Bird J. Electrical Circuit Theory and Technology

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708 Electrical Circuit Theory and Technology

The area of a hysteresis loop plotted for a ferromagnetic material is

80 cm

, the maximum ﬂux density being 1.2 T. The scales of B and H

are such that 1 cm D 0.15 T and 1 cm D 10 A/m. Determine the loss

due to hysteresis if 1.25 kg of the material is subjected to an alter-

nating magnetic ﬁeld of maximum ﬂux density 1.2 T at a frequency

of 50 Hz. The density of the material is 7700 kg/m

[0.974 W]

4 The cross-sectional area of a transformer limb is 8000 mm

and the

volume of the transformer core is 4 ð 10

. The maximum value

of the core ﬂux is 12 mWb and the frequency is 50 Hz. Assuming the

Steinmetz constant is 1.6, the hysteresis loss is found to be 250 W.

Determine the hysteresis loss when the maximum core ﬂux is 9 mWb,

the frequency remaining unchanged. [157.8 W]

5 The hysteresis loss in a transformer is 200 W when the maximum ﬂux

density is 1 T and the frequency is 50 Hz. Determine the hysteresis

loss if the maximum ﬂux density is increased to 1.2 T and the

frequency reduced to 32 Hz. Assume the hysteresis loss over this

range to he proportional to B



1.6

. [171.4 W]

6 A hysteresis loop is plotted to scales of 1 cm D 0.004 T and 1 cm D

10 A/m and has an area of 200 cm

. If the ferromagnetic circuit for

the loop has a volume of 0.02 m

and operates at 60 Hz frequency,

determine the hysteresis loss for the ferromagnetic specimen.

[9.6 W]

Eddy current loss

7 In a magnetic circuit operating at 60 Hz, the eddy current loss is

25 W/m

. If the frequency is reduced to 30 Hz with the ﬂux density

remaining unchanged, determine the new value of eddy current loss

per cubic metre. [6.25 W/m

]

8 A transformer core operating at 50 Hz has an eddy current loss of

150 W/m

and the core laminations are 0.4 mm thick. The core is

redesigned so as to operate with the same eddy current loss but at

a different voltage and at 200 Hz frequency. Assuming that at the

new voltage the ﬂux density is half of its original value and the

resistivity of the core remains unchanged, determine the necessary

new thickness of the laminations [0.20 mm]

9 An inductor core has an eddy current loss of 25 W and a hysteresis

loss of 35 W when operating at 50 Hz frequency. Assuming that the

ﬂux density remains unchanged, determine (a) the value of the losses

if the frequency is increased to 75 Hz, and (b) the total core loss if

the frequency is 50 Hz and the laminations are 2/5 of their original

thickness. [(a) P

D 52.5W,P

D 56.25 W (b) 36.6 W]

10 A transformer is connected to a 400 V, 50 Hz supply. The hysteresis

loss is 250 W and the eddy current loss is 120 W. The supply voltage

Magnetic materials 709

is increased to 1.2 kV and the frequency to 80 Hz. Determine the

new total core loss if the Steinmetz index is assumed to be 1.6

[2173.6 W]

11 The hysteresis and eddy current losses in a magnetic circuit are

5 W and 8 W respectively. If the frequency is reduced from 50 Hz

to 30 Hz, the ﬂux density remaining the same, determine the new

values of hysteresis and eddy current loss. [3 W; 2.88 W]

12 The core loss in a transformer connected to a 600 V, 50 Hz supply

is 1.5 kW of which 60% is hysteresis loss and 40% eddy current

loss. Determine the total core loss if the same winding is connected

to a 750 V, 60 Hz supply. Assume the Steinmetz constant to be 1.6

[2090 W]

Separation of hysteresis and eddy current losses

13 Tests to determine the total loss of the steel core of a coil at different

frequencies gave the following results:

Frequency (Hz) 40 50 70 100

Total core loss (W) 40 57.5 101.5 190

Determine the hysteresis and eddy current losses at (a) 50 Hz and

(b) 80 Hz. [(a) 20 W; 37.5 W (b) 32 W; 96 W]

14 Explain why, when steel is subjected to alternating magnetization

energy, losses occur due to both hysteresis and eddy currents.

The core loss in a transformer core at normal ﬂux density was

measured at frequencies of 40 Hz and 50 Hz, the results being 40 W

and 52.5 W respectively. Calculate, at a frequency of 50 Hz, (a) the

hysteresis loss and (b) the eddy current loss.

[(a) 40 W (b) 12.5 W]

15 Results of a test used to separate the hysteresis and eddy current

losses in the core of a transformer winding gave the following results:

Total core loss (W) 48 96 160 240

Frequency (Hz) 40 60 80 100

If the ﬂux density is held constant throughout the test, determine the

values of the hysteresis and eddy current losses at 50 Hz.

[20W;50W]

16 A transformer core has a total core loss of 275 W at 50 Hz and

600 W at 100 Hz, the ﬂux density being constant for the two tests.

(a) Determine the hysteresis and eddy current losses at 75 Hz. (b) If

the ﬂux density is increased by 40% and the lamination thickness is

increased by 20% determines the hysteresis and eddy current losses

at 75 Hz. Assume the Steinmetz index to be 1.6

[(a) 375 W; 56.25 W (b) 642.4 W; 190.5 W]

Assignment 12

This assignment covers the material in chapters 36 to 38.

The marks for each question are shown in brackets at the end of

each question.

1 A voltage waveform represented by

v D 50sin ωt C20sin



3ωt C





C 5sin



5ωt C





volts

is applied to a circuit and the resulting current i is given by

i D 2.0sin



ωt 





C 0.462sin3ωt

C 0.0756sin5ωt  0.71 amperes.

Calculate (a) the r.m.s. voltage, (b) the mean value of voltage, (c) the

form factor for the voltage, (d) the r.m.s. value of current, (e) the

mean value of current, (f) the form factor for the current, (g) the total

active power supplied to the circuit, and (h) the overall power factor.

(24)

2 The value of the current i (in mA) at different moments in a cycle are

given by:

 degrees 0 30 60 90 120 150 180

i mA 50 75 165 190 170 100 150

 degrees 210 240 270 300 330 360

i mA 210 185 90 10 35 50

Draw the graph of current i against  and analyse the current into

it’s ﬁrst three constituent components, each coefﬁcient correct to 2

decimal places. (30)

3 The cross-sectional area of a transformer limb is 8000 mm

and the

volume of the transformer core is 4 ð 10

. The maximum value

of the core ﬂux is 12 mWb at a frequency of 50 Hz. Taking the Stein-

metz index as 1.6, the hysteresis loss is found to be 80 W. Determine

the value of the hysteresis loss when the maximum core ﬂux is 9 mWb

and the frequency is 50 Hz. (6)

4 The core of an inductor has a hysteresis loss of 25 W and an eddy

current loss of 15 W when operating at 50 Hz frequency. Determine

(a) the values of the losses if the frequency is increased to 70 Hz, and

(b) the total core loss if the frequency is 50 Hz and the laminations

are made three quarters of their original thickness. Assume that the

ﬂux density remains unchanged in each case. (10)

39 Dielectrics and

dielectric loss

At the end of this chapter you should be able to:

ž understand electric ﬁelds, capacitance and permittivity

ž assess the dielectric properties of materials

ž determine dielectric loss, loss angle, Q-factor and dissipation

factor of capacitors

39.1 Electric ﬁelds,

capacitance and

permittivity

Any region in which an electric charge experiences a force is called an

electrostatic ﬁeld. Electric ﬁelds, Coulombs law, capacitance and permit-

tivity are discussed in Chapter 6—refer back to page 55. Summarizing

the main formulae:

Electric ﬁeld strength, E

volts/metre

Capacitance C

farads

Electric ﬂux density, D

coulombs/metre

D "

= "

Relative permittivity "

ﬂux density in material

ﬂux density in vacuum

The insulating medium separating charged surfaces is called a dielectric.

Compared with conductors, dielectric materials have very high resistivities

(and hence low conductance, since  D 1/). They are therefore used to

separate conductors at different potentials, such as capacitor plates or

electric power lines.

For a parallel-plate capacitor, capacitance C

A.n − 1/

39.2 Polarization

When a dielectric is placed between charged plates, the capacitance of

the system increases. The mechanism by which a dielectric increases

capacitance is called polarization. In an electric ﬁeld the electrons and

atomic nuclei of the dielectric material experience forces in opposite

712 Electrical Circuit Theory and Technology

directions. Since the electrons in an insulator cannot ﬂow, each atom

becomes a tiny dipole (i.e., an arrangement of two electric charges of

opposite polarity) with positive and negative charges slightly separated,

i.e., the material becomes polarised.

Within the material this produces no discernible effects. However,

on the surfaces of the dielectric, layers of charge appear. Electrons are

drawn towards the positive potential, producing a negative charge layer,

and away from the negative potential, leaving positive surface charge

behind. Therefore the dielectric becomes a volume of neutral insulator

with surface charges of opposite polarity on opposite surfaces. The result

of this is that the electric ﬁeld inside the dielectric is less than the elec-

tric ﬁeld causing the polarization, because these two charge layers give

rise to a ﬁeld which opposes the electric ﬁeld causing it. Since electric

ﬁeld strength, E D V/d, the p.d. between the plates, V D Ed. Thus, if E

decreases when the dielectric is inserted, then V falls too and this drop

in p.d. occurs without change of charge on the plates. Thus, since capac-

itance C D Q/V, capacitance increases, this increase being by a factor

equal to ε

above that obtained with a vacuum dielectric.

There are two main ways in which polarization takes place:

(i) The electric ﬁeld, as explained above, pulls the electrons and nucleii

in opposite directions because they have opposite charges, which

makes each atom into an electric dipole. The movement is only small

and takes place very fast since the electrons are very light. Thus, if

the applied electric ﬁeld is varied periodically, the polarization, and

hence the permittivity due to these induced dipoles, is independent

of the frequency of the applied ﬁeld.

(ii) Some atoms have a permanent electric dipole as a result of their

structure and, when an electric ﬁeld is applied, they turn and tend

to align along the ﬁeld. The response of the permanent dipoles is

slower than the response of the induced dipoles and that part of

the relative permittivity which arises from this type of polarization

decreases with increase of frequency.

Most materials contain both induced and permanent dipoles, so

the relative permittivity usually tends to decrease with increase of

frequency.

39.3 Dielectric strength

The maximum amount of ﬁeld strength that a dielectric can withstand

is called the dielectric strength of the material. When an electric ﬁeld is

established across the faces of a material, molecular alignment and distor-

tion of the electron orbits around the atoms of the dielectric occur. This

produces a mechanical stress which in turn generates heat. The produc-

tion of heat represents a dissipation of power, such a loss being present in

all practical dielectrics, especially when used in high-frequency systems

where the ﬁeld polarity is continually and rapidly changing.

A dielectric whose conductivity is not zero between the plates of a

capacitor provides a conducting path along which charges can ﬂow and

Dielectrics and dielectric loss 713

thus discharge the capacitor. The resistance R of the dielectric is given

by R D l/a, l being the thickness of the dielectric ﬁlm (which may be

as small as 0.001 mm) and a being the area of the capacitor plates. The

resistance R of the dielectric may be represented as a leakage resistance

across an ideal capacitor (see Section 39.8 on dielectric loss). The required

lower limit for acceptable resistance between the plates varies with the use

to which the capacitor is put. High-quality capacitors have high shunt-

resistance values. A measure of dielectric quality is the time taken for

a capacitor to discharge a given amount through the resistance of the

dielectric. This is related to the product CR.

Capacitance, C /

area

thickness

and

area

thickness

thus CR is a characteristic of a given dielectric. In practice, circuit design

is considerably simpliﬁed if the shunt conductance of a capacitor can be

ignored (i.e. R !1) and the capacitor therefore regarded as an open

circuit for direct current.

Since capacitance C of a parallel plate capacitor is given by C D

A/d, reducing the thickness d of a dielectric ﬁlm increases the

capacitance, but decreases the resistance. It also reduces the voltage

the capacitor can withstand without breakdown (since V D Q/C). Any

material will eventually break down, usually destructively, when subjected

to a sufﬁciently large electric ﬁeld. A spark may occur at breakdown which

produces a hole through the ﬁlm. The metal ﬁlm forming the metal plates

may be welded together at the point of breakdown.

Breakdown depends on electric ﬁeld strength E (where E D V/d), so

thinner ﬁlms will break down with smaller voltages across them. This

is the main reason for limiting the voltage that may be applied to a

capacitor. All practical capacitors have a safe working voltage stated on

them, generally at a particular maximum temperature. Figure 39.1 shows

the typical shapes of graphs expected for electric ﬁeld strength E plotted

against thickness and for breakdown voltage plotted against thickness. The

shape of the curves depend on a number of factors, and these include:

(i) the type of dielectric material,

(ii) the shape and size of the conductors associated with it,

(iii) the atmospheric pressure,

(iv) the humidity/moisture content of the material,

(v) the operating temperature.

Dielectric strength is an important factor in the design of capacitors as

well as transformers and high voltage insulators, and in motors and gener-

ators. Dielectrics vary in their ability to withstand large ﬁelds. Some

typical values of dielectric strength, together with resistivity and rela-

tive permittivity are shown in Table 39.1. The ceramics have very high

relative permittivities and they tend to be ‘ferroelectric’, i.e., they do not

lose their polarities when the electric ﬁeld is removed. When ferroelectric

effects are present, the charge on a capacitor is given by Q D CV C

(remanent polarization). These dielectrics often possess an appreciable

Figure 39.1

714 Electrical Circuit Theory and Technology

TABLE 39.1 Dielectric properties of some common materials

Relative Dielectric

Resistivity,  permittivity, strength

Material (m) ε

(V/m)

Air 1.0 3 ð 10

Paper 10

3.7 1.6 ð 10

Mica 5 ð 10

5.4 10

–10

Titaniumdioxide 10

100 6 ð 10

Polythene >10

2.3 4 ð 10

Polystyrene >10

2.5 2.5 ð 10

Ceramic (type 1) 4 ð 10

6–500 4.5 ð 10

Ceramic (type 2) 10

–10

500–1000 2 ð 10

–10

negative temperature coefﬁcient of resistance. Despite this, a high permit-

tivity is often very desirable and ceramic dielectrics are widely used.

39.4 Thermal effects

As the temperature of most dielectrics is increased, the insulation

resistance falls rapidly. This causes the leakage current to increase, which

generates further heat. Eventually a condition known as thermal avalanche

or thermal runaway may develop, when the heat is generated faster than

it can be dissipated to the surrounding environment. The dielectric will

burn and thus fail.

Thermal effects may often seriously inﬂuence the choice and

application of insulating materials. Some important factors to be

considered include:

(i) the melting-point (for example, for waxes used in paper

capacitors),

(ii) aging due to heat,

(iii) the maximum temperature that a material will withstand without

serious deterioration of essential properties,

(iv) ﬂash-point or ignitability,

(v) resistance to electric arcs,

(vi) the speciﬁc heat capacity of the material,

(vii) thermal resistivity,

(viii) the coefﬁcient of expansion,

(ix) the freezing-point of the material.

39.5 Mechanical

properties

Mechanical properties determine, to varying degrees, the suitability of a

solid material for use as an insulator: tensile strength, transverse strength,

shearing strength and compressive strength are often speciﬁed. Most

solid insulations have a degree of inelasticity and many are quite brittle,

thus it is often necessary to consider features such as compressibility,

Dielectrics and dielectric loss 715

deformation under bending stresses, impact strength and extensibility,

tearing strength, machinability and the ability to fold without damage.

39.6 Types of practical

capacitor

Practical types of capacitor are characterized by the material used

for their dielectric. The main types include: variable air, mica, paper,

ceramic, plastic, titanium oxide and electrolytic. Refer back to Chapter 6,

Section 11, page 69, for a description of each type.

39.7 Liquid dielectrics

and gas insulation

Liquid dielectrics used for insulation purposes are reﬁned mineral oils,

silicone ﬂuids and synthetic oils such as chlorinated diphenyl. The prin-

cipal uses of liquid dielectrics are as a ﬁlling and cooling medium for

transformers, capacitors and rheostats, as an insulating and arc-quenching

medium in switchgear such as circuit breakers, and as an impregnant of

absorbent insulations— for example, wood, slate, paper and pressboard,

used mainly in transformers, switchgear, capacitors and cables.

Two gases used as insulation are nitrogen and sulphur hexaﬂuoride.

Nitrogen is used as an insulation medium in some sealed transformers

and in power cables, and sulphur hexaﬂuoride is ﬁnding increasing use

in switchgear both as an insulant and as an arc-extinguishing medium.

39.8 Dielectric loss and

loss angle

In capacitors with solid dielectrics, losses can be attributed to two causes:

(i) dielectric hysteresis, a phenomenon by which energy is expended

and heat produced as the result of the reversal of electrostatic stress

in a dielectric subjected to alternating electric stress—this loss is

analogous to hysteresis loss in magnetic materials;

(ii) leakage currents that may ﬂow through the dielectric and along

surface paths between the terminals.

The total dielectric loss may be represented as the loss in an additional

resistance connected between the plates. This may be represented as either

a small resistance in series with an ideal capacitor or as a large resistance

in parallel with an ideal capacitor.

Series representation

The circuit and phasor diagrams for the series representation are shown

in Figure 39.2. The circuit phase angle is shown as angle . If resistance

is zero then current I would lead voltage V by 90

, this being the case

of a perfect capacitor. The difference between 90

and the circuit phase

angle  is the angle shown as υ. This is known as the loss angle of the

capacitor, i.e.,

loss angle, d = .90

− f/

Figure 39.2 (a) Circuit

diagram (b) Phasor diagram

716 Electrical Circuit Theory and Technology

For the equivalent series circuit,

tanυ D

i.e., tan υ D

1/ωC



D R

ωC

Since from Chapter 28, Q D

ωCR

then

tand = R

39.1

Power factor of capacitor,

cos D

since X

³ Z

when υ is small. Hence power factor D cosf ≈ R

i.e.,

cosf ≈ tand

39.2

Dissipation factor, D is deﬁned as the reciprocal of Q-factor and is an

indication of the quality of the dielectric, i.e.,

D =

= tand

39.3

Parallel representation

The circuit and phasor diagrams for the parallel representation are shown

in Figure 39.3. From the phasor diagram,

tanυ D

V/R

V/X

i.e.,

tand =

39.4

Power factor of capacitor,

cos D

V/R

V/Z

Figure 39.3 (a) Circuit

diagram (b) Phasor diagram

Dielectrics and dielectric loss 717

since X

³ Z

, when υ is small. Hence

power factor D cos ³

ωC

i.e.,

cosf ≈ tand

(For equivalence between the series and the parallel circuit repre-

sentations,

³ C

D C and R

ωC

from which R

³ 1/R

)

Power loss in the dielectric D VI cos. From the phasor diagram of

Figure 39.3

cosυ D

V/X

VωC

or I D

VωC

cosυ

Hence power loss D VI cos D V



VωC

cosυ



cos

However, cos  D sinυ (complementary angles), thus

power loss D V



VωC

cosυ



sinυ D V

ωC tanυ

(since sin υ/ cosυ D tanυ)

Hence

dielectric power loss = V

!C tand

39.5

Problem 1. The equivalent series circuit for a particular capacitor

consists of a 1.5  resistance in series with a 400 pF capacitor.

Determine for the capacitor, at a frequency of 8 MHz, (a) the loss

angle, (b) the power factor, (c) the Q-factor, and (d) the dissipation

factor.

(a) From equation (39.1), for a series equivalent circuit,

tanυ D R

ωC

D 1.52 ð8 ð 10

400 ð 10

12

 D 0.030159

Hence loss angle, d D arctan0.030159 D 1.727

or 0.030 rad.

(b) From equation (39.2), power factor D cos ³ tanυ D 0.030

hence Q D

tanυ

0.030159

D 33.16