BASIC ELECTROTECHNOLOGY.pdf

REED'S

\SIC

ELECTROTECHNOLOGY

FOR

ENGINEERS

BY C

E D M U N D

G . R . K R A A L

C E n g . . D F . H . (Hons.), M.I.E.E., M.1.Mar.E. K e . . - .f

-.

.

.-, . Elecrrlcal Engmeering ond Radio Depvrtment

t'irst Edition - 1965 j'rrt, Edition - 1973 (SI u n ~ t s )

PREFACE

FIRST EDITION

This book is intended to cover the basic theoretical work in the syllabuses for Electrotechnology in Part B of the Department of Trade and Industry Examinations for Second and First Class Engineers and also Principles of Electricity and Electrical Engineering for Marine Engineer Cadets of the Alternative Training Scheme for Marine Engineers.

I t follows a similar pattern a s the previous yoiumes in this series which has already proved so successful, giving emphasis on first principles, referring to numerous illustrations. providing worked examples within the text, and supplying many problems for the student t o attempt o n his own. The subject matter has been treated in the order and in the manner in which it would be taught at a college and the book is thuscomplementary to lecture notes taken at such a college.

The typical examination questions at the end provide the student with the opportunity of'fin;illy testing himself thoroughly hcl'orc;~~lcrnplinp r h c c x ; ~ m ~ n ; ~ t i o n . 1:ullyworkctl-or11 slcp-by-step solutions are given to cvcry p1.0blc111. t11115 \ ) C I I I ~ I ~ I I . I I C I I I ; I I . I ~ , I I S C -

ful to the engineer at sea without a college tutor a t hand.

The author wishes to acknowledge the assistance given

by

h ~ s College colleagues, M r . J . W . Powell for drawing the diagrams and M r . T . E. Fox for assisting with the proof reading. Acknow- ledgement is also made to the Controller of Her Majesty's Stationery

Office

for permission to reproduce and use the speci- men questions from "Examination of Engineers in the Mer- cantile Marine" as are made available by the Department of Trade and Industry.

PREFACE

This revision has been undertaken to meet the requirement of metrication and the up-dating of the Department of Trade and Industry examinations. In this connection, an additional chapter No.

15

has been devoted to the introduction of electronics and some extra material has been added to.preceding chapters.

As for the first edition, the author has been assisted by his colleagues Messrs Powell and Fox. Their help is gratefully

acknowledged. #-

CONTENTS

~ I { A P T F ! W I---THE ELECTRIC CIRC(JIT. EI~ECTRICAI P A C ; I .

'TERMS

Circuit conditions,

Ohm's

I ~ I w .

Scsic:, and parallcl circui~s. Kirchhull's laws. Internal resistance of supply source. Electromotive force and terminal p.d. or voltage. The series-parallel circuit. Ammeters and voltmeters. Range ex-

tension of ammeters and voltmeters 1-1 7 CHAPTER 2-THE ELECTRIC CIRCUIT (CONTINUED).

ELECTRICAL UNITS

The SI system. Mechanical units of force, work and energy, power. Elec- trical units of current, quantity, volt- age and resistance. Examples relating mechanical and elecirical eneigy. Effi-

ciency. Grouping of e l l s . . . . 18-36 CHAPTER 3- CONDUCTORS AND INSULATORS

Resistance of a conductor,-variation with dimensions and material. Varia- tion of conductor resistance with tem- perature. Temperature coefficient of resistance. Resistance of an insulator. --variation with dimensions and material. Variation 01' insulation re- sistance with temperature. Resistance of a semi-conductor,-variation with temperature. Heat and electrical energy. Relations between mechanical and heat energy. Relation between

electrical and heat energy .

.

. . 37-57

CHAPTER LELECTROCHEMISTRY

Electrolysis. Electrolytic cells. Volt- ammeters (water, and copper). Quanti- tative laws of electrolysis ( ~ a r a d a ~ ' ~ ) . The electro-chemical equivalent, chem- ical equivalent, valency and atomic weight. Back e.m.f. of electrolysis. Resistance of electrolytes. Power ex- pended during electrolysis. Primary and secondary cells. The simple voltaic cell,--cell e.m.f. Electromotive series. Polarisation. The primary cell,-

CHAPTER L C o n t inued PAGE Daniell (energy and e.m.f.). The Le-

clanche cell (wet and dry types). The secondary cell,--capacity and effici-

ency. Charging procedure

.

.

.

.

58-86

CHAPTER %MAGNETISM. ELECTROMAGNETISM Natural and artificial magnets. The magnetic field.-flux and flux-density

.

Molecular theory of magnetism Elec- tromagnetism. Fields due to long, straight, current-carrying conductor, loop and solen~id,-introduction of

itn iron core. Force on a current- carrylrtg conductor in a magnetic field,

units of ampere, flux-density and flux. T h e magnetic circuit, magnetising force or magnetic field strength. Mag- netising force of a current-carrying conductor. Permeability of free space

(Po>.

. .

. .

.

.

. .

. .

87-110

CHAPTER ~ E L E C T R O M A G N E ~ C INDUCTION

Flux-linkages. Faraday's and Lenz's laws of electromagnetic induction. Static induction,+.m.f. of self and mutual induction. Dynamic induction, -magnitude of e.m.f. The weber. Direction of induced e.m. f.-Flem- ing's right-hand rule. The simple magneto-dynamo. The simple d.c. generator, commutation, and practical

requirements,-windings. A.C. and

d.c. theory,- introduction

.

.

.

. 111-137 CHAPTER 7- BASIC A.C. THEORY

The a.c. waveform. Representati'on o f sinusoidal alternating quantities,- trigonometrical and phasor represen- t:ttion. Addition ilnd subtraction of i l l - ternating quantities,-graphical and m:~thcm;~tic;~l mcthotls R o o t mc;ln square and average values. 1;orlil

factor. Peak factor .

.

. . . . 138-- 158

CHAPTER 8-THE D.C. GENERATOR

D.C. machine construction,-field sys- tem and armature. D.C. armature winding :\rr:lngcmcnts. The d.c. gcn- erator,-4.rn.f. equation, no-load

characteristics. Associated rnuynctic P A G E circuit effects. Generator characteris-

tics. Types of d.c, generator,-per- IllilYldIlt mayscl ul~d scpurutcly-cxcitcd types. The shunt-connected generator, -theory of sclf-excitation, 7'he mag- netisation curve or O.C.C. and crhical resistance. Load characteristic. The series-connected generator, self-exci- tation and load characteristic. The compound-connected generator.Types

of connection. Load characteristic

. .

158-1 9 1 CHAPTER 9-THE A.C. CIRCUIT (CONTINUED)

Impedance, inductance, inductive re- actance. Circuits with pure resistance, pure inductance and resistance and in- ductance in series,-power factor,-- true and apparent power. C a p ~ i t a n c e , capacitive reactance. Circuits with pure capacitance, and resistance and capa- citance in series. The series circuit,- inductive impedances in series and in- ductive and capacitive impedances in series. The general series circuit,-

resonance

.

.

. .

. .

. . 192-222 ('lIAPT1:R 10 TFIE D.C. MOTOR

I>ircctinn o f f;>rc.c, -l'lclil~rig's Icft- hand rule. Magnitude of force. Back e.m.f. of a motor. Voltage, current and speed equations. Specd control- ling factors. Types of d.c. motor,- shunt, series and compound. The power and torque equations. Torque controlling factors. Motor characteris- tics. The shunt motor,--electrical characteristics (speed and torque), mechanical characteristic. The series motor, - electrical characteristics (speed and torque), mechanical charac- teristic. The compound motor,-xlec- trical characteristics (speed and torque), mechanical characteristics. Cumulative and differential connec- tion of fields,--strength of shunt and series fields. Motor starters. Speed

CHAPTER 1 ~ A . C . C ' I K C ~ I T S ( C O N T I N L E D ) A N I ~ S Y S T E M S PAGE Power in the a.c. circuit. A c t ~ v e and

reactive components. The parallel cir- cuit. Inductive impedances in parallel. Inductive and capacitive impedances in parallel. Parallel resonance. Power- factor improvement, advantages of p.f. improvement. k W , k V A and kVAr.

Power-factor improvement ( k V A

method). Polyphase working,-three- phase systems. Star o r Y connectiqn, -use of the neutral. Balanced and un- balanced loads. Delta o r A (mesh) con- nection. Three-phase power Three-

phase k V A , k

W

and

k

VAr .

.

. .

247-275

I I I A I - I I I( I 2 I , I - I ; ( , I R O M A C ~ N I ~ I . I S M ( C O N T I N L J I ~ ) Permeability of free space

01,).

Mag- netising force due to a long, straight, current-carrying conductor, inside a solenoid and inside a toroid. Ferro- magnetism. Relative permeability

01,).

The B-H o r magnetisation curve. Ab- solute permeability

01).

Relilctance (S). The composite magnetic circuit,- series and parallel arrangement. Mag- netic fringing and leakage. Iron iosses, -the hysteresis loop. hysteresis and eddy-current losses. Pull of an elec-

tremagnet

.

.

.

.

. .

. . 276- 300 CHAPTER 1 3-THE ELECTRON THEORY, BASIC

ELECTRONICS AND ELECTROSTATICS Constitution of matter. The structure of the atom. Current flow as electron movement, ionisation. Electric field. The electroscope. Potential difference. Electrostatic charging,- induction. Distribution 01' charge. Electrostatic fields o f force. Electrostatic flux. Elec- I ri(, l v ) l r t \ ! i : ~ I 'l'lir (,:I p ; ~ ( . i t o r ( ' ; I ~ ; I ( . I

tor jystcn1s.- \crle\ ; ~ n d par:~llcl corl- ncction. c;\p:~citor currcnr Encrg! stored in an electr~c field or dielectric Relative and absolute permittivity (s, and s). Permittivity of free space (e,,).

CHAPTER I L M I S C E L L A N E ~ U S CIRCUIT CONDITIONS A N D METHODS 01- SOLUTION. SPECIAL

APPLICATIONS

D.C. rrctwork s. Applic~tionu or Kircli- hoffs laws. Maxwell's circulating cur- rent theorem. Thc super-position of current theorem. Conductance, suscep- tance and admittance,-- the series and parallel circuit. The Wheatstone bridge. Measurement of resistance. Temperature measurement. The po- tentiometer. The thermocouple. Tem- perature measurement, compensating leads o r cables, instrumentation . .

CHAPTER 1 5--ELECTRONI(.S

Thermionic devices. Electron emission. The vacuum diode. static characteris- tic, dynamic characteristic,- load line, diode as a rectifier. Ionisation. Dis- charge lamps, The fluorescent lamp (low-pressure). The fluorescent lamp (high-pressure). Cold-cathode lamps. The cathode-ray oscilloscope. The cathode-ray tube (C.R.T.), operation, focussing, deflection. Time-base. Solid- state devices. Semiconductors. Basic

tllcory, co-v;~lcrit 13onding. ~ ' o n d i ~ c t i o t ~ control, in11 insic conductivity, im- purity (extrinsic) conductivity, N-type germanium. P-type germanium, ioni- sation. The P-N junction. The junction diode, forward bias, reverse bias, diode characteristic. Rectifier opera- tion, static and dynamic operation

. .

SOLUTIONS TO PRACTICE EXAMPLES . . SECOND CLASS EXAMINATION QUESTIONS SOLUTIONS TO S1:COND CLASS QUESTIONS

FIRST CLASS EXAMINATION QUESTIONS SOLUTIONS TO FIRST CLASS QUESTIONS

THE S.I. SYSTEM

PREFIXES, SYMBOLS, MULTIPLES A N D SUB-MULTIPLES

PREFIX SYMBOL UNITS MULTIPLYING FACTOR

tetra T x 1012 gigs G x 109 mega M x lo6 kilo k lo3 milli m micro C1 x nano n pica

P

x 10- "

Examples 1 megawatt (MW) =

I

x lo3 kilowatts (kW) 1 x lo6 watts (W) 1 kilovolt (kV) = 1 x

lo3

volts (V)

1 milliampere (mA) = 1 x ampere (A) 1 microfarad (pF) = 1 x farad (F)

PHYSICAL, QUANTITIES (ELECTRICAL), SYMBOLS A N D UNITS

The table has been compiled from recommendations in B.S. 1991 and the List of symbols and Abbreviations issued by the I.E.E.

QUANTITY

..

SYMBOL UNIT

Force Work or Energy Torque Power '1-~nlc Angular vcloclt

>

Speed lilectric

chnrgc

Potential difference ( p . d . ) F newton joule W o r newton metre T newton metre P watt I jccond

to (omega) radians per wcond N rcvolut~ons per minute n revolutions per second

Q

coulomb C' volt ABBREVI- ATION OF UNIT AFTER NUMERICAL VALUE

N

J Nm Nm W S radis revls

QUANTITY SYMBOL UNIT

Electromotive

form

E

volt

(e.m.f.)

Current I a w p r e

Resistance R ohm

Resistivity (specific p (rho) ohm metre resistance)

Conductance G siemens

Magnetomotive force F ampere-tun)

(m.m.0

Magnetic field H ampere-turn

strength per metre or

ampere per metre Magnetic flux @ (phi) weber *

Magnetic flux B tesla

density

Reluctance S ampere-turn or

ampere per weber

Absolute permea- p, (mu) henry per metre Lility of free space

Absolute permea-

bility P henry per metre

Relative _Pk

-

permeability

Self inductance L henry

Mutual inductance M henry

Reactance X o h

Impedance Z ohm

Frequency

f

hertz

Capacitance C farad

Absolute permittivity 6, farad per metre of free space (epsilon)

Absolute permittivity E farad per metre

Relative permittivity E , - (dielectric constant,

specific inductive capacity)

Electric field strength, I _{volt per metre}

electric force

Electric flux 'P (psi) coulomb

ABBREVI- ATION OF UNIT A R F R NUMERICAL. VALUE

v

A R (omega) R m Atlm

Electric flux density, D coulomb per C/mZ electric displacement square metre

Active power P watt W

Reactive power

Q

volt ampere VAr

reactive

Apparent power S volt ampere VA

Phase difference

4

(phi) degree G Power factor (p.f.) cos 4 -

CHAPTER 1

THE

ELECTRIC CIRCUIT

ELECTRICAL TERMS

A circuit can be defined as the path taken by an electric current. Such a current will flow through a circuit if (i) a source of electrical energy such as a battery or generator is connected and (ii) the circuit is continuous or conducting throughout its

whole length. The diagram (Fig 1) represents a simple circuit in which current is flowing. It shows the source, from

which

energy is transmitted through the medium of the current, the conducting path or cables along which the current flows and the 'load'. The load is the point at which the energy is required to be released o r work is to be done through the agency of the current flowing.

SWITCH

t

I-'-\

H E A T I N G CELL E L E M E N T

-

A -AIROW SHOWS DIRECTION O f C U R R E N T F L O W W H E N S W I T C H I S CLOSED Fig 2 C A B L E -CURRENT-+

I

LO*. I

I

4 - C U R R t N T

-

C A B L E

I

Fig 1 '

The conditions of Fig 1 are better represented by a circuit diagram as in Fig 2, which illustrates the energy source as a chemical cell, the conducting path as the leads or wires and the load as a heating element. A switch is shown as a pivoted link which, when opened, interrupts the corUinuity of the circuit and thus stops current flowing.

Consideration of the simple circuit introduces more funda- mental terms and the practical units as used in electrical engineering. Flow of electricity o r current is the result of a pressure built u p within the energy source which manifests itself, at the circuit connecting-points o r terminals, as a pressure difference. One terminal, called the positive, is considered to be at a higher pressure o r potential than the other terminal, called the negative. A potential cli8erenc.e (p.d.) is said to exist between these terminals. The direction of current is from the positive (

+

ve) terminal through the circuit external to the energy source, back

to

the ncgativc (- vc) tcrmin;~l and thence through the source to the +ve terminal. Thus for the load, current is from +ve to - ve terminal, but for the energy source in the form of a cell, battery or generator, current is I'rom the -vc to the +vc

terminal.

The electrical pressure generated by the energy source is termed its electromotive force (e.rn.f.). The symbol used is E, and e.m.f. is measured as a voltage. The unit is the Volt, which will be defined later, but any voltage value can be represented by the letter V attached to the numerical value. Thus a vo!tage of two hundred and twenty volts would be written a s 220V. For reasons which will be explained when the mathematics of the circuit is considered. the whole generated e.m.f. of 3 cell, battery o r generator does not appear a t the terminals, when current is flowing. The p.d. across the terminals is also measured in tcrms of the potential o r voltage dropped round the external circuit. The symbol used for the terminal p.d. is V and it is measured as a voltage. ie it? volts.

CIRCUIT LAWS

1. For any circuit, current strength is found to be proportional to the voltage appfied-%iosS its- ends. Current strength- is d ~ Y i h - e ~ S j r i i i 7 i a n d ~ s measured in Ainperes. The ampere will be defined later by consideration of the electromagnetic effect of current Row, but uny currcnt valuc can bc rcprcscntcd by the letter A appended to the numerical value. Thus 200A mcllnri two hundred nrnpcrcs.

Any electrical circuit is found to oll'er opposition to lhe Ilow of curmit. This opposition is termed the rc,.vistrrrtc~t1 of the circuit and is denoted by the symbol R. The priictical unit of resistance is the Ohm, but any value is rcprcsentcd by thc Grcck Icttcr cirpit;rl

fJ

(omega) appended to the numeric;il viill~e. Thus 10000 rncans one thousand ohms. The ohni can ht. dcfinetl in terms of ~ I I C v o l t i111cl i1111l~r'c I ~ I U S : il 1 c s i 5 1 o t I \ . \ \ C I L;II\IC' 01' O I I C 011111

THE ELECTRIC CIRCUIT: ELECTRICAL TERMS 3 resistance, if a current of one ampere passes through it when a potential difference of one volt is applied across I& ends. An alternative definition appears in Chapter

2.

2.

The current in any circuit, for a constant voltage,

is

found to vary inversely with the resistance; for instance, the greater the resistance, the smaller the current and vice versa.

OHM'S LAW

The relationships set out above, are summarised by the first law of the electrical circbit, which is called Ohm's law and can be expressed thus:-the current in a circuit is directly pro- portional to the voltage and inversely proportional to the resistance. This can be written

Current = Voltage Resistance V (volts) or I (amperes) = - -R (ohms) V Other forms are V = IR or

R

= -

I

When using the Ohm's law formula it is essential to pay due regard to the correct magnitudes of the units used. Reference should be made to the appropriate table of conversions.

Example 1. A voltage of 6V is applied across a 300R resistor. Find the current which will flow.

1' 6

/ = - = = 0.02A or 20 x = 20mA.

R 300

Example 2. A current of lOmA passes through a

60kR

resistor. Find the voltage drop across the ends of the resistor.

V = IR = (10 x x (60 x lo3) = 600V. SERIES A N D PARALLEL CIRCUITS

Study of the electrical circuit shows that in its simplest form it may be built up as (i) a series circuit, (ii) a parallel circuit. For converilence, reiis'lance 'is consihereh 10 b e concentrated ' ~ n a resistor, as for Fig 2, or in more than one resistor; while the con~lecting leads are assumed to have negligible resistance, unless

a

definite resistance value for these is stated. Similarly the cell, battery o r generator can be assumed to have no resistance unless otherwise stated.

The diagram (Fig 3) shows a Series Circuit. It will be seen that only one current path is provided and that the same current passes through all the resistors. The current is thus common for such a circuit but the applied potential is dropped progressively as the current flows along the circuit.

A P P L I E D P D

V V O L T S

1

1

I

AMPS

-

Fig 3

The diagram (Fig 4) shows a Parallel Circuit. Here the main current is made up of a number of branch currents, but the ;~pplicd p.d. 1s tllc s;\rnc o r common l o r all branches. It will bc noted that at an), junction point there is no accumulation 01' current. ir the tot;~l current entering that point is the same as the

lotitI currcrll I C ~ I V I I I K I ~ I C potrll. S I I I ~ ~ ~ C li~ws hitscd o n lhc v o l l i ~ ~ c

conditions for the series circuit and the current conditions 1'0s the parallel circuit have been evolved which allow the solution of associated problems for these simple circuits, and also for those of the more complicated series-parallel arrangements and electrical networks.

-

*

Fig 4 KIRCHHOFF'S LAWS

1. VOLTAGE LAW. The sum of the potential o r voltage drops taken round a circuit must be equal to the applied potential difference.

Thus for Fig 3 C',

+

V ,

+

V:, = V

2. CURRENT LAW . The current flowing away from a junction point in ii circuit must C ~ L I ; I I tllc curscnl flowing into that point.

Thus Ibr Fig 4 I ,

+

I 2 -t l3 = I

'1'11~ above I;Iw\ '11 c ~ 1 x 1 1 L O C I C C I L I C C \ ~ I I I ~ T I C S ~ I . I I ~ L I ~ ~ I C 1'01. 1 1 1 ~ '

series and parallel circuits in tcrms of' the equivalent resistances of the circuits.

THE SERIES CIRCUIT. For Flg 3. let I amperes be the common current 'flowing round the clrcult. Then from Ohm's law, the voltage dropped across resistor- R , is C', volts = I R , . Similarly the voltage dropptxi :tc.~.ors K , is C', = I/?, and so on. It' R I S

THE ELECTRIC ( IKC I I I T . L : F; TKIC A L rL.KMS

- - 5

taken as the equivalent resistance of the \thole clrcult then as I ' is llic iipplicd volti~gc i i ~ i i l I ( \ V I I I L I I ~ o ~ ~ ~ ~ L ~ c I O V ~ I L - ~ L I I \ , : I ~ C * I I I

resistance, we can write C' = IR.

llsirlg Kircl~hoITs volt;\gc

I;IW

tlicn C ' = C',

+

I.', 4- C',

o r I R = I R ,

+

I R ,

+

I R , = I ( R ,

+

R ,

+

R,)

giving finally R = R ,

+

R,

+

R ,

THE PARALLEL CIRCUIT. For Fig 4, let V volts be the common voltage applied t o all the parallel branches and let i t cause a total main current of I amperes. Voltage V would also cause a current of I amperes through an equivalent circuit of resistance R ohms.

v

T h u s I = - and using Kirchhoffs current law then R

v

But for branch 1 V = I l R l o r I i z - Rl

v

Similarly for branch 2 I - - and so on - R ,

Thus I = I,

+ I,

+ 1; can be written

1 - 1 1 1

-

- - + . - +

R R , Rz R ,

N o r t , . 'l'hc reciprocal ol' rrsis~ancc i \ I ' ~ ~ c q u c r ~ ~ l y rcl'crrccl 10 : t \

I

Conducranc,e. Symbol G = - R

The unit is the Sirtncns and the symbol used is S So for a parallel circuit G = G , 4 G ,

+

G , ere.

Example 3. Three resistors of values 2, 4 and 8 ohms are con- nected in series across a supply of 42 volts. Find the current taken from the supply and the voltage dropped across each resistor.

Here R = R ,

+

R ,

+

R , = 2 t 4

+

8 = 14R 42

So supply current = - = 3A 14

Voltage dropped across 2R resistor = 3 x 2 = 6V Voltage ,, ), 4Q , = 3 x 4 = 12V Voltage ,, 7 , 8 R = 3 x 8 = 24V

Check. 6V

+

12V

+

24V = 4 2 i (the applied voltage). Example 4. The above resistor: are connected In parallel across the same supply voltage, fit1 the total current and the current in each branch.

X 4 2

o r H = - = 1.1452 and I = -- = 36.75A

7 1.14

The current in Branch I =

7

= 21A The ,, ,

.

2 =

-4:

= 1 0 . 5 ~

The ,,

.

3 = = 5.25A

Chcck. 21A

+

10.5A

+

5.25A = 36.75A (the total supply current).

INTI'KNAI. Kt:SISfAN('li 0 1 : SIJPIJ[.\' SOUK('li

l i p to now the energy supply source has been considered as h:~ving nepligihlc rcsistancc. I n practicc a cell, hi~ttery o r gcncr-

i r ~ o r . , I I O rllallsr I I O W wcll dcsrgnrcl, 11i1s all internal rcsistirncr

which results in an internal voltage drop when current is being supplied. Thus the emf generated appears at the supply terminals, shown as A R in the diagram (Fig 5), only when the circuit switch is open, ir the cell is on open circuit (O.C.). When current I is supplied. an internal voltage drop given by

/ R i

occurs,

'Ri

being the internal resistance of the cell. The potential difference V appearing at the energy source terminals is E -

IR,.

Thus V is

less than the generated e.m.f. E by the p.d. required to drive the current through the resistance of the cell itself.

.

K

Fig 5

I~~L~I:('TKOMOTIVli FOKCE A N D TERMINAL PD OK VOLTAGE

I n Fig 5.

R i

represents the internal resistance of the cell as \ l i o w n crttcrn;~l l o ~ l i c ccll itsclf. Tlii:; is di;lgri~mmiitic only; \ornctlmca r'csistance is not shown, being written 2s a figure O I ~ I ! hcaidc the ccl! e.ni.f. If given, however, internal resistance

I T I L I ~ ( be taken into account. The foregoing paragraph describes

the l'undamental dill'erence between Electromotive Force E, and Potcntiirl Difference 1'. On open-circuit (O.C.) the terminal p . d . c>S;~n energy source eclualsthe e.m.f. generated; but 'on load', ic.

THE ELECTRIC CIRCUIT : 1II.IiC:THIC:AL. TliKMS

- -- - - -- - . - . .- - . . - . 7

minus the internal voltage drop. This can bc surnrn:~risetl

mathematically thus:

O n Open Circuit (O.C.) V = E

On

Load V = '5.- IR,

We can also deduce that since V = IR where R is the load resistance

Then IR = E - I R i and

E

= IR

+

I R i or E = I(R

+

R , ) Expressed another way :

On O.C. Cell terminal voltage = E O n Load Cell terminal voltage = E - I R i

= Circuit terminal voltagc

= IR

Problems can be treated as a simple serle:, clrcult, if' E i h wed

as the circuit voltage and R i is included in the series resistance. Example 5. A battery of e.1n.f. 42V and internal resistance 7 0 is used to supply the series circuit of Example 3, ie three resistors of 2 , 4 and 8R in series. Find the current and the terminal voltage and by how much the cell voltage 'sits down' when supplying the load,.

Note. I t is appropriate a t this stage to explain that a battery is an arrangement of more than one cell. The methods of connect- ing the cells will be discussed in Chapter 2, but for the present a battery can be considered to be an arrangement of cells in series. Thus the battery c.m.f. is the sum of the ccll e.m.f.s and the battcry intcrnal ~.cbistii~lcc is thc sun1 ol' 111c ccll intcrrlal rcbib- tances.

4 2 V O L T S +

.

7 O H M S 1 A M P S

,z+---

4

+

1

Fig 6

External resistance of the circuit = 2

+

4

+

8 = 14R Battery resistance = 7R

Total resistance of circuit = 14

+

7 = 2 1 R Circuit current = = 2A

Terminal voltage = 14 x 2 7 28V

Voltage drop in cell = 7 x 2 = 14V So terminal voltage = 42 - 14 = 28V Cell voltage 'sits down' 14 volts.

THE SERIES-PARALLEL CI'.CUIT. AS the ritle implies, circu~ts may

solution of' associated problems, though not straightforward, f'ollow\ a logical sequence of operiition based on the methods

u5c.d I'or rllc sltnplc scrics ;rnd parallel arrangement of resistors.

I t cannot be too strongly urged that here is a case of 'practice maklng perfect' and the reader should work a variety of appro- priate problems. A method of solution for any particular prob- lem will thus readily become apparent once its form is recognised and both tlmc and labour will ultimately be saved. The more dillicult ne~work problems are solved by reverting to funda- m e n t ; ~ ] principles such ils the direct application of Kirchhoff s I;~ws 01. using ;r 'I'licorctn or Mctliod ol' SoIu1ion b;~scd on IIic\c

laws. Such problems will be introduced at a very much later st;~gc in this book. when the render should feel sufficiently confident to appreciate the fincr points of theory and may cvcn en.joy the methods and techniques necessary to allow solution. The diagram (Fig 7a) shows a straightforward series-parallel circuit. I t consists of a series circuit made up of two sections, each comprising o f a group of resistors in parallel. Since the main circuit o r supply current may require to be found together with the current in 'each resistor, solutiori can only be obtained by a simplification of the problem. It will be noted thzt the parallel groups o r banks of resistors have been called,sections A and €3 respectively. The voltage dropped across these sections is not known and since such knowledge is essential, procedure would be as follows.

Fig 7a

' I llc crrcurt I:, si~l~plilicd by linclrng tlic cquivi~lent resistance

\slues R , and R B of the parallel banks from

OnCr. K,, and

K,,

h;~vc heen found the total supply current can hc ohtnined sincc. :IS 1s shdwn in the diagram (Fig 7b). thc

'IHE ELECTRIC CIR(.IJI I I:L.I:('I~KI(.,'AL rEKMS 9

current i.; I -

v

and the volt;~gc drop.; ;+crow g r o t r p A

f t A 1 and B are

C', - I R A and L',, - I K , ,

I--

S U P P L Y V O L T A G E V V O L T S

Fig 7b

Once V A and V B are known, the Individual currents in each resistor can be found by reverting to the original circuit.

V A

v

A

r

Thus I, = -- and I, = - Also I -

5

etc.

R I R2 -

R4

I t is not intended that the above method of solution should be memorised. I t is given to explain the solution of Example 6 and to illustrate the step-by-step procedure. There is no short-cut for problems of the series-parallel type. The reader should work only with the data given and should not make any assumptions. Methods of solution using proportions for currents or voltages

;~croh\ p:~r:~llcl o r hcricb S C C ~ I O I I S of' [tic circi~it iirc d i s c o ~ ~ r ; ~ ~ c ( l ~

slnctl seldo~n I Q pructlce art. !I,c resistance ratios s ~ m p l e . and only good can come from a d h e r ~ n g to and following the straight- forward, though mmcrimc\ rnore teciious methods.

Example 6: A circuit IS built u p from five resistors. Kesistors of

values 40, 6R and 8R are connected in parallel to form a group, whilst resistors of

3R

and 6 0 are connected in parallel to form

rLT-'--j

I 6 OHMS 8 OHMS C--r 6 OHMS

B

another group. The two parallel groups of resistors are con- nected in series across a 10V supply. Find the voltage dropped iicross each parallel group, the main supply current and the current in each resistor.

Let R, be the equivalent resistance of the first group.

Simil:trly Ict R , bc the equivalent resistance of the second group.

1 1 1 2 + 1 3

Tticn - - -

Nu - 3 + 6 0 6

For the equivalent series circuit total resistance is R o r R = R A _{. -} -t R, _- = 1.85

+

2 = 3.85Q

v

10

Main supply current I = - = - = 2.6A R 3.85.

Voltage drop across RA o r the first parallel group = 1.85 x 2.6 = 4.8V

., ,, ,,

R E

o r the second

,,

,, = 2 x 2.6 = 5.2V

Check. Total supply voltage is (4.8

+

5-2) = 10V. 4-8

Current in

BR

resistor = - = 1.2A 4

Check. Total current is (1.2

+

0.8

+

0.6) = 2.6A 5.2

Similarly: Current in 3R resistor =

-

= 1.73A

3 .

,, ,, 6R ,, - 5'2 ₆ 0.87A

C ' h r c k . 'I'otal current IS (1.73

+

0.87) = 236A.

Example 7. A battery of e m f . 42V and internal resistance 7R feeds a circuit consisting of three resistors connected in parallel. Tlic resistors have values of 2R, 4R and 8R. Find the battery current, the battery terminal voltage and the current in each

The circuit can now be considered to have a total resistance of S.14R made up from 1.14R and 7R in series.

-

VOLTS + 7 0 H M S

-1

- - -

-

- - - + -I

42 r

The battery current I is given by = 5.16A 8.!"

2 OHMS

4 O W I,

-

The termlnal voltage will be 5.16 x 1.14 = 5.88V Or ,, ,) 9 9 , , 4 2 - ( 7 x 5.16) = 42 - 36.12 = 5.85V 5'88 = 2.94A Current I, in 2Q resistor 1s -- 2 Current I, in 4 0 ,. 51.88 = 1 . 4 7 ~ ' 4 " I Current I, in

8R

,, 5 , 8 8 - - 0 . 7 4 ~ " 8 8 0tIHS

7

Fig 9

Check. Total current I = 5.16A. A M M E T E R S A N D VOLTMETERS

These are the primary instruments used for electrical work and the diagram (Fig lo), shows how they are connected into the circuit. Ammeters are used for measuring current and volt- meters for measuring potential difference o r voltage. Both instruments operate on the same principle, but ammeters must

be of very low resistance since they are in series with the load and must be responsible for negligble voltage drop. Voltmeters on the other hand must be of high resistance, since they may be connected across points which may be at a high potential difference. For most circuit purposes, the ammeter is considered to have negligible resistance and the voltmeter t o have infinite resistance, ie to take n o current.

In Fig 10 a generator is shown as the energy source, S may

be a singk-pole o r double-pole switch, as is shown here, and R is the load resistance. .As ;I practical example, the generator may li:ivc ; I I I 111tcr11:)l I . ~ ~ I S ~ ; I I I C C 01' 0 , 0 2 f l , t l ~ c c;~l)lc

I C ; I ~ S

I I L I ) have a total resistance of 0.03R and K may have a value of' Sf2 I f ttlc gcllcr;~tor is .;ct ((7 320V o n opcn-circuit. icz with

tllc switch open, then when the switcll I S closed a current 01' 220 -- -

220

= 43.56A would flow round the circuit.

5

+

0.02

+

0.03 5.05

The terminal voltage of the generator would 'sit down' to 220 - (43.56 x 0.02) volts = 220 - 0.87 = 219.13V. T h s would be shown by the voltmeter, while the ammeter would show 43.56A. If the voltmeter w a s disconnected and then con- nected directly across R it would indicate 219.13 - (43.56 x 0-03) volts = 219.13 - 1.3 = 217.83V. The voltage drop in the cables would be 1.3V. It will be seen that the example of a simple l s t r i b u t i o n system has been worked as a simple series circuit and that the instrume~lts perform their required functions. The ammeter show2 the series circuit current, whilst the vollmeter indicates the potential d r o p across any particular portion of the circuit. I t also can record the e.m.f. built up by the generator when the switch is open, since this is the only condition when the e.m.f. appears at the terminals of the energy source.

RANGE EXTENSION OF AMMETERS A N D VOLTMETERS

For practical work it may not be possible to pass all the circuit current through the ammeter. I t may be difficult to construct a suit;iblc instrument bec:~usc of' size or otlicl. limitations, and in ordcr to introduce a certain amount of standardisation. i t may

I>c caslrr. lo LISC tI1c ~ I I I ~ I I I C I C I . w i [ l l ;I . V / I I I I I I in O I . C ~ C I . lo I I I C ; I S I I I . C the circuit current. A shunt 1s a spec~ally constructed resistor ol' low ohmic value and, in order to make an ammeter capable of measuring a current greater than that which can be passed through i t . a parallel arrangement of thc ammeter and the shunt is used. The ammeter is designed to carry a definite but small rr:~ction of tiic main current : ~ n d the rest of the current is made lo by-pass the ammeter through thc shunt, which is accurately

made a n d set to :I dcfinitc l-cs~sl;lncc \ ; I I I I ( . I t 15 C : I I I I ~ ~ : I I C ~ I \ 4 7 1 1 1

the ammeter instrument and r~iust ;rl~iagh

bc used w ~ t h

i t Tlit'

calibrated leads between Instrument ;rnd .;ll~rnt I'orrn p;irt of tlic ;Irrangcmenl ; ~ n d nluhl not hc cut 01. ~ I . I ~ > S ~ I I I . I ~ C ~ 1'01. 13) j71cces 0 1

ordinary copper wlre. T h e diagram ( F i g I I ) shows tile normal Lrrrangernent o f instrument and \bun! and tlie e.\ample hlious the form of' cnlcul:itlon necessary I t will tx 5ct.n that tlic calculation follow^ the piittern set I'or parallel rcsl<t;incc C I ~ . ~ L I I I \

S H U N T Rw

RM IKLUOES RE518 TANCE OF _P

INSTRUMENT L LEADS Fig I I

Example 8 . Calculate the reslstance of' a shunt r e q u ~ r e d to operate with a milliammeter. ivh~ch p i e s full-scale deflection for a current of 15mA a n d which has a reslstance of 512. (,Vorc.. 5 R can be taken to include the resistance o f the c o n n e c t ~ n g leads. since n o specific mcntion of lead I-eb~stance h;is been made ) TI-ie cornbin;ltion o t ' m c t c r ;lnd sI111nt is rcquirrcl 1 0 rc;lcl c l t r r c n l , ~ 1 1 1 1

to IOOA.

Voltage d r o p across instrument when g i ~ , i n g full-scale det1c.c- tion = current causing full-scale deflection x resistance 01'

instrument circuit

= I , R , = ( 1 5 5 = 75

= 0.075V o r 751nV

N o w the voltage d r a p across the i n s t r u ~ n e n t is tlie hame a s the voltage d r o p across the shunt o r Is, x R,, = 0.075 volta.

But the shunt current I,, would be 100 - meter current

= 100 - 0.015 = 99.985A

= 0.000 75112

I t is important to note the low resistance u l u e of the shunt which is designed to carry the curr_{ent without 'heating up' The}

shunt I S usually mounted on the switchboard, behind the a n i - meter a n d in the main current circuit. Tlie 'light' calibrated le;ldj are coiled to take u p any 'slack' and then brought out to tllc instrument. Thus the a m m e t e r m a y be marked 0-100 arnyxrcs. but in actual fact only a minute current, some 15mA. passes

through the instrument itself. The remainder and by far the largest proportion of the current, passes through the shunt. The reason for always using the instrument with its own calibrated shunt and leads is thus obvious.

T o measure voltages higher than that for which the instrument movement is designed a series or range resistor must be used. This resistor is designed to drop the excess voltage and dissipates a certain amount of heat; It consists of special fine-gauge wire wound on a porcelain spool or on a mica card, the whole being moi~nted inside a ventilated case. Here again the arrangement n l a y bc mounted behind the switchboard, if i t cannot be con- tained in the case of the instrument. Thin leads for carrying the small instrument current connect the range resistor unit and the instrument l o ~ h c nuin supply terminals, usually through fuses. Thus the voltmeter may be scaled 0-250 volts, but in fact only 0.075V may be dropped across it, when full-scale deflection occurs. By far the major voltage drop occurs across the range resistor, which is always high in ohmic value: thousands of ohms. This fact should be noted. The diagram (Fig 12) shows the arrangement and Example 9 shows how the value of an appropriate range resistor is calculated.

VOLTAGE TO @E

t 7 Y 2 .

---j

Fig 12

Example 9. Calculate the resistance of the range resistor required to be placed in series with the instrument of Example 8 to make it into a voltmeter reading

0-250V.

(The instrument has a resistance of 5Q and gives full-scale deflection with a current o f I 5 m A ) .

T h e current through the complete voltmeter circuit must be limited to 15mA, otherwise the instrument will be 'burnt out'.

the series or range resi\tor R,, muit h a \ ? ;I \ : ~ l l l c of ( I h h h f , - 5 J o l l l n h I0 OOlil

The actual 'movement' or working unlt 01' ;in amrnstcr 01. voltmeter is much t h e same ;lnd i t is tllc usc 01' :I sliunl 01' rangc

resistor which decides whether current o r voltage can be measured. Multi-purpose portable test instruments are available which can make a range of measurements. A range switch is provided which makes the appropriate connection of shunt or range resistors.

Consider an instrument movement in which 15mA at a p . d . o f 75 1 0 - ~ 75mV gives full-scale deflection. Its resistance =

15 x lo-' = 5R.

If a voltage range &15V is required, the instrument circuit resistance must be I 5 = lOOOR and a range resistor of

15 x 10- ,.

1000 - 5 = 9 9 5 a must be swltched in.

Similarly if a voltage range of 0-150V is required the range resistor must be 150 - 5 = 1 0 0 0 0 - 5 = 9995R.

15 x l o - '

If a current range 0-5A is required, a shunt is used whose value can be obtained thus:

P.D. across shunt = p.d. across instrument movement for full-scale deflection

0 r

..

= 75111v -- 75 X 10

'

~ 0 1 1 s

The current through the shunt = 5 - 0.015 = 4.98%

CHAPTER 1 PRACTICE EXAMPLES

1. A circuit is made up from four resistors of value 2R, 4R, 5R and 10R connected in parallel. If the current is 8.6A, find the voltage drop across the arrangement and the current in each resistor.

2. One resistor group consists of 452, 6R and 852 connected

in pi~~.;rllcl ; ~ n d it second group consists of 3R and 6R in parallel. The two groups are connected in series across a 24V supply. Calculate (a) the circuit current, (b) the p.d. across d i t ~ l l ~ I . L ) L I I ) , (c) fllc curre111 in cuch resistor.

3. If the resistor arrangement of Q1 is connected to a 12V battery of internal resistance 0.6552, find the circuit current and the battery terminal voltage. Find also, the current in the 5R resistor.

4. A moving-coil instrument has a resistance of 10R a n d requires a current of l5mA to give full-scale deflection. Calculate the resistance value of the resistor necessary to enable it to be used to measure (a) currents .up to 25A, (b) voltages up to 500V.

5 . Two.+esistors of 60k52 and

40kR

value are connected in series acFoss a 240V supply and a voltmeter having a resist- ance value of 40kR is connected across the 40kR resistor. What is the reading on the voltmeter?

6. When a 10R resistor is connected across a battery, the current 1s measured to be 0.18A. If similarly tested with a 2 5 0 resistor, the current is measured to be 0.08A. Find the c.11l.l'. ol' ~ h c bnttcry and its intcrni~l resistancc. Neglect thc resistance of the ammeter used to measure the current.

7 , ' I wo groups ol' resistors A and B are conncctcd in scrlcs. Group A consists of four resistors of values 2R, 4R, 6R and 8R connected in parallel and group B consists of two resistors of values 10Q and 15R in parallel. If the current i n the 4R resistor is 1,5A, calculate ( a ) the current in each of tlic rornaininp ~.t.sistors, ( h i !hc. supply voltage, (c) the v o l t ; ~ ~ c ( I r o p i ~ c r o s ~ 1 hc g r o t ~ p s A i ~ n d R .

X I lic \ o l t ; ~ g c of' ; I d c , generator. wl-icrl \ i l p p l > . ~ n g ; I current 01' 7 5 t l ( ( 1 ; I l ~ ~ i ~ d , I > I ~ ~ C ; I > L I I C ~ I [LI 1~ IOS SL' . I [ I I I C > \ + I I L ' ~ \ -

board. At the load. the voltage recorded is 10SV ;lnd ~ , h c n the Ioxi is switched off' the volt;~ge rises 10 I I O V . 1:11id ~ l i c internal 'resistance of the generator, the resistance of thk supply cables and estimate the fault current if a 'short- circuit' of negligible resistance occurred at the load terminals.

9. The ammeter on a swltchboaid. scaled 0-300A is acci- dentally damaged. The associated shunt is marked 300A. 150mV A small ammeter, scaled 0-1A with a resistance of 0.12R, is available, and the possibility of using this is con- sidered. Find if such an arrangement is possible, and if so, how it could be achieved using surplus resistors which are

also available. t

10. Five resistors AB. BC. C D , D E and EA are connected to form a closed ring ABCDEA. A supply of 90V is connected across A D , A being positive. The following is known about the resistors: A B is IOR, BC is of unknown value R , ohms, CD is of unknown value R , ohms, DE is 6R and EA is 9 0 . A high resistance voltmeter (taking negligible current) when connected across RE reads 34V with H positive and when c.orl~icc.~cd :IL.I.O.;~ ( ' 1 : r c ; ~ t I \ OV w ~ l l i I ! I I O ~ I I I L . ~ 1 , 1 1 1 ~ 1 1 1 1 ~ ' valucs 01' K , 411d K,, thc current in branch ABCD and the main supply current.

CHAPTER 2

THE

ELECTRIC CIRCUIT (CONTINUED):

ELECTRICAL UNITS

All engineering studies stress theneed for units and an intro- duction to some of these will have been made when the subjects of mechanics and heat were being covered. Units allow measure- ments to be taken and calculations to be made. They are essential to the derivation of formulae from the basics of theory and enable presentations of related principles to be evolved. Thus

for

electrical

engineering, even at the Chapter 1 stage, the ampere, volt and ohm were considered, and although these units have yet to be defined, their importance in rel~tion to the basic electric circuit will have been appreciated. The student will also recognise these units as being amongst those in common every-day usage. If, however, the impression has been given that the study of electrotechnology will involve the knowledge of an entire range of new units, then it is stressed that this is not the case. The whole modern concept of engineering technology is based on the universal adoption of SI units and, since some of these have been encountered in earlier work, it will not be long before, in this study of electrical engineering, common ground is being covered and the relevant relationships with associated units, already treated from the mechanical engineering aspect, are being stressed.

Before proceeding with any further study of units of the S1 system, it w o u l d a e useful to introduce a historical note and consider the situation in engineering as it has developed. Towards the end of the last century two systems of units began to be employed in engineering; the British or foot-pound- second (fps) system and a metric or centirnetre-gramme-second (cgs) system. The British or Imperial system had no merits since all units of the same kind, such as those of length, area, volume ther.

indeed-there

were also ric and horsepower which were ntly defined. The metric system t to industry and commerce but tages and it was adopted prior ical circles. In 1873 the British nt of Science selected the centi- units of !ength and mass for

h;~sc-unit of rime and the tio option of the second. for this purpose, gave the c e n t i m e t r e - g r a n i ~ ~ ~ c - s e c o ~ ~ d (cgb) systrln.

The metric system, in the cgs form, was adopted to a Iiirpt' cxtcnt li)s electrical cngiuccsing in thc curly dirys 01' dcvclop- ment. The system had the advantage that all the same kind of quantities are multiples of ten and it was also international. The sizes of the absolute unit of the centimetre and the gramme were found to give rise to difficulties for the desired electrical units which became either too large or too small for practical work- ing. The use of these absolute units for essential formulae in engineering work also proved difficult and thus more workable or practical units had to be devised. Such practical units were to include the volt, ampere and ohm. In about 1900 practical measurement in metric units began to be based on the metre. kilogramme and the second and the aforementioned electrical practical units.These constituted the unrationalised MKS system.

The next developmeqt came from a fact, which was re- peatedly pointed out over the first half of the present century, that a system of units could be devised to make the practical units of the volt, ampere and ohm the absolute units of such a system. If, in addition, suitable adjustments of certain constants encountered in electromagnetism and electrostatics are accepted, then a more workable system of units would result. This system was known as the rationalised MKS system and its adoption

W A S ~.ccommcndcd by tllc Internation:~l L i l ~ ~ t r ~ t ~ c l ~ ~ ~ i ~ i ~ l Ci)111- mission of 1950. The change to the MKS caused some little

inconvenience to the older electrical engineer and necessitated the revision of many of the better known works of reference and text-books. The student was required to appreciate however. that the new system did not upszt the course of learning in any way and that, if anything, the 11nits introduced made matters easier and formulae more manalr,eable.

Prior to 1970, conditions d;3 exist when both the older Imperial and the newer MKS sydems of units were in use. The latest extension of.metric units .nto all branches of commerce and industry has enabled enginiering to evolve the SI system, the units of which are used throc ghout this book. Thus from the electrical viewpoint, it can be faid that the SI system is the rationalised M K S system with :nits in all the other fields of measurement being fully metrica..cd.

T H E SI S 'STEM

All measurement consists in cc mparison with some standard or unit. The three fundamental un ts are those of length, of mas.-

and of time. In the SI system the metre is taken as the funda- mental unit of length, the kilogramme as the unit of mass and the second as the unit of time. From the fundamental units, can be built u p the derived units, which can be further classified as mechanical or electrical units. Thus Force I S a derived mechani- cal unit involving a fundamental unit and a derived unit, ie mass and acceleration. For the SI system, a unit of force, called the

Nekton, has been introduced. Velocity is similarly a derived

unit involving distance a n d - time. So also is acceleration a derived unit, involving velocity and time. Both velocity and acceleration are mechanical units. The ampere is really a derived unit involving force and length but as stated previously it is used as a fundamental electrical unit. Other electrical units are the volt i ~ n d t h o ohm which ore nlso dcrivcd units, but thc Joule and Wutt, although used principally in the past in connection with electrical engineering, are really derived mechanical units and will be defined under this heading.

Once the units are recognised and understood, the reader is advised to discontinue their classification as mechanical o r electrical units and to accept them as general engineering units. This applies particularly to the units of work and power. Both the mechanical and electrical engineering fields are concerned with common practical appliances or associated problems and a ready use of the appropriate units, with a correct appraisal of the magnitudes of the quantities involved, is essential.to the modern engineer.

MECHANICAL* UNITS

The fundamental units require little definition since they are accepted standards. Thus the metre is the absolute standard, taken as the distance between two marks on a certain metal bar. Similarly the kilogramme is the mass of an accepted standard lump of'metal. The time unit is the second, which is defined as

I

of a mean solar day.

86

400

Most of the principal SI derived units will already have been introduced to the student but a revision is made here to allow an extension into the fieid of' electrical engineering.

UNIT O F F O R C E

THE NEWTON. This is theforce required to accelerate a mass of' one kilogramme at the rate of one metre/second2. It has been fbund that the force of gravity acting on a mass of I kg is 9.81 ncwtons ; ~ n d , since thc force on ;I body due to tlic ciirth's

THE ELECTRIC CIRCUIT FLFCTKICAI, ONIT? 2 1 -- -- - -- - - - - -- . - -- - attraction is termed weight, i t follows that the weight of 1 k ~ l o - gramme is 9.81 newtons.

The symbol for force is F but any value in newtons can be represented by the letter N after the numerical value.

Thus: lkg = 9.81N. UNIT OF WORK AND ENERGY

THE JOULE. This can be defined as the irork done or energ1

stored when a force of one newton acts through a d~stance of one metre in the direction of the force.

The symbol for work or energy is W but any value In joules can be represented by the latter J after the numerical value.

From the definition, it follows that a force of F newtons. acting through a distance of s metres, does F x s newton metres of work o r Fs joules.

Thus: W (joules) = F (newtons) x s (metres) * UNIT OF POWER

THE WATT. Power is the rate at which work is done or energj is converted and the unit is the watt. A watt is the power resulting, when a joule of energy is expended in a second.

The symbol for power o r rate of doing work is P but any value in watts can be represented by the letter W after the numerical value.

The definition can be more generally written as I + ' (Joulc5)

P

( w n t t s ) = -

r (seconds) Other tbrms are U' = Pr or 1 = - CI '

P

The jaule and watt are the units orignally used in electrical engineering and they will be encountered constantly in electrical problems. Example 10 is set out here to serve as an introduction to electrolmechanical relationships.

Example 10. A pump is required to lift 1200 litres of water through 10 metres in 6 minutes. Calculate the work done in joules and the power rating of the pump. Assume 1 litre of' water to have a mass of' 1 kilogramme.

Work done = force of gravity x distance lifted = (1200 x 9.81) : . 10 = 12 x 9.81 x l C 3 newton metres = 1 17.72 x 10.' = 1 1 7 720N1n or 1 17 7103 work done - 1 7 720 - 1 I 772 - 327W Power = - -- - - time 6 x 6 0 . 36

Note. In the above problem no account has been taken of

machine efficiency. This will be introduced in due course, but for the example, the practical rating figure of the electric motor, assuming this to be the means of driving the pump, would be much larger.

ELECTRICAL UNITS

The same fundamental units are used as for the mechanical units namely: the metre, kilogramme and s e ~ o n d . The primary derived unit is the ampere, which has been adopted as the basic electrical unit

of

current and ;IS

a

I'ourth fundamental unit. Before considering the definition for the ampere, it is necessary at this stage, to describe two associated effects, which would be observed when a current flowed in a circuit.

(1) If the resistance of the circu~t was concentrated in a short length of conductor, then a temperature rise of the wire in this region would txnoted, showing a conversion of energy into heat.

(2) If the circuit was supplied through two wires laid to- gether, then especially if the current is large and the wires flexible, a mechanical effect would be noted. When the current is switched on, the wires would be observed to move and this electromagnetic effict, as it is called, has been used to define the ampere for the SI system. The factors governing the magnitude and direction of the force on the wires will be described in the chapter on Electromagnetism.

UNIT OF CURRENT

THE AMPERE. This is that current which, when maintained in each of two infinitely long, straight, parallel conductors situated in a vacuum and separated by a distance of one metre between centres, produces on each conductor a force of 2 x lo-' newton per metre length of conductor.

As stated in Chapter 1, the symbol for current is I and any value in amperes is represented by the letter A after the nwneri- cal value. The reader is reminded that practical circuit currents may rilngc from thousands of amperes to minute values of' milli-amperes and attention is drawn to the Table of Prefixes of' Magnitudes as given at the front of this book. Full consideration must be given to the correct use of the abbreviation which follows the numerical value.

When

a

current flows for n given tir~le, a quantity of electricity is said to be conveyed round the circuit. The quantity which passes ciin

bc

sliown to

be

related to thc work done in the circuit,

THE ELECTRIC CIRCUIT : ELECTRICAL UNITS - - - - - - - -

23 but before this relationship is considered further, it is necessary to dcfinc qu:~ntity ofclcctricity in terms of current : ~ n d timr UNIT 0 1 . QUANTIT)'

THE COULOMB. The usual unit-sometimes called the ampere second. For practical purposes a larger unit, for everyday electrical engineering is used. This is the Ampere hour as used in connection with the capacity of batteries and for accumulator charging.

The symbol for quantity of electricity is Q and any value in coulombs can be represented by the letter C after the numerical value. Any value in ampere hours is represented by the letters

A h. after the numerical value. Since the quantity ~f electricity which is conveyed round a circuit would vary with the strength of the flow of electricity and with time, a simple definition for the coulomb can be deduced thus:

, A coulomb is the quantity of electricity cdnveyed by a steady current of one ampere flowing for a time of one second.

Thus Q (coulombs) = I (amperes) x 1 (seconds) or Q (ampere hours) = I (amperes) x t (hours)

From the above, the following can be deduced:

1 ampere hour = 1 ampere x 1 hour = 1 ampere x 3600 seconds = 3600 ampere seconds = 3600 coulombs. Thus

1A h = 3600C.

Ex:~mplc I I . C'onsidct. I~?c:~mplr 5, w l ~ c ~ c :I h:~ttcr.v of' c 1 n . f

42V and ~nternal resistance 752 is used to supply a clrcult of' three resistors 2, 4 and 8 0 in series. If the current is switched on for 30 minutes, find the quantity of electricity which would have been conveyed.

Total resistance of circuit = 7

+

2

+

4

+

8 = 2152 42

Circuit Current = -- = 2A ; 1

Quantity of Electricity = current x time in seconds = 2 x 30 2 60 = 3600C or Quantity of Electricity = current x time in hours

- - 2 x + = 1 A h .

The passage of an electric current results in energy being expended. This energy may appear as the work done by the rotation of an electric motor, as the action of heating up a furnace element o r as the agency responsible for t h e electrolytic dissociation of a salt solution. The relation between conveying a quantity of electricity round a circuit by an applied voltage and the resulting work done can be used to derive the units of voltage

and resistance in terms of the coulomb and the joule which have already been defined.

UNIT OF VOLTAGE

THE VOLT. This is the unit of electromotive force and potential difference and can be defined as the potential difference required between two points in a circuit, if one joule of work is to be done when passing one coulomb of electricity between the points. As stated in Chapter 1 , the symbol for voltage or e.m.f. is V

and any value in volts is represented by the letter V after the numerical value. In accord with the remarks made concerning the representation of current, the reader's attention is drawn to the Table of Prefixes of Magnitudes, and to the correct use of the Abbreviations.

From tho dofinition set out tibovc it is stated that the

work

done by part of an electrical circuit = quantity of electricity conveyed x voltage applied across that part of the circuit.

Thus : fl Goules) = Q (coulombs) x V (volts)

or W (joules7 = Z (amperes) x t (seconds) x V (volts).

v 2 t

Other forms are W =

or W = It(IR) = 1 2 ~ t

Example 12. Consider Example 1 1 . A battery of e.m.f. 42V

- $ and internal resistance 7 0 is used to supply a circuit of three

resistors, 2, 4 and 8 0 in series. If the current is switched on for 30 minutes, find the energy converted (as heat) by each resistor and inside the battery &self.

Circuit current was found to be 2 amperes Using form W = 12Rt then energy converted in

2 ohm resistor = 22 x 2 x ' 30 x 60 = 14400 joules 4 ohm resistor = 22 x 4 x 30 x 60 = 28 800joules 8 ohm resistor = 2* x 8 x 30 x 60 = 57 600joules 7 ohm battery = 2' x 7 x 30 x 60 = 50400 joules Total energy converted by the circuit =

14 400

+

78 800

+

57 600

+

50 400 = 151 200 joules

( ' l / t ~ X . The total energy converted by the entire circuit may be

Ii)utltl 1'1 orll Cj' =- I.'// joulcs

= 42 x 2 x 30 x 60 = 151 200 joules.

The definitions of' Power and Energy have already been con- \~dered, but i t would be as well to summarise the points of' Importance, n:rnlely t h a t power is the rate at which work is done

1,'ro111 c i ~ * ( I i ~ ~ - t i o r ~ ~ qct 0111 : I \ N T \ C 11' - 1'11 : I I I ~ I 1 1 l o l l o \ ~ ~ \ ! I I . I I P = VI o r P (watts) = V (volts) x I (amperes).

The iibove is a most importi~nt reli~tinnship. I t c i ~ n ;11so k expressed in the following f'orms:

v Z

P = 12R o r P = -

R

The attention of the reader is drawn to the following which must also be known.

W

Since P = - o r W = Pt itfollows that

t

joules = watts x seconds and that a joule is one watt second. Now a joule is a small unit of energy and for practical purposes a much larger electrical unit of energy is used. This is the kilort,irtr

holrr, abbreviated to k W h and is also know^ as the commerci:il unit of electricity o r more commonly as 'a unit'.

Since one kilowatt hour = one kilowatt x one hour = lO!N watts x 3600 seconds So one kilowatt hour = 3 600

000

joules.

Example 13. A 220V electric fire is rated at 2kW. Find the current taken when the fire is switched o n and also how much it would cost to use the fire for 5 hours with electricity being charged at 0 . 6 ~ per unit.

X I OOU

Current taken = L7 = Y4YA 220

Electricity used = 2 x 5 = 10 kW h = 10 units Cost = 10 x 0.6 = 6p.

UNIT OF RESISTANCE

T H E O H M . This was defined in Chapter 1 as the unit of' sesisl- ance and in terms of the volt and ampere thus:-a resistor has a value of one o h m resistance, if one ampere passes through i t

when a potential difference of one volt is applied across ~ t s ends Now that the relations between the ampere, volt, joule and watt have been defined. i t is possible to give a l'ust1ie1- definition t'or the ohm which is associated with power o r energy dissipation Thus the ohm can be defined as:--that resistance which when one ampere passes thrclugh it produces power at the rate. of one watt. Alternatively, the ohm is that resistance in which a current of one ampere flowing 'or one second generates a joule of energy. For a resistor the ensrgy produced by current flow appears a heat. and the followin: is of' importance.

Since P = F I and I/ = I R then P = ( 1 R ) I