mywbut.com Lesson 11 Study of DC transients in R-L-C Circuits

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1 mywbu.om esson Sudy of DC ransiens in R--C Ciruis

2 mywbu.om Objeives Be able o wrie differenial equaion for a d iruis onaining wo sorage elemens in presene of a resisane. To develop a horough undersanding how o find he omplee soluion of seond order differenial equaion ha arises from a simple R C irui. To undersand he meaning of he erms (i) overdamped (ii) riiallydamped, and (iii) underdamped in onex wih a seond order dynami sysem. Be able o undersand some erminologies ha are highly linked wih he performane of a seond order sysem... Inroduion In he preeding lesson, our disussion foused exensively on d iruis having resisanes wih eiher induor ( ) or apaior ( C ) (i.e., single sorage elemen) bu no boh. Dynami response of suh firs order sysem has been sudied and disussed in deail. The presene of resisane, induane, and apaiane in he d irui inrodues a leas a seond order differenial equaion or by wo simulaneous oupled linear firs order differenial equaions. We shall see in nex seion ha he omplexiy of analysis of seond order iruis inreases signifianly when ompared wih ha enounered wih firs order iruis. Iniial ondiions for he irui variables and heir derivaives play an imporan role and his is very ruial o analyze a seond order dynami sysem... Response of a series R--C irui due o a d volage soure Consider a series R C irui as shown in fig.., and i is exied wih a d volage soure V s. Applying KV around he losed pah for >, di() Ri() v() = V s (.) The urren hrough he apaior an be wrien as

3 mywbu.om dv () i () = C Subsiuing he urren i () expression in eq.(.) and rearranging he erms, dv() dv() C RC v () = Vs (.) The above equaion is a nd -order linear differenial equaion and he parameers assoiaed wih he differenial equaion are onsan wih ime. The omplee soluion of he above differenial equaion has wo omponens; he ransien response vn() and he seady sae response vf(). Mahemaially, one an wrie he omplee soluion as v() vn() v f () ( Ae α α = = Ae ) A (.3) Sine he sysem is linear, he naure of seady sae response is same as ha of foring funion (inpu volage) and i is given by a onsan value A. Now, he firs par vn() of he oal response is ompleely dies ou wih ime while R > and i is defined as a ransien or naural response of he sysem. The naural or ransien response (see Appendix in esson-) of seond order differenial equaion an be obained from he homogeneous equaion (i.e., from fore free sysem) ha is expressed by dv() dv() dv() R dv() C RC v () = v () = C dv() dv() R a b v () = (where a, b and = = = ) (.4) C The haraerisi equaion of he above homogeneous differenial equaion (using he d d operaor α =, α = and v () ) is given by α R a b α = C α α R = (where a =, b = and = ) (.5) C and solving he roos of his equaion (.5) one an find he onsans α and α of he exponenial erms ha assoiaed wih ransien par of he omplee soluion (eq..3) and hey are given below. R R b b α = = a ; C a a (.6) R R b b α = = a C a a R where, b = and =. C The roos of he haraerisi equaion (.5) are lassified in hree groups depending upon he values of he parameers R,, and C of he irui. 3

4 mywbu.om R Case-A (overdamped response): When >, his implies ha he roos are C disin wih negaive real pars. Under his siuaion, he naural or ransien par of he omplee soluion is wrien as α α v () = A e A e (.7) n and eah erm of he above expression deays exponenially and ulimaely redues o zero as and i is ermed as overdamped response of inpu free sysem. A sysem ha is overdamped responds slowly o any hange in exiaion. I may be noed ha he exponenial erm A e α akes longer ime o deay is value o zero han he erm A e α. One an inrodue a faor ξ ha provides an informaion abou he speed of sysem response and i is defined by damping raio R Aual damping b ( ξ ) = = = > (.8) riial damping a C R Case-B ( riially damped response): When =, his implies ha he roos C of eq.(.5) are same wih negaive real pars. Under his siuaion, he form of he naural or ransien par of he omplee soluion is wrien as R vn() = ( A A ) e α (where α = ) (.9) where he naural or ransien response is a sum of wo erms: a negaive exponenial and a negaive exponenial muliplied by a linear erm. The expression (.9) ha arises from he naural soluion of seond order differenial equaion having he roos of haraerisi equaion are same value an be verified following he proedure given below. R The roos of his haraerisi equaion (.5) are same α= α= α= when R R = = and he orresponding homogeneous equaion (.4) C C an be rewrien as dv() R dv() v ( ) = C dv() dv() or α α v ( ) = d dv() dv() or αv() α αv() = df dv () or α f = where f = α v () 4

5 mywbu.om The soluion of he above firs order differenial equaion is well known and i is given by f = Ae α dv () Using he value of f in he expression f = α v () we an ge, dv() α α dv() α d αv() Ae e e αv() A e α v () = A Inegraing he above equaion in boh sides yields, v () = A A e α n = = ( ) ( ) In fa, he erm Ae α (wih α = zero as R ) deays exponenially wih he ime and ends o. On he oher hand, he value of he erm A e α (wih α = equaion (.9) firs inreases from is zero value o a maximum value A e R R ) in a a ime = = = and hen deays wih ime, finally reahes o zero. One an α R R easily verify above saemens by adoping he onep of maximizaion problem of a single valued funion. The seond order sysem resuls he speedies response possible wihou any overshoo while he roos of haraerisi equaion (.5) of sysem having he same negaive real pars. The response of suh a seond order sysem is defined as a riially damped sysem s response. In his ase damping raio R Aual damping b ( ξ ) = = = = (.) riial damping a C R Case-C (underdamped response): When <, his implies ha he roos of C eq.(.5) are omplex onjugaes and hey are expressed as R R ; R R α = j = β jγ α= j = β jγ. The C C form of he naural or ransien par of he omplee soluion is wrien as α α ( β jγ) ( β jγ) v () = Ae Ae = Ae A e n γ (.) β = e Bos( γ ) Bsin ( γ ) where B= A A ; B= j( A A) For real sysem, he response v () n mus also be real. This is possible only if A and A onjugaes. The equaion (.) furher an be simplified in he following form: β e K β = e ( A A ) os( γ ) j( A A ) sin ( ) sin ( γ θ ) (.) 5

6 mywbu.om where β = real par of he roo, γ = omplex par of he roo, B K = B B and =. Truly speaking he value of Kand θ an be θ an B alulaed using he iniial ondiions of he irui. The sysem response exhibis osillaion around he seady sae value when he roos of haraerisi equaion are omplex and resuls an under-damped sysem s response. This osillaion will die down wih ime if he roos are wih negaive real pars. In his ase he damping raio R Aual damping b ( ξ ) = = = < (.3) riial damping a C Finally, he response of a seond order sysem when exied wih a d volage soure is presened in fig... for differen ases, i.e., (i) under-damped (ii) over-damped (iii) riially damped sysem response. 6

7 mywbu.om Example:.. The swih S was losed for a long ime as shown in fig..3. Simulaneously a =, he swih S is opened and S is losed Find ( a) i ( ); ( b) v ( ); dv ( ) () i R (); ( d) v( ); ( e) i( ) ; ( f) Soluion: When he swih S is kep in posiion for a suffiienly long ime, he irui reahes o is seady sae ondiion. A ime =, he apaior is ompleely harged and i as as a open irui. On oher hand,. he induor as as a shor irui under seady sae ondiion, he urren in induor an be found as 5 i ( ) = 6= A 5 Using he KC, one an find he urren hrough he resisor i ( R ) = 6 = 4A and subsequenly he volage aross he apaior v ( ) = 4 5 = vol. Noe a = no only he urren soure is removed, bu Ω resisor is shored or removed as well. The oninuiy properies of induor and apaior do no permi he urren hrough an induor or he volage aross he apaior o hange insananeously. Therefore, a = he urren in induor, volage aross he apaior, and he values of oher variables a = an be ompued as i( ) = i( ) = A; v( ) = v ( ) = vol. Sine he volage aross he apaior a = is vol, he same volage will appear aross he induor and he 5 Ω resisor. Tha is, v( ) = vr( ) = vol. and hene, he urren ( i R ( )) in 5Ω resisor = = 4 A. Applying KC a he boom erminal 5 7

8 mywbu.om of he apaior we obain ( ) = (4 ) = 6 A and subsequenly, i dv( ) i( ) 6 = = = 6 vol./ se. C. Example:.. The swih S is losed suffiienly long ime and hen i is opened a ime = as shown in fig..4. Deermine dv () di () dv () () i v ( ) ( ii) ( iii) i ( ), and ( iv) ( v) when R = R = Ω. 3 = = = Soluion: A = (jus before opening he swih), he apaior is fully harged and urren flowing hrough i oally bloked i.e., apaior as as an open irui). The volage aross he apaior is v( ) = 6 V = v( ) = v bd ( ) and erminal b is higher poenial han erminal d. On he oher branh, he induor as as a shor irui (i.e., volage aross he induor is zero) and he soure volage 6V will appear aross he 6 resisane R. Therefore, he urren hrough induor i( ) = = A = i( ). Noe a 3 =, ( ) = (sine he volage drop aross he resisane R = Ω = v = 6V ) v ad and v ( d ) = 6V and his implies ha v a ( ) = 6V = volage aross he induor ( noe, erminal is ve erminal and induor as as a soure of energy ). Now, he volage aross he erminals b and ( v( ) ) = v ( ) v ( ) = V. The following expressions are valid a = dv dv C = i ( ) = A = vol/ se. (noe, volage aross he apaior will = = bd 3 d ab 8

9 mywbu.om dv derease wih ime i.e., = vol /se ). We have jus alulaed he volage aross he induor a = = as di() di() 6 va ( ) = = 6V = = A/ se..5 = = dv ( ) dv ( ) di ( ) = = 3 = 35 vol /se. Now, R ( ) Example:..3 Refer o he irui in fig..5(a). Deermine, (i) di ( ) i( ), i ( ) and v( ) (assumed v () = ; i () = ) (ii) dv( ) and (iii) i( ) i ( ) and v( ) Soluion: When he swih was in off posiion i.e., < i() = i ()=, v()= and v C()= The swih S was losed in posiion a ime = and he orresponding irui is shown in fig.5 (b). (i) From oninuiy propery of induor and apaior, we an wrie he following expression for = - - i() = i()=, v()= v()= i( ) = v ( ) = v( )= i ( ) 6= vol. 6, 9

10 mywbu.om (ii) KC a poin a 8 = i ( ) i( ) i( ) A =, he above expression is wrien as 8 = i() i() i() i ( ) = 8 A We know he urren hrough he apaior i () an be expressed as dv () i()=c dv ( ) i()=c dv ( ) = 8 = vol./se.. 4 Noe he relaions dv ( ) = hange in volage drop in 6Ω resisor = hange in urren hrough 6Ω ( ) resisor 6 = 6 di di ( ) = 6 = 3 amp./se. Applying KV around he losed pah b--d-b, we ge he following expression. v () = v () v() A, = he following expression v( ) = v( ) i( ) di( ) di( ) = v() v() = = = di ( ) di = ( ) dv( ) and his implies = = v/se = =

11 mywbu.om Now, v () = Ri() also a = dv( ) di ( ) di ( ) = R = = vol / se. (iii) A = α, he irui reahed is seady sae value, he apaior will blok he flow of d urren and he induor will a as a shor irui. The urren hrough 6Ω and Ω resisors an be formed as 8 6 i( ) = = = 5.333A, i ( ) = =.667 A 8 3 ( = ) 3 vol. v Example:..4 The swih S has been losed for a suffiienly long ime and hen i is opened a = (see fig..6(a)). Find he expression for (a) v (), (b) i (), > for induor values of () i =.5 H ( ii) =.H ( iii) =.H and plo v () vs and i () vs for eah ase. Soluion: A = (before he swih is opened) he apaior as as an open irui or blok he urren hrough i bu he induor as as shor irui. Using he properies of induor and apaior, one an find he urren in induor a ime = as i( ) = i( ) = = A (noe induor as as a shor irui) and volage aross he 5 5Ω resisor = 5= vol. The apaior is fully harged wih he volage aross he 5Ω resisor and he apaior volage a = is given by v( ) = v( ) = vol. The irui is opened a ime = and he orresponding irui diagram is shown in fig..6(b). Case-: =.5 H, R = Ω and C = F e us assume he urren flowing hrough he irui is around he losed pah is i () and apply KV equaion

12 mywbu.om di() dv() d v() dv () Vs = Ri() v() Vs = RC C v () (noe, i () = C ) dv() R dv() Vs = v () (.4) C The soluion of he above differenial equaion is given by v () = v () v () (.5) n f The soluion of naural or ransien response v is obained from he fore free equaion or homogeneous equaion whih is dv() R dv() v () = C (.6) The haraerisi equaion of he above homogeneous equaion is wrien as R α α = (.7) C The roos of he haraerisi equaion are given as R R α R R = =.; α = =. C C and he roos are equal wih negaive real sign. The expression for naural response is given by v () = A A e α (where α = α= α= ) (.8) n ( ) The fored or he seady sae response vf () is he form of applied inpu volage and i is onsan A. Now he final expression for v () is ( ) α ( ) n( ) v () = A A e A = A A e A The iniial and final ondiions needed o evaluae he onsans are based on v ( ) = v ( ) = vol; i ( ) = i ( ) = A (Coninuiy propery). (.9)

13 mywbu.om A = ; v () A e A A A = = = A A= (.) dv () Forming (from eq.(.9)as dv () α α = α ( A A) e Ae = ( A A) e Ae dv () = A A A A = (.) = dv( ) dv( ) (noe, C = i( ) = i( ) = = vol/ se. ) I may be seen ha he apaior is fully harged wih he applied volage when = and he apaior bloks he urren flowing hrough i. Using = in equaion (.9) we ge, v( = ) A A= Using he value of A in equaion (.) and hen solving (.) and (.) we ge, A = ; A=. The oal soluion is ( ) ( ) v () = e = e ; dv () i () = C = ( ) e e = ( ) e (.) The irui responses (riially damped) for =.5 H are shown fig..6 () and fig..6(d). Case-: =. H, R = Ω and C = F I an be noed ha he iniial and final ondiions of he irui are all same as in ase- bu he ransien or naural response will differ. In his ase he roos of haraerisi equaion are ompued using equaion (.7), he values of roos are α=.563; α = The oal response beomes α α v () = Ae Ae A = Ae Ae A (.3) dv () α α = α Ae αae = Ae.563 Ae (.4) Using he iniial ondiions( v ( dv ( ) ) =, = vol /se.) ha obained in ase- are used in equaions (.3)-(.4) wih A= ( final seady sae ondiion) and simulaneous soluion gives A =.3; A =.3 3

14 mywbu.om The oal response is v ( ) =.3e.3e dv () i ( ) = C =.4e.4e (.5) The sysem responses (overdamped) for =. H are presened in fig..6() and fig..6 (d). Case-3: = 8. H, R = Ω and C = F Again he iniial and final ondiions will remain same and he naural response of he irui will be deided by he roos of he haraerisi equaion and hey are obained from (.7) as α= β jγ =.63 j.43; α= β jγ =.63 j.4 The expression for he oal response is β v() = vn() vf() = e Ksin( γ θ) A (.6) β (noe, he naural response v () n = e Ksin( γ θ ) is wrien from eq.(.) when roos are omplex onjugaes and deail derivaion is given here.) dv () β = Ke β sin ( γ θ) γ os( γ θ) (.7) Again he iniial ondiions ( v ( dv ( ) ) =, = vol /se.) ha obained in ase- are used in equaions (.6)-(.7) wih A= (final seady sae ondiion) and simulaneous soluion gives K= 4.3; θ = 8.98 ( deg ree ) The oal response is β 63 v ( ) = e Ksin γ θ = e 4.3sin ( ) ( ) 63 ( ) v ( ) = 4.3e sin (.8) dv () 63 i ( ) = C = e.999*os( ).6sin ( ) The sysem responses (under-damped) for = 8. H are presened in fig..6() and fig..6(d). 4

15 mywbu.om 5

16 mywbu.om Remark: One an use = and = in eq.. or eq..5 or eq..8 o verify wheher i saisfies he iniial and final ondiions ( i.e., iniial apaior volage v ( ) = vol., and he seady sae apaior volage v ( ) = vol. ) of he irui. Example:..5 The swih S in he irui of Fig..7(a) was losed in posiion suffiienly long ime and hen kep in posiion. Find (i) v( )(ii) i () for if C is (a) 9 F (b) 4 F () 8 F. Soluion: When he swih was in posiion, he seady sae urren in induor is given by i ( ) = = A, v ( ) = i ( ) R = = vol. Using he oninuiy propery of induor and apaior we ge - - i ( ) = i ( ) =, v ( ) = v ( ) = vol. The swih S is kep in posiion and orresponding irui diagram is shown in Fig..7 (b) Applying KC a he op junion poin we ge, v () i ()i ()= R 6

17 mywbu.om v () dv () C i ()= R di () d i () C. i ()= R [noe: () di () v = ] d i () di () i ( )= (.9) RC C or The roos of he haraerisis equaion of he above homogeneous equaion an obained for C= F C RC RC α = = = C RC RC α = = = 3. Case-( ξ =.6, over damped sysem) : C= F, he values of roos of haraerisi 9 equaion are given as α=.5, α = 3. The ransien or neural soluion of he homogeneous equaion is given by i()=ae Ae (.3) To deermine A and A, he following iniial ondiions are used. A = ; - i ( )=i ( )= A A (.3) = A A - di () v()=v()=v()= = = A -.5 e -3. A e (.3) = [ -.5A-3A ] = - 3A- 6A Solving equaions (.3) and (,3) we ge, A= 6.66, A= The naural response of he irui is 8 5 i = e e = 6.66e 6.66e

18 mywbu.om di 5 3 = e e v( ) = v () = e - 8e dv () i ( ) 3. e e 3.33e 33.33e = = ( ) = ( ) Case- ( ξ =.77,under damped sysem) : For equaion are α =. j.= β j γ C= F, he roos of he haraerisi 4 α =. j.= β j γ The naural response beomes β i()=k e sin( γ θ ) (.33) Where k and θ are he onsans o be evaluaed from iniial ondiion. A =, from he expression (.33) we ge, ( ) i = k sinθ = k sinθ (.34) di() β β = k β e sin( )e os( ) γ θ γ γ θ (.35) = = Using equaion (.34) and he values of β and γ in equaion (.35) we ge, = k( β snθ γ os θ) = kos θ (noe: β =, γ = and k sinθ = ) (.36) From equaion (.34 ) and (.36 ) we obain he values of θ and k as - o an θ = θ = an = 6.56 and k.36 = sinθ = The naural or ransien soluion is - o i () =.36 e sin 6.56 ( ) di() = v [ ] β () = k β sin ( γ θ) γ os ( γθ) e o o = 44.7 os ( 6.56 ) - sin ( 6.56 ) e dv () d o o i ( ) = = 44.7 { os ( 6.56 ) - sin ( 6.56 ) e 4 =.36 os( 6.56) e - 8

19 mywbu.om Case-3( ξ =, riially damped sysem) : For C= F; he roos of haraerisi 8 equaion are α= ; α = respeively. The naural soluion is given by i () = A A e α (.37) ( ) where onsans are ompued using iniial ondiions. A = ; from equaion (.37) one an wrie i ( ) = A A = di() = A αe αa e A e = α α α ( α) = α α = A A e αae = di() = v ( ) = = ( A A) A= 3 = The naural response is hen ( ) i ( ) = 3 e di () d = 3 ( ) e di () = v () = [ 6e ] dv () d i () = = ( 6) e = e 3e 8 Case-4 C= F 3 ( ξ =,over damped sysem ): For Following he proedure as given in ase- one an obain he expressions for (i) urren in induor i () (ii) volage aross he apaior v () i( ).5e =.5e di() ( ) 44.8 = v = e 4.8 e dv () d i ( ) = = 44.8 e 4.8 e =.837e.9e..3 Tes your undersanding (Marks: 8) T.. Transien response of a seond-order d nework is he sum of wo real exponenials. [] 9

20 mywbu.om T.. The omplee response of a seond order nework exied from d soures is he sum of response and response. [] T..3 Ciruis onaining wo differen lasses of energy sorage elemens an be desribed by a order differenial equaions. [] T..4 For he irui in fig..8, find he following [6] dv() dv() di() di() ( a) v( ) ( b) v( ) ( ) ( d) ( e) ( f) (Ans. ( a) 6 vol. ( b)6 vol. ( ) V / se. ( d) V / se. ( e) amp/ se. ( f) 3 amp./ se. ) T..5 In he irui of Fig..9, Find,

21 mywbu.om dvr( ) dv( ) ( a) vr( ) and v( ) ( b) and ( ) vr( ) and v( ) [8] (Assume he apaior is iniially unharged and urren hrough induor is zero). (Ans. ( a) V, V ( b) V, Vol./ Se. ( )3 V, V ) T..6 For he irui shown in fig.., he expression for urren hrough induor i ( ) = 3 e for is given by ( ) Find, ( a ) he values of, C ( b ) iniial ondiion ( ) (Ans. ( ) v () he expression for v () >. ( a) = H, C = F ( b) v( ) = V ( ) v( ) = e V. ) [8] 8 T..7 The response of a series RC irui are given by v ( ) =.3e.3e i ( ) =.8e.8e where v() and i() are apaior volage and induor urren respeively. Deermine (a) he supply volage (b) he values R,, C of he series irui. [44] (Ans. ( a) V ( b) R = Ω, =. H and C = F ) T..8 For he irui shown in Fig.., he swih S was in posiion for a long ime and hen a = i is kep in posiion.

22 mywbu.om Find, ( a) i ( ); ( b) v ( ); ( ) v ( ); ( d) i ( ); [8] R Ans. ( a) i ( ) = A ; ( b) v ( ) = 4 V; ( ) v ( ) = 4 V ( d) i ( ) = A R T..9 For he irui shown in Fig.., he swih S has been in posiion for a long ime and a = i is insananeously moved o posiion. Deermine i () for and skeh is waveform. Remarks on he sysem s response. [8] 7 7 (Ans. i () = e e amps. ) 3 3 T.. The swih S in he irui of Fig..3 is opened a = having been losed for a long ime.

23 mywbu.om Deermine (i) v( ) for (ii) how long mus he swih remain open for he volage v( ) o be less han % o is value a =? [] ( ) (Ans. (i) ( i) v ( ) = 6 4 e ( ii).75se. ) T.. For he irui shown in Fig..4, find he apaior volage v( ) and induor urren i () for all ( < and ). [] Plo he wave forms v( ) and i () for. 5 5 (Ans. v () = e sin(.5 ); i( ) = 5( os(.5 ) sin(.5 ) ) e ) T.. For he parallel irui RC shown in Fig..5, Find he response 3

24 mywbu.om of i() and v() respeively. [] (Ans. ( ) i( ) = 4 4e amps.; v( ) = 48 e vol. ) 4

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