2014-G12-Physics-E

EEXEXQ ) 16 ALTERNATING CURRENT Learning Objectives At the end of this chapter the students will be able to: Understand and describe time period, frequency, the peak and root mean square values of an alternating current and voltage. 2. Know and use the relationship for the sinusoidal wave. 3. Understand the flow ofA .C . through resistors, capacitors and inductors. 4. Understand how phase lags and leads in the circuit. 5. Apply the knowledge to calculate the reactances of capacitors and inductors. 6. Descnbe impedance as vector summation of resistances. Know and use the formulae ofA .C . power to solve the problems. 8. Understand the function of resonant circuits. 9. Appreciate the principle of metal detectors used for security checks. Describe the three phase A .C. suppty. Become familiar with electromagnetic spectrum (ranging from radio waves toy rays). 12, Know the production, transmission and reception of electromagnetic waves. W * have read in the last chapter that an A .C . generator produces alternating voltage/current. Now a days most of the electrical energy is produced by A .C . generators using water power or huge steam turbines. The main reason for the world wide use ofA .C . is that it can be transmitted to long distances easily and at a very low cost. 16.1 ALTERNATING CURRENT Alternating current (A C .) is that which is produced by a voltage source whose polarity keeps on reversing with time (Fig.16.1 a.b). In Fig.16.1 (a), the terminal A of the source is positive with respect to terminal B and it remains so during a time interval 0 to T / 2 . At t = T / 2 . the terminals change their polarity. Now A becomes negative with respect to B (Fig.16.1 b). This state continues during the time interval T / 2 to T. after which terminal A again becomes positive with respect to B and the next cycle starts. As a result of this change of polarity, the direction of the current flow in the circuit also changes. During the time 0 - T/ 2 . it flows in one direction and during the interval T / 2 - T i n opposite direction (Fig. 16.1 a.b). This time interval T III

during which the voltage source changes its polarity once is known as period 7 of the alternating current or voltage. Thus an alternating quantity is associated with a frequency /given by T h e most com m on source of alternating voltage is an A C . generator which has been described in the previous chapter. T h e output V o f this A C . generator at an y instant is given by V0si.n y2 it x I, (1 6 2 ) where 7 is period of the rotation of the coil and is equal to the period ofA .C . and ^ = 2 s / = w is angular frequency of rotation of the coil. Th u s =<oJ is the angle 9 through which the coil rotates m time t. Eq.16.2 shows that the value of alternating voltage V is not constant It changes with timef.f 19 1*1 W h en I « 0. o - y *f is 0 and V is zero W h en t = 7/4. 0 = x — a — and Vattains its m axim um value V at this T 42 instant. At I = 7/2. 0 * n and V is zero. At this instant V changes its polarity and becom es negative henceforth. 37 and V = ■Vc and finally at the end of tho Whenf =— . 0 4/7 \C p E cycle w hen I = 7. 0 = 2s and V = 0. T h e variation of V with */2 \ 3V 2 time f and 0 is shown in F ig . 16.2 (a .b). Th is graph between voltage and time is known a s waveform of alternating voltage. It can bo seen that it is a sine curve. T h u s the output voltage of a n A .C . generator varies sinusoidally with time. In our daily life we are mostly dealing with this type of voltage, s o w e will consider it in deta il (b) 1. In sta ntane o us valueFig. 16.2 T h e value of voltage or current that exists in a circuit at any instant of timo t m easured from som e reference point is known a s its instantaneous value. It can have any value between plus maximum value * V , and negative maximum 112

value -V„ ar>d is denoted by V. T h e entire waveform shown inFig. 16.2 is actually a set of all the instantaneous values thatexist during a period T. Mathematically, it is given byV = V .s in 6 = V .sin <of ............. (1 6 .3 ) 2- , V = V .s i n 'y x ‘ =V .sin2*ff 2. Peak value Do You Know?It is the highest value reached by the voltage or current in one ' - . • 0 7 /.cycle. F o r exam ple, voltage shown in Fig. 16.2 has a peakvalue of V.. A .C Waveform 3. Peak to Peak ValueIt is the sum of the positive and negative peak values usuallywntten as p-p value. Th e p-p value of the voltage waveformshown in Fig. 16.2 is 2 V.. R o o t M e a n S q u a r e (r m s 1V a Iu cIf we connect a n ordinary D .C . am m eter to m easure V-alternating current, it would m easure its value as averaged §over a cycle. It can be seen in Fig. 16.2 that the average value &of current and voltage over a cycle is zero, but the power }delivered during a cycle is not ze ro because power is I 'R and |the values of /‘are positive even for negativo values of /. Th u sthe average value of / 'is not zero and is called the m eansquare current. T h e alternating current or voltage is actuallymeasured by squaro root of its m ean square value known asroot m ean square (rm s ) value. FI9- 1 « JLet us com pute the average value of V over a cycle. F ig .16.3shows a n alternating voltage and the w a y its V values vary.Note that the values of V 1 are, positive on the negative halfcycle also. A s the graph of V is symmetrical about tho line2 Vot so for thisfigure the m ean or the average value of V is ’ Vo. T h e root m ean square value of V is obtained by taking 113

th e s q u a re roo t o f V * 12 T h e re fo re . v~ * J ? mT2 m07K (16-4)Similarly / _ = ^ =0.7/oMost of the alternating current and voltage meters arecalibrated to read rms values. When we speak of A .C . meterreading, we usually mean rms values unless statedotherwiseE x a m p le 16.1 : A n A .C . voltmeter reads 250 V. What is itspeak and instantaneous values if the frequency of alternatingvoltage is 50 H z?S o lu tio n :rms valuo of alternating voltage ■ V*. ■ 250 VIts peak value V. is given by tho relationor K = J 2 V -. = >fex 250 V = 353.5 VAngular froquoncy o>= 2itf = 2xitx50H z = 100nHzTherefore, instantaneous value is given by V = V.sin o>/ = 353.5 sin (lO O x f) VPhase of A.C.W e have seen that the instantaneous value of the alternatingvoltage is given by V = V.sin t»for V = V.sinOThis angle 0 which specifies the instantaneous value of thealternating voltage or current is known as its phase. InFig. 16.2 (b). we can say that the phase at the points A . B. C . 0and E is 0. x / 2. x. 3x / 2 and 2x respectively because theseangles are the values of 0 at these points. Th u s each point onthe A .C . waveform corresponds to a certain phase. 1 14

T h e phase at the positive peak is x/2 = 90’ and it is F lg .1 M 3x/2 = 270’ at tho negative peak. T h e points whore the Flfl. i t s waveform crosses the time axis correspond to phase 0 and x. F*g. 16.$ Phase Lag and Phase LeadIn practice, the phase difference between two alternatingquantities is more important than their absolute phases.Fig 1 6 4 shows two waveforms 1 and 2. Th e phase angles ofthe waveform 1 at the points A. B. C . D and E have been shownabove the axis and those of waveform 2 below the axis. At thepoint B. the phase of 1 is x 12 and that of 2 is 0. Stmiarly it canbe seen that at each point the phase of waveform 2 is less thanthe phase of waveform 1 by an angle of x / 2 W e s a y th a tA .C . 2is lagging behind A .C , 1 by an angle of x / 2 It means that ateach instant, the phase ofA C . 2 is less than the phase ofA C . 1by it / 2. Similarly it can be seen m Fig. 16.5. that the phase ateach point of the waveform of A C . 2 is greater m an that of Awaveform 1 by an angle s /2. In mis case, it is said that A .C . 2 isleading the A C . 1 by n / 2. It means mat at each instant of time,the phase ofA .C . 2 is greater than that of 1 by x /2.Phase lead and lag between two alternating quantities isconveniently shown b y representing the two A C . quantitiesas vectors. V e cto r R epresentation of an A lte rn a tin g Q ua ntityA sinusoidally alternating voltage or current can begraphically represented by a counter clockwise rotatingvector provided it satisfies the following conditions.1. Its length on a certain scale represents the peak or rm s value of m e alternating quantity.2. It is in the horizontal position at m e instant when the alternating quantity is zero and is increasing positively.3. Th e angular frequency of m e rotating vector is the sam e as the angular frequency to of the alternating quantity.Fig. 16.6 (a ) shows a sinusoidal voltage waveform leading analternating current waveform by x /2. T h e sam e fact has beenshown vectorially in Fig. 16.6 (b). Here vector 01 representsthe peak or rms value of the current which is taken as thereference quantity. Similarly O V represents m e rms or peakvalue of the alternating voltage which is leading the current by90°. Bom vectors are supposed to be rotating in m e counter 115

c lo c k w is e d ire c tio n a t the a n g u la r fre q u e n c y o> o f the two alternating quantities. Fig . 16.6 (b ) s h o w s the position of v o lta g e a n d c u rre n t v e c to r a t f = 0 . _______________________________ 16.2 A .C . C IR C U IT S T h e basic circuit elem ent in a D .C . circuit is a resistor (R ) w hich controls the current o r voltage an d the relationship be tw een them is g iv e n b y O h m 's la w that is V = I R . In A .C . circuits, in additio n to resistor R , tw o n e w circuit elem ents nam ely IN D U C T O R (L ). and C A P A C IT O R (C ) b e co m e relevant. T h e current a n d vo lta g e s in A .C . circuits are controlled b y three elem ents R . L a n d C . W e w ould study the r e s p o n s e o f a n A . C . circuit w h e n it is e x cite d b y a n a lte rn ating v o lta g e ._____________________________________________ 16.3 A .C . T H R O U G H A R E S IS T O R F ig . 16.7 (a ) s h o w s a resistor of resistance R connected with an alternating voltage source. A t a n y tim e t the potential difference a c ro s s the term inals of the resistor is g iven by V = V ,.s in c o f ................ (1 6 .5 ) w h e re V . is the pe ak valu e of the alternating voltage. T h e current / flow ing through the circuit is g iven b y O h m 's law V ' - R ft sin<of or / = /„sinw f (1 6 .6 ) Ysl w h e re / is the instantaneous current a n d /, = ft is the i p e a k v a lu e of th e c u rre n t. It fo llo w s fro m E q s .1 6 .5 a n d 16.6 (e) that the instantaneous values of both voltage and current are sin e functions w h ich va ry with tim e (F ig , 1 6 .7 b ). T h is figureFlfl. 16.7 s h o w s that w h e n v o lta g e rise s, the c u rre n t a ls o rises. If the voltage falls, the current also d o e s s o - both pa ss their m axim um a n d m inim um va lu e s at the s a m e instant. T h u s in a purely resistive A .C . circuit, instantaneous va lu e s of voltage a n d current are in p h a se . T h is b e haviour is s h o w n graphically in F ig .16.7 (b ) and vectonalty in F ig . 16.7 (c ). Fig . 16.7 (c ) s ho w s V an d I vectors for resistance. T h e y are dra w n parallel b e c a u s e there is n o p h a se difference betw een the m . T h e o p p o s itio n to A .C . w h ich the circuit 116

presents is the resistance*= 7 (16.7)T h o instantaneous power in the resistance is given byp a l’ R ^ V Ix V '/ R ............. (16.8)P is in watts. V is in voits. I is in amperes and R Is in ohms. It isvery important to note that the Eq. 16.8 holds only when thecurrent and voltage are in phase.16.4 A.C. THROUGH A CAPACITORAlternating current can flow through a resistor, but it is notobvious that how it can flow through a capacitor. This can bedemonstrated by the circuit shown in Fig. 16.8. A low powerbulb is connected in series with a 1 uF capacitor to supplymains through a switch. W hen the switch is closed, the bulblights up showing that the current is flowing through thecapacitor. Direct current cannot flow through a capacitorcontinuously because of the presence of an insulatingmedium between tho platos of the capacitor. N ow let us seehow does A .C . flows through a capacitor. T h e current flowsbecause the capacitor plates are continuously charged,discharged and charged the other way round by thealternating voltage (Fig. 16.9 a). Th e basic relation betweentho charge q on a capacitor and the voltage V across itsplates i.e. q = C Vholds at every instant. If V = V0sin u>/is theapplied alternating voltage, the charge on the capacitor atany instant will be given byq = C V » C V . sinwf ............. (16.9)Since C. V, are constants, it is obvious that q will vary the sameway as applied voltage i.e.. V a nd ga re in phase (Fig. 16.9b). pTh e current /is the rate of change ofq with time i.e.. / . A^tS o the value of /at any instant is the corresponding slope ofthe p /curve. At O when q = 0. the slope is maximum, so /isthen a maximum. From O to A. slope of the q -t curvedecreases to zero. S o /is zero atN. F ro m A to B th e slope ofthe q - 1curve is negative and so /is negative from N to R. Inthis way the curve P N R S T gives the variation of currentwith time.

Referring to the Fig. 16.2 (b ) itcan be seen that the phase at Ois zero and the phase at the upper maximum is x / 2. S o inFig.16.9 (b ) the phase of V at O is zero but the current at thispoint is maximum so its phase is x / 2. Thus, the current isleading the applied voltage by 90* or x / 2. Now consider thepoints A and N. Th e phase of alternating voltage at A is x / 2but the phase of current at N is x. Again the current is leadingthe voltage by 90° or x /2 Similarly by comparing the phaseat the pair of points (B. R ). (C . S ) and (D . T ) it can be seen thatat all these points the current leads tho voltage by 90’ or n/2.This is vectorialty represented in Fig. 16.9 (c).Roactance of a capacitor is a measure of the oppositionoffered by the capacitor to the flow of A .C . It is usuallyrepresented by X c . Its value is given by (16.10)whore VL, is the rms value of the alternating votage across thecapacitor and / „ is the rms value of current passing through thecapacitor. The unit of reactance is ohm. In case of capacitor 2x1fC <o1C (16.11)According to Eq. 16.11. a certain capacitor will have a largrreactanco at low frequency. So the magnitude of th«opposition offered by it will be large and the current in th«circuit will be small. O n the other hand at high frequency. th<roactance will be low and the high frequency current througltho same capacitor will be largeE x a m p le 16 .2: A 100 uF capacitor is connected to aralternating voltage of 24 V and frequency 50 H z. Calculate (a ) Th e reactance of the capacitor, and (b ) Th e currdnt in tho circuitS o lu tio n : 1(a ) Reactance of the capacitor Xc * 1 2 x 3.14 * 50 s\"’ x 100 x 10Xc = 3 1 .8 -— * - 31.8 ft c Cs ’118

(b ) From the equation Xf-= —v _ , V -. 24 V , 7CA 075A16.5 A.C. THROUGH AN INDUCTORA n inductor is usually in the form of a coil or a solenoid woundfrom a thick wire so that it has a large value of self inductanceand has a negligible resistance. W e have already seen howself inductance opposos changes of current. So when analternating source of voltage is applied across an inductor, itmust oppose the flow ofA .C . which is continuously changing(Fig. 16.10). Let us assume that the resistance of the coil isnegligible. W e can simplify the theory by considering first, thecurrent and then finding the potential difference across theinductor which will cause this current. Suppose the current is/ = /.sin 2 Jiff. If L is the inductance of the coil, the changingcurrent sets up a back emf in the coil of magnitude M ‘ • 'L uTo maintain the current, the applied voltage must be equal tothe back e.m.f. T h e applied voltage across the coil must,therefore, be equal toSince L is a constant. V is proportional to — . Fig. 16.10 (b ) Afshows how the current / varies with time. The value ofA/ / Af is given by the slope of the / - f curve at the variousinstants of time. At O, the value of the slope is maximum, so themaximum value of V equal to V, occurs at O and is representedby O P (Fig. 16.10 b). From O to A the slope of I - f graphdecreases to zero so the voltage decreases form V, to zero atQ. From A to B. the slope of the / - f graph is negative, so thevoltage curve goes from Q to /?. In this way the voltage isrepresented by the curve P Q R S T corresponding to currentcurve O A B C D . By comparing the phases of the pair of points(O , P). (A, Q ). (B. R), (C . S ) and (D. T ). it can be seen that thephase of the current is always less than the phase of voltage by90'’ or x / 2 i.e.. current lags behind the applied vottago by 90*or x/2 or the applied voltage leads the current by 90* or n / 2. 119

T h is is vectorially shown in Fig. 16.10(c) Inductive reactance is a m easure of the opposition offered by the inductance coil to the flow ofA .C . It is usually denoted by X L. *= (16.12) If V ^ s is rms value of the alternating voltage across an inductance and l„.x. the rms value of the current passing through it. the value of X L is given byIiltnrirstinp Information X u”* 2 x f L <oL (16.13) T h e reactance of a coil, therefore, depends upon the frequency of the A .C . and the inductance L. It is directly proportional to both /and L . L is expressed in henry, /in hertz, and X in ohms. It is to be noted that inductance and capacitance behave oppositely as a function of frequency. If / is low X L is smaH but X c is large. F o r high /. X L is large but X c is small. T h e behaviour of resistance is independent of frequencyIrrXKtcn t f « m tires So Rofem ng to Fig. 16.10 (b). it can be seen that no power isptv«sriio-r>tm-waewniddeinvdu«nsS«rtyy. functfon* in dissipated in a pure inductor. In the first quarter of cycle both V and / are positive so the power is positive, which means energy is supplied to inductor. In the second quarter. V is positive but /is negative. N ow power is negative which implies that energy is returned by the inductor. Again in third quarter, it receives energy but returns the sam e amount in the fourth quarter. Th u s, there is no net change of energy in a complete cycle. Since an inductor coil does not consum e energy, the .coil is often employed for controlling A .C . without consumption ofe ne rm ^S uchan in du cten cecaliskTO w na scM ^^^^ im p e d a n c e H H B H H H H H W e already know that resistance R offers opposition to tho flow of current. In case of A .C . a n inductance L or a capacitance C also offer opposition to the flow of A .C . which is m easured by reactances X L and X t respectively. A n A .C . circuit m ay consist of a resistance R , a n inductance L , a capacitance C or a combination of these elements. Th e com bined effect of resistance and reactancos in such a circuit is known as im pedance and is denoted by 2 . It is measured by the ratio of the rm s value of the apphod voltage to the rm s value of Moulting A C . Th u s (1 6 1 4 ) 120

It is also ex p re sse d in o h m s.E x a m p le 1 6 .3 : W h e n 10 V are applied to an A .C . circuit, thecurrent flowing in it is 1 0 0 m A . F in d its im pedance.S olution :rm s value of applied voltage = 10 Vrms value of current = / „ = 1 0 0 m A = 100 x 10' A v -. 1° v in n oIm podanceZ* u ~ 100 x 1 0 ’ A “16.7 R - C A N D R - L S E R IE S C IR C U ITSConsider a series network of resistance R and a capacitor Cexcited b y a n alternating voltage (F ig .16.11 a ). A s R and Care in series, the sam e current w ou ld flow through each ofthem. If /^, is the value of current, the potential differencea cross the resistance R w ou ld b e / „ . R a n d it w ou ld be inphase with current / T h o vector diagram of tho voltage andcurrent is sho w n in Fig.16.11 (b ). Taking the current asreference, the potential difference L . R across tho resistanceis represented by a line along th e current line becausepotential d ro p R is in phase wixh current. T h e potentialdifference a c ro s s the capacitor will b e / _ X c = / „. / <->C. A s <•)this voltage la gs the current b y 9 0 \". s o the line representing Flfl 16.11the vector / _ / coC is dra w n at right angles to the current line(Fig.16.11 b).T h e appltod v d ta g e V that will s e n d the current / in thecircuit is obta-ned by the resultant of the vectors /_. R and / ■ i.e.. «CImpedance Z < t»cy (1 6 1 5 )It can b e s e e n in Fig. 16.11 (b ) that th e current and the appliedvoltage a re not in ph a se . T h e c u n e n t leads the appliedvoltage by an angle 0 such that 12!

w8 = tan (16.16)E q .16.15 suggests that we can find the im pedance of aseries A C . circuit by vector addition. T h e resistance R isrepresented by a horizontal line in the direction of currenlwhich is taken as reference. T h e reactance X c - — is wCshow n by a line lagging the R - line by 9<r (Fig. 16.11 c ). Th eim pedance 2 of the circuit is obtained by the vectorsummation of resistance and reactance. Fig. 16.11 (c ) isknown as impedance diagram of the circuit. T h e an gle whichthe lino representing the impedance 2 makes with R linegives the phase difference between the voltage and current.In Fig.16.11(c). the current is leading the voltage applied byan angle N o w we will calculate the impedance of a R - L series circuit by drawing its impedance diagram . Fig.16.12 (a ) sho w s an R - L series circuit excited by a n A .C . source o f frequency o>. Th e current is token a s reference, so rt is represented by a horizontal line. Resistance R is drawn along this line because the potential drop / „ R is in phase with current. A s the potential across the inductance V L = / _ X t = (<»L) leads(•) the current by 90*. so the vector line of reactance X L = o>L is draw n at right angle to P i n e (F ig . 16.12 b). T h e im pedance Z of the circuit is obtained by the vector sum of R and <»L lines. Thus Z - vR* ♦ (<»Lf T h e angle o = tan' 1— w hich Z makes with R line gives the phase difference between the applied voltage and current. In this case the voltago leads the current by 0*. B y comparing the .m pedance diagrams of R ■ C and L - R circuits it can be seen that the vectcr linos of reactances Xc and X L are directed opposite to tach other with R as reference. 16.8 PO W ER IN A .C . CIR CU ITS T h e expression for power is P = V C ./ .,. Th is expression is true in case of A C . d ro its , only when V a n d / are in phase as in • 122

case of a purely resistive drey*. W e have already seen that thepow er dissipation in a pore inductive or in a pure capacitancedrcuit is zero. In these cases the current lags or leads thoapplied vottago by 90* and com ponent of applied voltagevector V along the current vector is ze ro (Fig. 16.9 c and16.10 c ). In A . £ . drcuit the phase difference between appliedvoltage t/and the current / _ is 0 (Fig.16.11 b and 16.12 b). Th ecom ponent of V along current / _ is cosQ. Actually it is thiscom ponent of voltage vector which is in phase with current. S othe power dissipated in A C . drcuit P = L x V „ cos0 .............. (16.17)T h e factor cosG is kno w n a s po w e r factor.E x a m p le 1 6 .4 : At what frequency will an inductor of1.0 H have a reactance of 500 O ? Do You Know?S o lu tio n : L = 1.0H . X t «5 0 0 fi XL* of. = 2 ntL - 500 O 5000 5000 / c 2xfL 2 x x 1 .0 H =80HzE x a m p le 1 6 .5 : A n iron core coil of 2 .0 H and 50 0 is placed A <M*ctor « u M d to local*in series with a resistance of 4 5 0 0 . A n A .C . supply of 100 V. tum*0 m*ttic6t*ct»50 H z is connected across the drcuit. Find (i) the currentflowing in the coil, (ii) phase angle betw een the current andvoltage.S olu tio n :Resistance = R » 5 0 0 * 4 5 0 0 = 5 0 0 0Inductance = L = 2.0 HSupply voltage = V _ 3 100 VFrequency =f=50H zT h e reactance = X l = o>f. = 2 x f/.Im pedance = 2 x 3.14 x 50 s ' x 2.0 H = 628 0 « Z = y R 2 » ( « of.)2 = V(500 O )’ ♦ (628 O )2 = 803 O Current / „ V - 100 V = 0.01245A = 12.45 m A Z . 803Q .(cof.Phase difference 0 = tan 123

f 628 Q 'l [ s o o n ] ' 5' 5'E x a m p le 1 6 .6 : A circuit consists of a capacitor of 2 p F and aresistance of 1000 Q connected in series. A n alternatingvoltage of 12 V ar>d frequency 50 H z is applied. Find (I) thocurrent in tho circuit, and (ii) the average pow er supplied.S o lu tio n : Resistance = R = 1000 O Capacitance = C = 2 p F = 2X10\"*F Frequency = /= 50 H z !_ Reactance c j r.fC ------------------------- --------------------- t— = 1592 n 2 x 3 .1 4 x 5 0 s x 2 x 10 FImpedance Z f e * * ( X c f ->/(1OOOn), * (150 2fr)2 = 1 8 8 0 0Current = U w 1 2 V = 0.0064 A = 6.4 m A ■~ 18800 ZPhaso Difference 0 - tan ' 1 R tan 11.1000 f l != 57.87° )Avorage power = V _ /„, cosO = 12 V x 0.0064 A x 0.532 = 0.04 WConsider a R - L - C series circuit which is excited by analternating voltage sourer; w hoso frequency coukJ be varied(F ig. 16.13 a). T h e imped an co diagram of the circuit is shownin Fig. 16.13 (b). A s expla ined earlier, the inductive reactanceX t = m L and capacitor reactance xV c = 1 are directedopposite to each other. W h en the frequency ofA C . source isvery small * c ^ jS m uch greater than X L * t»L. S o thocapacitance dom inates at low frequencies and the circuit 124

b e h a v e s like a n R - C circuit. A t high frequencies X t = «>L ism u ch greater than X c = ~ . In this c a s e the inductance * -uL •■dom inates a n d the circuit b e h a v e s like R - L circuit. In * -= cbetw een the se freq uen cie s there will b e a freq uen cy ta, at (b>w hich X L = X c . T h is condition is called resonance. T h u s atresonance the inductive reactance being equal and oppositeto capacitor reactance, cancel each other and the im pedancediagram assum es the form (Fig. 16.13 c). T h e value of theresonance frequency can be obtained b y putting a’L a ^ c («)or 21 or 1 FI9 16.13 Tlc = LCor f, 2 7 \ i C (1 6 .1 8 )T h e following are the properties of the series resonance,i) T h e reson ance frequency is given by '-'d t c Flfl. 16.14 T h e im pedance of the circuit at resonance is resistive so the current and voltage are in phase. T h e pow er factor is 1 .i:i) T h o im p e d a n c e of the circuit is m in im um a t this froq uon cy a n d it is equal to R .iv ) If the am plitude of the s o u rce voltage V , is constant, the current is a m axim um at the resonance frequency an d its v a lu e is V0/ R . T h e variation of current with the frequency is show n in Fig. 16.14.v ) A t re s o n a n c e V L, the voltage d ro p across inductance and V c the voltage drop across capacitance m ay be m uch larger than the source voltage.16.10 P A R A L L E L R E S O N A N C E C IR C U ITF ig 16 15 s h o w s a n L - C parallel circuit It is excited b y analternating source of voltage w hose frequency could beva rie d T h e inductance cod L has a resistance r w h ich isnegligibly small T h e capacitor draw s a leading current 125

LC whereas the cotl draws a tagging current T h e circuit resonates at a frequency <•>= to, which makes X t = X c . so thatrtg H I S the two branch currents are equal but opposite Hence, they 'i t 1» 1« cancel out with the result that the current draw n from the supply is zero In actual practice, the current is not zero but has a m inimum value due to small resistance r of the coil Properties of parallel resonant circuits ar^ i) Resonance frequency is ^ ~ z I—llv7l-7V2s '■) At the resonance frequency, the orcuil impedance is maximum. It is resistive, in) A t the resonance the current is minimum and it is in phase with the applied voltage. S o the power factor «s one. T h e variation of current with the frequency of the source is shown in Fig. 16.16. iv) At resonance, the branch currents /L and /c m ay each be larger than the source current /,. E x a m p le 1 6 .7 : Find the capacitance required to construct a resonance circuit of frequency 1000 k H z with an inductor of 5mH. S olution: Resonance frequency f * 1000 kHz L = 5 m H = 5 x 10 3 H C=? Resonance frequency = I ■ 1 4 **<*£. 4 x(3 .1 4 )* x (10* 8 ' f xx 5 x 1 0 * H 5 0 9 p F 1 W o have already studied that an A .C . generator consists of a coil with a pair of slip rings. A s the cod rotates an alternating voltage is generated across the slip rings. In a three phase A .C . generator, instead of one coil, there are three coils inclined at 1 2 0 *to each other, each connected to its own pair of slip rings. W h e n this combination of three coils rotate in the magnetic field, each coil generates an alternating voltage •across its ow n pair of slip rings. Th u s , three alternating voltages are generated. Th e phase difference between these voltages is 120°. It m eans that when voltage across the first 126

jair of slip rings is zero, having a phase of 0 . the voltagejcross the second pair of slip rings would not be zero but itvill have a phase of 120*. Similarly at this instant the voltagejenerated across the third pair will havo a phase 240'. Th is isshown in Fig. 16.17. T h e m achine, instead of having sixerm inals. two for each pair of slip rings, has only fourerm inals because the starting point of all the three coils has ax m m o n junction which is often earthed to the shaft of thejenerator and the other three ends of the coils are connectedx> three separate terminals o n the m achine. Th e se four:erminals along with the lines and coils connected to them are>hown in Fig .16.18. T h e voltage across each of linesconnected to terminals A . B . C and the neutral line is 230 V.Because of 1 2 0 ” phase shift, the voltage across any two liness about 400 V. T h e main advantage of having a three phaseiupp ly is that the total load of the house or a factory is dividedn three parts, so that none of the line is over loaded. If heavyoad consisting of a number of air conditioners and motorsate., is supplied pow er from a singlo phase supply, its voltages likely to drop at full load. Moreover, the throe phase supplyalso provides 400 V which can be used to operate som especial appliances requiring 400 V for their operation.16.12 PRINCIPLE O F M ETA L D E TE C TO R SA coil and a capacitor are electrical components whichtogother can produce oscillations of current. A n L ■C circuitbehaves just like an oscillating m ass - spring system. In thiscase energy oscillates between a capacitor and an inductor.Th e circuit is called an electrical oscillator. Tw o suchoscillators A and B are used in the operation of a com m ontype of metal detector (F ig. 16.19). In the absence of anynearby OKMHoronxHB OicAMy o-oMA 127

metal object, the inductances L A and Lj, are the sam e andhence the resonance frequency of the two circuits is alsosame. W hen the inductor B. called the search coil comesnear a metal object, its inductance decreases andcorresponding oscillator frequency increases and thus abeat note is heard in the attached speaker. S uch detectorsare extonsivcly used not only for various security checks butalso to locate buried metal objects.16.13 CH OK EIt is a coil which consists of thick copper wire w ound closely ina large num ber of turns over a soft iron laminated cores. Th ismakes the inductance L of the coil quite large whereas itsresistance R is ve ry small. Th u s it consum es extrem ely smallpower. It is used in A .C . circuits to limit current with extremelysmall wastage of energy as compared to a resistance or arheostat16.14 E L E C TR O M A G N E TIC W AVESIt is a very important class of w aves which requires nom edium for transmission and which rapidly propagatesthrough vacuum.In 1864 British physicist Ja m e s Clark Maxwell formulated aset of equations known as Maxwell's equations whichexplained the various electromagnetic phenom ena.According to these equations, a changing magnetic fluxcreates an electric field and a changing electric flux creates amagnetic field. C onsider a region of space A B as show n inFig. 16.20. Suppose a change of magnetic flux is taking placethrough it. Th is changing magnetic flux will set up a changingelectric flux in the surrounding region. T h e creation of electricfield in the region C D will cause a change of electric fluxthrough it due to which a m agnetic field would be set u p in thespace surrounding C D and so on. T h u s each field generatesthe other and the whole package of electric and magneticfields will m ove along propelling itself through space. Suchm oving electric and m agnetic fields are know n aselectromagnetic waves Th e electric field, magnetic field andthe direction of their propagation are mutually orthogonal(Fig. 16.21). It can be seen in this figure that theelectromagnetic waves are periodic, hence they have awavelength /. which is given b y the relation c = f). w here ( isthe frequency and c is the speed of the w ave. In free spacethe speed of electromagnetic w aves is 3 x 10* m s'1.128

Depending upon the values of wavelength and frequency,the electromagnetic w ave s have been classified into differenttypes of waves a s radiowaves, microwaves, infrared rays,visible light etc. Fig. 16.22 shows the complete spectrum of ^^ spectrum _^—rui-o -o in c y TV and M u u illK .liP W P ra*«6on X-rays ray* A U «no*JO rado ceso '«• > w ter icr ,0 ' 10\" 10\" 10\" 10\" Iff10* I f f Iff tO* W (irewe1)0' 10* 1 0 - 10\" 10 d e o t w n g wavelengei Fig. t i n T h o electromagnet.: * p e«rumElectromagnetic w aves from the low radio waves to high Tit-bitsfrequency gamma rays. Shake an electrically c h a r g e16.15 PRINCIPLE O F GEN ERATIO N, object to and Fro. and you produce TRANSMISSION AND RECEPTION olectrcm agnet w a ve s OF ELECTROMAGNETIC WAVESW e have seen that electromagnetic waves are generatedwhen electric or magnetic flux is changing through a certainregion of space. A n electric charge at rest gives rise to aCoulom b's field which do es not radiate in space because nochange of flux takes place in this type of field. A chargemoving with constant velocity is equivalent to a steadycurrent which generates a constant magnetic field in the

surrounding space, but such a field also does not radiate out because no changes of magnetic flux are involved. Th u s only chance to generate a w ave of moving field is when we accelorate the electrical charges. Oo You Know? A radio transmitting antenna provides a good example ofMCond. they produced redo were* generating electromagnetic w aves by acceleration ofhaving frequency 94 tttt charges. T h e piece of wire along which charges are m ade to accelerate is known as transmitting antenna (F ig. 16.23). It is charged b y an alternating source of potential of frequency t and time period T. A s the charging potential alternates, the charge on the antenna also constantly roverses. For exam ple if the top has ♦ q charges at an y instant, then after time 77 2 the charge on it will be - q. S uch regular reversal of charges on the antenna gives rise to an electric flux that constantly changes with frequency f. Th is changing electric flux sets up an electromagnetic w a v e which propagates out in space aw ay from the antenna. T h e frequency with which the fields alternate is always equal to the froqucncy of the source generating them. These electromagnetic waves which are propagated out in space from antenna of a transmitter are known as radio w aves. In free space these w aves travel with the speed of light. Suppose these waves impinge on a piece of wire (Fig. 16.24). T h e electrons in the wire m ove under the action of the oscillating electric field which give rise to an alternating voltage across the wire. T h e frequency of this voltage is the sam e as that of the w ave intercepting the wire. This wire receiving the wave is known a s receiving antenna. A s the electric field of the w ave is very w eak at a distance of m any kilometres from tho transmitter, the voltage that appears across the receiving antenna is very small. E a ch transmitter propagates radio w aves of one particular frequoncy. So w hen a num ber of transmitting stations operate simultaneously, we have a num ber of radio waves of different frequencies in the space. Th u s the voltage that appears across a receiving antenna placed in space is usually due to the radio waves of large num ber of frequencies. T h e voltage of one particular frequency can be picked up by connecting an inductance L and a variable capacitor C in parallel with one end of the receiving antenna (Fig. 16.24). If one adjusts the value of the capacitor so that the natural frequency of L ■ C circuit is the sam e as that of the transmitting station to be picked up. the circuit will resonate 130

under the driving action of the antenna. Consequently, the F o r Your Inform ation L - C circuit will build u p a large respon se to the action of only' that ra d io w a v e to w h ic h it is tu n e d . In y o u r r a d io r e c e iv e r set m m sm m w h e n y o u c h a n g e stations yo u actually adjust the value o f C . carrier wave 16.16 M O D U L A TIO N a A AI S pe ech and m usic etc. are transmitted h undred o f kilom etres ampMudo modiXatod wave a w a y b y a radio transmitter. T h e s ce n e in front of a television Fig. 16Z5 ca m e ra is also sent m a n y kilom etres a w a y to view ers. In all the se uses, the carrie r of the p ro g ra m m e is a high fre q u e n cy VAIUUM (Ml radio w a ve . T h e inform ation i.e.. light, soun d or other d a ta is im p re ss e d o n the rad io w a v e a n d is c a rrie d a lo n g w ith it to the t r. destination. o fmiiMV miUlAVAVAV/vlMi nIfMnMmAi M odulation is the proce ss o f com binin g the lo w freq uen cy frequency mmtuuted wave signal with a high frequency radio w a v e called carrier w ave . T h e resultant w a v e is called m odulated carrier w a ve . T h e low frequency signal is know n as m odulation signal. M odulation is ach ie ve d b y ch a n g in g the am plitude or the freq uen cy o f the carrier w a ve in accordance with the m odulating signal. T h u s w e have tw o types of m odulations w hich are 1. Am plitude m odulation (A .M ), 2. Frequency m odulation (F .M ) A m p litu d e M o d u la tio n In this t yp e of m o d u la tio n the a m p litu d e o f th e c a rrie r w a v e is increased o r dim inished as the am plitude of the supe rposing m odulating signal increases a n d decreases. Fig. 16.25 (a ) represents a high frequency carrier w a v e of constant am plitude and frequency. Fig. 16 .25(b) represents a low o r au dio frequency signal of a sine w aveform . F ig .16.25 (c ) sho w s the result obtained b y m odulating thccarrier w a ve s with the m odulating w ave. T h e A .M . transmission frequencies ran ge from 54 0 k H z to 1600 kH z. F re q u e n c y M o d u la tio n In this type o f m odulation the fre q u e n cy o f the carrier w a v e is increased or dim inished as the modulating signal amplitude «c re a s e s or decreases but the carrier w a ve amplitude rem ains constant. Fig . 16.26 show s frequency m odulation. T h e frequency of the m odulated carrier w a v e is highest (point H ) w h e n the signal am plitude is a t its m a x im u m positive v a lu e a n d is a t its lowest freq uen cy (point L ) w h e n signal am plitude h as m axim um negative. W h e n the signal amplitude is zero, the carrier fre q u e n cy is a t its norm al fre q u e n cy f„131

Th e F.M. transmission frequencies are much higher and ranges between 88 M H z to 108 M Hz. F.M. radio waves are affected less by electrical interference than A. M. radio waves and hence, provide a higher quality transmission of sound. However, they have a shorter range than A M waves and are less able to travel around obstacles such as hSs and large buildings. am Alternating current is that which is produced by a voltage source w hose polarity keeps on reversing with time.• T h e time interval during which the voltage source changes its polarity once is known as period T of the alternating current or voltage. T h e value of voltage or current that exists in a circuit at an y instant of time measured from som e reference point is known as its instantaneous value. T h e highest value reached by the voltage or current in one cycle is called the peak value of the voltage or current. T h e sum of positive and negative peak values is called peak to peak value and is written as p -p value. T h e root mean square value (rm s ) is the square root of the average value of V ‘or /’. T h e angle 0 which specifies the instantaneous value of the alternating voltage or current, gives the phase lag or phase lead of o n e quantity over the other. A n inductor is usually in the form of a coil or a solenoid wound from a thick wire so that it has a large value of self inductance and has negligible resistance. T h e com bined effect of resistance and reactance in a circuit is known as impedance and is denoted b y Z . Choke is a coil which consists of thick copper wire wound closely in a large number of turns over a soft iron laminated core.• Electrom agnetic w aves are thoso which require no medium for transmission and rapidly propagate through vacuum. Modulation is the process of com bining tho low frequency signal with a high frequency radio w ave, called carrier waves. T h e resultant wave is called modulated carrier wave. I r l 'i m w Asinusoidal current has rms value of 10 A W hat is the maxim um or peak value? 132

16 2 N a m e the d e vic e that will perm it flow o f direct current but op p o s e the flow of alternating current permit flow of alternating current but not the direct current.16 .3 H o w m a n y tim es per second will a n incan de scen t la m p reach m axim um brilliance when connected to a 50 H z source?16.4 A circuit contains a n iron-cored inductor, a switch and a D .C . source arran ged in series. T h e switch is closed and after an interval reopened. Explain w h y a spark jum ps across the switch contacts?16 5 H o w does doubling the frequency affect the reactance of an inductor a capacitor?16 6 In a R - L circuit, will the current la g o r le ad the voltage? Illustrate yo u r an sw er by a vector diagram.16.7 A choke cod placed in series with an electric lam p an A C . circuit causes the lam p to be com e dim . W h y is it so? A variable capacitor ad ded in series m this circuit m a y be adjusted until the lamp glow s with norm al brilliance. Explain, how this is possible? Explain the conditions under which electrom agnetic w aves are produced from a y source?16 9 H o w the reception of a particular rad io station is selected o n yo u r rad io set?16 10 W h a t is m e a n t b y A .M . and P.M .? G 23EEJ A n alternating current is represented b y the equation / = 2 0 sin 100 itf. C o m p u te its frequency and the m axim um and rms values of current. (A n s :5 0 H z .2 0 A . 14 A )16 2 A sinusoidal A .C . has a m axim um valu e of 1 5 A W h a t are its rm s va lu e s ? If the time16 4 is recorded from the instant the current is ze ro and is becom ing positive, w hat is the instantaneous value of the current after 1 /3 00 s. given the frequency is 50 H z. (A n s : / ,= 1 0 .6 A . Instantaneous current = 13.0 A ) Find the value of the current and inductive reactance w hen A .C . voltage of 220 V at 5 0 H z is passed through an inductor of 10 H . (A n s : / _ = 0.07 A . X L= 3140 O ) A circuit h a s an inductance of 1/n H a n d resistance of 2 0 0 0 O . A 5 0 H z A . C . is supplied to it. C a lcula te the reactance a n d im p e d a n ce offered b y the circuit. (A n s : X L= 100 O . Z = 2002.5 0 )16 5 A n inductor of pure inductance 3/n H is co n nected in series with a resistance of 4 0 O . F in d (i) the peak value of the current tho rm s value, a n d (Hi) the ph a se difference betw een the current a n d the applied voltage V = 3 5 0 sin(1 OOit t). (A n s : (i) 1.16 A . (ii) 0.81 A . (iii) 82.4‘ ) A 10 m H , 20 O coil is connected across 240 V a n d 180 I z H z source. H o w m uch p o w e rd o e s it dissipate? (A n s:2 7 7 8 W ) 133

Find the value of the current flowing through a capacitance 0 .5 p F w hen connected to a source of 150 V at 50 H z. (A n s :/ nm= 0 .0 2 4 A ) A n alternating so u rc e of em f 12 V a n d freq uen cy 5 0 H z is applied to a capacitor of capacitance 3 p F in series with a resistor of resistance 1 k O , Calculate the phase angle. (A n s : 46.7*) W h a t is the resonant frequency of a circuit which includes a coil o f inductance 2.5 H and a capacitance 40 p F ? (A n s : 15.9 H z)16 1 A n inductor of inductance 150 p H is connected in parallel with a variable capacitor w hose capacitance can bo changed from 500 p F to 2 0 pF. Calculate the m axim um frequency and m inim um frequency for w hich the circuit can be tune d. (A n s : 2.91 M H z. 0.58 M H z) 134

PHYSICS OF SOLIDSLearning ObjectivesAt the end ofthis chapter the studentswill bo able to: Distinguish between the structure of crystalline, glassy, amorphous and polymeric solids. Understand the idea oflattice. Appreciate that deformation is caused by a force and that, in one dimension, the deformation can be tensileor compressive Define and use the terms Young's modulus, bulk modulus and shear modulus. Describe an experiment todetermine elastic limit and yield strength. Distinguish between elasticand plastic deformation ofa material. Synthesize and deduce the strain energy ina deformed material from the area under the forceextension graph Describe the energy bands in sofcds. Classify insulators, conductors, semi-conductors on the basisof energy bands. Distinguish between intrinsic and extrinsic semiconductors. Explain how electrons and holes flowacross a junction. Descnbe superconductors. Distinguish between dia. para and ferro magnetic materials. Understand and describe the conceptof magnetic domains in a material. Know the Curie point. Classify hard and soft ferro magnetic substances. Understand hysteresis and hysteresis loss.I \ ^ aterials have specific uses depending upon their characteristics and properties, suchas hardness, ductility, malleability, conductivity etc. What makes steel hard, lead soft, ironmagnetic and copper electrically conducting? Itdepends upon the structure - the particularorder and bonding of atoms in a material. This due has made it possible to design andcreate materials with new and unusual propertiesfor use in modem technology. 17.1 C L A S S IF I C A T IO N O F S O L ID SCrystalline SolidsIn crystalline solids there is a regular arrangement of molecules. The neighbours of everymolecule are arranged in a regular pattern that isconstant throughout the crystal. There is.thus an ordered structure in crystalline solids. 135

T h e vast majority of solids, e.g.. metals such as copper, iron and zinc, ionic com pounds such as sodium chloride, ceramics such as zirconia are crystalline. Th e arrangement of molecules, atoms or ions within all types of crystalline solids can bo studied using various X -ra y techniques. It should be noted that atom s, molecules o r ions in a crystalline sobd are not static. F o r exam ple, each atom in a metal crystal vibrates about a fixed point with an amplitude that increases with rise in temperature. It is the average atom ic positions which are perfectly ordered over large distances. T h e cohesive forces between atoms, molecules or ions in crystalline solids maintain the strict lo ng-range order inspite of atomic vibrations. For every crystal, however, there is a temperature at w hich the vibrations be com o so great that the structure suddenly breaks up. and the solid melts. Th e transition from solid (order) to liquid (disorder) is. therefore, abrupt o r discontinuous. E v e ry crystalline solid has a definite melting point.Q ta ily and cryWaftrw k M h MXI A m o rp h o u s o r G la s s y Solids*nd k x v '* nO* order Th e word am orphous m eans without form or structure. Th u s Fo r Your Information in am orphous solids there is no regular arrangem ent of m olecules like that in crystalline solids. W e can. therefore, say that am orphous solids are m ore like liquids with the disordered structure frozen in.'♦ • » + + # f +. F o r exam ple ordinary glass, w hich is a solid at ordinary temperaturo. has no regular arrangement of molecules. O n♦ ♦ ♦ »♦ ♦ ♦ ♦ + + < heating, it gradually softens into a paste like state before it be com es a very viscous liquid at alm ost 800°C. T h u sTran*rr»**ion B * O o m U c r o g n p n <rf am orphous solids are also called glassy solids. This type ofm * atomic totte* of a gold crystal solids have no definite melting point. P o lym e ric S olid sFor Your Information Polym ers m ay be said to be m ore or less solid materials with a structure that is intermediate between order and disorder. T h e y can be classified as partially or poorly crystalline solids. Polymers form a large group of naturally occurring and synthetic materials. Plastics and synthetic rubbers are termed •Ploymers' because they are formed by polymerization reactions in which relatively simple molecules are chemically com bined into m assive long chain molecules, or th re e dimensional\" structures. Th e s e materials have rather low specific gravity com pared with even the lightest of metals, and

yet exhibit good strength-to-weight ratio. For Your InformationPolymers consist wholly or in part of chemical combinations o f ,carbon with oxygen, hydrogen, nitrogen and other metallic o r 'non-metallic elements. Polythene, polystreno and nylon etc..are examples of polymers. Natural rubber is composed in thepure state entirely of a hydrocarbon with the formula (C ,H ,L .Crystal LatticeA crystalline solid consists of three dimensional pattern thatrepeats itself over and over again. Th is smallest threedimensional basic structure is called unit cell. Th e wholestructure obtained by the repetition of unit cell is known ascrystal lattice. F o r example, the pattern of NaCI particleshave a cube shape. Th e cube shape of the sodium chloride isjust one of several crystal shapos. In a cubic crystal all thesides meet at right angles. Other crystal shapes have comersin which one or more of the angles are not right angles.17.2 M ECH ANICAL PRO P ER TIES OF SOLIDSE eform ation in Solids O If w e hold a soft rubber ball in our hand and then squeeze it. F the shape or volume of the bail will change. However, H we stop squeezing the ball, and open our hand, the ball will \ return to its original spherical shape. Th is has been illustrated schematically in Fig. 17.1. •:m v Similarly, if we hold two ends of a rubber string in our hands, O and move our hands apart to some extent, the length of the string will increase under iho action of the applied force Ongmal rubber bafl exerted by our hands. Greater the applied force larger will be Sbueejed rubber b o i subjected the increase in length. Now on removing the applied force, force F by the hand the string witl return to its original length. From these Rubber be* after removng fore* examples, it is concluded that deformation (i.e.. change in shape, length or volum e) is produced when a body is subjected to some external force. In crystalline solids atoms are usually arranged in a certain order. These atoms are held about their equilibrium position,which depends on the strength of the inter-atomic cohesive force between them. W hen external force is applied on such a body, a distortion results because of the displacement of the atoms from their equilibnum position and the body is said to be in a state of stress After the removal of external force. 137

Urtf c*« i a N> outwtrd the atoms return to their equilibrium position, and the body*tr*Ww>g tore* regains its original shape, provided that external applied force was not too great. The ability of the body to return to its original shape is called elasticity. Fig.17.2 illustrates deformation produced in a unit cell of a crystal subjected to an external applied force. Stress and Strain The results of mechanical tests are usually expressed in terms of stress and strain, which are defined in terms of applied force and deformation. Stress It is defined as the force applied on unit area to produce any change in the shape, volume or length of a body. Mathematically itis expressed as (' 7” The SI unit of stress (o ) is newton per square metre (N m 7). which is given the name pascal (Pa). Stress may cause a change in length, volume and shape. When a stress changes length, it is called the tensile stress, when itchanges the volume itis called the volume stress and when itchanges the shape itis called the shear stress.Un4 o*< under irMard Strain is a measure of the deformation of a solid when stress is applied to it. In the case of deformation in one dimension,■pphedtorte strain is defined as the fractional change in length. If Af is theUrwtc»l removing change in length and I is the original length (Fig. 17.3 a).then strain is given byeppted lore* S ,^ C h a n g e i n ^ j ^ .................... Rb- 172 Original length (/) Since strain is ratio of lengths, it is dimensionless. and therefore, has no units. If strain c is due to tensile stress o, it is called tensile strain, and if it is produced as a result of compressive stress o. it is termed as compressive strain. In case when the applied stress changes the volume, the change in volume per unit volume is known as volumetric strain (Fig. 17.3 b). Thus 138

Volumetric strain: AV F__ V,N o w referring to Fig. 17.3 (c ). w h e n the opposite faces of arigid c u b e are subjected to shear stress , the shear strainproduced is given by Aa tano (1 7 .3 )r »■H ow ever, for small values of angle 0. m easured in radian,tan 0 * 0 . so thatr- 9 (1 7 .4 )Elastic C onsta ntsExperim ents have revoalod that the ratio of stress to strain isa constant for a given material, provided the external appliedforce is not too great. T h is ratio is called m odulus of elasticity,and can be mathematically described asM odulus of Elasticity* ^StreaSin^ - (1 7 .5 )S ince strain is a dim ensw nloss quantity, the units o f m odulus Ft0.i7aof elasticity are the sam e a s those of stress, i.e.. N m : or Pa. j W r o puled along <ts tengti byIn the case of linear deformation, the ratio of tensile (or stretching (ortn Fcom pressive) stress o (= F I A ) to tensile (or com pressive) CySnder sUyected to compresrsne •crcoFstrain e = A * U is called Y o un g ’s m odulus Cube subjected to sheenng lorts FrA (17.6) mFor three dimensional deformation, when volum e is involved, Thor* a re seven, drflerent crystal*then the ratio of applied stress to volumetric strain is called system s b ated on the geom etricalBulk modulus. arrangem ent 0< the» atom * and the mil HftotQicmttncil Cotec «!* «■ ■ T re e n * (1 7 .7 )w he re A V is the ch a n ge in original volum e V.H ow ever, w h e n the sh ear stress t = ( F I A ) a n d sh ear strainy(= tan 0 ) are involved, then their ratio is called shear modulus.

Elastic constants for som e of the materials are given in Table 17.1.Table17.1 Elastic constants E la stic Lim it a n d Yield S tre ngth for som o materials In a tensile test, metal wire is extended at a specifiedM*t«n«i l ^ I f deformation rate, and stresses generated in the wire during doformation are continuously m easured by a suitable 10‘Nm* 10’Nm* electronic device fitted in the m echanical tosting machine. 30 Force-etongation diagram or stress-strain curve is plotted| Menrxum 70 70 60 automatically on X -Y chart recorder. A typical stress-strain curve for a ductile material is show n in Fig. 17.4.le o~ 15 - 36 • In the initial stage of deformation, stress rs increased linearlyBr»*s 91 61 44 with the strain till w e reach point A on the stress-strain curve Th is is called proportional hmit (o„) It is defined as theConcrete 25 • 450 greatest stress that a material can endure without losingCopper 110 140 23 straight line proportionality between stress and strain 3 Hooke's law which states that the strain is directlyDiamond 1120 540 56 proportional to stress is obeyed in the region O A From A to 0 B. stress and strain are not proportional, but nevertheless, ifGUM 55 31 64 the load is rem oved at any point between O and B. the curve 150 will be retraced and the material will return to its original| lea 14 6 length In the region O B . the matenal «s said to be elastic Th e 0 point B * caned the yield point T h e value of stress at B is[ lend 15 7.7 known as elastic limit o . If the stress is increased beyond the yield stress or elastic limit of the matenal. the specimen| Mercury 0 27 becom es permanently changed and does not recover its original shape or dvnension after the stress is rem oved Thisf Sioel 200 160 kind of behaviour is called plasticity 390 200[ Tungsteni ia«-■- - 0 22typical duello meter*! T h e region of plasticity is represented by the portion of the curve from B to C . the point C in Fig 17 4 represents the ultxnate tensile strength ( U T S ) o „ of the matenal T h e U T S is defined as the maximum stress that a matenal can withstand and can be regarded as the n o m n a i strength of the material O n c e point C corresponding to U T S is crossed the matenal breaks at point D. responding the fracture stress ( a ) Substances w hich undergo plastic deformation until they break, are known as ductile substances. Lead, copper and w rought iron are ductile. O th er substances which break just after the elastic limit is reached, are know n as brittle substances. G la s s and high carbon steel are brittle. E x a m p l e 1 7 .1 : A steel w ire 12 m m in diameter is fastened 140

to a log and is then pulled by tractor. Th e length of steel wirebetween the log and the tractor is 11 m. A force of 10.000 N isrequired to pull the log. Calculate (a ) the stress in the wire and(b ) the strain in the wire, (c ) How much does the wire stretchwhen the log is pulled? ( E = 200 x 10’ Nm *)S o lu tio n : F 10.000 N(a) A s tensile stress ^ \ 4 x ^ 04 m f = 88.46 x 10*Nm '= 88.46 MPa(b ) A/Also T h e tensile strain e 1 I(c) Stress 88.46 x 104 N m 'J 200x10* Nm Strain Strain Strain 88.46 x106 Nm * 4.4 x 1 0 4 200x10® Nm 2 Af Now using the relation Strain = — . w e get M = 4 .4 x 1 0 ‘*x11 m = 4 .8 4 x 1 0 ,m = 4.84m mStrain E n e rg y in D eform ed M aterialsConsider a wire suspended vertically from one end. It is T h » a ■ macfrna u»*0 to rrvaatigatestretched by attaching a weight at the other end. W e can vanat wtfi ft* tore*tta w a n g itincrease the stretching force by increasing the weight. Bynoting the extension I of the wire for different values of thestretching force F . a graph can be drawn between the force Fand the extension I (Fig.17.5). If the elastic limit is notexceeded, the extension is directly proportional to force F . A sthe force F stretches the wire, it does some amount of workon wire which is equal to product of force F and the extensionI Suppose w e are required to find the amount of the workdone when the extension is I,. Let the force for this extensionbe F,. Fig.17.5 shows that the force F does not remainconstant in producing the extonsion l „ it varies uniformlyfrom 0 to F,. In such a situation the work is calculated bygraphical method.Suppose at some stage before the extension I, is reached, 17.Sthe force in the wire is F and that the wire now extends by avery small amount Ax. Th e extension Ax is so small that theforce F m ay be assum ed constant in ax. so the work done inproducing this small extension is F x Ax. In the figure it can be 141

seen that it is represented by the area of the shaded strip. Inthis w a y the total extension I , can be divided into very smallextensions and the work done during each of these smallextensions would be gtven by the area of the strips (Fig. 17.5).S o the total work d o ne in producing the extension I , is the sumarea of all these strips which is equal to area between thegraph and the axis on which extension has been plotted uptoI - 1 , . In this case it equals to area of the triangle O A B . Work done ■ Area of A O AB 1 =2 OAxAB ■ jA x F , .............. (1 7 .9 )T h is is the am ount of en ergy stored in the wire. It is the gain inthe potential en ergy of the m olecules due to theirdisplacement from their m ean positions. E q.17.9 gives theenergy in joules w hen F is in newton and I in metres.Eq.17.9 can also be expressed in terms of m odulus ofelasticity E . If A is the area of cross-section of the wire a n d Lits total length then cL A t, E A x iyor F ,» L ■Substituting the va lu e of F , in E g . 17.9 w e haveWork do ne: 1 E A x t ? (1 7 .1 0 ) 2 l <■T h e area m ethod is quite a general on o. F o r exam ple if theextension is increased from I , to*,, the am ount of w ork doneby the stretching force w ould be given by the area o f thetrapozium A B C D (F ig . 17.5). It is also valid for both the linear(elastic) and the non-linear (non-elastic) parts of the force-extension graph. If the extension occurs from O to G(Fig. 17.5). this work done would be the area of O H G17.3 ELECTR ICAL PROPERTIES OF SOLIDST h e fundamental electrical property of a solid is its responseto an applied electric field, i.e.. its ability to conduct electriccurrent. Th e electrical behaviours of various materials are142

diverse. S o m e are very good conductors, e.g.. metals with Do You Know? GteM » Jtoo kncrnn M totdconductivities of the order o f 10 ’ ( O m ) ’.A t the other extreme, bo cauM o m o l*cU »» \"• «ragU «rtysom e solids, e .g.. w ood, diam ond etc.. have very low •rrangad m n • Iq u 4 bul fae<J inconductivities ranging between 10'** and 1 0 ” (O m ) ’. these t w n M M poabontare called insulators. Solids with intermediate conductivities,generally from 104to 104(O m ) \ are termed semiconductors,e.g.. silicon, germ anium etc. T h e conventional free electrontheory based o n Bohr model of electron distnbution in anatom failed to explain com pletely the vast diversity in theelectrical behaviour of these three types of materials.O n the other hand, energy band theory based o n wavem echanical model has been found successful in resolvingthe problem.Energy Band TheoryElectrons of a n isolated atom are bound to the nucleus, andcan only have distinct energy levels. However, when a largenum ber of atom s, say N . are brought close to one another toform a solid, each en ergy level of the isolated atom splits intoN sub-levels, called states, under the action of the forcesexerted by other atom s in the solid. Th e s e permissibleenergy statos are discrete but so closely spaced that theyappear to form a continuous e n e rg y band. In between twoconsecutive permissible energy bands, there is a range ofenergy states which cannot be occupied by electrons. Theseare called forbidden en ergy states, and its range is term ed asforbidden energy gap.T h e electrons in the outermost shell of an atom are calledvalence electrons and the energy band occupying theseelectrons is know n as valence band. It is obviously thehighest occupied band. It m a y be either com pletely filled orpartially filled with electrons and can never be em pty. T h eband a b o ve the valonco band is called conduction ba nd. Inconduction band, electrons m ove freely and conduct electriccurrent through solids. Th a t is. w h y the electrons occupyingthis band are known a s conductive electrons or freeelectrons. A n y electron leaving the valence band isaccom m odated by this band. It m a y be either em pty orpartially filled with electrons. T h e bands below the valenceband are norm ally com pletely filled and as such play no partin the conduction procoss. Th u s , while discussing theelectrical conductivity w e will consider only the valence andconductionbands.143

_I_n__s_u__la__t_o_r__s___ Insulators are those materials in which valence electrons are bound very tightly to their atom s and are not free. In term s of en ergy bands, it m eans that an insulator, as show n in Fig. 17.6 has an empty conduction band (no free electrons) a full valence band a large energy gap (several e V ) between themfCooSuctxxi C o n d u c t o r s Conductors are those w hich have plenty of a»n.j free electrons for electrical conduction. In term s of energy bands, conductors are those materials in w hich valence and VMnoe conduction bands largely overlap each other (Fig. 17.7). Band T h e re is no physical distinction betw een the tw o bands w hich ensures the availability of a large num ber of free electrons. S e m i c o n d u c t o r s In te rm s of e n o rg y b a n d s , sem iconductors are those materials which at room temperature have 0 ) partially filled conduction band (ii) partially filled valence band (iii) a very narrow forbidden energy gap (of the order of 1 eV ) between the conduction and valence bands (Fig. 17.8). A t 0 K . there are no electrons in the conduction band and their valence band are com pletely filled. It m ea ns at 0 K . a piece of G e or Si is a p o rte d insulator. H ow ever, withCondocton band increase in temperature, som e electrons posses sufficient en ergy to ju m p across the small en ergy g a p from valen ce to conduction band. T h is transfers so m e free electrons in the*w o w tosiddan» » « 9K> c o n d u d io n bands and creates so m e vacan cie s of e lo d ro n s in the valence band. Th e vacancy of ele d ro n in the valenceVWcnoeBand band is know n as a hole. It be haves like a positive charge. Th u s at room temperature. G e or Si crystal becomes a semiconductor. Intrinsic an d E x trin sic S e m i-c o n d u c to r A sem i-conductor in its extrem ely pure form is know n as intrinsic sem i-conductor. Th e electrical behaviour of se m ic o n d u d o r is extremely sensitive to the purity of the material. It is substantially changed on introducing a small am ount of impurity into the pure s e m i-c o n d u d o r lattice. T h e process is called doping, in w hich a small num ber of atom s of som e other suitable elem ents are a d ded as impurity in the ratio of 1 to 10*. T h e doped s e m i-c o n d u d in g m atenals are called extrinsic sem i-condudors. Pure elem ent of silicon and germ anium are intrinsic 144

semi-conductors. These semi-conductor elements have <•>atom s with four valence electrons. In solid crystalline form,the atoms of these elements arrange themselves in such a (*»pattern that each atom has four equidistant neighbours Ftg. 17.10F»g. 17.9 sh ow s this pattern along with its valence electrons.E a c h atom with its four valence electrons, shares an electron • • • • ••from its neighbours. T h is effectively allocatos eight electrons • • • • ••in the outerm ost shell of e a c h atom w hich is a stable state.Th is sharing of electrons between two atoms creates •• • • ••covalent bonds. D ue to these covalent bonds electrons are •• • • ••b o und in their respective shells. 17.11W h en a silicon crystal is doped with a pcntavatent element,e.g.. arsenic, antimony or phosphorous etc.. four vale^coelectrons of the impurity atom form covalent bond with thefour ne»ghbouring Si atom s, while the fifth valence electronprovides a free electron in the crystal. S uch a doped orextrinsic sem i-conductor is called n-type semi-conductor.Fig. 1 7 .1 0 (a ) illustrates silicon crystal lattice doped with apentavalent im purity such as phosphorous. Th ephosphorous atom is called a donor atom because it readilydonates a free electron, w hich is therm ally excited into theconduction band.O n the other hand, w hen a silicon crystal is d o p -.d with atrivalent element, e.g.. aluminium, boron, gallium or indiumetc.. threo valence electrons of the impurity atom formcovalent bond with the three neighbouring S i atoms, whilethe o n e missing electron in the covalent bond with the fourthneighbouring Si atom , is called a hole w hich in fact is vacancyw here an electron can be accom modated. Such a semi­conductor is called p-type semi-conductor. Fig. 17.10 (b)illustrates silicon crystal lattice doped with alum inium . T h ealum inium atom is called a n acceptor atom b e cause it is easyfor the aluminium ion core to accept a valence electron from anearby silicon atom , thus creating a hole in the valenco bond.E le c tric a l C o n d u c tio n b y E le c tro n s a n d H o le s inS em icon ducto rsC onsider a sem i-conductor crystal lattice, o .g .. G e o r S i asshown in Fig. 17.11. T h o circles represent the positiveioncores of Si or G e atoms, and the blue dots are valenceelectrons. These electrons are bound by covalent bond.H ow ever, at room tem perature they have thermal kineticmotion w hich, in case of so m e electrons, is so vigorous that 145

A 0 .,t A©. A©. A©. the covalent bond is unable to keep them bound. In suchB ® B@o B©. B®. cases the electrons break the covalent bond and get them selves free leaving a vacant seat for an electron, i.e.. acO - cO ! C0 ©# Jf c® . hole. T h u s w he never a covalent bond is broken, an electron-0 ©<•>. 0 ®« » . (C) hole pair is created. Both the electrons and the holes m ove in 0 ®W )o the sem i-conductor crystal lattice a s explained below. © Electron C onsider a ro w of Si atom s in crystal lattice. S upp o se a hole is present in the valence shell of atom A . A s hole is a k * 0 -+ deficiency of electron, so the core of atom A would have a not “H k positive charge (Fig. 17.12 a). Th is attracts an electron from a neighbouring atom say B . T h u s the electron m o ve s from B to A and the hole (* v e ch a rg e ) shifts to B (Fig. 17.12 a .b ). N ow an electron is attracted from C to B and a hole is created at C (Fig. 17.12 b.c) and positive charge appears at C . This process is repeated between the a tom s C a n d D with the result that the electron m oves from D to C and the hole (♦ve charge) appears at D (Fig. 17.12 c . d ) . Th u s w e notice that if a hole is present in a n y vale nce shell, it cannot stay there but it m o ve s from o n e atom to other with the electron m oving in opposite direction. S eco n dly w e notice that the appearance of hole is accom panied by a positive chargo. Th u s a moving hole is equivalent to a moving positive chargo.t' In this exam ple w e have considered a special case in which the electron and the hole are m oving in a straight line. . Actually their motion is random b e cause positively charged * core of the atom can attract a n electron from a n y of its neighbouring atoms. Th u s , in sem i-conductors there are two kinds of charge carriers; a free electron (- e ) and a hole (+ e). W h e n a battery is connected to a semi-conductor, it establishes an electric field across it due to w hich a directed flow of electrons and holes takes place. T h e electrons drift towards the positive end, w hereas the holes dnft towards the negative end of the semi-conductor (Fig. 17.13) T h e current / flowing through the sem i-conductor is earned by both electrons and holes. It m ay be noted that the electronic current and the hole current add up together to give the current /. 17.4 S U P E R C O N D U C T O R S There are som e materials w hose resistivity becom es zero below a certain temperature Tf called critical temperature as shown in rosistivity-temperaturo graph in Fig. 17.14. Below this temperature, such materials are called superconductors. 14 6

Th e y offer no resistance to electric current and are. therefore, Oo You Know?perfect conductors. O nce the resistance of a material dropsto zero, no energy is dissipated and the current, once Sop or conductor* a rc a to y s r m atestablished, continues to exist indefinitely without the source cortam tam peraturo*. conductofanemf. H o c a -o ry m T i n o r**i*tanc*Th e first superconductor was discovered in 1911 by For Your InformationKmaerimgh O m e s when it was observed that electncalresistance of m ercury disappears suddenly as the Magnate Rosonanca «magng (M RI)temperature is reduced below 4.2 K. Som e other metals such u*o* strong m agnate f**j producadas aluminium ( Tt = 1.18 K). tin ( T , - 3.72 K ). and lead ( Tt=7.2 K ) b y super c o n d u c in g m a to n ah loralso become superconductors at very low temperatures. In scanning com putor p ro co svn g1986 a new class of ceramic materials was discovered that pro duces the im age identifyingbecomes superconductor at temperatures as high as 125 K. tumor* and W tam od bA ny superconductor with a critical temperature above 77 K.the boiling point of liquid nitrogen, is referred as a hightemperature superconductor.Recently a complex crystalline structure known asYttrium barium copper oxide (YB a,Cu,. O ,) have beenreported to become superconductor at 163 K or -110 *C byProf. Yao Lian's Lee at Cambridge University. Perhaps oneday even room temperature superconductor will bodeveloped and that day will be a new revolution in electricaltechnologies. Superconductors have many technologicalapplications such as in magnetic resonance imaging (M R I).magnetic levitation trains, powerful but small electric motorsand faster computer chips.1 7 .5 M A G N E T IC P R O P E R T IE S O F S O L ID SFrom the study of the magnetic fields produced by bar Long Bar magnatmagnets and moving charges, i.e., currents, it is possible totrace the origin of the magnetic properties of the material. It isobserved that the field of a long bar magnet is like the fieldproduced by a long solenoid carrying current and the field ofa short bar magnet resembles that of a single loop(Fig. 17.15). Th is similarity between the fields produced bymagnets and currents urges an enquiring mind to think thatall magnetic effects m ay be due to circulating currents (i.e..moving charges); a view first held by Ampere. Th e idea wasnot considered very favourably inAmpere's time because theStructure of atom was not known at that time. Taking intoconsideration, the internal structure of atom, discoveredthereafter, the Am pere's view appears to be basically correct.Th e magnetism produced by electrons within art atom can147

M jg n c t field of ■ c u rre n t lo o p arise from two motions. First, each electron orbiting the nucleus behaves like an atom ic sized loop of current that FIB 17.15 generates a small m agnetic field; this situation is similar to the For Y our Information field created by the current loop in Fig. 17.15 (d ). Secondly each electron possesses a spin that also gives rise to a Atojdmuve magnetic field. T h e net magnetic field created by the electrons withm an atom is due to the com bined field createdS q u id s ( o r su p o r-co n d o ctio g by their orbital and spin motions. S ince there are a num ber ofquartern rto d o re rv o d e vtc o i) *f# electrons in an atom, their currents or spins m ay be sousod to detect v e ry t m k m a gn otc oriented o r aligned a s to cancel the m agnetic effects mutuallyM ed such a* proOucod b y tho br*r> o r strengthen the effects of each other. A n atom in which there is a resultant m agnetic field, behaves like a tiny magnet and is called a m agnetic dipoie. T h e m agnetic fields of the atoms are responsible for. the magnetic behaviour of the substance m ade u p of these atom s. Magnetism is, therefore, due to the spin and orbital motion of the electrons surrounding the nucleus and is thus a property of all substances. It m ay be mentioned that the charged nucleus itself spins giving rise to a m agnetic field. H ow ever, it is m uch w eaker than that of the orbital electrons. T h u s the source of m agnetism of an atom is the electrons. Accepting this view of magnetism it is concluded that it is impossible to obtain an isolated north pole. T h e north-pole is m erely one side of a current loop. T h e othor side will afw ays be present as a south pole and these cannot be separated. T h is is an experimental reality. T w o cases arise w hich have to be distinguished. In the first case, the orbits and the spin axes of the electrons in an atom are so oriented that their fields support each other and the atom be haves like a tiny m agnet. Substances with such atoms are called paramagnetic substances. In second type of atom s there is n o resultant field as the m agnetic fields produced by both orbital and spin motions of the electrons might add upto zero. Th e se are called diamagnetic substances, for exam ple the atom s of water, copper, bismuth and antimony. However, there are som e solid substances e.g.. Fe. Co. Ni. C h rom ium dioxide, and Alnico (a n iron aluminium - nickel - cobalt a lloy) in w hich the atom s co-operate with each other in such a w a y so as to exhibit a strong m agnetic effect. T h e y are called, ferromagnetic substances. Ferromagnetic materials are of great interest for electrical engineers Recent studies of ferromagnetism h a v e show n that there exists in ferromagnetic substance small regions called 'domains'. T h e dom ains are of macroscopic size of the order 148

of millimetres or less but large enough to contain 10’' to 10’' f ig 17.ie Magnetic doman* watoms. Within each domain the magnetic fields of all the an unmagnetirad lorrcc-jgnetspinning electrons are parallel to one another i.e.. eachdomain is magnetized to saturation. Each domain behavos asa small magnet with its own north and south poles. Inunmagnetised iron the domains are oriented in a disorderlyfashion (Fig.17.16). so that the net magnetic effect of asizeable specimen is zero. W hen the specimen is placed inan external magnetic field as that of a solenoid. the domainsline up parallel to lines of external magnetic field and theentire specimen becomes saturated (Fig. 17.17). Th ecombination of a solenoid and a specimen of iron inside it thusmakes a powerful magnet and is called an electromagnet.Iron is a soft magnetic material. Its domains are easily rig 17.17oriented on applying an external field and also readily roturnto random positions when the field is removed. This isdesirable in an electromagnet and also in transformers.Domains in steel, on the other hand, are not so easilyoriented to order. T h e y require very strong external fields, butonce oriented, retain the alignment. Th u s steel makes a goodpermanent magnet and is known as hard magnetic materialand another such material is a special alloyAlnico V.Finally, it must be mentioned that thermal vibrations tend to Do You Know?disturb the orderliness of the domains. Ferromagneticmaterials preserve the orderliness at ordinary temperaturos. LUgnotc mad* owl c<organ*: tromnt*When heated, they begin to lose the# orderliness due to the c © » bo m od n otftoo andincreased thermal motion. Th is procoss begins to occur at a components. >n computers. mobJeparticular temperature (different for different materials) called phono*. T V » motor* generator* andCune temperature. Above the Curie temperature iron isparamagnetic but not ferromagnetic. Th e Curie temperature d .iu storage donees O curt* can make use ol ceramic magnet* tnat do notfor iron is about 750 °C. conducts <M*f*ie>iyH ysteresis Loop 17.I tTo investigate a ferromagnetic material, a bar of that materialsuch as iron is placed in an alternating current solenoid. Whenthe alternating current is at its positive peak value, it fullymagnetises the specimen in one direction and when the currentis at its negative peak, itfully magnetises itin opposite direction.Thus as the alternating current changes from its posrtivo peakvalue to its negative peak value and then back to its positivepeak value, the specimen undergoes a complete cycle ofmagnetization Th e flux density versus the magnetizationcurrent of the specimen for the various values of magnetizingcurrent of the solenoid is plotted by a C R O (Fig.17.18).149