Conductors and insulators

5

Properties of

Conductors and Insulators

UNIT 1 PROPERTIES OF CONDUCTORS AND INSULATORS

Structure 1.1 Introduction

Objectives

1.2 Classification of Materials 1.2.1 Conductors 1.2.2 Insulators 1.2.3 Semiconductors

1.3 Resistance 1.3.1 Resistivity of Material 1.3.2 Effect of Temperature on Resistance 1.3.3 Voltage and Current Division in Resistive Circuits

1.4 Inductance

1.5 Capacitance

1.6 Summary

1.7 Answers to SAQs

1.1 INTRODUCTION

Conductors, semiconductors and dielectrics or insulators are the most common materials used in electronic and electrical devices. In this unit, the characteristics and properties of the above materials are being presented. Here you will get acquainted with various definitions, applications, types and properties of conductors, insulators and semiconductors. In this unit, semiconductor properties are being described in short. You will learn about semi-conductor in detail in the Block 2. In next section of this unit we will acquaint you with the important basic electrical properties such as resistance, inductance and capacitance of a material. Concepts of resistance, inductance and capacitance will be discussed in this unit with sufficient theory and examples.

Objectives After studying this unit, you should be able to

• explain the basic definitions of conductors and insulators,

• understand the capacitance, types and properties of conductors and insulator,

• define the concept of resistance,

• define the self and mutual inductance,

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

• define the concept of capacitance, and

• distinguish between inductive and capacitive reactance.

1.2 CLASSIFICATION OF MATERIALS

Most common materials used in electrical and electronic devices are conductors, insulators or dielectric and semiconductors. In this section, the characteristics and properties of the above materials are presented. In electrical devices where conductors and insulators are most commonly used material will be discussed in detail.

6

1.2.1 Conductors Electrical

Copper, silver, aluminium are some common examples of conductors. We can define conductor as given below.

Definitions

(a) A material is said to be a conductor if it allow electricity or current through it freely or more simply.

(b) A material is defined as a conductor if the width of the forbidden band in energy band structure is zero, that means there is overlap in conductor and valence bands.

Conduction Band

Valence Band

Figure 1.1 : Conductors Energy Band Structure

(c) A medium is defined as conductor if σωε

is very high. Here σ is

conductivity, ω is angular frequency, and ε is permittivity.

Applications

Some of the applications of conductors are given here. Conductors are used to deliver energy from the source to the load, to make antennas, connectors. They are used in all types of transmission lines in capacitors, PCB layouts, in bread boards, in switches in fuses, in batteries, in circuit breakers, ICs, etc.

Properties

Some important properties of conductors are given below :

(a) Conductor has small resistance.

(b) Voltage drop across a conductor is small, the power dissipation is small.

(c) If the resistance of the conductor is small, current is high.

(d) Free electrons concentration is very large in good conductors (≈ 1028 electrons/m3).

(e) Conductors do not absorb electromagnetic energy. They reflect electromagnetic waves.

(f) Their temperature coefficient is positive.

(g) Electric and magnetic field does not exist in a conductor.

(h) Depth of penetration of electromagnetic waves is zero in conductors.

(i) The size of the round conductor wire in terms of its diameter is specified by gage members.

(j) If gage number is high, the diameter is small and for a thin wire, the resistance is high for given length.

(k) Conductors have always one electron in outer orbit.

7

Properties of

Conductors and Insulators

1.2.2 Insulators A material can be defined as an insulator if

(a) it is a poor conductor of electricity,

(b) it has high resistance of the order of megaohms. Insulator is also called as dielectric material,

(c) width of forbidden band is very large in energy band structure,

(d) its σωε

is very small where σ is conductivity, ε is permittivity, ω is angular

frequency (ω = 2π f).

Various types of insulator are : air, vaccum, wax, paper, sandy earth, bakelite, pure water, wood, plastics, mica, glass, rubber, oil, porcelain, etc.

Conduction

Band

Forbidden

Band

Valence Band

Figure 1.2 : Energy Band Structure for Insulator

Applications Some of the important applications of insulators are :

(a) They are used in capacitors, they are used to store electric charge and to store electrostatic energy.

(b) Insulators protect from electric shock, and they are also used for good propagation of electromagnetic (EM) waves.

Properties Some of the important properties of insulator (or dielectrics) are as under :

(a) Resistance of a insulator is very high, conductivity is very small. (b) It does not conduct electricity. (c) Insulators have 8 electrons in the outer orbit. (d) The concentration of free electron is very small (~ 107 electrons/m3). (e) Insulators allow penetration of electromagnetic (EM) waves. (f) In good insulators alternation of EM waves is negligible.

1.2.3 Semiconductors Definitions

A material is called semiconductor if (a) it conducts electricity moderately.

(b) Its forbidden band gap is very small (≈ 1 eV) energy band structure is shown in Figure 1.3.

Forbidden Band

Conduction Band

8

Electrical EG ≈ 1 eV

Valence

Band

Figure 1.3 : Energy Band Structure in Semiconductors

(c) its wσε

is neither small nor large.

Some typical semiconductor materials are : Graphite, Germanium and Silicon. Applications

Semiconductors are used in diodes, bipolar junction, transistor construction, JFETs, MOSFTs, UJTs and SCRs, amplifiers, oscillators, microwave devices, etc.

Properties Some of the important properties of semiconductors are given below :

(a) Conductivity of a semiconductor lies between that of conductors and insulators.

(b) Pure semiconductor behaves like insulator at low resistance. (c) Its conductivity increases with increase in temperature. (d) Semiconductors have always 4 electrons in its outer orbit. (e) Concentration of free electron lies between that of conductors and

insulators. So far, you have studied about the various properties and characteristics of materials which are classified as conductors, insulators and semiconductors. In next sections you will learn about the properties like resistance, inductance and capacitance of materials.

1.3 RESISTANCE A resistor is the most basic electrical and electronics component that offers a specific amount of electrical resistance to the flow of current when conducted in the circuit. Resistance is the property of any substance due to which it opposes the flow of current through it and convert electrical energy into heat energy. It has the same role in electric circuit as that of friction in mechanical system. This opposition is basically due to the molecular structure of the substance. When electrons flow through any substance then they collide with the other molecules or atoms of the substance. In each collision, some energy is dissipated in the form of heat. So we can say due to resistance some energy is wasted in the form of heat (which is given by I2 Rt). The resistance is defined as the ratio of voltage and current in any circuit and its unit is ohm (Ω).

ohm ( )VRI

= Ω

Reciprocal of resistance is known as the conductance (G) and its unit is Siemen mho ( ).

∴ 1 mhoIGR V

= = .

On the basis of their resistance, substances may be classified as good conductor, semiconductor and bad conductor. Good Conductor

9

Properties of

Conductors and Insulators

Materials with low resistance and high conductance are known as the good conductors of electricity. Example

Metals (like copper, aluminum, silver etc.), acids and electrolytes. Semiconductor

Materials which are bad conductors at low temperature and good conductors at high temperature are classified as the semiconductors. Such materials are partially conducting, but also has properties of an insulator. The amount of current conduction that can be supported can be varied by “doping” the material with appropriate materials, which results in the increased presence of free electrons for current flow. They have medium resistance (between good and bad conductors) at room temperature. Example

Germanium, Silicon, GaAs. Bad Conductors or Insulators

Materials which offer very high resistance to flow of electricity are known as the bad conductors of electricity. They are normally used as the insulator in electrical machines. Example

Mica, Glass, Paper, Rubber, Wood, Bakelite etc.

1.3.1 Resistance Law and Resistivity of Material Resistance of any material is directly proportional to the length of material and inversely proportional to its area of cross-section (assuming cylindrical material of length l and cross-section area A).

i.e. R l∝

1RA

∝

∴ lRA

∝

or lRA

= ρ

where l = length in metre,

A = area of cross section in metre2, and

resistivity of material (specific resistance). =ρ

It is defined as the resistance between the opposite faces of a metre cube of any material.

ARl

ρ =

2metre= ohm ×

metre

= Ohm-metre

Reciprocal of resistivity is known as conductivity or the specific conductance. It is denoted by ‘σ’ and its unit is Ω/metre.

1σ =

ρ

10

Electrical . mho/metrlG

A= e

Conductors has its conductivity very high or in other words, the resistivity of conductor is very small. Insulator has its conductivity very low or resistivity very high. Semiconductor has its conductivity lies between that of conductor and insulator.

1.3.2 Effect of Temperature on Resistance In ideal conditions, resistance is the constant element of the circuit. But as the current flows, heat is produced, and temperature is increased. With increase in temperature following effects are observed :

(a) Resistance of the metal conductors increases with increase in temperature. The Resistance-temperature graph is a straight line and shown in Figure 1.4.

oC (Temperature)

R (Resistance)

Rt

Ro

O

Figure 1.4 : Resistance-Temperature Curve for Metals

(b) Resistance of alloy also increases with increase in temperature. But this increment is relatively slow and irregular.

(c) Resistance of semiconductors decreases with increase in temperature. At very low temperature, they acts like insulators but at high temperature they show the property of conductors.

(d) Resistance of electrolytes and insulators (bad conductors) decreases with increase in temperature.

Temperature Coefficient of Resistance

Let R0 is resistance at any initial temperature t0 and Rt is the resistance at higher temperature t. Then increment in resistance (Rt – R0) is directly proportional to initial value of resistance R0 and increment in temperature (t – t0).

i.e. 0 0 0 0( )tR R R t t− = α −

where α0 is the temperature coefficient of resistance referred to temperature t0

00

0 0( )tR R

R t t−

α =−

If initial temperature t0 = 0oC then 0

0

tR RR t−

α =×

. Unit of temperature coefficient is

per oC.

Resistance at any temperature is defined as

0 (1 )tR R t= + α (here t0 = 0oC)

11

Properties of

Conductors and Insulators

The value of temperature coefficient of resistance (α) is not constant. Its value depends upon the initial temperature on which the increment of resistance is based.

If the value of α at t1oC is α1 then its value at t2

oC will be

2

2 11

11 ( )t t

α =+ −

α

The temperature coefficient of resistance ‘α’ is positive for metals and alloys. α is negative for semiconductors, electrolytes and insulators.

Example 1.1

A platinum coil has a resistance of 3.146 Ω at 40oC and 3.767 Ω at 100oC. Find the resistance at 0oC and the temperature-coefficient of resistance at 40oC.

Solution

0 0 0 0( )tR R R t= + α − t

0

Take t0 = initial temperature = 0o.

0 0tR R t= + α R

0 0(1 )R t= + α

100 0 0(1 100 )R R= + α at t = 100oC . . . (i)

and 40 0 0(1 40 )R R= + α at t = 40oC . . . (ii)

From Eqs. (i) and (ii)

o00

0

1 1003.767 0.00379 per C3.146 1 40

+ α= ⇒ α =

+ α

From Eq. (i)

03.767 (1 100 0.00379)R= + ×

∴ 0 2.732,R =

o40

2 10

1 1 1 per C1 1 304( ) (40 0)0.00379

t tα = = =

+ − + −α

1.3.3 Voltage and Current Division in Resistive Circuits The equivalent resistance of series combination of resistances is the summation of all resistances connected in series.

eq 1 2 3R R R R= + +

R1 R3

I

V

V2V1 V3

R2

Figure 1.5 : Resistances in Series

In series circuit, current remains same through all the resistances while the voltage is divided in all resistances.

1 2V V V V= + + 3

12

Electrical 3 1 2IR IR IR= + +

1 2 3 e( )I R R R I R q= + + =

Voltage Division Rule

Since in series circuit, shown in Figure 1.5, current is given by equation

eq

VIR

=

then 11 1 .

eq

RV IR VR

= =

22 2 .

eq

RV IR VR

= =

33 3 .

eq

RV IR V

R= =

Figure 1.6 shows the resistance connected is parallel.

I

V

R1

R3

R2

I1

I3

I2

Figure 1.6 : Parallel Resistors

In parallel circuits, current is divided while the voltage remains same across all parallel branches.

eq 1 2 3

1 1 1 1R R R R

= + +

2 3 1 3 1 2

1 2 3

R R R R R RR R R+ +

=

∴ 1 2 3eq

1 2 2 3 3 1

R R RR

R R R R R R=

+ +

It is better to represent the circuit in conductance form.

i.e. eq 1 2 3G G G G= + +

Current Division Rule

Current, 11

VIR

=

But, . eqV I R=

∴ eq 2 31

1 1 2 2 3 3. .

R R RI I I1R R R R R R R

= =+ +

13

Properties of

Conductors and Insulators

Similarly 1 32

1 2 2 3 3 1.

R RI I

R R R R R R=

+

And, 1 23

1 2 2 3 3 1. R RI I

R R R R R R=

+ +

Voltage and Current Division Rules for Combination of Two Resistances Voltage Division Rule

1 21 2

1 2 1 2. , .R RV V V V

R R R= =

+ + R

I

V

R1 R2

V2V1

Figure 1.7 : Series Resistances

Current Division Rule

2 11 2

1 2 1 2. , .R RI I I I

R R R= =

+ + R

R1

R2

I

V

I1

I2

Figure 1.8 : Parallel Resistances

1.3.4 Types of Resistors On the Basis of Value

Fixed Resistors

Their values remain constant. Fixed resistors are commonly made of carbon composition.

R

Figure 1.9 : Fixed Resistor

Tapped Resistors

There are tapping on the resistor so that any desired value can be achieved.

Figure 1.10 : Tapped Resistor

Variable Resistor

14

If the value of resistor can be changed with the help of movable contact then it is known as the variable resistor (like rheostat).

Electrical

Figure 1.11 : Variable Resistor

On the basis of the resistance type or fabrication technique adopted for its manufacture, fixed resistance are classifies as

(a) Carbon Composition Resistors

(b) Carbon Film Resistors

(c) Metal Film Resistors

(d) Wirewound Resistors

In carbon composition resistors the resistance element is a slug of resistive material (carbon in this case) whereas in film resistors, it is a relatively thin layer of resistive material painted or deposited on a ceramic or glass-tube. The resistive material is carbon in carbon-film and metal, metal-alloy or metal oxide in case of metal film resistors. In a wirewound resistor, resistive element in a length of wire would on an insulating core in one or more than one layers.

Wirewounds resistors are preferred power carbon composition resistors in low resistance, low noise applications.

Significant advantages of carbon composition resistors includes :

(a) Extremely small inherent inductance and capacitance.

(b) Do not fail catastrophically.

(c) They can withstand higher voltage than film resistors of standard configuration.

Metal film resistors have the advantages of both carbon film and precision wirewound resistors without having disadvantages of either.

There are different varieties of variable resistors and they can be classified as

(a) Potentiometers or simply pots,

(b) Trimmers or Presets, and

(c) Rheostats.

The resistance value of variable resistors, as name suggests, can be varied over the specified resistance range.

Non-linear Resistor

A non-linear resistor is the one, the current through which does not vary according to Ohm’s law. It varies as some function of voltage in case of Voltage Dependent Resistors (also called varistors) or the body temperature in case of thermistors, VDRs are essentially non-linear resistors with negative coefficient. There are thermistors with negative temperature coefficient as well as thermistors with positive temperature coefficient.

Example 1.2

Find the current through each element of the given network and also find potential difference across 15Ω resistor.

15

15Ω

5Ω

100V 10Ω

10Ω

Properties of Conductors and

Insulators

Figure for Example 1.2

Solution

The circuit can be reduced in the following form :

Figure

Total current i in the circuit is obtained by

100 100 4.76 Amp5 10 6 21

ViR

= = = =+ +

So the current through 5Ω and 10Ω resistor will be i = 4.76 Amp. Now, current through 10 resistor in parallel branch is determined by current division rule :

Ω

21

1 2

ri ir r

= ×+

154.76 2.85 Amp10 15

= × =+

Similarly, current through 15Ω resistor is

21 2

10i ir r

= ×+

104.76 1.9 Amp10 15

= × =+

∴Potential difference across 15 resistor will be Ω

2 15V i= × 1.9 15 28.5 volts= × =

Example 1.3 Find the equivalent resistance of network across the source terminals and determine the current drawn from the source.

4Ω100V 6Ω

2Ω

2Ω

2Ω4Ω

Figure for Example 1.3

16

Solution Electrical

After solving both the parallel branches, we get

eq 4 0.67 2.9 7.067R = + + = Ω

Figure

∴ The current drawn from the source

eq

100 14.16 Amp7.067

VIR

= = = .

SAQ 1

(a) A tungsten filament has a temperature of 2,050oC and a resistance of 500 Ω when taking normal working current, calculate the resistance of the filament when it has a temperature of 25oC. Temperature coefficient at 0oC is 0.005/oC.

(b) Find equivalent resistance for the following passive network.

Figure for SAQ 1

1.4 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 can be expressed by VL or e) is given by

LdNdtφ

ν = . . . (1.1)

17

Properties of

Conductors and Insulators

where N = number of turns in the coil, and

φ = flux per turn (in weber)

From Eq. (1.1)

( )L

d N ddt dtφ ψ

ν = = . . . (1.2)

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, . . . (1.3) Liψ =

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. (1.3) to Eq. (1.2)

LdiLdt

ν = . . . (1.4)

In a similar manner, we can define the mutual inductance. Consider two isolated coils shown in Figure 1.12, in which coil 1 is carrying the current i1.

1 2

N2N1

i1

φ12

φ11

Figure 1.12 : Two Isolated Coils with Current i1

Let current i1 is alternating current and producing flux φ1 (per turn).

Then 12111 φ+φ=φ . . . (1.5)

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. (1.4))

Similarly, we can write the expression for e2

12

die Mdt

= . . . (1.6)

where M = mutual inductance between coils 1 and 2 (in henry)

e2 can also be written as :

122 2

de Ndtφ

= . . . (1.7)

18

On equating Eqs. (1.6) and (1.7) Electrical

1 12

di dM Ndt dt

2φ=

⇒ 122

1

dM Ndiφ

=

Assuming the zero initial conditions

122

1M N

iφ

= . . . (1.8)

Now consider the reverse case, in which we give the supply to the secondary coil and determining the induced emf in primary coil (Figure (1.13)).

N2N1

i2

22

e1

21

Figure 1.13 : Two Isolated Coils with Current i2

22212 φ+φ=φ . . . (1.9)

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

= . . . (1.10)

(Due to self inductance L2 of the coil). Induced emf in coil 1 (due to mutual inductance M between the two coils) assuming permeability of mutual flux path is constant

21

die Mdt

=

also 211 1

de Ndtφ

=

or 2 21

di dM Ndt dt

1φ=

211

2

dM Ndiφ

= . . . (1.11)

Assuming the zero initial conditions

211

2M N

iφ

= . . . (1.12)

Multiply Eq. (1.9) and (1.12)

2 12 212 1

2M N N

i i⎛ ⎞φ φ⎛ ⎞= ⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠ . . . (1.13)

19

Properties of

Conductors and Insulators

Now we define the coefficient of coupling

12 21

1 2K φ φ= =

φ φ . . . (1.14)

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. (1.14)

12 1 21 2andK Kφ = φ φ = φ

Put these values of φ 21 and φ 12 in Eq. (1.13)

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= ± . . . (1.15)

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

1.4.1 Inductances in Terms of Reluctance Self Inductance

l

I

N

Figure 1.14 : Solenoid

Figure 1.14 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

= . . . (1.16)

Flux linked per turn : BA Hφ = = μ A . . . (1.17)

[ ]B H= μ∵

where μ = permeability. From Eq. (1.3)

LIψ =

or NLI Iψ φ

= =

20

Put value of from Eq. (1.17) φElectrical

( )N HALIμ

=

Put value of H from Eq. (1.16)

N N IALl Iμ

=

2N Alμ

=

2N

lA

=⎛ ⎞⎜ ⎟μ⎝ ⎠

2N

S= . . . (1.18)

where S = reluctance of magnetic circuit in AT/wb

lA

=μ

. . . (1.19)

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

S= . . . (1.20)

where K = Nagoka’s constant which depends on the ratio of length to diameter of solenoid. By analogy, formula for M is,

1 2 HenryN NMS

= . . . (1.21)

where N1 = no. of turns of coil 1 N2 = no. of turns of coil 2

S = reluctance 0 1

lA

=μ μ

So, the inductance of a coil depends upon the number of turns, the relative permeability of the core material, area of cross-section of the coil and length of the coil.

1.4.2 Types of Inductor Inductors can be categorised as :

(a) Fixed inductors, (b) Variable inductors, (c) Preset inductors, and (d) Chokes.

Fixed Inductors Fixed inductors offer a fixed inductance when connected in an electrical circuit. Figure 1.15(a) shows the circuit representation of fixed inductors. Fixed inductors are wound in such a fashion that the number of turns do not charge and if it uses any core of magnetic material, iron core of ferrite, its position with respect to the winding remains fixed. Coils wound on toroidal cores are an example of fixed inductors (Figure 1.15(b)).

21

Properties of

Conductors and Insulators

Variable Inductors

A variable inductor is the one whose inductance can be charged over a specific range as and when required (Figure 1.15(c)). The inductance is usually varied by changing the position of the core with respect to winding.

Preset Inductors

Preset inductors (Figure 1.15(d)) is a variable inductor that is varied and than set for the most optimum performance. An application of a preset inductor is in Intermediate Frequency Transformers (IFTs) used in radio and television receivers where the inductance is varied by moving the core to get the desired IF response curve.

Chokes

A choke is usually an inductor wound on a core (Iron or Ferrites core) that is used to impede the flow of alternating current or pulsating direct current. Such chokes are known as smoothing chokes. Chokes are also used as coupling component for coupling the signal from one stage to another. Figure 1.15(e) shows circuit representation of a choke.

(a)

(b)

(c)

(d)

(e)

Figure 1.15

1.4.3 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 1.16 : Series Inductors

22

Electrical Self induced emf in 1 1

diA 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

= − + + ) . . . (1.22)

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

diLdt

= − . . . (1.23)

From Eqs. (1.22) and (1.23) we get

1 2( 2L L L M )= + + . . . (1.24)

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

A Be1 e2

e1’ e2’i

Figure 1.17 : Inductor in Series Opposition

Now, 1 1 1anddi die L e Mdt dt

′= − =

2 2 2anddi die L e Mdt dt

′= − =

Total induced emf 1 2( 2di L L Mdt

= − + − )

∴ Equivalent inductance 1 2 2L L L M= + − . . . (1.25)

Inductors in Parallel

A B

L1

L2

i1

i2

Mi

Figure 1.18 : 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.

23

Properties of

Conductors and Insulators

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−

=+ −

. . . (1.26)

Case 2

When the two fields oppose each other, then

21 2

1 2 2L L ML

L L M−

=+ +

. . . (1.27)

Derivations of the above equations will be explained in Unit 2 in detail.

1.4.4 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 . . . (1.28)

Example 1.4

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.

Solution

Figure 1.19 is shown as follows :

0.1 m 0.04

Figure 1.19

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

24

Electrical = 0 1 2

wb1000m

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 1.5

If a coil of 150 turns is linked with a flux of 0.01 wb when carrying a current of 10 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]0.1

L − −= −

= 30 volt.

Example 1.6

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

Solution

For M-Positive :

Leq = L1 + L2 + 2M

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

25

Properties of

Conductors and Insulators

L1

M

L2

L eq.

Leq.

Figure 1.20

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)

20.8 0.4 2 0.15L= + + ×

∴ L2 = 0.1 Henry

The coefficient of coupling 1 2

MkL L

= = 0.75

SAQ 2 (a) 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.

(b) 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.

(c) Find Leq for the arrangements shown in Figure 1.21.

2Hi

4H

1H

5H

i

8H

M=2H

(b)

(a)

Figure 1.21

26

Electrical 1.5 CAPACITANCE

Next to resistor, capacitors are most constantly used electrical and electronic component. In the most basic form where a capacitor is constituted by two parallel plates separated by an insulating (or dielectric) medium, today capacitors are fabricated in a variety of different types with each type having its own characteristics is making it more suitable for a specific area of applications. Based on the type of dielectric used, fixed capacitors can be categorised as

(a) Paper capacitors, (b) Ceramic capacitors, (c) Plastic film capacitors, like polyester capacitors, polystyrene capacitors, etc., (d) Air dielectric capacitors, and (e) Oil filled capacitors and so on.

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 1.22 : Capacitor

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

= = . . . (1.29)

1.5.1 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 1.23 : Capacitor with Dielectric

27

Properties of

Conductors and Insulators

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) . . . (1.30)

1.5.2 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 1.24 : Capacitors in Series

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

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=

+ + . . . (1.31)

Capacitors in Parallel

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

∴ 1 2Q Q Q Q= + + 3

3

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

or 1 2C C C C= + + . . . (1.32)

V

C3+Q3

C2+Q2

C1+Q1

Figure 1.25 : Parallel Capacitors

28

Energy Stored in a Capacitor Electrical

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

. . . Joule=dvvC dt Cv dvdt

s

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 . . . (1.33)

also 21

2QWC

= .

Example 1.7

Three capacitors C1, C2 and C3 have capacitance 20 μf, 15 μf, 30 μf, 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 1.26

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 − −= = × × = ×

29

Properties of

Conductors and Insulators

(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 1.27

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 1.8

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

−×=

Capacitance with dielectric slab :

02

2

r AC

dε ε

=∑

30

Electrical

Air

0.5 cm Figure 1.28

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 1.29

Example 1.9

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 1.30

Total capacitance C = C1 + C2

= 50 + 100 = 150 μF

Energy loss 212

Cv=

6 21 (150 10 ) (250)2

−= × 4.687= joules

31

Properties of

Conductors and Insulators

SAQ 3 (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.

C1 = 1 μF

200V

C2 = 2 μF

C3 = 4 μF

Figure 1.31

(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 :

Voltage gradient across each dielectric.

1.6 SUMMARY

This unit provides the basic properties and characteristics of the most common materials which are classified as conductor, semiconductor and insulator.

In this unit, we have introduced you with various definitions, applications and properties of conductors, insulators or dielectric and semiconductors. Here we have described semiconductors properties in short. These have been discussed in detail in the first unit of Block 2.

In next sections of this unit you have seen acquainted with the basic elements of electrical engineering such as resistors, inductors and capacitors. Important electrical properties of materials such as resistance, inductance and capacitance have been discussed in detail.

Concepts of magnetic flux and self and mutual inductance, energy stored in inductor and a capacitor have also been discussed in this unit.

Solved examples and self assessment question (SAQs) have been given for practicing on these topics.

32

Electrical 1.7 ANSWERS TO SAQs

SAQ 1 (a) 0 0tR R t= + α R

At t = temperature = 2050oC Rt = 500

∴ 500 = R0 (1 + 2050 α) . . . (i) at t = 25oC

Rt = R0 (1 + 25 α) . . . (ii) Now, dividing Eqs. (ii) by (i)

1 25 5001 2050tR + α

= ×+ α

But α = 0.005 per oC

1 25 0.005 500 501 2050 0.005tR + ×

= × =+ ×

Ω

50tR = Ω

(b)

Figure for Answers to SAQ 1

Req in the parallel combination of 2 and 15 Ω resistors.

eq2 15 302 15 17

R ×= = Ω

+.

SAQ 2

(a) (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

=

33

Properties of

Conductors and Insulators

= 1 × 200 = 200 Volt

(b) 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

×

Area of coil, 2 21 (3 10 )

4A −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.

(c) (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

34

Figure for Answer to SAQ 1(d) Electrical

Coils are connected in series opposition so

Leq = L1 + L2 – 2M

= 5 + 8 – 2(2)

= 9 H

SAQ 3

(a) 0 Faradr AC

dε 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

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

35

Properties of

Conductors and Insulators

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