∠ 0o is taken as reference.

87.84o

I

VR

VC

V=100VV V V= +R C

2.16o

Figure 2.70 : Phasor Diagram

SAQ 6 (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.

2.12 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.

82

2.12.1 Series Resonance Electricity

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 . . . (2.72)

In the series RLC circuit of Figure 2.71, the circuit current I is given by

V volts, f Hz

VR

I

CLR

VL VC

Figure 2.71 : Series Resonance Circuit

VI AZ

=

where Z represents the equivalent impedance of the circuit. 1Z R j L

j C= + ω +

ω

1R j L jC

= + ω −ω

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 2.72 : 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 2.73.

83

VL

VC

VR

V


(VL – Vc)

Figure 2.73 : 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 = = = . . . (2.73)

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

or 20 0

1 1i.e. rad/ secLC LC

ω = ω =

i.e. 01 Hz

2f

LC=

π . . . (2.74)

Voltage vector diagram at resonance is shown in Figure 2.74. VL

VC

I V

Figure 2.74 : Voltage Vector Diagram at Resonance (VL = Vc)

Impedance Curve

Figure 2.75 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 2.75 : Impedance Curve

84

Angle Curve Electricity

Angle φ (between voltage and current) is given by

1 ( )tan L cX XR

− −φ = . . . (2.75)

Here, φ is known as the power factor angle. Figure 2.76 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 2.76. As ω approaches zero the angle of Z approaches – 90.

0

High R

Low R+90o

-90o

φ

ω ω0

Figure 2.76 : Angle Curve

Admittance Curve Admittance Y is reciprocal of impedance. Its variation with frequency is shown in Figure 2.77.

1YZ

= . . . (2.76)

Low R

Adm

ittan

ce

High R

w0 Figure 2.77 : Admittance Curve

Resonance Curve Figure 2.78 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.

85

I

I0

I / 20


ω (rad/sec) ω1 ω0 ω2

Figure 2.78 : 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. . . . (2.77)

or BW = f2 – f1 in Hz . . . (2.78)

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

0 1 2 0 1and 2f f fω = ω ω = . . . (2.79)

2.12.2 Parallel Resonance Figure 2.79 represents a parallel resonating circuit where a coil is connected in parallel with a capacitor C and the combination is connected across a AC voltage source of variable frequency. Figure 2.80 represents the vector diagram of the given circuit.

R L

Coil

CIC

IL

IV

AC .f Hz

Figure 2.79 : A Typical Parallel Resonating Circuit

IL

I cosL φ

I sinL φφ

IC

V

Figure 2.80 : 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,

86

Electricity 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

= . . . (2.80)

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=

Also 20

0

1 . , i.e.LLZ L Z

C C= × ω = =ω L

LC

. . . (2.81)

[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 = − = − . . . (2.82)

i.e. 20

12

Lf RL C

= −π

. . . (2.83)

If the resistance of the coil be neglected,

01 1

2 2Lf

L C LC= =

π π . . . (2.84)

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.]

87


or, 2L

LZ LCZR R CRΩ = = = L

LZC

⎡ ⎤=⎢ ⎥

⎣ ⎦Q . . . (2.85)

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 2.81 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 2.81

BBC = ωC and BLB =1/ωL are the capacitive and inductive susceptance. Net susceptance B = BBC – BL B

= ωC – 1/ωL . . . (2.86) 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 2.82 : 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.

88

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

Electricity

(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 2.16

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 2.83

Also find the expression for resonant current.

Solution

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 2.17

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

89


R

5j -3j

4

Figure 2.84

Solution

Admittance of first branch

1 21 5 mho ( )

5 25R jY

R j R−

= =+ +

J

Admittance of second branch

21 4 3 mho ( )

4 3 25jY

j+

= =−

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−

+ =+

25 3

2525R=

+

⇒ = 4.082 ΩR

Example 2.18

A circuit shown in Figure 2.85 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

90

C

110 V, 50 Hz

0.4 H5Ω

i

Electricity

Figure 2.85

(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

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

ω π= =

2 50 0.5

4π × ×=

= 25.1327

Example 2.19

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 2.86

91


Solution

(a) Let capacitance = C farad

Resonant frequency, f0 = 400 Hz

Capacitive reactance, 1 1 12 400 800CX

C C= = =ω π × × π × C

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

=+

(ΘApplied voltage V = 25 V)

RL = 30 Ω

SAQ 7 (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 2.87 to be resonant at all frequencies.

Figure 2.87

92

Electricity 2.13 POWER, POWER FACTOR AND QUALITY FACTOR

2.13.1 Power We know that power is defined as the time rate of change of energy and is given by P.

dwPdt

= . . . (2.87)

dw dqdq dt

× v i= × . . . (2.88)

Energy is given by

joulesw P dt= ∫ . . . (2.89)

Power in AC

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

where Z = impedance (ohm)

Active Power

It is given by P = VI cos θ Watt . . . (2.90)

It is the power, which is actually consume in any circuit. Its unit is watt.

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.

It is not consumed in the circuit but it flows from source to load and then load to source.

Power Triangle

All the three power S, P and Q can be represented in a triangle, called the power triangle as shown in Figure 2.88.

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 2.88 : Power Triangle

Power can also be represented in complex form.

93


S = P + jQ . . . (2.91)

Magnitude 2 2S P Q= +

2.13.2 Power Factor

It may be defined as

• Cosine of the angle of lead or lag

• The ratio ResistanceImpedance

RZ=

• The ratio True Power Watts=Apparent Power Volt amperes

WVA

=

2.13.3 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

or 1cos

ZQR

= =φ

2 2

LR XR+

=

If R is very less as compared to XL, then

LX LQR R

ω= = .

Example 2.19

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

220 0 ;60Hz∠ o

i

0.2H15Ω

Figure for Example 2.19

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

94

Electricity

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.

2.14 SUMMARY

This unit gives the different applications of electric current, detailed discussion about the self and mutual inductance and capacitance. Unit provides the magnetic effects of electric current and relations are derived for long straight conductor, solenoid, parallel conductors and coils. For developing the concept of magnetic effect of electric current two basic laws – Biot-savart law and Ampere’s circuital law – are given. Biot-savart law gives the relation of magnetic flux density B

uror magnetizing force H

uurin terms of electric current I.

Similarly, Amperes circuital law provides the relation for work done or mmf in terms of current. Also, concept of magnetic flux and inductance (self and mutual) are discussed in detail. Equivalent inductance for parallel and series combination is derived. As inductor stores energy in the form of magnetic field so the formula for it is (1/2 Li2 joules) also determined in this unit.

95


Important applications of electricity has been discussed by introducing three important machines in this unit. The main stress is given to principle, construction, working and important characteristics of DC generator, AC generator and transformer. In the next section of the unit, we discuss the AC circuit with R, L and C with single phase AC supply. We define inductive and capacitive reactance (XL and XC), impedance, impedance triangle, power and power factor. Relation for current and its lagging and leading effect for AC circuit are also determined. Series and parallel resonance along with resonant characteristics are also presented at the end.

Sufficient examples and SAQs are given for better understanding of each topic.

2.15 ANSWERS TO SAQs

SAQ 1 (a) ψ = Li

Nφ = Li

150 0.01 0.15 Henry10

NLiφ ×

= = =

emf induced die Ldt

= −

[ ]10 100.15 300 Volt

0.01− −

=− =

(b) (i) Self inductance, 2

ReluctanceNL =

2

4(100)10

=

= 1 Henry

(ii) Energy Stored, 212

E Li=

21 1 (5) 12.5 Joule2

= × × =

(iii) Induced emf diLdt

=

= 1 × 200 = 200 Volt (c) O O O O O O O O solenoid

O O O O O O 3 cm coil O O O O O O O O O O O O O O 60 cm Coil turns = 50; Solenoid turns = 2500; and I = 2 Amp. Flux density due to solenoid :

0NIB Hl

=μ =

20 2

2500 2 wb/m60 10−

×= μ

×

96

Electricity Area of coil, 2 2

1 (3 10 )4

A −π= × ×

4 29 10 m4

−π= × ×

Flux linked with coil, 1 1BAφ =

402

2500 29 10 wb

460 10−

−μ × × π

= × × ××

Mutual inductance, 2 1NMIφ

=

N2 = 50

∴ M = 0.185 mH. (d) (i)

2Hi

4H

1H

(a) Coils are connected in series so

Leq = L1 + L2 + 2M = 2 + 4 + 2(1) = 8 H

(ii)

5H

8H

2H

i

Figure for Answer to SAQ 1(d)

Coils are connected in series opposition so

Leq = L1 + L2 – 2M = 5 + 8 – 2(2)

= 9 H

SAQ 2

(a) 0 Faradr ACd

ε ε=

A = area of plate, d = distance between the plates 12 4

38.85 10 6 11 10

2 10C

− −

−× × × ×

=×

= 29.2 pF

(b) Ceq = C1 + C2 + C3 = 7 μF

Total charge, Q = Ceq ν

= 7 × 10− 12 × 200

= 14 × 10− 10 Coulomb

97


Charge across C1 = Q1 = C1 ν

= 10− 6 × 200 = 2 × 10− 4 C Charge across C2 = Q2 = C2 ν

= 2 × 10− 6 × 200 = 4 × 10− 4 C Charge across C3 = Q3 = C3 ν

= 4 × 10− 6 × 200

= 8 × 10− 4 C (c) The composite capacitance is given by

1 2 3

0

31 2

r r r

AC

dd d

ε=⎡ ⎤⎢ ⎥+ +ε ε ε⎢ ⎥⎣ ⎦

12 4

3 3 38.85 10 (100 10 )

10 10 103 4 6

− −

− − −× × ×

=

+ +

1111.8 10−= ×

= 118 pF

Total charge, Q = CV

= 11.8 × 10− 11 × 2000 = 23.6 × 10− 8 C

Flux density

8

6 24

23.6 10 23.6 10 C/m100 10

QDA

−−

−×

= = = ××

Potential gradient, 1

10

8.9 kV/cmr

DE = =ε ε

2

20

6.67 kV/cmr

DE = =ε ε

33

04.44 kV/cm

r

DE = =ε ε

SAQ 3

(a) Total lamp load

P1 = Number of lamps × Voltage of each lamp

= 60 × 40 = 2400 W

Terminal voltage, V = 100V

Load current, 1 2400 24 Amp100L

PIV

= = =

Shunt field current, 100 2 Amp50sh

sh

VIR

= = =

(i) Total armature current, Ia = IL + Ish

= 24 + 2

= 26 Amp

98

(ii) Current per path, 26 6.5 A4

aC

II

A= = =

Electricity

(iii) Generated emf, Eg = V + Ia ra + Brush Drop

= 100 + 26 × 0.1 + 2 × 1

= 104.6 Volt

(b) Load, PL = 25 × 1000 = 25000 Watt

Load voltage, VL = 500 V

Load current, 50 AmpLL

L

PIV

= =

Shunt field current, 500 2.5 Amp200sh

sh

VIR

= = =

Armature current, Ia = IL + Ish

50 + 2.5 = 52.5 Amp

Generated emf, Eg = V + Ia ra + Is Rse + Brush Drop

= 500 + 52.5 × 0.03 + 52.5 × 0.04 + 1

= 504.675 volt

SAQ 4

(a) P = 6, N = 1000 rpm

Frequency of generated emf : 6 1000 50 Hz120 120PNf ×

= = =

(b) Generated emf per phase

EP = 2.22 Kp Kd f φ Zp

= 2.22 kw f φ ZP [Kw = Kp kd]

Kw = winding factor

ZP = No of conductors/phase

Total No. of conductors = 72 × 10 = 720

Conductors per phase, 720 2403PZ = =

Frequency, 8 750 50 Hz120 120PNf ×

= = =

∴ EP = 2.22 (0.96) × 50 × 0.1 × 240

= 2557 volt

SAQ 5

(a) (i) At full load, 125 1000 8.33 Amp

3000I ×= =

1 2

2 1

I NI N

=

∴ 12 1

2

500 8.33 104.15 Amp40

NI IN

= × = × =

99

Current, Electricityand Its Effect(ii) Secondary emf, 2

2 11

NE EN

= ×

40 3000 240 V500

= × =

(iii) Using relation, E1 = 4.44 N1 f φm

3000 = 4.44 × 500 × 50 × φm

φm = 0.027 wb

(b) 1200 1000 60.6 Amp

3300I ×= =

2200 1000 833.3 Amp

240I ×

= =

E2 = 4.44 N2 f φm

240 = 4.44 × 80 × 50 × φm

φm = 13.51 mwb

Now 1 1

2 2

N EN E

=

⇒ 1 1100 turnsN =

SAQ 6

(a)

i

Z

ν Figure for Answer to SAQ 6(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

100

Current, I = 12 Amp, Electricity

Power, P = 720 Watts

Current resistance = R Ω

P = I2 R

2 2720 5

(12)PRI

= = = Ω

But P = VI cos φ

P.f. 720cos 0.35294170 12

PVI

φ = = =×

φ = 69.33275o

Impedance 170 14.166712

VI

= = = Ω

Reactance = XL

tan LXZ

θ =

∴ XL = 37.556

= ω L

∴ 99.61 mHLXL = =ω

SAQ 7

(a)

I

R=1.06Ω L C

125V, 220.6 Hz Figure for Answer to SAQ 7(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= = =

101


(b) R=40Ω L=0.75 H C

250V, 50 Hz Figure for Answer to SAQ 7(b)

Resonance frequency, f0 = 55 Hz

Let 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.

current

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