Module 3. R-L & R-C Transients. Version 2 EE IIT, Kharagpur

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1 Module 3 - & -C Transiens

2 esson 0 Sudy of DC ransiens in - and -C circuis

3 Objecives Definiion of inducance and coninuiy condiion for inducors. To undersand he rise or fall of curren in a simple series circui excied wih dc source. Meaning of ' Time Consam ( τ )' for circui and explain is relaionship o he performance of he circui. Energy sored in an inducor Definiion of capaciance and Coninuiy condiion for capaciors. To undersand he rise or fall volage across he capacior in a simple series C circui excied wih dc source. Meaning of ' Time Consam ( τ )' for C circui and explain is relaionship o he performance of he circui. Energy sored in a capacior.0. Inroducion So far we have considered dc resisive nework in which currens and volages were independen of ime. More specifically, Volage (cause inpu) and curren (effec oupu) responses displayed simulaneously excep for a consan muliplicaive facor (V = I ). Two basic passive elemens namely, inducor ( ) and capacior ( C ) are inroduced in he dc nework. Auomaically, he quesion will arise wheher or no he mehods developed in lesson-3 o lesson-8 for resisive circui analysis are sill valid. The volage/curren relaionship for hese wo passive elemens are defined by he derivaive (volage across he inducor () di () v =, where i () =curren flowing hrough he d dvc inducor ; curren hrough he capacior C () () i = C, v ( C ) = volage across he d capacior) or in inegral form as i() = v() d i(0) or vc() i() d v (0) = C C 0 0 raher han he algebraic equaion (V = I) for all resisors. One can sill apply he KC, KV, Mesh-curren mehod, Node-volage mehod and all nework heorems bu hey resul in differenial equaions raher han he algebraic equaions ha we have considered in resisive neworks (see ession-3 o lesson-8). An elecric swich is urned on or off in some circui (for example in a circui consising of resisance and inducance), ransien currens or volages (quickly changing curren or volage) will occur for a shor period afer hese swiching acions. Afer he ransien has ended, he curren or volage in quesion reurns o is seady sae siuaion (or normal seady value). Duraion of ransien phenomena are over afer only a few micro or milliseconds, or few seconds or more depending on he values of circui parameers (like,, and C ).The siuaion relaing o he sudden applicaion of dc volage o circuis possessing resisance ( ), inducance ( ), and capaciance (C ) will

4 now be invesigaed in his lesson. We will coninue our discussion on ransiens occurring in a dc circui. I is needless o menion ha ransiens also occur in ac circui bu hey are no included in his lesson..0. Significance of Inducance of a coil and dc ransiens in a simple - circui Fig.0. shows a coil of wire forming an inducance and is behavior is o resis any change of elecric curren hrough he coil. When an inducor carries curren, i produces a cerain amoun of magneic flux ( Φ ) in he core or space around i. The produc of he magneic flux ( Φ ) and he number of urns of a coil (an inducor) is called he flux linkage of he coil. Considering he physical fac ha he volage across he coil is direcly proporional o he rae of change of curren hrough he inducor and i is expressed by he equaion di() e() emf = e() = = (0.) d di()/ d where is he consan of proporionaliy called inducance of coil and i is measured in vol s econd = henry (H ). The direcion of induced emf is opposie o ha of curren ampere increases or decreases (enz s aw) di() e () = (0.) d e us assume ha he coil of wire has 'N ' urns and he core maerial has a relaively high permeabiliy (or magneic pah relucance is very low), so ha he magneic flux ( Φ ) produced due o curren flowing hrough he coil is concenraed wihin he core area. The basic fundamenal principle according o Faraday, he changing flux hrough he coil creaes an induced emf ( e ) and i is expressed as dφ() e () = N (0.3) d

5 In words, Faraday s law saes ha he volage induced in a coil (inducor) is proporional o he number of urns ha he coil has, and also o he rae of change of he magneic flux passing hrough is coils. From equaions (0.) and (0.3), one can wrie he following relaion N dφ() change in flux linkage N Φ = = = (0.4) di change in curren I The inducance of a coil can also be defined as flux ( Φ ) linkage per uni of curren flowing hrough he coil and i is illusraed hrough numerical example. Example-.0.: Consider wo coils having he same number of urns N. One coil is wrapped in a nonmagneic core (say, air) and he oher is placed on a core of magneic maerial as shown in fig.0.. Calculae he inducances of boh coils for same amoun of curren flowing hrough hem. Case-A: Nonmagneic maerial 4 N Φ 00 (0.5 0 ) Inducance of nonmagneic maerial = = = = 5mH I Case-B: Magneic maerial N Φ 00 (0.05) Inducance of magneic maerial = = = 5H ( Noe: > ) I.0.. Inducance calculaion from physical dimension of coil A general formula for he inducance of a coil can be found by using an equivalen Ohm ' s law for magneic circui and he formula for relucance. This opic will be discussed in deail in esson-. Consider a solenoid-ype elecromagne/oroid

6 wih a lengh much greaer han is diameer (a leas he lengh is en imes as grea as is diameer). This will produce an uniform magneic field inside he oroid. The lengh l of a oroid is he disance around he cener axis of is core, as indicaed in fig.0.3 by doed line. Is area A is he cross-secional area of he oroid, also indicaed in ha figure. Appling ampere-circuial law for magneic circui (see esson-) one can wrie he following relaion NI NI= Hl H= A/ m (0.5) l We know, flux is always given by he produc of flux densiy ( B ) and area ( A ) hrough which flux densiy ids uniform. Tha is, NI Φ= B A= μ H A = μ A ( noe, μ = μ0 μ r ) (0.6) l where B = μ H and H is he uniform field inensiy around he mean magneic pah lengh l. Subsiuing he equaion (0.6) ino he defining equaion for inducance, equaion (0.4) gives NΦ μ N I A μn A = = = (0.7) I Il l emark-: The expression (0.7) is derived for long solenoids and oroids, compuaion of inducance is valid only for hose ypes.

7 .0.. Coninuiy condiion of Inducors The curren ha flows hrough a linear inducor mus always be a coninuous. From he expression (0.), he volage across he inducor is no proporional o he curren flowing hrough i bu o he rae of change of he curren wih respec o di() ime,. The volage across he inducor ( v ) is zero when he curren flowing hrough d an inducor does no change wih ime. This observaion implies ha he inducor acs as a shor circui under seady sae dc curren. In oher words, under he seady sae condiion, he inducor erminals are shored hrough a conducing wire. Alernaing curren (ac), on he oher hand, is consanly changing; herefore, an inducor will creae an opposiion volage polariy ha ends o limi he changing curren. If curren changes very rapidly wih ime, hen inducor causes a large opposiion volage across is erminals. If curren changes hrough he inducor from one level o anoher level insananeously i.e. in d 0 sec., hen he volage across i would become infinie and his would require infinie power a he erminals of he inducor. Thus, insananeous changes in he curren hrough an inducor are no possible a all in pracice. emark-: (i) The curren flowing hrough he inducor canno change insananeously (i.e. i(0 ) jus righ before he change of curren = i(0 ) jus righ afer he change of curren). However, he volage across an inducor can change abruply. (ii) The inducor acs as a shor circui (i.e. inducor erminals are shored wih a conducing wire) when he curren flowing hrough he inducor does no change (consan). (iii) These properies of inducor are imporan since hey will be used o deermine boundary condiions..0.3 Sudy of dc ransiens and seady sae response of a series - circui. Ideal Inducor: Fig.0.4 shows an ideal inducor, like an ideal volage source, has no resisance and i is excied by a dc volage source. V S

8 The swich S is closed a ime = 0 and assumed ha he iniial curren flowing hrough he ideal inducor i (0) jus before closing he swich is equal o zero. To find he sysem response ( i () vs ), one can apply KV around he closed pah. KV V S i () di() = 0 di() VS d d (0.8) Vs Vs V di() = d i() = i(0) i = (noe i (0) = 0 ) (0.9) () s 0 0 Equaion (0.9) implies ha he curren hrough inducor increases wih increase in ime and heoreically i approaches o infiniy as bu in pracice, his is no really he case. eal or Pracical inducor: Fig.0.5 shows a real or pracical inducor has some resisance and i is exacly equal o he resisance of he wire used o wind he coil. e us consider a pracical inducor is conneced in series wih an exernal resisance and his circui is excied wih a dc volage as shown in fig.0.6(a). V S

9 Our problem is o sudy he growh of curren in he circui hrough wo sages, namely; (i) dc ransien response (ii) seady sae response of he sysem. D.C Transiens: The behavior of he curren (( i )); charge ( q ( )) and he volage ( v ( )) in he circui (like ; C: C circui) from he ime ( (0 ) ) swich is closed unil i reaches is final value is called dc ransien response of he concerned circui. The response of a circui (conaining resisances, inducances, capaciors and swiches) due o sudden applicaion of volage or curren is called ransien response. The mos common insance of a ransien response in a circui occurs when a swich is urned on or off a raher common even in an elecric circui Growh or ise of curren in - circui To find he curren expression (response) for he circui shown in fig. 0.6(a), we can wrie he KV equaion around he circui di() = = (0.0) ( ) VS i() v() 0 VS i() d where V is he applied volage or forcing funcion, is he resisance of he coil, is S he exernal resisance. One can combine he resisance of coil o he exernal resisance in order o obain a simplified form of differenial equaion. The circui configuraion shown in fig. 0.6(a) is redrawn equivalenly in fig.0.6(b) for our convenience. The equaion (0.0) is he sandard firs order differenial equaion and is soluion can be obained by classical mehod. The soluion of firs or second order differenial equaion is briefly discussed in Appendix (a he end of his lesson-0). The following relaion gives he soluion of equaion (0.0) α i () = i() i () = Ae A (0.) n f Here, in ( ) is he complemenary soluion/naural soluion of differenial equaion (0.0). I is also someimes called as ransien response of sysem (i.e. he firs par of

10 response is due o an iniial condiion of he sysem or force free response). The second par i () of eq. (0.) is he paricular inegral soluion/force response or seady sae f response of he sysem due o he forcing funcion ( f ( ) = VS ) or inpu signal o he series circui. I may be noed he erm A provide us he seady sae soluion of he firs order differenial equaion while he forcing funcion (or inpu o he sysem) is sep funcion (or consan inpu). More specifically, for a linear sysem, he seady sae soluion of any order differenial equaion is he same naure of forcing funcion ( f ( )) or inpu signal bu differen in magniude. We have lised in abular form he naure of seady sae soluion of any order differenial equaion for various ypes of forcing funcions (see in Appendix). To ge he complee soluion of eq. (0.0), he values of α, A and A are o be compued following he seps given below: Sep-: How o find he value of α? d Assigning V S = 0 and inroducing an operaor α = in eq.(0.0), we ge a d characerisic equaion ha will provide us he numerical value of α. This in urn, gives us he ransien response of he sysem provided he consan A is known o us. The Characerisic equaion of (0.0) is α = 0 α =. Sep-: How o obain he consans A and A? I may be noed ha he differenial eq. (0.0) mus be saisfied by he paricular inegral soluion or seady sae soluion i (). The value of i () a seady sae condiion (i.e. f ) can be found ou using he eq.(0.) and i is given below. Using final condiion ( ) dif () VS = if () (0.) d (noe: a seady sae ( ) i () = A= cons an from eq. (0.)) f f da V A A d V = = s (0.3) S Using iniial condiion ( =0) Case-A: Assume curren flowing hrough he inducor jus before closing he swich S (a = 0 ) is i (0 ) = 0. ( 0 ) (0 ) (0 ) i = i = i = A A (0.4) Vs 0 = A A A= A= Using he values of α, A and A in equaion (0.), we ge he curren expression as

11 V () S i e = (0.5) The able shows how he curren i() builds up in a - circui. Acual ime () in sec Growh of curren in inducor (Eq.0.5) = 0 i (0) = 0 τ Vs = = i( τ ) = 0.63 = τ Vs i( τ ) = = 3τ Vs i(3 τ ) = = 4τ Vs i(4 τ ) = 0.98 = 5τ Vs i(5 τ ) = Noe: Theoreically a ime he curren in inducor reaches is seady sae value bu in pracice he inducor curren reaches 99.3% of is seady sae value a ime = 5 τ (sec.). The expression for volage across he exernal resisance (see Fig. 0.6(a)) V s = v = i() = e (0.6) The expression for volage across he inducor or coil V s vcoil () = vinducor () = VS v () = V ( ) S e (0.7) Graphical represenaion of equaions (0.5)-(0.7) are shown in Fig.0.7 for differen choices of circui parameers (i.e., & )

12

13 Case-B: Assume curren flowing hrough he inducor jus before closing he swich S (a = 0 ) is i(0 ) = i0 0. Using equaions (0.3) and (0.4), we ge he values of V A = S VS and A i(0) A i0 = =. Using hese values in equaion (0.), he expression for curren flowing hrough he circui is given by V () S i e = ie 0 (0.8) The second par of he righ hand side of he expression (0.8) indicaes he curren flowing o he circui due o iniial curren of inducor and he firs par due o he V S forcing funcion applied o he circui. This means ha he complee response of he circui is he algebraic sum of wo oupus due o wo inpus; namely (i) due o forcing funcion V (ii) due o iniial curren i S 0 hrough he inducor. This implies ha he superposiion heorem is also valid for such ype of linear circui. Fig.0.8 shows he i 0

14 response of inducor curren when he circui is excied wih a consan volage source and he iniial curren hrough inducor is. i 0 V S emark-3: One can also solve his differenial equaion by separaing he variables and inegraing. Time consan ( τ ) for exponenial growh response (τ ): We have seen ha he curren hrough inducor is represened by V () S i e = when a series circui is excied by a consan volage source ( ) and an iniial curren hrough he inducor i 0 is assumed o be zero. Furher i may be noed ha he curren hrough he inducor (see fig.0.7) increases as ime increases. The shape of V S

15 VS growing curren before i reaches o a seady sae value enirely depends on he parameers of circui (i.e. & ) ha associaed wih he exponenial erm e. As grows larger and larger he ransien, because of is negaive exponenial facor, diminishes and disappears, leaving only he seady sae. Definiion of Time Consan ( ) variable or signal (in our case eiher curren ( τ of Circui: I is he ime required for any i ()) or volage ( ) ( i.e he ime a which he facor e 00 ( e ) 00= 63. v or v ) o reach 63.% in eq.(0.5) becomes %) of is final value. I is possible o wrie an exac mahemaical expression o calculae he ime consan ( τ ) of any firs-order differenial equaion. e is he ime required o reach 63.% of seady-sae value of inducor curren (see fig. 0.6(a)) and he corresponding ime expression can be obained as V S V S i ( ) = 0.63* = e 0.63 = e = e = e = = (sec.) τ The behavior of all circui responses (for firs-order differenial equaion) is fixed by a single ime consan τ (for circuiτ = ) and i provides informaion abou he speed of response or in oher words, i indicaes how firs or slow he sysem response reaches is seady sae from he insan of swiching he circui. Observe he equaion (0.5) ha he smaller he ime consan (τ ), he more rapidly he curren increases and subsequenly i reaches he seady sae (or final value) quickly. On he oher hand, a circui wih a larger ime consan (τ ) provides a slow response because i akes longer ime o reach seady sae. These facs are illusraed in fig.0.7(a). In accordance wih ( ) convenience, he ime consan of an exponenial erm () 0 ( a say p p e ) = is he reciprocal of he coefficien a associaed wih he in he power of exponenial erm. emark-4: An ineresing propery of exponenial erm is shown in fig. 0.7(a). The ime consan τ of a firs order differenial equaion may be found graphically from he response curve. I is necessary o draw a angen o he exponenial curve a ime = 0 and mainained he same slope unil i inersecs he seady sae value of curren curve a P poin. A perpendicular is drawn from he poin P o he ime axis and i inersecs he ime axis a = τ (see fig. 0.7(a)). Mahemaically, his can be easily verified by considering he equaion of a sraigh line angen o he curren curve a = 0, given by y= m where m is he slope of he sraigh line, expressed as

16 V S d e di() V S m = = 0= = 0 = (0.9) d d Here, we designaed he value of ime required o reach y VS from 0 o unis, assuming a consan rae (slope) of growh. Thus, VS VS (sec.) τ (0.0) I is ofen convenien way of approximaing he ime consan ( ) response curve (see fig.0.7(a) for curve-) Fall or Decay of curren in a - circui τ of a circui from he e us consider he circui shown in fig. 0.9(a). In his circui, he swich S is closed sufficienly long duraion in posiion. This means ha he curren hrough he inducor VS VS reaching is seady-sae value ( I = = = I0 ) and i acs, as a shor circui i.e. he volage across he inducor is nearly equal o zero since resisance. If he swich S is opened a ime =0 and kep in posiion for > 0 as shown in fig. 0.9(b), his siuaion is referred o as a source free circui.

17 Since he curren hrough an inducor canno change insananeously, he curren hrough he inducor jus before ( i (0 ) and afer ( i (0 ) opening he swich S mus be same. Because here is no source o susain he curren flow in inducor, he magneic field in inducor sars o collapse and his, in urn, will induce a volage across he inducor. The polariy of his induced volage across he inducor is jus in reverse direcion compared o he siuaion ha occurred during he growh of curren in inducor (i.e. when he swich S is kep in posiion ). This is illusraed in fig. 0.9(b), where he volage induced in inducor is posiive a he boom of he inducor erminal and negaive a he op. This implies ha he curren hrough inducor will sill flow in he same direcion, bu wih a magniude decaying oward zero. Appling KV around he closed circui in fig. 0.9(b), we obain di() i() = 0 (0.) d The soluion of he homogeneous (inpu free), firs-order differenial equaion wih consan coefficiens subjec o he iniial (boundary) inducor curren (iniial condiion, VS i(0 ) = i(0 ) = = I ) is given by i () = i() = Ae α (0.) n where α can be found from he characerisic equaion of eq.(0.) described by α = 0 α= (0.3) VS A ime = 0, he iniial condiion i(0 ) = i(0 ) = is used in equaion (0.) o compue he consan A and i is given below. VS i(0) = A A= = I

18 Using he values of A and α in equaion (0.), we ge final expression as V S i () = e for 0 (0.4) A skech of i () for 0 is shown in fig.0.0. Here, ransien has ended and seady sae has been reached when boh curren in inducor i () and volage across he inducor including is inernal resisance are zero. Time Consan ( τ ) for exponenial decay response: For he source free circui, i is he ime τ by which he curren falls o 36.8 percen of is iniial value. The iniial condiion in his case (see fig. 0.9(a) is considered o be he value of inducor s curren a he momen he swich S is opened and kep in posiion. Mahemaically, τ is compued as V S V S i ( ) = = e = τ = (0.5)

19 Alernaively, he ime consan for an exponenial decay response of a circui may be compued graphically by adoping he seps (see equaions (0.9) and (0.0)) as discussed before. In fig.0.0, a angen is drawn o he exponenial decay curve a ime = 0 and mainained he same slope unil he sraigh line inerceps ime axis a ime = τ. Approximaely, he value ofτ can hus be found direcly from graphical represenaion of exponenial decay curve Energy sored in an inducor e us urn our aenion o power and energy consideraion for an inducor. The insananeous power absorbed by he inducor is expressed by produc of he curren hrough inducor i () and he volage across i v (). di() p () = vi ()() = i () (0.6) d Since he energy is he produc of power and ime, he energy absorbed by an inducor over a period is expressed as

20 di() () () () ( 0) d 0 0 W = p d = i d = i i (0.7) where we selec he curren hrough inducor a ime 0 = is i ( ) =0. Then, we have W = i () and from his relaion we see ha he energy sored in an inducor is always non-negaive. A any consequen ime a which he curren is zero, no energy is sored in he inducor. The ideal inducor ( = 0Ω ) never dissipaes energy, bu only sores. In rue sense, a physical or pracical inducor dissipaes a very small amoun of sored energy due o is small series resisance. Example-.0. Fig.0. shows he plo of curren i () hrough a series circui when a consan forcing funcion of magniudevs = 50 V is applied o i. Calculae he values of resisance and inducance. Soluion: From fig.0. one can easily see ha he seady sae curren flowing hrough he circui is 0 A and he ime consan of he circui τ = 0.3 sec. The following relaionships can be wrien as VS 50 i = seady sae 0 5 = = Ω and τ = 0.3 = =.5 H 5 Example-.0.3 For he circui shown in Fig.0., he swich S has been closed for a long ime and hen opens a = 0.

21 Find, (i) (0 ) (ii) i (0 ), i (0 ) (iii) v ab (vii) ix ( ) for > 0 x i x (0 ) (iv) v ab (0 ) (v) ix ( = ) (vi) vab ( = ) Soluion: When he swich S was in closed posiion for a long ime, he circui reached in seady sae condiion i.e. he curren hrough inducor is consan and hence, he volage across he inducor erminals a and b is zero or in oher words, inducor acs as shor circui i.e., (i) v (0 ab ) = 0 V. I can be seen ha he no curren is flowing hrough 6 Ω resisor. The following are he currens hrough differen branches jus before he swich S is opened i.e., a = 0. 0 i (0 0 x ) = = 4 A and he curren hrough 0Ω resisor, i 0 Ω (0 ) = = A. The 5 0 algebraic sum of hese wo currens is flowing hrough he inducor i.e., (ii) i (0 ) = 4 = 6 A. When he swich S is in open posiion The curren hrough inducor a ime = 0 is same as ha of curren (0 ), since inducor canno change is curren insananeously.therefore, he curren hrough is given by i (0 ) = i (0 ) = 6A. x i i x (0 ) Applying KV around he closed loop a = 0 we ge, 0 ix(0 ) = vab(0 ) = vab(0 ) vab(0 ) = 0V The negaive sign indicaes ha inducor erminal b as ve erminal and i acs as a source of energy or mahemaically, vba (0 ) = 0V. A seady sae condiion ( ) he curren hrough inducor is consan and hence inducor acs as a shor circui. This esablishes he following relaions: 0 vba ( = ) = 0V and ix ( = ) = = 4A (0.8) 5

22 The circui expression ix ( ) for 0 can be obained using he KV around he closed pah (see fig.0.). KV equaion: dix () VS ix() 5 = 0 d dix () ix () 5 = VS (0.9) d The soluion of firs order differenial equaion due o forcing funcion and iniial condiion is given by i () = A e A (0.30) x Iniial and final condiions are: (i) A = 0, i (0) = i (0 ) = i (0 ) = 6 A (ii), x curren hrough inducor i ( = ) = 4A (see Eq. 0.8). Using iniial and final condiions equaion (0.30) we ge, A = 6 A and A= 4 A= From equaion (0.30), we ge he final expression as i () = 4 e for 0. Example:.0.4 The swich S is closed in posiion sufficienly long ime and hen i is kep in posiion as shown in fig.0.3. Compue he value of v and i (i) he insan jus prior o he swich changing; (ii) he insan jus afer he swich changes. Find dil () also he rae of change of curren hrough he inducor a ime = 0 ie..,. d = 0 x 5 Soluion: We assume ha he circui has reached a seady sae condiion when he swich was in posiion. Noe, a seady sae he inducor acs as shor circui and volage across he inducor is zero.

23 A = 0, he curren hrough and he volage across he inducor are 0 i (0 ) = 0 = 5 A and v (0 ) = 0V respecively. When he swich is kep in 0 0 posiion, curren hrough he inducor canno change insananeously bu his is no rue for he volage across he inducor. A = 0, one can wrie he following expressions: i (0 ) = 5A and (0 ) = = 00V ( b is more ve poenial han a v ( ) erminal). Noe ha he sored energy in inducor is dissipaed in he resisors. Now, he rae of change of curren hrough inducor a ime = 0 is obained as dil( ) dil( ) 00 = 00V = = 5 amp./ sec. d d 4 = 0 = 0 Example:.0.5 Fig. 0.4(a) shows ha a swich S has been in posiion for a long ime and is moved in posiion a ime = 0. Find he expression v () for 0. Soluion: When he swich S is in posiion, he curren hrough inducor (using he fundamenal propery of inducor currens) a seady sae condiion (see fig.0.4(b)) is 6 I = 6= 3A I(0 ) = I(0 ) = 3 A (0.3) 6 6

24 The circui for he swich S is in posiion is shown in fig.0.4 (c). The curren in inducor can be compued using following wo differen mehods. Mehod-: Using Thevenin s heorem Conver he par of a circui conaining independen sources and resisances ino an equivalen Thevenin s volage source as shown in fig.0.4.(d).

25 Using he KV around he closed pah is di () 9 i ( ) = 5 (0.3) d The soluion of he above equaion is given by 9 i () = A e A = i () i f () (0.33) where, in( ) if () n = complemenary/naural/ransien soluion of eq.(0.3) = paricular/ seady sae/final soluion of eq.(0.3) The consans A and A are compued using he iniial and final condiions of he circui when he swich is kep in posiion. A ime = 0, i (0) = i (0 ) = 3 = A A (0.34) A ime, he curren in inducor reached is seady sae condiion and acs as a shor circui in a dc source nework. The curren hrough inducor is 5 i( = ) = = amp. = if = A (0.35) 3 6 Using he above wo equaions in (0.33), one can obain he final volage expression for volage v () across he erminals a and b as 9 vab( ) = v( ) = 5 i ( ) 3 = e = e V Mehod-: Mesh curren mehod Assign he loop currens in clockwise direcions and redrawn he circui as shown in Fig. 0.4(e). The volage across he erminals a and b can be obained by solving he following loop equaions.

26 oop-: 0 6 i() 6 ( i() i() ) = 0 0= i() 6 i() i() = ( 0 6 i() ) (0.36) oop-: () () 6 () di di i 6 ( i() i() ) 0 6 i() i() 0 d = d = (0.37) Using he value of i () in equaion (0.37), we ge di () 9 i ( ) = 5 (0.38) d To solve he above firs order differenial equaion we mus know inducor s iniial and final condiions and heir values are already known (see, i(0 ) = i(0 ) = 3 A and 5 i ( = ) = = amp.). The soluion of differenial equaion (0.38) provides an 3 6 expression of curren i () and his, in urn, will give us he expression of i (). The volage across he erminals a and b is given by 9 di () vab = 0 6 i( ) = 6 i( ) = e V d where, i(), i() can be obained by solving equaions (0.38) and (0.36). The expressions for i () and hence i () are given below: 9 i ( ) =.445 e and i() = ( 0 6 i() ) = 9..5e

27 .0.4 Capacior and is behavior Fig.0.5 shows a capacior consiss of wo pieces of meal (he plaes) separaed from each oher by a good insulaor (he dielecric), wih wo wires (he leads) aached o he meal plaes. A baery is conneced across he capacior o ranspor charge from one plae o he oher unil he capacior charge volage buildup is equal o he baery volage V. The volage across he capacior depends on how much charge was deposied on he plaes and also how much capaciance he capacior has. In oher words, here is a relaionship beween he volage (V ), charge ( Q ) and capaciance ( C ), hey are relaed wih a mahemaical expression as Q ( coulumb ) C = (0.39) V ( vol) where Q = magniude of charge sored on each plae, V = volage applied o he plaes and he uni of capaciance is in Farad. Alhough he capaciance C of a capacior is he raio of charge per plae o he applied volage bu i mainly depends on he physical dimension of he capacior. If he area of he plaes is larger, he more would be he amoun of charge sored over he surface of he plaes, resuling higher value of capaciance. On he oher hand, if he spacing d beween he plaes is closer, accumulaes more charge over he parallel plaes and hus increases he value of he capaciance. The qualiy of dielecric maerial has an effec on capaciance beween he plaes. The good qualiy of dielecric maerial indicaes ha higher he permiiviy, resuling greaer he capaciance. The value of capaciance can be expressed in erms physical parameers of capacior as ε A ε A 0 ε r C = = where A is he area of each plae, d is he disance beween he plaes, d d ε 0 (= ) is he permiiviy of free-space, ε r = relaive permiiviy of dielecric maerial and C is he capaciance in Farad. I is imporan o noe ha when he applied

28 volage across he capacior exceeds a cerain value he dielecric maerial breaks down and loses i insulaion propery Coninuiy condiion of capaciors To find he curren-volage relaionship of he capacior, one can ake he derivaive of boh sides of Eq.(0.39) dvc () dq() dvc () C = = i() i () = C (0.40) d d d The volage-curren relaion can also be represened by anoher form as vc() = i() d vc( 0) C where v ( ) c 0 is volage across he capacior a ime 0. I can 0 be seen ha when he volage across a capacior is no changing wih ime, or, in oher words, he capacior is fully charged and he curren hrough he capacior is zero (see Eq.0.40). This means ha he capacior resembles as an open circui and blocks he flow of curren hrough he capacior. Equaion (0.40) shows ha an insananeous ( Δ= 0 ) change in capaciance volage mus be accompanied by an infinie curren ha requiring an infinie power source. In pracice, his siuaion will no occur in any circuis conaining energy soring elemens. Thus, he volage across he capacior (or elecric charge q() ) canno change insananeously (.., ie Δ = 0), ha is we canno have any disconinuiy in volage across he capacior. emark-5 (i) The volage across and charge on a capacior canno change insananeously (i.e. v c (0 ) jus righ before he change of volage = v c (0 ) jus righ afer he change of volage). However, curren hrough a capacior can change abruply. (ii) The capacior acs as an open circui (i.e., when he capacior is fully charged) when volage across he capacior does no change (consan). (iii) These properies of capacior are imporan since hey will be used o deermine boundary condiions Sudy of dc ransiens and seady sae response of a series -C circui. Ideal and real capaciors: An ideal capacior has an infinie dielecric resisance and plaes (made of meals) ha have zero resisance. However, an ideal capacior does no exis as all dielecrics have some leakage curren and all capacior plaes have some resisance. A capacior s leakage resisance is a measure of how much charge (curren) i will allow o leak hrough he dielecric medium. Ideally, a charged capacior is no supposed o allow leaking any curren hrough he dielecric medium and also assumed no o dissipae any power loss in capacior plaes resisance. Under his siuaion, he model as shown in fig. 0.6(a) represens he ideal capacior. However, all real or pracical capacior leaks curren o some exend due o leakage resisance of dielecric medium. This leakage resisance can be visualized as a resisance conneced in parallel

29 wih he capacior and power loss in capacior plaes can be realized wih a resisance conneced in series wih capacior. The model of a real capacior is shown in fig. 0.6(b). In presen discussion, an ideal capacior is considered o sudy he behavior of dc ransiens in Ccircui Charging of a capacior or Growh of a capacior volage in dc circuis e us consider a simple series Ccircui shown in fig. 0.7(a) is conneced hrough a swich S o a consan volage source. V S

30 The swich S is closed a ime = 0 (see fig. 0.7(a)). I is assumed ha he capacior is iniially charged wih a volage v (0) c = v0 and he curren flowing hrough he circui a any insan of ime afer closing he swich is i (). The KV equaion around he loop can be wrien as dvc () VS = i() vc() VS = C v c () (0.4) d The soluion of he above firs-order differenial equaion (0.4) due o forcing funcion V is given by s v () = v () (naural response/ransien response) v () (seady-sae response) c cn = A e α A (0.4) The consans A, and A are compued using he iniial and boundary condiions. The value of α is obained from he characerisic equaion given by (see in deail in Appendix) Cα = 0 α = C Eq. (0.4) is hen rewrien as C v () = A e A c A seady sae, he volage across he capacior is v ( ) = v = A which saisfy he original differenial equaion (0.4). i.e., dvcf da V = S C vcf C A A V d d = S Using he iniial condiion (a = 0 ) in equaion (0.43), we ge cf c c f (0.43)

31 0 C v (0) = v = A e A A = v A= v V c The values of A, S A, and Eq. (0.43) ogeher will give us he final expression for capacior volage as C C C vc() = ( v0 VS) e VS vc() = VS e v0 e (0.44) Thus, v0 < 0 v () c = C C vc() = VS e v0 e > 0 esponse of capacior volage wih ime is shown in fig Special Case: Assume iniial volage across he capacior a ime = 0 is zero i.e., vc (0) = v0 = 0. The volage expression for capacior a any insan of ime can be wrien from Eq.(0.44) wih vc (0) = v0 = 0. Volage across he capaciance Volage across he resisance vc() = V S e C C (0.45) v () = V v () = V e (0.46) S c S v C Charging curren hrough he capacior () V i= = S e (0.47) Charge accumulaed on eiher plae of capacior a any insan of ime is given by C C q () = Cvc() = CVS e = Q e (0.48) where Q is he final charge accumulaed in he capacior a seady sae ( i.e., ). Once he volage across he capacior ( ) is known, he oher quaniies (like, vc v (), i (), and q () ) can easily be compued using he above expressions. Fig. 0.9(a) shows growh of capacior volage vc ( ) for differen choices of circui parameers (assumed ha he capacior is iniially no charged). A skech for q () and i() is shown in fig. 0.9(b).

32

33 Following he definiion given in secion.0.3., ime consan of each of he exponenial expressions described in Eqs o 0.48 may be found as τ = C (for C circui) Discharging of a capacior or Fall of a capacior volage in dc circuis Fig. 0.7(b) shows ha he swich S is closed a posiion for sufficienly long ime and he circui has reached in seady-sae condiion. A = 0 he swich S is opened and kep in posiion and remains here. Our job is o find he expressions for (i) volage across he capacior ( v c ) (ii) volage across he resisance ( v )(iii) curren (()) i hrough he capacior (discharging curren) (iv) discharge of charge ( q( )) hrough he circui.

34 Soluion: For < 0, he swich S in posiion. The capacior acs like an open circui o dc, bu he volage across he capacior is same as he supply volage V S. Since, he capacior volage canno change insananeously, his implies ha vc(0 ) = vc(0 ) =V S When he swich is closed in posiion, he curren i () will flow hrough he circui unil capacior is compleely discharged hrough he resisance. In oher words, he discharging cycle will sar a = 0. Now applying KV around he loop, we ge dvc () C vc () = 0 (0.49) d The soluion of inpu free differenial equaion (0.49) is given by v () = A e α (0.50) c where he value of α is obained from he characerisic equaion and i is equal o α =. The consan A is obained using he iniial condiion of he circui in C Eq.(0.50). Noe, a = 0 ( when he swich is jus closed in posiion ) he volage across he capacior vc( ) = V S. Using his condiion in Eq.(0.50), we ge 0 C vc(0) = VS = Ae A = V S Now he following expressions are wrien as Volage across he capaciance Volage across he resisance C v () = V e (0.5) c S C v () = v () = V e (0.5) c S v C Charging curren hrough he capacior () V i= = S e (0.53) An inspecion of he above exponenial erms of equaions from (0.5) o (0.53) reveals ha he ime consan of C circui is given by τ = C (sec.) This means ha a ime = τ, he capacior s volage v c drops o 36.8% of is iniial value (see fig. 0.0(a)). For all pracical purposes, he dc ransien is considered o end afer a ime span of 5τ. A such ime seady sae condiion is said o be reached. Plos of above equaions as a funcion of ime are depiced in fig. 0.0(a) and fig. 0.0(b) respecively.

35

36 .0.5 Energy sored in a capacior The ideal capacior does no dissipae any of he energy supplied by he source. I sores energy in he form of an elecric field beween he conducing plaes. e us consider a volage source VS is conneced o a series C circui and i is assumed ha he capacior is iniially uncharged. The capacior volage ( vc ( )) and curren ( ic ( )) waveforms during he charging period are shown in fig.0. (see he expressions (0.45) and (0.47)) and insananeous power ( pc( ) = vc( ) i( ) ) supplied o he capacior is also shown in he same figure.

37 e us consider he insananeous power supplied o he capacior is given by pc( ) = vc( ) i( ) (0.54) Now, he energy supplied o he capacior in d second is given by dvc() Δ w= pc() d= vc() C d= Cvc() dvc () (0.55) d Toal energy supplied o he capacior in seconds is expressed as vc () = v q c c c vc (0) = 0 ( ) w() = C v () dv () = C v = ( Joules) (0.56) C (Noe iniial volage across capacior is zero and q() is he charge accumulaed on each plae a a ime ). When he capacior is fully charged, is erminal volage is equal o he source volage. The amoun of energy sored in capacior in he form of elecric field is given by Q W = C VS = ( Joules ) (0.57) C V S

38 where Q is he final charge accumulaed on each plae of he capacior a seady sae ( i.e., ) i.e., when he capacior is fully charged. Example:.0.6 The swich S shown in fig..0. is kep open for a long ime and hen i is closed a ime = 0. Find (i) (0 ) (ii) (0 ) (iii) (0 ) (iv) (0 ) (v) = 0 v c dvc () (vi) find he ime consans of he circui before and afer he swich is closed dy (iv) vc ( ) v c i c i c Soluion: As we know he volage across he capacior vc ( ) canno change insananeously due o he principle of conservaion of charge. Therefore, he volage across he capacior jus before he swich is closed (0 ) = volage across he capacior jus afer he swich is closed v c (0 ) = 40 V (noe he erminal a is posiively charged. I may be noed ha he capacior curren before he swich S is closed is i (0 c ) = 0 A. On he oher hand, a = 0, he curren hrough 0Ω resisor is zero bu he curren hrough capacior can be compued as vc (0) 40 i (0 c ) = = = 6.66 A (noe, volage across he capacior canno change 6 6 insananeously a insan of swiching). The rae of change of capacior volage a ime = 0 is expressed as v c dvc( ) dvc(0 ) ic(0 ) 6.66 C = ic (0) = = =.665 vol/ sec. d d C 4 = 0 Time consan of he circui before he swich was closed = τ = C= 0 4 = 40sec. Time 0 6 consan of he circui afer he swich is closed is τ = Th C= 4= 5sec. (replace 0 6 he par of he circui han conains only independen sources and resisive elemens by an

39 equivalen Thevenin s volage source. In his case, we need only o find he Thevenin resisance Th ). Noe: When he swich is kep in closed posiion, iniially he capacior will be in discharge sae and subsequenly is volage will decrease wih he increase in ime. Finally, a seady sae he capacior is charged wih a volage 40 vc ( = ) = 6 = 5V (heoreically, ime required o reach he capacior volage a 0 6 seady value is 5τ = 5 5= 75 sec. ). Example:.0.7 The circui shown in Fig.0.3 has been esablished for a long ime. The swich is closed a ime = 0. Find he curren (i) i(0 ), i(0 dvde ), i3 (0 ), and (ii) a seady sae he volage across he capaciors, d = 0 i ( ), i ( ) and i ( ). 3 Soluion: (i) A = 0 no curren flowing hrough he circui, so he volage a poins b and d are boh equal o 50 vol. When he swich S closes he capacior volage remains consan and does no change is volage insananeously. The curren i (0 ) hrough a bbranch mus hen equal o zero, since volage a erminal b is equal o v b (0 ) = 50 vol., curren hrough b c is also zero. One can immediaely find

40 50 ou he curren hrough c e equal o i (0 ) = = A. Appling KC a poin c, 50 i (0 ) 3 = A which is he only curren flow a = 0 around he loop d c e d. Noe he capacior across d e branch acs as a volage source, he change of capacior dvde volage = i 6 3 (0 ) = kvol/ sec. d = 0 (ii) a seady sae he volage across each capacior is given 50 = 50 = vol. 50 A seady sae curren delivered by he source o he differen branches are given by 50 i ( ) = = A; i ( ) = A and i 3 ( ) = 0A 50 Example:.0.8 The circui shown in fig. 0.4(a) is swiched on a ime = 0. How long i akes for he capacior o aain 70 % of is final volage? Assume he capacior is iniially no charged. Find also he ime consan ( τ ) of he circui afer he swich is closed. The circui conaining only resisive elemens and independen curren source (i.e., nonransien par of he circui) is convered o an equivalen volage source which is shown in fig.0.4(b).

41 Fig.0.4(c) shows he capaciorc is conneced across he Thevenin s volage erminals a and b in series wih Thevenin s resisance Th. The parameers of Thevenin s volage source are compued below: VTh 00 = 00 = 50V and Th = = 75Ω

42 Using KV around he closed pah, one can find he curren hrough he capacior and hence, he volage across he capacior. dvc () 50 = 75 i( ) vc( ) = 0.75 v c ( ) (0.58) d The soluion of he differenial equaion is given by C v () = A e A c (0.59) Using he iniial and boundary condiions of he circui, we obain he final expression of volage across he capacior vc ( ) as.33 ( e ) v () = 50 (0.60) c e is he ime required o reach he capacior volage sae) volage = 35= 50 e = 0.9sec. ( ) 70% of is final (i.e., seady Example:.0.9 The swich S of he circui shown in fig.0.5(a) is closed a posiion a = 0. Find volage vc( ) and curren ic ( ) expressions for 0. Assume ha he capacior is iniially fully uncharged (i.e.,. v (0) = 0 ). c (i) find he mahemaical expressions for vc ( ) and ic ( ) if he swich S is hrown ino posiion a = τ (sec.) of he charging phase. (ii) plo he waveforms obained in pars (i) o (ii) on he same ime axis for he volage vc( ) and he curren ic ( ) using he idenified polariy of volage and curren direcions.

43 Soluion: (i) The curren source is convered o an equivalen volage source and i is redrawn in fig.0.5(b) when he swich S is in posiion. KV around he closed pah: 40= 0 i () v(), where i () is in ma. c dvc () 40 = 0 C vc ( ) d (0.6) The volage expression across he capacior using he iniial and boundary condiions of he circui, one can wrie vc( ) as 3 6 C vc ( ) = 40 e = 40 e = 40( e ) (0.6) 0 40 vc ( ) 40 e 0 ic () = = = 4 e ( in ma) 0 0 (0.63) Noe ha he ime consan of he circui in par (i) is τ = C = 00 msec. (ii) The swich S is hrown ino posiion a = τ = 0.sec. and he corresponding circui diagram is shown in fig.0.5(c).

44 Noe, a ime = τ = 0.sec., he capacior is charged wih a volage = vc ( ) 0 0. ( τ = 0.) = 40 e = = 5.8V and a he same ime ( = τ = 0.sec. ) 0 he curren in capacior is 4 e = ( in ma) =. Considering he fig.0.5(c), one can wrie KV around he closed pah dvc () vc() C eq = 0 (0.64) d where = 4 6= 0kΩ and he capacior is now in discharging phase. eq The soluion of Eq.(0.64) can be found using he iniial and final volage of he capacior (iniial volage v ( = τ = 0.) = 5.8 V, v ( τ = ) = 0 V ) and i is given by c c ( τ ) eq C 0 ( τ ) v () = v ( τ = 0.) e = 5.8 e (0.65) c c Discharging curren expression is given by (noe, curren direcion is jus opposie o he assigned direcion and i is aken ino accoun wih a ve sign) 0 ( τ ) vc ( ) 5.8 e 0( τ ) ic () = = =.58 e ( in ma) (0.66) 0 eq (Noe, he above wo expressions are valid only for τ ) The circui responses for charging and discharging phases in (i) and (ii) are shown in fig.0.5 (d).

45 emark-6 Noe ha he curren hrough he capacior (see fig. 0.5(d)) can change insananeously like he volage across he inducor. Appendix-A.0.A Soluion of n h order linear ime invarian differenial equaion excied by forcing funcion. e us consider a linear ime invarian circui having several energy source elemens is described by he following dynamic equaion. n n n d x d x d x dx n n n 0 () n n iii = n n a a a a a x f d d d d where a, a, a, iii a, a 3 n n (0.A) are consan coefficiens associaed in he differenial equaion and hey are dependen on circui parameers (like,,, and C for elecric circui) bu independen of ime, f () is he forcing or driving funcion and x() is he soluion of differenial equaion or response of he sysem. We shall discussion he

46 soluion of differenial equaion resriced o second order differenial, say n = equaion (0.A). d x dx 0 a a a x= f() (0.A) d d The soluion of his differenial equaion provides he response of circui and i is given by x () = x () x () (0.A3) n f where xn ( ) is he naural response of circui, obained by seing f() = 0, and x f () is he forced response ha saisfies he original differenial equaion (0.A). By seing f() = 0 in equaion (0.A), as given in equaion (0.A4), he force free equaion is obained. d x dx 0 0 a a a x= (Homogeneous equaion) (0.A4) d d The soluion of such differenial equaion (or homogeneous equaion) is known as naural soluion or complemenary soluion or ransien soluion and i is denoed by x (). To ge he naural soluion xn ( ) of equaion (0.A4) he following seps are considered. e us use he following operaors d d d d α ; d d d d = = = α in equaion (0.A4) and resuls an equaion given by ( α α ) a a a x 0 = 0 Since x 0, he above equaion can be wrien as ( aα aα a0 = 0 ) (0.A5) which is known as characerisic equaion for a circui whose force free equaion is Eq.(0.A4). The naural or ransien soluion of Eq.(0.A4) is expressed by he exponenial erms as given below. α x () = A e A e (0.A6) n α where α and α are he roos of characerisic equaion (0.A5). The roos of second order characerisic equaion wih real coefficiens is eiher real or complex occur in conjugae pairs. The consans A and A are evaluaed from iniial or boundary condiions of circui. The principles of coninuiy of inducance curren and capaciance volage are used o esablish he required boundary condiions. n in

47 If xn ( ) is he naural or ransien soluion of unforced (or homogeneous) equaion differenial, i mus saisfy is own differenial equaion d xn dxn 0 n 0 a a a x = (0.A7) d d Furher, if x() = x () x () is he complee soluion of given differenial Eq.(0.A), i mus saisfy is own equaion n ( n f ) ( n f ) d x x d x x f ( n f ) a a a0 x x = f() (0.A8) d d Using he equaion (0.A7) in Eq.(0.A8), we ge d xf dxf 0 a a a x () f = f (0.A9) d d The above equaion implies ha x () is he forced soluion or seady sae soluion of f second order differenial equaion (0.A). Seady sae soluion of some common forcing funcions is lised in Table (assume a > 0, a > 0and a0 > 0). Table: Seady sae soluion x f () for any order differenial equaion excied by some common forcing funcion. Type of forcing funcion f () (inpu) Seady sae soluion x f () (oupu) f () = K (consan) x f () = A (consan) f () = K x () = A B f () = K a f () = Ke () f x f () = A B C a x f = Ae f () = sinb if () = Asinb Bcosb f () = cosb i () = Asinb Bcosb a f () e sinb = i () f = e ( Asinb Bcosb) = a i () = e a ( Asinb Bcosb) f () e cosb Coefficiens involve in he seady sae soluion can be found ou by using he boundary condiions of he circui. emark-7 (i) Eq. (0.A) is he differenial equaion descripion of a linear circui, superposiion may be used o find he complee soluion of a forcing funcion which is sum of naural and seady sae responses. (ii) Eq.(0.A6) is he naural soluion of force-free linear differenial equaion. Noe ha he consans α and α are he roos of he characerisic equaion (0.A5) and hey are enirely depending on he circui parameers. The roos of he characerisic equaion may be classified as f f

48 Case-: eal or Complex bu disinc The naural soluion of homogeneous equaion (0.A4) is given as x () A e α α = A e n Case-: oos are repeaed (i.e. α = α= α or mulipliciy of roos of order ) The naural soluion of homogeneous equaion (0.A4) is given as x () = β e α β n 0 Using iniial and final condiions of he circui, β 0 and β consans are compued. More discussions on hese issues can be seen in esson Tes your undersanding ( Marks: 70) T.0. Inducor ends o block curren bu pass curren. T.0. The basic fundamenal principle ha explains he acion of an inducor is known as law. T.0.3 Exponenial waveforms sar and finish T.0.4 A ransien approximaely always has a duraion of ime consans. T.0.5 Afer he firs ime consan, a ransien goes hrough % of is seady sae value. T hrough inducor canno change bu across he inducor can insananeously a he swiching phase. T.0.7 A simple series circui is excied wih a consan volage source, he speed of response depends on and of he circui. T.0.8 The energy sored in an inducor in he form of T.0.9 In a firs order circui if he resisor value is doubled, he ime consan is halved for an circui. T.0.0 An inducor acs as for a curren hrough i. T.0. Once a capacior has been charged up, i is able o ac like a T.0. If he spacing beween he plaes is doubled, he capaciance value is T.0.3 Afer a capacior is fully charged in a dc circui, i dc curren. T.0.4 The ime rae of change of capacior volage is represened by he angen line o he vc( ) -versus- curve. T.0.5 Immediaely afer a swich has been hrown, a capacior s mus mainain he same value ha excied jus before he swiching insan. T.0.6 A he insan of swiching, curren hrough he capacior insananeously.

49 T.0.7 A seady sae condiion in a dc circui, he capacior acs as an circui. T.0.8 A firs order circui wih a single resisor, if he resisor is doubled in value, he ime consan is also for an Ccircui. T.0.9 Time consan of a firs order sysem is he measure of response of he circui. T.0.0 The energy sored in a capacior in he form of [ 0] T.0. For he circui of fig.0.6, find (i) i (0 ), i (0 ) ( ), ( i = i = ) (iv) v (0 ), v ( = ). ab ab (ii) i(0 ), i (0 ) (iii) (Ans. ( i) 0, A ( ii).333 A, A ( iii) A, 0 A ( iv).33 V, 0 V ) [8] T.0. For he circui shown in Fig.0., he swich S has been opened for a long ime and hen closes a =0. Find, (i) v ab (0 ) (ii) i x (0 ) (iii) i x (0 ) (iv) v ab (0 ) (v) ix ( = ) (vi) vab ( = ) (vii) ix () for > 0 (Ans. [0] T.0.3 In he circui shown in fig.0.7, he swich was iniially open and no curren was flowing in inducor ( ). The swich was closed a = 0 and han re opened a = τ di () sec. A = 0, was 50 A / s. d

50 Find, (i) The value of (ii) Find he curren i () and volage vbc ( ) expressions for 0. Assume, no curren was flowing hrough he inducor a = 0 (i.e.,. i (0) 0 = ). (iii) Find he mahemaical expressions for i () and vbc ( ) if he swich S is reopened a = τ (sec.). (iv) Plo he waveforms obained in pars (ii) o (iii) on he same ime axis (ime in ms.) for he curren i () and he volage vbc ( ) considering he indicaed curren direcions and idenified polariy of volage across he b c erminals. i 40 ( ) =.5 e amp., v 40 ( ) = 5 e (Ans. (i) 0.3 H (ii) ( ) 40( ) τ (iii) i ( ) =.08 e, bc ( τ ) 40 v ( ) =.96 e, for τ.) [0] bc T.0.4 A seady sae condiion, find he values of I, I, I 3, I 4, I 5, V and for he circui shown in fig.0.8. V (Ans I = I = I = A, I = I = 0, V = 40V and V = 30V ) [6]

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