A.C. FUNDA- MENTALS. Learning Objectives

Transcription

1 C H A P T E R Learning Objectives Generation of Alternating Voltages and Currents Alternate Method for the Equations of Alternating Voltages and currents Siple Wavefors Cycle Different Fors of E.M.F. Equation Phase Phase Difference Root Mean Square (R.M.S.) Value Mid-ordinate Method Analytical Method R.M.S. Value of a Coplex Wave Average Value For Factor Crest or Peak Factor R.M.S. Value of H.W. Rectified A.C. Average Value For Factor of H.W. Rectified Vector Diagras Using R.M.S. Values Vector Diagras of Sine Waves of Sae Frequency Addition and Subtraction of Vectors A.C. Through Resistance, Inductance and Capacitance A.C. through Pure Ohic Resistance alone A.C. through Pure Inductance alone Coplex Voltage Applied to Pure Inductance A.C. through Capacitance alone A.C. FUNDA- MENTALS Alternating current circuits iproves the versatility and usefulness of electrical power syste. Alternating current plays a vital role in today s energy generation

2 454 Electrical Technology.. Generation of Alternating Voltages and Currents Alternating voltage ay be generated by rotating a coil in a agnetic field, as shown in Fig.. (a) or by rotating a agnetic field within a stationary coil, as shown in Fig.. (b). Fig.. The value of the voltage generated depends, in each case, upon the nuber of turns in the coil, strength of the field and the speed at which the coil or agnetic field rotates. Alternating voltage ay be generated in either of the two ways shown above, but rotating-field ethod is the one which is ostly used in practice... Equations of the Alternating Voltages and Currents Consider a rectangular coil, having N turns and rotating in a unifor agnetic field, with an angular velocity of ω radian/second, as shown in Fig... Let tie be easured fro the X-axis. Maxiu flux Φ is linked with the coil, when its plane coincides with the X-axis. In tie t seconds, this coil rotates through an angle θ ω t. In this deflected position, the coponent of the flux which is perpendicular to the plane of the coil, is Φ Φ cos ω t. Hence, flux linkages of the coil at any tie are N Φ N Φ cos ω t. According to Faraday s Laws of Electroagnetic Induction, the e..f. induced in the coil is given by the rate of change of flux-linkages of the coil. Hence, the value of the induced e..f. at this instant (i.e. when θ ω t) or the instantaneous value of the induced e..f. Fig.. is e d ( N ) volt N. d ( cos t) dt Φ dt Φ ω volt NΦ ω ( sin ωt) volt ω N Φ sin ωt volt ω N Φ sin θ volt...(i) When the coil has turned through 9º i.e. when θ 9º, then sin θ, hence e has axiu value, say E. Therefore, fro Eq. (i) we get E ω N Φ ω N B A f N B A volt...(ii) where B axiu flux density in Wb/ ; A area of the coil in f frequency of rotation of the coil in rev/second Substituting this value of E in Eq. (i), we get e E sin θ E sin ω t...(iii)

3 A.C. Fundaentals 455 Siilarly, the equation of induced alternating current is i I sin ω t...(iv) provided the coil circuit has been closed through a resistive load. Since ω f, where f is the frequency of rotation of the coil, the above equations of the voltage and current can be written as e E sin f t E sin t and i I T sin ft I sin t T where T tie-period of the alternating voltage or current /f It is seen that the induced e..f. varies as sine function of the tie angle ω t and when e..f. is plotted against tie, a curve siilar to the one shown in Fig..3 is obtained. This curve is known as sine curve and the e..f. which varies in this anner is known as sinusoidal e..f. Such a sine curve can be conveniently drawn, as shown in Fig..4. A vector, equal in length to E is drawn. It rotates in the counter-clockwise direction with a velocity of ω radian/second, aking one revolution while the generated e..f. akes two loops or one cycle. The projection of this vector on Y-axis gives the instantaneous value e of the induced e..f. i.e. E sin ω t. Fig..3 Fig..4 To construct the curve, lay off along X-axis equal angular distance oa, ab, bc, cd etc. corresponding to suitable angular displaceent of the rotating vector. Now, erect coordinates at the points a, b, c and d etc. (Fig..4) and then project the free ends of the vector E at the corresponding positions a, b, c, etc to eet these ordinates. Next draw a curve passing through these intersecting points. The curve so obtained is the graphic representation of equation (iii) above..3. Alternate Method for the Equations of Alternating Voltages and Currents In Fig..5 is shown a rectangular coil AC having N turns and rotating in a agnetic field of flux density B Wb./. Let the length of each of its sides A and C be l eters and their peripheral velocity v etre/second. Let angle be easured fro the horizontal position i.e. fro the X-axis. When in horizontal position, the two sides A and C ove parallel to the lines of the agnetic flux. Hence, no flux is cut and so no e..f. is generated in the coil. When the coil has turned through angle θ, its velocity can be resolved into two utually perpendicular coponents (i) v cos θ coponent-parallel to the direction of the agnetic flux and (ii) v sin θ coponent-perpendicular to the direction Fig..5 of the agnetic flux. The e..f. is generated due entirely to the perpendicular coponent i.e. v sin θ. Hence, the e..f. generated in one side of the coil which contains N conductors, as seen fro Art. 7.7, is given by, e N Bl v sin θ.

4 456 Electrical Technology Total e..f. generated in both sides of the coil is e BNl v sin θ volt...(i) Now, e has axiu value of E (say) when θ 9º. Hence, fro Eq. (i) above, we get, E B N l v volt. Therefore Eq. (i) can be rewritten as e E sin θ...as before If b width of the coil in eters ; f frequency of rotation of coil in Hz, then v bf E B N l b f f N B A volts...as before Exaple.. A square coil of c side and turns is rotated at a unifor speed of revolutions per inute, about an axis at right angles to a unifor agnetic field of.5 Wb/. Calculate the instantaneous value of the induced electrootive force, when the plane of the coil is (i) at right angles to the field (ii) in the plane of the field. (Electroagnetic Theory, A.M.I.E. Sec B, 99) Solution. Let the agnetic field lie in the vertical plane and the coil in the horizontal plane. Also, let the angle θ be easured fro X-axis. Maxiu value of the induced e..f., E f N B A volt. square coils Instantaneous value of the induced e..f. e E sin θ Now f /6 (5/3) rps, N, B.5 Wb/, A (i) In this case, θ º e (ii) Here θ 9º, e E sin 9º E Substituting the given values, we get e (5/3) V.4. Siple Wavefors The shape of the curve obtained by plotting the instantaneous values of voltage or current as the ordinate against tie as a abscissa is called its wavefor or wave-shape. Fig..6 An alternating voltage or current ay not always take the for of a systeatical or sooth wave such as that shown in Fig..3. Thus, Fig..6 also represents alternating waves. But while it is scarcely possible for the anufacturers to produce sine-wave generators or alternators, yet sine wave is the ideal for sought by the designers and is the accepted standard. The waves deviating fro the standard sine wave are tered as distorted waves. In general, however, an alternating current or voltage is one the circuit direction of which reverses at regularly recurring intervals.

5 A.C. Fundaentals Coplex Wavefors Coplex waves are those which depart fro the ideal sinusoidal for of Fig..4. All alternating coplex waves, which are periodic and have equal positive and negative half cycles can be shown to be ade up of a nuber of pure sine waves, having different frequencies but all these frequencies are integral ultiples of that of the lowest alternating wave, called the fundaental (or first haronic). These waves of higher frequencies are called haronics. If the fundaental frequency is 5 Hz, then the frequency of the second haronic is Hz and of the third is 5 Hz and so on. The coplex wave ay be coposed of the fundaental wave (or first haronic) and any nuber of other haronics. In Fig..7 is shown a coplex wave which is ade up of a fundaental sine wave of frequency of 5 Hz and third haronic of frequency of 5 Hz. It is seen that Fig..7 Fig..8 (i) the two halves of the coplex wave are identical in shape. In other words, there is no distortion. This is always the case when only odd haronic (3rd, 5th, 7th, 9th etc.) are present. (ii) frequency of the coplex wave is 5 Hz i.e. the sae as that of the fundaental sine wave. In Fig..8 is shown a coplex wave which is a cobination of fundaental sine wave of frequency 5 Hz and nd haronic of frequency Hz and 3rd haronic of frequency 5 Hz. It is seen that although the frequency of the coplex wave even now reains 5 Hz, yet : (i) the two halves of the coplex wave are not identical. It is always so when even haronics (nd, 4th, 6th etc.) are present. (ii) there is distortion and greater departure of the wave shape fro the purely sinusoidal shape. Soeties, a cobination of an alternating and direct current flows siultaneously through a circuit In Fig..9 is shown a coplex wave (containing fundaental and third haronic) cobined Fig..9 with a direct current of value I D. It is seen that the resultant wave reains undistorted in shape but is raised above the axis by an aount I D. It is worth noting that with reference to the original axis, the two halves of the cobined wave are not equal in area..6. Cycle One coplete set of positive and negative values of alternating quantity is known as cycle. Hence, each diagra of Fig..6 represents one coplete cycle.

6 458 Electrical Technology A cycle ay also be soeties specified in ters of angular easure. In that case, one coplete cycle is said to spread over 36º or radians..7. Tie Period The tie taken by an alternating quantity to coplete one cycle is called its tie period T. For exaple, a 5-Hz alternating current has a tie period of /5 second..8. Frequency The nuber of cycles/second is called the frequency of the alternating quantity. Its unit is hertz (Hz). In the siple -pole alternator of Fig. 4. (b), one cycle of alternating current is generated in one revolution of the rotating field. However, if there were 4 poles, then two cycles would have been produced in each revolution. In fact, the frequency of the alternating voltage produced is a function of the speed and the nuber of poles of the generator. The relation connecting the above three quantities is given as f PN/ where N revolutions in r.p.. and P nuber of poles For exaple, an alternator having poles and running at 3 r.p.. will generate alternating voltage and current whose frequency is 3/ 5 hertz (Hz). It ay be noted that the frequency is given by the reciprocal of the tie period of the alternating quantity. f /T or T /f.9. Aplitude The axiu value, positive or negative, of an alternating quantity is known as its aplitude... Different Fors of E.M.F. Equation The standard for of an alternating voltage, as already given in Art.., is e E sin θ E sin ω t E sin f t E sin T t By closely looking at the above equations, we find that (i) the axiu value or peak value or aplitude of an alternating voltage is given by the coefficient of the sine of the tie angle. (ii) the frequency f is given by the coefficient of tie divided by.. For exaple, if the equation of an alternating voltage is given by e 5 sin 34t then its axiu value of 5 V and its frequency is f 34/ 5 Hz. Siilarly, if the equation is of the for e I ( R + 4 ω L ) sin ω t, then its axiu value is E I ( R + 4 ω L ) and the frequency is ω/ or ω/ Hz. Exaple.. The axiu values of the alternating voltage and current are 4 V and A respectively in a circuit connected to 5 Hz supply and these quantities are sinusoidal. The instantaneous values of the voltage and current are 83 V and A respectively at t both increasing positively. (i) Write down the expression for voltage and current at tie t. (ii) Deterine the power consued in the circuit. (Elect. Engg. Pune Univ.) Solution. (i) In general, the expression for an a.c. voltage is v V sin (ω t + φ) where φ is the phase difference with respect to the point where t.

7 A.C. Fundaentals 459 Now, v 83 V ; V 4 V. Substituting t in the above equation, we get 83 4 (sin ω + φ) sin φ 83/4.77 ; φ 45º or /4 radian. Hence, general expression for voltage is v 4 (sin 5 t + /4) 4 sin ( t + /4) Siilarly, at t, sin (ω + φ) sin φ.5 φ 3º or /6 radian Hence, the general expression for the current is i (sin t + 3º) sin ( t + /6) Fig.. (ii) P VI cos θ where V and I are rs values and θ is the phase difference between the voltage and current. Now, V V/ 4/ ; I / ; θ 45º 3º 5º (See Fig..) P (4/ ) (/ ) cos 5º 3864 W Exaple.3. An alternating current of frequency 6 Hz has a axiu value of A. Write down the equation for its instantaneous value. Reckoning tie fro the instant the current is zero and is becoing positive, find (a) the instantaneous value after /36 second and (b) the tie taken to reach 96 A for the first tie. Solution. The instantaneous current equation is i sin ft sin t Now when t /36 second, then (a) i sin ( /36)...angle in radians sin ( 8 /36)...angle in degree sin 6º 3.9 A (b) 96 sin 8 6 t...angle in degree or sin (36 6 t) 96/ t sin.8 53º (approx) t θ/f 53/ second... Phase By phase of an alternating current is eant the fraction of the tie period of that alternating current which has elapsed since the current last passed through the zero position of reference. For exaple, the phase of current at point A is T/4 second, where T is tie period or expressed in ters of angle, it is / radians (Fig..). Siilarly, the phase of the rotating coil at the instant shown in Fig.. is ω t which is, therefore, called its phase angle. Fig.. Fig..

8 46 Electrical Technology In electrical engineering, we are, however, ore concerned with relative phases or phase differences between different alternating quantities, rather than with their absolute phases. Consider two single-turn coils of different sizes [Fig.. (a)] arranged radially in the sae plane and rotating with the sae angular velocity in a coon agnetic field of unifor intensity. The e..fs. induced in both coils will be of the sae frequency and of sinusoidal shape, although the values of instantaneous e..fs. induced would be different. However, the two alternating e..fs. would reach their axiu and zero values at the sae tie as shown in Fig.. (b). Such alternating voltages (or currents) are said to be in phase with each other. The two voltages will have the equations e E sin ω t and e E sin ω t.. Phase Difference Now, consider three siilar single-turn coils displaced fro each other by angles α and β and rotating in a unifor agnetic field with the sae angular velocity [Fig..3 (a)]. Fig..3 In this case, the value of induced e..fs. in the three coils are the sae, but there is one iportant difference. The e..fs. in these coils do not reach their axiu or zero values siultaneously but one after another. The three sinusoidal waves are shown in Fig..3 (b). It is seen that curves B and C are displaced fro curve A and angles β and (α + β) respectively. Hence, it eans that phase difference between A and B is β and between B and C is α but between A and C is (α + β). The stateent, however, does not give indication as to which e..f. reaches its axiu value first. This deficiency is supplied by using the ters lag or lead. A leading alternating quantity is one which reaches its axiu (or zero) value earlier as copared to the other quantity. Siilarly, a lagging alternating quantity is one which reaches its axiu or zero value later than the other quantity. For exaple, in Fig..3 (b), B lags behind A by β and C lags behind A by (α + β) because they reach their axiu values later. The three equations for the instantaneous induced e..fs. are (Fig..4) e A E sin ω t...reference quantity Fig..4 e B E sin (ωt β) e C E sin [ωt (α + β)]

9 A.C. Fundaentals 46 In Fig..4, quantity B leads A by an angle φ. Hence, their equations are e A E sin ω t...reference quantity e B E sin (ωt φ) A plus (+) sign when used in connection with phase difference denotes lead whereas a inus ( ) sign denotes lag..3. Root-Mean-Square (R.M.S.) Value The r..s. value of an alternating current is given by that steady (d.c.) current which when flowing through a given circuit for a given tie produces the sae heat as produced by the alternating current when flowing through the sae circuit for the sae tie. It is also known as the effective or virtual value of the alternating current, the forer ter being used ore extensively. For coputing the r..s. value of syetrical sinusoidal alternating currents, either id-ordinate ethod or analytical ethod ay be used, although for syetrical but nonsinusoidal waves, the idordinate ethod would be found ore convenient. A siple experiental arrangeent for easuring the equivalent d.c. value of a Fig..5 sinusoidal current is shown in Fig..5. The two circuits have identical resistances but one is connected to battery and the other to a sinusoidal generator. Watteters are used to easure heat power in each circuit. The voltage applied to each circuit is so adjusted that heat power production in each circuit is the sae. In that case, the direct current will equal I/ which is called r..s. value of the sinusoidal current..4. Mid-ordinate Method In Fig..6 are shown the positive half cycles for both syetrical sinusoidal and non-sinusoidal alternating currents. Divide tie base t into n equal intervals of tie each of duration t/n seconds. Let the average values of instantaneous currents during these intervals be respectively i, i, i 3... i n (i.e. id-ordinates in Fig..6). Suppose that this alternating current is passed through a circuit of resistance R ohs. Then, Fig..6

10 46 Electrical Technology Heat produced in st interval.4 3 i Rt/n kcal (ä /J /4.4 3 ) Heat produced in nd interval.4 3 i Rt/n kcal : : : : : : : : : : : : : : : Heat produced in nth interval.4 3 i n Rt/n kcal Total heat produced in t seconds is.4 3 i + i i n Rt kcal n Now, suppose that a direct current of value I produces the sae heat through the sae resistance during the sae tie t. Heat produced by it is.4 3 I Rt kcal. By definition, the two aounts of heat produced should be equal..4 3 I Rt.4 3 i Rt + i i n n I i + i i n i + i i n I n n square root of the ean of the squares of the instantaneous currents Siilarly, the r..s. value of alternating voltage is given by the expression V v v v n n.5. Analytical Method The standard for of a sinusoidal alternating current is i I sin ωt I sin θ. The ean of the squares of the instantaneous values of current over one coplete cycle is (even the value over half a cycle will do). i dθ ( ) i dθ The square root of this value is Hence, the r..s. value of the alternating current is I idθ I sin θ d θ Now, cos θ sin θ sin θ cos θ I I sin θ I ( cos θ) dθ θ 4 4 (put i I sin θ) I I 4 I I.77 I Hence, we find that for a syetrical sinusoidal current r..s. value of current.77 ax. value of current The r..s. value of an alternating current is of considerable iportance in practice, because the aeters and volteters record the r..s. value of alternating current and voltage respectively. In

11 A.C. Fundaentals 463 electrical engineering work, unless indicated otherwise, the values of the given current and voltage are always the r..s. values. It should be noted that the average heating effect produced during one cycle is I R ( I/ ) R I R.6. R.M.S. Value of a Coplex Wave In their case also, either the id-ordinate ethod (when equation of the wave is not known) or analytical ethod (when equation of the wave is known) ay be used. Suppose a current having the equation i sin ωt + 6 sin (3ωt /6) + 4 sin (5ωt + /3) flows through a resistor of R oh. Then, in the tie period T second of the wave, the effect due to each coponent is as follows : Fundaental... (/ ) RT watt 3rd haronic... (6/ ) RT watt 5th haronic... (4/ ) RT watt Total heating effect RT [(/ ) + (6/ ) + (4/ ) ] If I is the r..s. value of the coplex wave, then equivalent heating effect is I RT I RT RT [(/ ) + (6/ ) + (4/ ) ] I [(/ ) + (6/ ) + (4/ ) ] 9.74 A Had there been a direct current of (say) 5 aperes flowing in the circuit also*, then the r..s. value would have been ( / ) (6/ ) (4/ ) 5.93 A Hence, for coplex waves the rule is as follows : The r..s. value of a coplex current wave is equal to the square root of the su of the squares of the r..s. values of its individual coponents..7. Average Value The average value I a of an alternating current is expressed by that steady current which transfers across any circuit the sae charge as is transferred by that alternating current during the sae tie. In the case of a syetrical alternating current (i.e. one whose two half-cycles are exactly siilar, whether sinusoidal or non-sinusoidal), the average value over a coplete cycle is zero. Hence, in their case, the average value is obtained by adding or integrating the instantaneous values of current over one half-cycle only. But in the case of an unsyetrical alternating current (like half-wave rectified current) the average value ust always be taken over the whole cycle. (i) Mid-ordinate Method i With reference to Fig..6, I av + i in n This ethod ay be used both for sinusoidal and non-sinusoidal waves, although it is specially convenient for the latter. (ii) Analytical Method The standard equation of an alternating current is, i I sin θ I idθ I av sin θ d θ ( ) * The equation of the coplex wave, in that case, would be, i 5 + sin ωt + 6 sin (3ωt /6) + 4 sin (5ωt + /3) (putting value of i)

12 464 Electrical Technology I I I I twice the axiu current / cos θ + ( ) I av I /.637 I average value of current.637 axiu value Note. R.M.S. value is always greater than average value except in the case of a rectangular wave when both are equal..8. For Factor r..s. value.77 I It is defined as the ratio, K f.. (for sinusoidal alternating currents only) average value.637 I In the case of sinusoidal alternating voltage also, K f.77 E.637 E. As is clear, the knowledge of for factor will enable the r..s. value to be found fro the arithetic ean value and vice-versa..9. Crest or Peak or Aplitude Factor It is defined as the ratio K a axiu value I.44 (for sinusoidal a.c. only) r..s. value I / E For sinusoidal alternating voltage also, K a.44 E/ Knowledge of this factor is of iportance in dielectric insulation testing, because the dielectric stress to which the insulation is subjected, is proportional to the axiu or peak value of the applied voltage. The knowledge is also necessary when easuring iron losses, because the iron loss depends on the value of axiu flux. Exaple.4. An alternating current varying sinusoidally with a frequency of 5 Hz has an RMS value of A. Write down the equation for the instantaneous value and find this value (a).5 second (b).5 second after passing through a positive axiu value. At what tie, easured fro a positive axiu value, will the instantaneous current be 4.4 A? (Elect. Science-I Allahabad Univ. 99) Solution. I 8. A, ω 5 rad/s. The equation of the sinusoidal current wave with reference to point O (Fig..7) as zero tie point is i 8. sin t apere Since tie values are given fro point A where voltage has positive and axiu value, the equation ay itself be referred to point A. In the case, the equation becoes : i 8. cos t (i) When t.5 second i 8. cos.5...angle in radian 8. cos angle in degrees 8. cos 45º A...point B Fig..7

13 (ii) When t.5 second i 8. cos cos 5º 8. ( / ) A.C. Fundaentals 465 A...point C (iii) Here i 4.4 A cos 8 t cos 8 t or 8 t cos (.5) 6º, t /3 second...point D Exaple.5. An alternating current of frequency 5 Hz has a axiu value of A. Calculate (a) its value /6 second after the instant the current is zero and its value decreasing thereafter (b) how any seconds after the instant the current is zero (increasing thereafter wards) will the current attain the value of 86.6 A? (Elect. Technology. Allahabad Univ. 99) Solution. The equation of the alternating current (assued sinusoidal) with respect to the origin O (Fig..8) is i sin 5t sin t (a) It should be noted that, in this case, tie is being easured fro point A (where current is zero and decreasing thereafter) and not fro point O. If the above equation is to be utilized, then, this tie ust be referred to point O. For this purpose, half tie-period i.e. / second has to be added to /6 second. The given tie as referred to point O Fig..8 becoes 7 + second 6 6 i sin 8 7/6 sin º / 5 A...point B (b) In this case, the reference point is O sin 8 t or sin 8, t.866 or 8, t sin (.866) 6º t 6/8, /3 second Exaple.6. Calculate the r..s. value, the for factor and peak factor of a periodic voltage having the following values for equal tie intervals changing suddenly fro one value to the next :, 5,,, 5, 6, 5,,, 5,, 5, V etc. What would be the r..s value of sine wave having the sae peak value? Solution. The wavefor of the alternating voltage is shown in Fig..9. Obviously, it is not sinusoidal but it is syetrical. Hence, though r..s value ay be full one cycle, the average value has necessarily to be considered for half-cycle only, otherwise the syetrical negative and positive half-cycles will cancel each Fig..9 other out.

14 466 Electrical Technology Mean value of v r..s. value V (approx.) 965 V Average value (half-cycle) 3 V r..s. value For factor average value Peak factor 6/3 (approx.) 3 R.M.S. value of a sine wave of the sae peak value V. Alternative Solution If t be the regular tie interval, then area of the half-cycle is (5t + t + t + 5t) + 6t 3t, Base t Mean value 3t/t 3 V. Area when ordinates are squared (5t + t + 4t + 5t) + 36t 965t, Base t Mean height of the squared curve 965t/ t 965 r..s. value V Further solution is as before. Exaple.7. Calculate the reading which will be given by a hot-wire volteter if it is connected across the terinals of a generator whose voltage wavefor is represented by ν sin ωt + sin 3ωt + 5 sin 5ωt Solution. Since hot-wire volteter reads only r..s value, we will have to find the r..s. value of the given voltage. Considering one coplete cycle, R.M.S. value V vdθ where θ ωt or V ( sin θ + sin 3θ + 5 sin 5θ) dθ ( sin θ + sin 3θ + 5 sin 5θ +. sin θ. sin 3θ +.5. sin 3θ. sin 5θ + 5..sin 5θ. sin θ) dθ ,5 V 6,5 6 V Alternative Solution The r..s. value of individual coponents are (/ ), (/ ) and (5/ ). Hence, as stated in Art..6, V V + V + V (/ ) + (/ ) + (5/ ) 6 V 3.. R.M.S. Value of H.W. Rectified Alternating Current Half-wave (H.W.) rectified alternating current is one whose one half-cycle has been suppressed i.e. one which flows for half the tie during one cycle. It is shown in Fig.. where suppressed half-cycle is shown dotted.

15 A.C. Fundaentals 467 As said earlier, for finding r..s. value of such an alternating current, suation would be carried over the period for which current actually flows i.e. fro to, though it would be averaged for the whole cycle i.e. fro to. R.M.S. current I I idθ sin θ d θ I 4 I sin θ θ 4 ( cos θ) dθ I I 4 4 I I Fig...5I.. Average Value of H.W. Rectified Alternating Current For the sae reasons as given in Art.., integration would be carried over fro I av I id θ sin θ d θ I I cos I θ (ä i I sin θ).. For Factor of H.W. Rectified Alternating Current r..s. value I/ For factor.57 average value I / Exaple.8. An alternating voltage e sin 34t is applied to a device which offers an ohic resistance of Ω to the flow of current in one direction, while preventing the flow of current in opposite direction. Calculate RMS value, average value and for factor for the current over one cycle. (Elect. Engg. Nagpur Univ. 99) Solution. Coparing the given voltage equation with the standard for of alternating voltage equation, we find that V V, R Ω, I / A. For such a half-wave rectified current, RMS value I / / 5A. Average current I / / 3.8 A ; For factor 5/ Exaple.9. Copute the average and effective values of the square voltage wave shown in Fig... Solution. As seen, for < t <. i.e. for the tie interval to. second, v V. Siilarly, for. < t <.3, v. Also tie-period of the voltage wave is.3 second. T. V av v dt dt T.3 V T vdt T (.) 6.67 V.3. dt (4.) 33.3; V.5 V.3.3 Fig..

16 468 Electrical Technology Exaple.. Calculate the RMS value of the function shown in Fig.. if it is given that for < t <., y ( e t ) and. < t <., y e 5(t.) Solution. Y.. { }. y dt + y dt... { } t 5( t ( e ) dt ( e.) ) dt t t ( t.) ( + e e ) dt e dt. +. { }.. 5{ } t.5. t. ( t t e + e + e.). { } ( 5..5e +.e ) (.5 +.) + (.e ) (.) Y Exaple.. The half cycle of an alternating signal is as follows : It increases uniforly fro zero at º to F at αº,reains constant fro αº (8 α)º, decreases uniforaly fro F at (8 α)º to zero at 8º. Calculate the average and effective values of the signal. (Elect. Science-I, Allahabad Univ. 99) Solution. For finding the average value, we would find the total area of the trapeziu and divide it by (Fig..3). Area Δ OAE + rectangle ABDE (/) F α + ( α) F ( α) F average value ( α) F / RMS Value Fro siilar triangles, we get F y θ α or y F This gives the equation of the signal over the two triangles OAE and DBC. The signal reains constant over the angle α to ( α) i.e. over an angular distance of ( α) α ( α) F α Su of the squares θ α d θ + F ( α) F ( 4α/3). θ The ean value of the squares is F 4 4 ( α) F( α 3 3) 4 α r..s. value F 3 Fig.. Fig..3 Exaple.. Find the average and r..s values of the a.c. voltage whose wavefor is given in Fig..4 (a) Solution. It is seen [(Fig..4 (a)] that the tie period of the wavefor is 5s. For finding the average value of the wavefor, we will calculate the net area of the wavefor over one period and then find its average value for one cycle. A V s, A 5 V s

17 A.C. Fundaentals 469 Net area over the full cycle A + A V s. Average value V s/5s V. Fig..4 (b) shows a graph of v (t). Since the negative voltage is also squared, it becoes positive. Average value of the area 4 V s + 5 V s 45 V s. The average value of the su of the square 45 V /- s/5s 9V rs value 9 V 9.49 V. Fig..4 Exaple.3. What is the significance of the r..s and average values of a wave? Deterine the r..s. and average value of the wavefor shown in Fig..5 (Elect. Technology, Indore Univ.) Solution. The slope of the curve AB is BC/AC /T. Next, consider the function y at any tie t. It is seen that DE/AE BC/AC /T or (y )/t /T or y + (/T)t This gives us the equation for the function for one cycle. T T Y av ydt ( t) dt T T + Fig..5 T T T. dt. t. dt t 5t T T T T T T Mean square value ( ) T ydt + t dt T or RMS value 7/3 5. Exaple.4. For the trapezoidal current wave-for of Fig..6, deterine the effective value. (Elect. Technology, Vikra Univ. Ujjain, Siilar Exaple, Nagpur Univ. 999) Solution. For < t < 3T/, equation of the current can be found fro the relation T 3 t t T T t t dt T + + T T t 7 T + 3T + 3 Fig..6

18 47 Electrical Technology I I i or i t 3 T / 3T.t When 3T/ < t < 7T/, equation of the current is given by i I. Keeping in ind the fact that ΔOAB is identical with ΔCDE, RMS value of current T/ I dt 3 T/ 7 T/ idt+ 3T/ I 3 T 3T 5 3 T/ 7 T/ tdt+ I dt I 3T/ I (3/5). I.775 I Incidentally, the average value is given by { } 3 T / 7 T / 3 T / I 7 T / Iac idt Idt tdt I dt T T/ T 3T 3 T/ 3 / T I 7 T/ t 7 + I t T 3T 3 T/.I Exaple.5. A sinusoidal alternating voltage of V is applied across a oving-coil aeter, a hot-wire aeter and a half-wave rectifier, all connected in series. The rectifier offers a resistance of 5 Ω in one direction and infinite resistance in opposite direction. Calculate (i) the readings on the aeters (ii) the for factor and peak factor of the current wave. (Elect. Engg.-I Nagpur Univ. 99) Solution. For solving this question, it should be noted that (a) Moving-coil aeter, due to the inertia of its oving syste, registers the average current for the whole cycle. (b) The reading of hot-wire aeter is proportional to the average heating effect over the whole cycle. It should further be noted that in a.c. circuits, the given voltage and current values, unless indicated otherwise, always refer to r..s values. E / V (approx.) ; I / 55.5/5 6. A Average value of current for positive half cycle A Value of current in the negative half cycle is zero. But, as said earlier, due to inertia of the coil, M.C. aeter reads the average value for the whole cycle. (i) M.C. aeter reading 3.96/.98 A Let R be the resistance of hot-wire aeter. Average heating effect over the positive half cycle is I. R watts. But as there is no generation of heat in the negative half cycle, the average heating effect over the whole cycle is 4 I R watt. Let I be the d.c. current which produces the sae heating effect, then I R 4 I R I I / 6./ 3. A. Hence, hot-wire aeter will read 3. A (ii) For factor r..s value ; Peak factor average value.98 ax. value 6. r..s. value 3. Exaple.6. Find the for-factor of the wave for given in fig. [Nagpur University Noveber 99, Siilar exaple, Sabalpur University]

19 Solution. For-factor RMS value Average value Average value of the current /4 4 (5/4) t dt 5 ap Let RMS value of the current be I ap 4 I 4 (.5 t).dt.5.5 t 3 4 (/3) ( ) A.C. Fundaentals 47 Fig..7 Thus I ap, Hence, for factor Exaple.7. A half-wave rectifier which prevents current flowing in one direction is connected in series with an a.c. aeter and a peranent-agnet oving-coil aeter. The supply is sinusoidal. The reading on the a.c. aeter is A. Find the reading given by the other aeter. What should be the readings on the aeters, if the other half-wave were rectified instead of being cut off? Solution. It should be noted that an a.c. aeter reads r..s. value whereas the d.c. aeter reads the average value of the rectified current. As shown in Art.. fro H.W. rectified alternating current, I I / and I av I / As a.c. aeter reads A, hence r..s. value of the current is A. I / or I A I av / A reading of d.c. aeter. The full-wave rectified current wave is shown in Fig..8. In this case ean value of i over a coplete cycle is given as i dθ I sin θ dθ ( cos θ) d θ θ sin θ I I I I I / / 4.4 A a.c. aeter will read 4.4 A Now, ean value of i over a coplete cycle Fig..8 I sin d I I sin I d cos.73 A This value, as ight have been expected, is twice the value obtained in the previous case. d.c. aeter will read.73 A. Exaple.8. A full-wave rectified sinusoidal voltage is clipped at / of its axiu value. Calculate the average and RMS values of such a voltage.

20 47 Electrical Technology Solution. As seen fro Fig..9, the rectified voltage has a period of and is represented by the following equations during the different intervals. < θ < /4 ; v V sin θ /4 < θ < 3/4 ; v V /.77 V 3/4 < θ < ; v V sin θ /4 3 /4 V av vdθ+ vdθ+ vdθ /4 3 /4 /4 3 /4 { Vsin θdθ+.77 Vdθ+ V sin θdθ} /4 3 /4 V cos θ +.77 θ + cos θ ( ).54 V 4 / 3 /4 V { Vsin θdθ+ (.77 V) d θ+ V sin θdθ }.34V /4 3 /4 V.584 V Exaple.9. A delayed full-wave rectified sinusoidal current has an average value equal to half its axiu value. Find the delay angle θ. (Basic Circuit Analysis, Nagpur 99) Solution. The current wavefor is shown in Fig..3. I av I I sin d ( cos + cos θ) Now, I av I / I I ( cos + cos θ) cos θ.57, θ cos (.57) 55.5º Fig..3 Exaple.. The wavefor of an output current is as shown in Fig..3. It consists of a portion of the positive half cycle of a sine wave between the angle θ and 8º. Deterine the effective value for θ 3º. (Elect. Technology, Vikra Univ. 984) V /4 3 /4 { /4 3 /4} Solution. The equation of the given delayed halfwave rectified sine wave is i I sin ωt I sin θ. The effective value is given by I idθ or I sin. /6 I θ d θ /6 I I sin θ ( cos θ) dθ θ 4 4 or I /6 /6.4 I.4I.49 I Fig..9 Fig..3 Exaple.. Calculate the for factor and peak factor of the sine wave shown in Fig..3. (Elect. Technology-I, Gwalior Univ.) Solution. For < θ <, i sin θ and for < θ <, i. The period is.

21 Fig..3 I av { idθ+ dθ } { θ dθ } A.C. Fundaentals 473 sin 3.8 A I idθ sin θdθ 5 ; I 5A 4 for factor 5/ ; peak factor /5 Exaple.. Find the average and effective values of voltage of sinusoidal wavefor shown in Fig..33. (Elect. Science-I Allahabad Univ. 99) Solution. Although, the given wavefor would be integrated fro /4 to, it would be averaged over the whole cycle because it is unsyetrical. The equation of the given sinusoidal wavefor is v sin θ. Fig..33 V av sin θdθ cos θ 7. V /4 /4 V sin sin d ( cos ) d / /4 V 47.7 V Exaple.3. Find the r..s. and average values of the saw tooth wavefor shown in Fig..34 (a). Solution. The required values can be found by using either graphical ethod or analytical ethod. Graphical Method The average value can be found by averaging the function fro t to t in parts as given below : T Average value of (f) T f (t) dt (net area over one cycle) T Now, area of a right-angled triangle (/) (base) (altitude). Hence, area of the triangle during t to t.5 second is A ( Δ t) ( ) Siilarly, area of the triangle fro t.5 to t second is A ( Δ t) ( + ) Net area fro t to t. second is A + A + Hence, average value of f (t) over one cycle is zero. For finding the r..s. value, we will first square the ordinates of the given function and draw a new plot for f (t) as shown in Fig..34 (b). It would be seen that the squared ordinates fro a parabola. Area under parabolic curve base altitude. The area under the curve fro t to 3

22 474 Electrical Technology t.5 second is ; A ( Δ t) Fig..34 Siilarly, for t.5 to t. second A ( Δ t) Total area A + A 4 T +, r..s. value f ( t ) dt average of f ( t ) T r..s. value 4/3.5 Analytical Method The equation of the straight line fro t to t in Fig..34 (a) is f (t) 4 t ; f (t) 6 t 6 t + 4 T T Average value (4t ) dt 4t t T T T T 3 (6t 6t 4) dt 6t 6t 4t T T 3 r..s value Exaple.4. A circuit offers a resistance of Ω in one direction and Ω in the reverse direction. A sinusoidal voltage of axiu value V is applied to the above circuit in series with (a) a oving-iron aeter (b) a oving-coil aeter (c) a oving-coil instruent with a full-wave rectifier (d) a oving-coil aeter. Calculate the reading of each instruent. Solution. (a) The deflecting torque of an MI instruent is proportional to (current). Hence, its reading will be proportional to the average value of i over the whole cycle. Therefore, the reading of such an instruent : θdθ+ sin sin θdθ sin 4 sin 6 5. A (b) An MC aeter reads the average current over the whole cycle. Average current over positive half-cycle is A Average current over positive half-cycle is A Average value over the whole cycle is (6.37.7)/.55 A (c) In this case, due to the full-wave rectifier, the current passing through the operating coil of the instruent would flow in the positive direction during both the positive and negative half cycles. reading ( )/ 3.8 A (d) Average heating effect over the positive half-cycle is I R

23 Average heating effect over the negative half-cycle is where I / A; I / A I Average heating effect over the whole cycle is ( R R ) A.C. Fundaentals / 6 R If I is the direct current which produces the sae heating effect, then I R 6 R I 6 5. A Exaple.5. A oving coil aeter, a hot-wire aeter and a resistance of Ω are connected in series with a rectifying device across a sinusoidal alternating supply of V. If the device has a resistance of Ω to the current in one direction and 5. Ω to current in opposite direction, calculate the readings of the two aeters. (Elect. Theory and Meas. Madras University) Solution. R.M.S. current in one direction is /( + ) A Average current in the first i.e. positive half cycle is /..9 A Siilarly, r..s. value in the negative half-cycle is /( + 5) /3 A Average value ( /3)/..3 A Average value over the whole cycle is (.9.3)/.3 A Hence, M/C aeter reads.3 A Average heating effect during the +ve half cycle I rs R I R R Siilarly, average heating effect during the ve half-cycle is ( /3) R R/9 Here, R is the resistance of the hot-wire aeter. Average heating effect over the whole cycle is 5R ( R + R 9 ) 9 If I is the direct current which produces the sae heating effect, then I R 5R/9 I 5/3.745 A Hence, hot-wire aeter indicates.745 A Exaple.6. A resultant current wave is ade up of two coponents : a 5A d.c. coponent and a 5-Hz a.c. coponent, which is of sinusoidal wavefor and which has a axiu value of 5A. (i) Draw a sketch of the resultant wave. (ii) Write an analytical expression for the current wave, reckoning t at a point where the a.c. coponent is at zero value and when di/dt is positive. (iii) What is the average value of the resultant current over a cycle? (iv) What is the effective or r..s. value of the resultant current? [Siilar Proble: Bobay Univ. 996] Solution. (i) The two current coponents and resultant current wave have been shown in Fig..35. (ii) Obviously, the instantaneous value of the resultant current is given by i (5 + 5 sin ωt) (5 + 5 sin θ) (iii) Over one coplete cycle, the average value of the alternating current is zero. Hence, the average value of the resultant current is equal to the value of d.c. coponent i.e. 5A (iv) Mean value of i over coplete cycle is i dθ (5 + 5 sin θ) dθ Fig..35 (5 + 5 sin θ+ 5 sin θ) dθ R

24 476 Electrical Technology cosθ ( ) sin θ+ 5 dθ ( sin θ.5 cos θ) dθ cos sin 37.5 R.M.S. value I A Note. In general, let the cobined current be given by i A + B sin ωt A + B sin θ where A represents the value of direct current and B the r..s. value of alternating current. The r..s. value of cobined current is given by I rns i dθ or I r..s. i dθ (A + B sin θ) dθ ( A + B sin θ+ ABsin θ) dθ (A + B B cos θ + AB sin θ dθ B sin A B ABcos A B AB AB A + B I rs ( A + B ) The above exaple could be easily solved by putting A 5 and B 5 / (because B ax 5) I rs ( ) 5 + 5/ 6. A Exaple.7. Deterine the r..s. value of a seicircular current wave which has a axiu value of a. Solution. The equation of a sei-circular wave (shown in Fig..36) is x + y a or y a x I rs + a + a rs a a ydx or I ( a x) dx a a Fig..36 a + a ( a dx x dx) a + a x 3 3 a ax a a a a a 3 a a I rs a /3.86 a Exaple.8. Calculate the r..s. and average value of the voltage wave shown in Fig..37. Solution. In such cases, it is difficult to develop a single equation. Hence, it is usual to consider two equations, one applicable fro to and an other for to illisecond. For t lying between and s, v 4, For t lying between and s, v 4t + 4 Fig..37

25 A.C. Fundaentals 477 v rs v dt + v dt V rs 4 dt ( 4t 4) dt + + V av t t t + + t V rs 4 4 ( 4 4) 4 t v dt + v dt dt + t + dt 4 t t + + 3/ volt volt Tutorial Probles No... Calculate the axiu value of the e..f. generated in a coil which is rotating at 5 rev/s in a unifor agnetic field of.8 Wb/ 3. The coil is wound on a square forer having sides 5 c in length and is wound with 3 turns. [88.5 V]. (a) What is the peak value of a sinusoidal alternating current of 4.78 r..s. aperes? (b) What is the r..s. value of a rectangular voltage wave with an aplitude of 9.87 V? (c) What is the average value of a sinusoidal alternating current of 3 A axiu value? (d) An alternating current has a periodic tie of.3 second. What is its frequency? (e) An alternating current is represented by i 7.7 sin 5 t. Deterine (i) the frequency (ii) the current.5 second after passing through zero, increasing positively. [6.76 A ; 9.87 V ; 9.75 A ; 33.3 Hz ; 8.8 Hz ; 49.7 A] 3. A sinusoidal alternating voltage has an r..s. value of V and a frequency of 5 Hz. It crosses the zero axis in a positive direction when t. Deterine (i) the tie when voltage first reaches the instantaneous value of V and (ii) the tie when voltage after passing through its axiu positive value reaches the value of 4.4 V. [(i) (.5 second (ii) /3 second)] 4. Find the for factor and peak factor of the triangular Fig..38 wave shown in Fig..38 [.55;.73] 5. An alternating voltage of sin 47 t is applied to a h.w. rectifier which is in series with a resistance of 4 Ω. If the resistance of the rectifier is infinite in one direction and zero in the other, find the r..s. value of the current drawn fro the supply source. [.5 A] 6. A sinusoidally varying alternating current has an average value of 7.4 A. When its value is zero, then its rate change is 6,8 A/s. Find an analytical expression for the sine wave. [i sin t] 7. A resistor carries two alternating currents having the sae frequency and phase and having the sae value of axiu current i.e. A. One is sinusoidal and the other is rectangular in wavefor. Find the r..s. value of the resultant current. [.4 A] 8. A copper-oxide rectifier and a non-inductive resistance of Ω are connected in series across a sinusoidal a.c. supply of 3 V (r..s.). The resistance of the rectifier is.5 Ω in forward direction and 3, Ω in the reverse direction. Calculate the r..s. and average values of the current. [r..s. value 5. A, average value 3. A] 9. Find the average and effective values for the waveshape shown in Fig..39 if the curves are parts of a sine wave. [7. V, 47.7V] (Elect. Technology, Indore Univ.)

26 478 Electrical Technology. Find the effective value of the resultant current in a wire which carries siultaneously a direct current of A and a sinusoidal alternating current with a peak value of 5 A. [4.58A] (Elect. Technology, Vikra Univ. Ujjain). Deterine the r..s. value of the voltage defined by e sin (34 t + /6) [6. V] (Elect. Technology, Indore Univ.). Find the r..s. value of the resultant current in a wire which carries siultaneously a direct current of A and a sinusoidal alternating current with a peak value of A. [.5 A] (Elect. Technology-I; Delhi Univ.) 3. An alternating voltage given by e 5 sin t is applied to a circuit which offers a resistance of 5 ohs to the current in one direction and copletely prevents the flow of current in the opposite direction. Find the r..s. and average values of this current and its for factor. [.5 A,.95 A,.57] (Elect. Technology, Indore Univ.) 4. Find the relative heating effects of three current waves of equal axiu value, one rectangular, the second sei-circular and the third sinusoidal in wavefor [: /, /] (Sheffield Univ. U.K.) 5. Calculate the average and root ean-square value, the for factor and peak factor of a periodic current wave have the following values for equal tie intervals over half-cycle, changing suddenly fro one value of the next. [, 4, 6, 8,, 8, 6, 4, ] (A.M.I.E.) 6. A sinusoidal alternating voltage of aplitude V is applied across a circuit containing a rectifying device which entirely prevents current flowing in one direction and offers a resistance of oh to the flow of current in the other direction. A hot wire aeter is used for easuring the current. Find the reading of instruent. (Elect. Technology. Punjab Univ.).3. Representation of Alternating Quantities It has already been pointed out that an attept is ade to obtain alternating voltages and currents having sine wavefor. In any case, a.c. coputations are based on the assuption of sinusoidal voltages and currents. It is, however, cubersoe to continuously handle the instantaneous values in the for of equations of waves like e E sin ωt etc. A conventional ethod is to eploy vector ethod of representing these sine waves. These vectors ay then be anipulated instead of the sine functions to Fig..4 Fig..39 achieve the desired result. In fact, vectors are a shorthand for the represen-tation of alternating voltages and currents and their use greatly siplifies the probles in a.c. work. A vector is a physical quantity which has agnitude as well as direction. Such vector quantities are copletely known when particulars of their agnitude, direction and the sense in which they act, are given. They are graphically represented by straight lines called vectors. The length of the line represents the agnitude of the alternating quantity, the inclination of the line with respect to soe axis of reference gives the direction of that quantity and an arrow-head placed at one end indicates the direction in which that quantity acts.

27 A.C. Fundaentals 479 The alternating voltages and currents are represented by such vectors rotating counter-clockwise with the sae frequency as that of the alternating quantity. In Fig..4 (a), OP is such a vector which represents the axiu value of the alternating current and its angle with X axis gives its phase. Let the alternating current be represented by the equation e E sin ωt. It will be seen that the projection of OP and Y-axis at any instant gives the instantaneous value of that alternating current. OM OP sin ωt or e OP sin ωt E sin ωt It should be noted that a line like OP can be ade to represent an alternating voltage of current if it satisfies the following conditions : (i) Its length should be equal to the peak or axiu value of the sinusoidal alternating current to a suitable scale. (ii) It should be in the horizontal position at the sae instant as the alternating quantity is zero and increasing. (iii) Its angular velocity should be such that it copletes one revolution in the sae tie as taken by the alternating quantity to coplete one cycle..4. Vector Diagra using R.M.S. Values Instead of using axiu values as above, it is very coon practice to draw vector diagras using r..s. values of alternating quantities. But it should be understood that in that case, the projection of the rotating vector on the Y-axis does not give the instantaneous value of that alternating quantity..5. Vector Diagras of Sine Waves of Sae Frequency Two or ore sine waves of the sae frequency can be shown on the sae vector diagra because the various vectors representing different waves all rotate counter-clockwise at the sae frequency and aintain a fixed position relative to each other. This is illustrated in Fig..4 where a voltage e and current i of the sae frequency are shown. The current wave is supposed to pass upward through zero at the instant when t while at the sae tie the voltage wave has already advanced an angle α fro its zero value. Hence, their equations can be written as Fig..4 i I sin ωt and e E sin (ωt + α) Sine wave of different frequencies cannot be represented on the sae vector diagra in a still picture because due to difference in speed of different vectors, the phase angles between the will be continuously changing..6. Addition of Two Alternating Quantities In Fig..4 (a) are shown two rotating vectors representing the axiu values of two sinusoidal voltage waves represented by e E sin ωt and e E sin (ωt φ). It is seen that the su of the two sine waves of the sae frequency is another sine wave of the sae frequency but of a different axiu value and phase. The value of Fig..4