# Module 4. Single-phase AC circuits. Version 2 EE IIT, Kharagpur

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1 Module 4 Single-phase A circuis ersion EE T, Kharagpur

2 esson 5 Soluion of urren in A Series and Parallel ircuis ersion EE T, Kharagpur

3 n he las lesson, wo poins were described:. How o solve for he impedance, and curren in an ac circui, consising of single elemen, / /?. How o solve for he impedance, and curren in an ac circui, consising of wo elemens, and /, in series, and hen draw complee phasor diagram? n his lesson, he soluion of currens in simple circuis, consising of resisance, inducance and/or capaciance conneced in series, fed from single phase ac supply, is presened. Then, he circui wih all above componens in parallel is aken up. The process of drawing complee phasor diagram wih curren(s and volage drops in he differen componens is described. The compuaion of oal power and also power consumed in he differen componens, along wih power facor, is explained. One example of series circui are presened in deail, while he example of parallel circui will be aken up in he nex lesson. Keywords: Series and parallel circuis, impedance, admiance, power, power facor. Afer going hrough his lesson, he sudens will be able o answer he following quesions;. How o compue he oal reacance and impedance / admiance, of he series and parallel circuis, fed from single phase ac supply?. How o compue he differen currens and also volage drops in he componens, boh in magniude and phase, of he circui? 3. How o draw he complee phasor diagram, showing he currens and volage drops? 4. How o compue he oal power and also power consumed in he differen componens, along wih power facor? Soluion of urren in -- Series ircui Series (-- circui O D E A - Fig. 5. (a ircui diagram The volage balance equaion for he circui wih, and in series (Fig. 5.a, is di v i i d d sin ersion EE T, Kharagpur

4 The curren, i is of he form, i sin ( ± As described in he previous lesson (#4 on series (- circui, he curren in seady sae is sinusoidal in naure. The procedure given here, in brief, is followed o deermine he form of curren. f he expression for i sin ( is subsiued in he volage equaion, he equaion shown here is obained, wih he sides (HS & HS inerchanged. sin ( cos ( (/ cos ( sin or sin ( [ (/ ] cos ( sin The seps o be followed o find he magniude and phase angle of he curren, are same as described here (#4. So, he phase angle is an [ (/ ] / and he magniude of he curren is / where he impedance of he series circui is [ (/ ] Alernaively, he seps o find he rms value of he curren, using complex form of impedance, are given here. The impedance of he circui is ± ( where, ( (/, and an ( an (/ 0 0 ± ( ( 0 ( (/ Two cases are: (a nducive >, and (b apaciive <. (a nducive is posiive, under he condiion ( > ( /. The curren lags he volage by (aken as posiive, wih he volage phasor aken as reference. The power facor (lagging is less han (one, as The complee phasor diagram, wih he volage drops across he n his case, he circui is inducive, as oal reacance ( ( / ersion EE T, Kharagpur

5 componens and inpu volage (OA, and also curren (O, is shown in Fig. 5.b. The volage phasor is aken as reference, in all cases. may be observed ha O ( i D [ i ( ] DA [ i ( ] OA ( i using he Kirchoff s second law relaing o he volage in a closed loop. The phasor diagram can also be drawn wih he curren phasor as reference, as will be shown in he example given here. The expression for he average power is cos. The power is only consumed in he resisance,, bu no in inducance/capaciance (/, in all hree cases. E (- A ( O. D nducive ( > Fig 5. (b Phasor diagram is posiive, under he condiion ( > ( /. The curren lags he volage by (posiive. The power facor (lagging is less han (one, as The complee phasor diagram, wih he volage drops across he componens and inpu volage ( OA, and also curren ( O, is shown in Fig. 5.b. The volage phasor is aken as reference, in all cases. may be observed ha O ( i D [ i ( ] DA [ i ( ] OA ( i using he Kirchoff s second law relaing o he volage in a closed loop. The phasor diagram can also be drawn wih he curren phasor as reference, as will be shown in he example given here. The expression for he average power is cos. The power is only consumed in he resisance,, bu no in inducance/capaciance (/, in all hree cases. n his case, he circui is inducive, as oal reacance ( ( / ersion EE T, Kharagpur

6 (b apaciive E O. D ( (- A apaciive ( < Fig 5. (c Phasor diagram is negaive, under he condiion ( < ( /. The curren leads he volage by, which is negaive as per convenion described in he previous lesson. The volage phasor is aken as reference here. The complee phasor diagram, wih he volage drops across he componens and inpu volage, and also curren, is shown in Fig. 5.c. The power facor (leading is less The circui is now capaciive, as oal reacance ( ( / han (one, as 0 90, being negaive. The expression for he average power remains same as above. The hird case is resisive, as oal reacance ( / is zero (0, under he condiion ( /. The impedance is 0 0. The curren is now a uniy power facor ( 0, i.e. he curren and he volage are in phase. The complee phasor diagram, wih he volage drops across he componens and inpu (supply volage, and also curren, is shown in Fig. 5.d. This condiion can be ermed as resonance in he series circui, which is described in deail in lesson #7. The magniude of he impedance in he circui is minimum under his condiion, wih he magniude of he curren being maximum. One more poin o be noed here is ha he volage drops in he inducance, and also in he capaciance,, is much larger in magniude han he supply volage, which is same as he volage drop in he resisance,. The phasor diagram has been drawn approximaely o scale. ersion EE T, Kharagpur

7 E (. - ( O -, (. OD D, A O esisive ( Fig. 5. (d Phasor diagram may be observed here ha wo cases of series (- & - circuis, as discussed in he previous lesson, are obained in he following way. The firs one (inducive is ha of (a, wih very large, i.e. / 0, which means ha is no here. The second one (capaciive is ha of (b, wih no being here ( or 0. Example 5. A resisance, is conneced in series wih an iron-cored choke coil (r in series wih. The circui (Fig. 5.a draws a curren of 5 A a 40, 50 Hz. The volages across he resisance and he coil are 0 and 00 respecively. alculae, (a he resisance, reacance and impedance of he coil, (b he power absorbed by he coil, and (c he power facor (pf of he inpu curren. A r D Fig. 5. (a ircui diagram Soluion (O 5 A S (OA 40 f 50 Hz π f The volage drop across he resisance ( O 0 ersion EE T, Kharagpur

8 The resisance, / 0/ 5 4 Ω The volage drop across he coil ( A 00 The impedance of he coil, r / 00 / 5 40 Ω From he phasor diagram (Fig. 5.b, A E D A A 40, 50 Hz Fig. 5.(b: Phasor Diagram OA O A (0 (40 (00 cos cos AO OA O The power facor (pf of he inpu curren cos ( lag The phase angle of he oal impedance, cos ( npu volage, S ( OA 40 The oal impedance of he circui, 3,000 57,600 ( r / 40 / 5 48 Ω ( r ( Ω The oal resisance of he circui, r 4 r Ω The resisance of he coil, r Ω The reacance of he coil, π f Ω 39.9 The inducance of he coil, 0.7 H 7 0 π f π 50 The phase angle of he coil, cos ( r / cos (.67 / 40.0 cos ( r ( Ω The power facor (pf of he coil, cos ( lag The copper loss in he coil Example 5. r W S 3 7 mh ersion EE T, Kharagpur

9 An inducive coil, having resisance of 8 Ω and inducance of 80 mh, is conneced in series wih a capaciance of 00 μf across 50, 50 Hz supply (Fig. 5.3a. alculae, (a he curren, (b he power facor, and (c he volages drops in he coil and capaciance respecively. Soluion f 50 Hz - A 80 mh H 00 μf Ω 6 E D Fig. 5.3 (a ircui diagram π f π rad / s l F S ( OA 50 The impedance of he coil, ( Ω The oal impedance of he circui, ( ( ( Ω The curren drawn from he supply, A ( The curren is, A The power facor (pf cos cos ( lead D Ω Ω 9.6 A. ( A. E. (- Fig. 5.3 (b Phasor diagram ersion EE T, Kharagpur

10 Please noe ha he curren phasor is aken as reference in he phasor diagram (Fig. 5.3b and also here. The volage drop in he coil is, θ 0 ( ( The volage drop in he capaciance in, θ 0 ( Soluion of urren in Parallel ircui Parallel circui The circui wih all hree elemens,, & conneced in parallel (Fig. 5.4a, is fed o he ac supply. The curren from he supply can be compued by various mehods, of which wo are described here. - Firs mehod Fig. 5.4 (a ircui diagram. The curren in hree branches are firs compued and he oal curren drawn from he supply is he phasor sum of all hree branch currens, by using Kirchoff s firs law relaed o he currens a he node. The volage phasor ( is aken as reference. All currens, i.e. hree branch currens and oal curren, in seady sae, are sinusoidal in naure, as he inpu (supply volage is sinusoidal of he form, v sin Three branch currens are obained by he procedure given in brief. v, or i v / ( / sin sin, i where, ( / d i Similarly, v d So, i is, i ( / v d (/ (sin d [ /( ] cos sin ( 90 cos ersion EE T, Kharagpur

11 where, / ( wih, from which is obained as, d i v / ( i d d d d v i cos cos ( sin ( 90 sin ( where, / ( c wih / ( Toal (supply curren, i is cos cos sin i i i i sin ( cos ( sin The wo equaions given here are obained by expanding he rigonomeric form appearing in he las erm on HS, ino componens of cos and sin, and hen equaing he componens of cos and sin from he las erm and las bu one (previous. cos and ( sin From hese equaions, he magniude and phase angle of he oal (supply curren are, ( ( l an (/ (/ (/ an an an where, he magniude of he erm (admiance of he circui is, Please noe ha he admiance, which is reciprocal of impedance, is a complex quaniy. The angle of admiance or impedance, is same as he phase angle, of he curren, wih he inpu (supply volage aken as reference phasor, as given earlier. Alernaively, he seps required o find he rms values of hree branch currens and he oal (suuply curren, using complex form of impedance, are given here. Three branch currens are 90 ; 0 (/ 90 ersion EE T, Kharagpur

12 Of he hree branches, he firs one consiss of resisance only, he curren, is in phase wih he volage (. n he second branch, he curren, lags he volage by 90, as here is inducance only, while in he hird one having capaciance only, he curren, leads he volage 90. All hese cases have been presened in he previous lesson. The oal curren is ± ( c where,, and ( an an The wo cases are as described earlier in series circui. (a nducive O D A E nducive ( > Fig. 5.4 (b Phasor diagram n his case, he circui being inducive, he curren lags he volage by (posiive, as >, i.e. / >, or < /.This condiion is in conras o ha derived in he case of series circui earlier. The power facor is less han (one. The complee phasor diagram, wih he hree branch currens along wih oal curren, and also he volage, is shown in Fig. 5.4b. The volage phasor is aken as reference in all cases. may be observed here ha ( OD ( D ( ( O ersion EE T, Kharagpur

13 The Kirchoff s firs law relaed o he currens a he node is applied, as saed above. The expression for he average power is cos /. The power is only consumed in he resisance,, bu no in inducance/capaciance (/, in all hree cases. (b apaciive The circui is capaciive, as >, i.e. > /, or > /. The curren leads he volage by ( being negaive, wih he power facor less han (one. The complee phasor diagram, wih he hree branch currens along wih oal curren, and also he volage, is shown in Fig. 5.4c. The hird case is resisive, as, i.e. / or /. This is he same condiion, as obained in he case of series circui. may be noed ha wo currens, and, are equal in magniude as shown, bu opposie in sign (phase difference being 80, and he sum of hese currens is zero (0. The oal curren is in ( phase wih he volage ( 0, wih, he power facor being uniy. The complee phasor diagram, wih he hree branch currens along wih oal curren, and also he volage, is shown in Fig. 5.4d. This condiion can be ermed as resonance in he parallel circui, which is described in deail in lesson #7. The magniude of he impedance in he circui is maximum (i.e., he magniude of he admiance is minimum under his condiion, wih he magniude of he oal (supply curren being minimum. O D A E apaciive ( < Fig. 5.4 (c Phasor diagram ersion EE T, Kharagpur

14 The circui wih wo elemens, say &, can be solved, or derived wih being large ( 0 or / 0. O D, ( Second mehod - esisive ( (/( ( /(- Fig. 5.4 (d Phasor Diagram efore going ino he deails of his mehod, he erm, Admiance mus be explained. n he case of wo resisance conneced in series, he equivalen resisance is he sum of wo resisances, he resisance being scalar (posiive. f wo impedances are conneced in series, he equivalen impedance is he sum of wo impedances, all impedances being complex. Please noe ha he wo erms, real and imaginary, of wo impedances and also he equivalen one, may be posiive or negaive. This was explained in lesson no.. f wo resisances are conneced in parallel, he inverse of he equivalen resisance is he sum of he inverse of he wo resisances. f wo impedances are conneced in parallel, he inverse of he equivalen impedance is he sum of he inverse of he wo impedances. The inverse or reciprocal of he impedance is ermed Admiance, which is complex. Mahemaically, his is expressed as As admiance ( is complex, is real and imaginary pars are called conducance (G and suscepance ( respecively. So, G. f impedance, wih being posiive, hen he admiance is where, 0 ( G E ( ersion EE T, Kharagpur

15 ; G Please noe he way in which he resul of he division of wo complex quaniies is obained. oh he numeraor and he denominaor are muliplied by he complex conugae of he denominaor, so as o make he denominaor a real quaniy. This has also been explained in lesson no.. The magniude and phase angle of and are / ( an ;, and G G an an ; To obain he curren in he circui (Fig. 5.4a, he seps are given here. The admiances of he hree branches are 90 ; The oal admiance, obained by he phasor sum of he hree branch admiances, is G ± 3 where, an ;, and G / ; / The oal impedance of he circui is G G G G ± The oal curren in he circui is obained as ± ± ± ( 0 0 where he magniude of curren is / The curren is he same as obained earlier, wih he value of subsiued in he above equaion. This is bes illusraed wih an example, which is described in he nex lesson. The soluion of he curren in he series-parallel circuis will also be discussed here, along wih some examples. ersion EE T, Kharagpur

16 Problems 5. alculae he curren and power facor (lagging / leading for he following circuis (Fig. 5.5a-d, fed from an ac supply of 00. Also draw he phasor diagram in all cases Ω 5 Ω 00-5 Ω - -0 Ω (a (b 00-5 Ω 5 Ω (c - -0 Ω Fig Ω - 0 Ω (d Ω 5. A volage of 00 is applied o a pure resisor (, a pure capacior, and a lossy inducor coil, all of hem conneced in parallel. The oal curren is.4 A, while he componen currens are.5,.0 and. A respecively. Find he oal power facor and also he power facor of he coil. Draw he phasor diagram. 5.3 A Hz supply is conneced o a lamp having a raing of 00, 00 W, in series wih a pure inducance,, such ha he oal power consumed is he same, i.e. 00W. Find he value of. A capaciance, is now conneced across he supply. Find value of, o bring he supply power facor o uniy (.0. Draw he phasor diagram in he second case..(a Find he value of he load resisance ( o be conneced in series wih a real volage source ( S S in series, such ha maximum power is ransferred from he above source o he load resisance. (b Find he volage was 8Ω resisance in he circui shown in Fig. (b..(a Find he Theremin s equivalen circui (draw he ck. beween he erminals A, of he circui shown in Fig. (a. ersion EE T, Kharagpur

17 (b A circui shown in Fig. (b is supplied a 40, 50Hz. The wo volages and (magniude only is measured as 60 and 5 respecively. f he curren, is measured as A, find he values of, and. Also find he power facor of he circui (--. Draw he complee phasor diagram. 3.(a Find he line curren, power facor, and acive (real power drawn from 3-phase, 00, 50Hz, balanced supply in he circui shown in Fig. 3(a. (b n he circui shown in Fig. 3(b, he swich, S is pu in posiion a 0. Find i e (, > 0, if v c (0-6. Afer he circui reaches seady sae, he swich, S is brough o posiion, a T. Find i c (, > T. Swich he above waveform. 4.(a Find he average and rms values of he periodic waveform shown in Fig. 4(a. (b A coil of mh lowing a series resisance of Ω is conneced in parallel wih a capacior, and he combinaion is fed from 00 m (0., khz supply (source having an inernal resisance of 0Ω. f he circui draws power a uniy power facor (upf, deermine he value of he capacior, qualiy facor of he coil, and power drawn by he circui. Also draw he phasor diagram. ersion EE T, Kharagpur

18 is of Figures Fig. 5. (a ircui diagram (-- in series (b Phasor diagram ircui is inducive ( l > (c Phasor diagram apaciive ( l < (d Phasor diagram esisive ( Fig. 5. (a ircui diagram (Ex. 5., Fig. 5.3 (a ircui diagram (Ex. 5., Fig. 5.4 (a ircui diagram (-- in parallel (b Phasor diagram ircui is inducive ( l (b Phasor diagram (b Phasor diagram (c Phasor diagram apaciive ( l > (d Phasor diagram esisive ( l < l ersion EE T, Kharagpur

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