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Module some Single-phase ALTERNATING CURRENT Circuits Type 2 EE IIT, Kharagpur Lesson 13 Representation of Sinusoidal Sign by a Phasor and Answer of Current in R-L-C Series Brake lines Version 2 EE IIT, Kharagpur Within the last lesson, two points were defined: 1 . What sort of sinusoidal volt quality waveform (ac) is generated? 2 . How the average and rms values of the regular voltage or perhaps current waveforms, are calculated? Some examples can also be described presently there.

In this lesson, the representation of sinusoidal (ac) voltage/current signals with a phasor will be explained. The polar/Cartesian (rectangular) form of phasor, as sophisticated quantity, can be described.

Lastly, the algebra, involving the phasors (voltage/current), is presented. Different mathematical operations ” addition/subtraction and multiplication/division, on several phasors, happen to be discussed. Keywords: Phasor, Sinusoidal signals, phasor algebra After going through this lesson, the students will be able to solution the following questions, 1 . Precisely what is meant by term, ‘phasor’ in respect of a sinusoidal signal? 2 . The right way to represent the sinusoidal ac electricity or current waveform by phasor? a few. How to write a phasor variety (complex) in polar/Cartesian (rectangular) form? 4.

How to conduct the businesses, like addition/subtraction and multiplication/division on several phasors, to obtain a phasor? This lesson varieties the background with the following lessons in the full module of single alternating current circuits, starting with the next lessons on the answer of the current in the stable state, in R-L-C series circuits. Emblems i or perhaps i(t) Immediate value from the current (sinusoidal form) I Im? Current (rms value) Maximum benefit of the current Phasor representation of the current Phase angle, say in the current phasor, with respect to the guide phasor I Same emblems are used for voltage or any other phasor. Representation of Sinusoidal Signal by a Phasor A sinusoidal volume, i. electronic. current, i (t ) = My spouse and i m bad thing? t, is definitely taken up for example. In Fig. 13. 1a, the length, OP, along the x-axis, represents the maximum value of the current I actually m, over a certain range. It is being rotated inside the anti-clockwise course at an slanted speed,?, and takes up a situation, OA after a short time t (or angle,? sama dengan? t, with the x-axis). The vertical discharge of OA is drawn in the right hand side of the over figure according to angle? It can generate a sine wave (Fig. 13. 1b), as OA is at an position,? with the x-axis, as stated before. The straight projection of OA along y-axis is usually OC = AB sama dengan Version two EE IIT, Kharagpur i (? ) = We m trouble?, which is the instantaneous worth of the current at any time capital t or position?. The perspective? is in lista., i. elizabeth.? =? capital t. The angular speed,? is in rad/s, i actually. e.? sama dengan 2? farreneheit, where farrenheit is the regularity in Hz or cycles/sec. Thus, my spouse and i = I actually m desprovisto? = I m bad thing? t = I meters sin 2? ft So , OP signifies the phasor with respect to the above current, i actually.

The line, OPERATIVE can be taken as the rms value, My spouse and i = My spouse and i m / 2, instead of maximum benefit, Im. Then a vertical output of OA, in degree equal to OPERATIVE, does not stand for exactly the fast value of I, but represents that with the scale factor of just one / two = 0. 707. The explanation for this range of phasor since given over, will be succumbed another lesson later with this module. Variation 2 EE IIT, Kharagpur Generalized circumstance The current can be of the kind, i (t ) = I m sin (? t? ) as proven in Fig. 13. 1d. The phasor representation of the current may be the line, OQ, at an angle,? could possibly be taken as negative), with the series, OP along x-axis (Fig. 13. 1c). One has to maneuver in clockwise direction to attend OQ by OP (reference line), though the phasor, OQ is believed to move in anti-clockwise course as offered earlier. After a time t, Z will be at an angle? with OQ, which is at an angle (?? sama dengan? t? ), with the series, OP along x-axis. The vertical output of Z along y-axis gives the fast value of the current, i = a couple of I sin (? to? ) sama dengan I m sin (? t? ). Phasor portrayal of Ac electricity and Current The volts and current waveforms receive as, v = a couple of V trouble? and i = 2 We sin (? +? ) It can be viewed from the waveforms (Fig. 13. 2b) of the two sinusoidal quantities ” voltage and current, that the voltage, Versus lags the current I, which means the positive optimum value of the voltage is reached previously by a great angle,?, in comparison with the positive maximum value with the current. In phasor explication as referred to earlier, the voltage and current are represented simply by OP and OQ (Fig. 13. 2a) respectively, the length of which are proportional to voltage, V and current, I actually in different weighing scales as relevant to each 1.

The volts phasor, OP (V) lags the current phasor, OQ (I) by the perspective?, as two phasors rotate in the anticlockwise direction mentioned previously earlier, although the angle? is also tested in the anticlockwise direction. In other words, the current phasor (I) potential clients the volts phasor (V). Version 2 EE IIT, Kharagpur Mathematically, the two phasors can be showed in extremely form, while using voltage phasor ( Sixth is v ) taken as reference, including V = V? zero 0, and I = We?. In Cartesian or rectangular form, these are generally, V = V? 0 0 = V + j zero, and I = I? = I cos? + j I trouble?, where, the symbol, j is given simply by j =?. Of the two terms in each phasor, the first one is termed as true or it is component in x-axis, even though the second one is imaginary or perhaps its component in y-axis, as proven in Fig. 13. 3a. The viewpoint,? is in level or lista.??? Phasor Algebra Before discussing the mathematical operations, just like addition/subtraction and multiplication/division, including phasors and also complex amounts, let us look into the two forms ” extremely and square, by which a phasor or complex amount is represented. It may be seen here that phasors can also be taken as complicated, as given above.

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Portrayal of a phasor and Modification A phasor or a intricate quantity in rectangular type (Fig. 13. 3) can be, A = ax & j a y Type 2 EE IIT, Kharagpur? where a by and a y are real and imaginary parts, of the phasor respectively. In polar contact form, it is indicated as A sama dengan A? a = A cos? a + j A trouble? a? where A and? a are size and phase angle of the phasor. From the two equations or movement, the procedure or rule of transformation coming from polar to rectangular kind is a x = A cos? a and a y sama dengan A desprovisto? a From the above, the guideline for modification from square to polar form is usually 2 two A sama dengan a by + a y and? = tan? 1 (a y as well as a back button ) The examples applying numerical ideals are given at the end of this lesson. Addition/Subtraction of Phasors Prior to describing the guidelines of addition/subtraction of phasors or intricate quantities, everyone should recollect the secret of addition/subtraction of scalar quantities, which may be positive or signed (decimal/fraction or fraction with integer). It may be explained that, for the two businesses, the quantities must be either phasors, or complex. The example of phasor is voltage/current, and that of complex amount is impedance/admittance, which will be discussed in the next lessons.

But one phasor and another complex quantity must not be used for addition/subtraction operation. For the operations, the two phasors or sophisticated quantities must be expressed in rectangular contact form as A = a by + l a con, B sama dengan bx & j n y If they are in polar form like a = A? a, N = M? b In such a case, two phasors are to be transformed to rectangle-shaped form by the procedure or perhaps rule provided earlier. The rule of addition/subtraction operation is that the two real and imaginary parts have to be separately treated because?? where c x = (a times n x ), c y = (a y b con ) Say, for addition, real parts must be added, thus also to get imaginary parts.

Same secret follows intended for subtraction. Following the result is usually obtained in rectangular type, it can be converted to polar one. It might be observed the six values of a’ s, b’ s and c’ s i9000 ” elements of the two phasors and the resultant one, are all signed scalar quantities, although in the case in point, a’ s i9000 and b’ s will be taken as great, resulting in great values of c’ s. Also the phase viewpoint? , h may lie in any with the four quadrants, though below the perspectives are inside the first sector only. This rule for addition could be extended to 3 or more amounts, as will probably be illustrated through example, which is given by the end of this lesson.

C = A B = (a back button bx ) + j (a y b sumado a ) = c back button + m c y?? Version a couple of EE IIT, Kharagpur The addition/subtraction procedures can also be performed using the quantities as?? phasors in polar form (Fig. 13. 4). The two phasors are A (OA) and N (OB). The find the sum C (OC ), a range AC is definitely drawn the same and parallel to DURCH. The line BC is the same and parallel to OA. Thus, C = OCCITAN = OA + AC = OA + HINSICHTLICH = A + N. Also, OC = HINSICHTLICH + BC = HINSICHTLICH + OA?? To obtain the difference D (OD), a collection AD is drawn the same and parallel to DURCH, but in opposite direction to AC or OB.

A line FACTORY is also drawn equal to DURCH, but in opposite direction to OB. Both AD and OE symbolize the phasor (? N ). The queue, ED is equal to OA. Thus, M = OD = OA + ADVERTISEMENT = OA? OB sama dengan A? N. Also Z = OE + IMPOTENCE =? OB + OA. The illustrations using numerical values receive at the end on this lesson.?? Multiplication/Division of Phasors Firstly, the process for multiplication is adopted. In this case zero reference has been made to the rule including scalar amounts, as everyone is familiar with these people. Assuming that both phasors are available in polar coming from as A = A? a and W = M? b.

Otherwise, they are being transformed via rectangular to polar kind. This is also valid for the process of split. Please note a phasor is usually to be multiplied with a complex amount only, to obtain the resultant phasor. A phasor is certainly not normally increased by another phasor, besides in exceptional case. Same is for section. A phasor is to be divided by a intricate quantity only, to obtain the resultant phasor. A phasor is definitely not normally divided simply by another phasor.?? To find the magnitude of the product C, the two magnitudes with the phasors have to be multiplied, whereas for stage angle, the phase sides are to added.

Thus, Edition 2 EE IIT, Kharagpur C = C? c = A? B sama dengan A? A? B? M = ( A? N )? (? a +? b )?? where C = A? B and? c =? a +? b? Take note that the same symbol, C is used for the product in cases like this.?? To split A. by B to discover the result G., the magnitude is received by label of the variation, and the phase is big difference of the two phase sides. Thus, G = Deb? d sama dengan? A? = B where D sama dengan A as well as B and? d =? a? m? A? a? A? sama dengan?? (? a? b ) B? n? B? If the phasors will be expressed in rectangular kind as A sama dengan a back button + m a sumado a and W = bx + l by right here A = (a 2 x? 2 + a y,? a = bronze? 1 (a y as well as a back button ) ) The beliefs of B are not provided as they can be obtained by substituting b’ h for a’ s. To obtain the product, C = C? c = A? W = (a x + j a y )? (bx + j m y ) = (a x bx? a sumado a b y ) & j (a x m y + a sumado a bx )?? Please note that j a couple of =? 1 ) The degree and phase angle of the result (phasor) are, C = (a x bx? a con b y ) & (a back button b con + a y bx ) two [ 1 2 2 ] sama dengan (a two x two + ay? ) (b 2 by 2 + b con = A? B, and )? c = color? 1?? a x n y + a con bx? a x bx? a sumado a b y?? The stage angle,? c =? a +? n = suntan? 1? a x w y + a y bx = tan? 1? a m? a m y y? x x?? ay? ax??? b? + tan? you? y? b? x? (a / a ) + (b y / bx )?? sama dengan tan? 1? y by?? 1? (a y as well as a by )? (b y / bx )?? The above the desired info is obtained simply by simplification.? To divide A by M to obtain G as D = dx + t dy =? A? = ax + j a y bx + j by? M To make simpler D, we. e. to obtain real and imaginary parts, both numerator and denominator, are to be multiplied by the complex conjugate of B, to be able to convert the? denominator in to real value only. The complex conjugate of N is Edition 2 EE IIT, Kharagpur

B 2. = bx + m b sumado a = W?? b In the complex conjugate, the indication of the fabricated part is negative, as well as the phase viewpoint is negative.? (a times + t a y )? (bx? j simply by ) sama dengan? a by bx + a y by? + j? a y bx? a by by??? G = dx + l dy sama dengan (bx & j simply by )? (bx? j by simply )? bx2 + by2? bx2 + by2??? The magnitude and phase viewpoint of the end result (phasor) are, [(a b D= x back button + a y n y ) + (a y bx? a times b con ) a couple of 1 2 2 (b 2 by +b two y ) ] = (a (b a couple of x two x two + ay 2 + by ) A =, and ) B? a y bx? a times b con?? d sama dengan tan? one particular? a m +a n? y sumado a? x x The period angle,? ay? ax?? bronze? 1? con? b? times? a n? a xby? = suntan? 1? con x? a b +a b sumado a y? times x??? deb =? a? b = tan? 1? The steps will be shown within brief, because detailed measures have been given earlier. Example? The phasor, A in the rectangular form (Fig. 13. 5) can be, A = A? a = A cos? a + t A sin? a sama dengan a times + t a con =? 2 + j 4 the place that the real and imaginary parts are a x =? two,? ay sama dengan 4 To transform the phasor, A into the polar type, the magnitude and period angle happen to be Version 2 EE IIT, Kharagpur two 2 A = a x & a con = (? 2) two + 4 2 = 4. 472? 4? sama dengan tan? one particular?? 116. 565 = 2 . 034 rad?? 2? Please be aware that? a is in the second quadrant, because real component is adverse and mythical part can be positive.? a = tan? 1?? ay? ax? Transforming the phasor, A in to rectangular contact form, the real and imaginary parts are a back button = A cos? a = some. 472? cos116. 565 sama dengan? 2 . zero a sumado a = A sin? a = some. 472? trouble 116. 565 = four. 0 Phasor Algebra?? One more phasor, W in rectangle-shaped form is definitely introduced besides the earlier one, A W = 6 + j 6 sama dengan 8. 325? 45 Firstly, let us take the addition and subtraction from the above two phasors. The sum and? difference receive by the phasors, C and D respectively (Fig. 13. 6). C = A+ B sama dengan (? 2 + t 4) +(6 + t 6) = (? 2 + 6) + t (4 & 6) = 4 & j 10 = 15. 77? 68. 2 D = A? B sama dengan (? two + l 4)? (6 + j 6) = (? a couple of? 6) & j (4? 6) sama dengan? 8? t 2 = 8. 246? 166. 0 It may be noted that pertaining to the addition and subtraction operations regarding phasors, they must be represented in rectangular contact form as given above. If any one of the phasors Version two EE IIT, Kharagpur??? is at polar type, it should be changed into rectangular form, for determining the effects as displayed.

If the two phasors are in polar form, the phasor picture (the plan must be drawn to scale), or perhaps the geometrical method can be used while shown in Fig 13. 6. The effect obtained using the diagram, while shown are exactly the same as acquired earlier. [ C (OC) = 10. 77,? COX = 68. 2, and M ( OD) = almost eight. 246,? DOX = 166. 0 ] Right now, the copie and section operations will be performed, using the above two phasors showed in polar form. In the event that any one of the phasors is in rectangular form, it might be transformed into polar form. Likewise note that the same symbols for the phasors are used here, as utilized earlier.

After, the method of both copie and department using rectangle-shaped form of the phasor rendering will be discussed.?? The resulting phasor C, i. e. the product from the two phasors is C = A? B = 4. 472? 116. 565? 8. 325? 45 = (4. 472? 8. 485)? (116. 565 + 45) = thirty seven. 945? 161. 565 sama dengan? 36 + j 12 The product of the two phasors in rectangular form is available as C = (? 2 & j 4)? (6 & j 6) = (? 12? 24) + l (24? 12) =? thirty six + t 12???? The actual result ( Deb ) obtained by the trademark A by B is usually D=? A? = W = 0. 167 + j zero. The above effect can be worked out by the method described previously, using the rectangular form of both phasors since D=? four. 472? 116. 565? four. 472? sama dengan?? (116. 565? 45) sama dengan 0. 527? 71. 565 8. 325? 45? 8. 485? A? = B 12 + j thirty eight = sama dengan 0. 167 + j 0. your five 72? a couple of + j 4 (? 2 + j 4)? (6? t 6) (? 12 + 24) + j (24 + 12) = = 6+ j6 ( 6th + t 6)? ( 6? j 6) 62 + sixty two The procedure for the elementary operations applying two phasors only, in both types of representation can be shown. It might be easily extended, for declare, addition/multiplication, employing three or even more phasors.

The simplification procedure with the scalar quantities, making use of the different fundamental operations, which can be well known, can be extended for the phasor quantities. This will provide in the study of air conditioning unit circuits being discussed in the following lessons. The background essential, i. electronic. phasor rendering of sinusoidal quantities (voltage/current), and algebra ” mathematical operations, such as addition/subtraction and multiplication/division of phasors or complex amounts, including transformation of phasor from rectangular to extremely form, and vice versa, has been discussed below.

The study of air conditioner circuits, beginning from series ones, will be referred to in the next couple of lessons. Version 2 EE IIT, Kharagpur Problems 13. 1 Use plasor technique to evaluate the manifestation and then find the statistical value in t sama dengan 10 ms. i ( t ) = one hundred and fifty cos (100t , 400 ) + 500 sin (100t ) + deb? cos 100t , 35 0 )? dt? ( 13. two Find the end result in both rectangular and polar forms, for the next, using sophisticated quantities: a few , j12 15? 53. 1 b) ( 5 , j12 ) +15? , 53. 1 a) 2? 30 , 4? 210 your five? 450 1? d)? a few? 0 &?. 2? 210 3 two? , 45? c)

Variation 2 EE IIT, Kharagpur List of Characters Fig. 13. 1 (a) Phasor manifestation of a sinusoidal voltage, and (b) Waveform Fig. 13. 2 (a) Phasor portrayal of volts and current, and (b) Waveforms Fig. 13. 3 Representation of any phasor, at rectangular and polar varieties Fig. 13. 4 Addition and subtraction of two phasors, equally represented in polar contact form Fig. 13. 5 Portrayal of phasor as an example, at rectangular and polar varieties Fig. 13. 6 Addition and subtraction of two phasors displayed in polar form, for example Version 2 EE IIT, Kharagpur

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