7. Capacitors and Inductors

Transcription

1 7. Capaciors and Inducors 7. The Capacior The ideal capacior is a passive elemen wih circui symbol The curren-volage relaion is i=c dv where v and i saisfy he convenions for a passive elemen The capacior is an open circui o dc The capaciance C is measured in farad (F) A capacior is formed from wo conducing plaes separaed by a hin insulaing layer he capaciance is given by C=εA/d, where ε is he permiiviy of he insulaing maerial beween he plaes, Capaciors can be bulky and ypical values range from pf o μf i A d

2 Deermine he curren i flowing hrough he capacior for he wo volage waveforms of if C = F. The capacior is an open circui o dc

3 Inegral Volage-Curren Relaionships The capacior volage may be expressed as dv= i C Then inegrae beween he imes 0 and and beween he corresponding volages v(0) and v(): v = i d v 0 C 0 When he capacior is iniially uncharged (a 0=, v( )=0) v = i d C Since i=dq/, we have v = q C where q() and v() represen insananeous values of he charge on eiher plae and he volage beween he plaes, respecively 3

4 4 Find he capacior volage ha is associaed wih he shown curren. The value of he capaciance is 5 μf v() = v = 0 d v C 0 ms v = 0.0d v 0 C 0

5 5 Energy Sorage The power delivered o he capacior p=vi=cv dv The energy sored in a capacior is p d =C 0 v v d v= C {v v 0 } v 0 Then wc w C 0 = C {v v 0 } where wc(0) and v(0) are he sored energy and he volage a 0 If v(0)=0, we have wc = C v

6 Find he maximum energy sored in he capacior of and he energy dissipaed in he resisor over he inerval 0 < < 0.5 s. The energy sored in he capacior is wc = C v =0. sin J The power dissipaed by he resisor ir = v =0 4 sin A R 4 6 p R=iR R= 0 0 sin W The energy dissiped in he resisor beween 0 < < 0.5 s wr = 0 pr = 0 0 sin J 6

7 7 7. The Inducor The ideal inducor is a passive elemen wih circui symbol The curren-volage relaion is v=l di where v and i saisfy he convenions for a passive elemen The uni of he inducance L is measured in Henry (H) The inducor is a shor circui o dc A physical inducor is consruced by winding a lengh of wire ino a coil, he inducance is given by L=μNA/l, where μ is he permeabiliy of he maerial inside he helix, N is he number of urns, A is he cross-secional area, l is he lengh Inducors can be bulky and ypical values range from μh o H

8 8 Given he waveform of he curren in a 3 H inducor, deermine he inducor volage and skech i. i = i = i = v =3 d i

9 Inegral Volage-Curren Relaionships The inducor curren may be expressed as di= v L Then inegrae beween he imes 0 and and beween he corresponding curren i(0) and i(): i = v d i 0 L 0 When he inducor is iniially uncharged (a 0=, i( )=0) i = v d L Energy Sorage The absorbed power delivered by he inducor p=vi=li di The energy is sored in he magneic field around he coil i wl wl 0 = L {i i 0 } wl = L i p d =L i d i 0 i 0 9

10 Find he maximum energy sored in he inducor of and he energy dissipaed in he resisor over he inerval 0 < < 6 s. The energy sored in he inducor is wl = L i =6sin J 6 The power dissipaed by he resisor p R=iR R=4.4sin W 6 The energy dissiped in he resisor beween 0 < < 6 s 6 6 wr = 0 pr = 0 4.4sin J 6 0

11 7.3 Inducance and Capaciance Combinaions Inducors in Series Apply KVL vs=v v vn di di di L LN di = L L LN =L The equivalen inducance vs=l eq di Leq =L L LN

12 Inducors in Parallel Apply KCL The equivalen inducance is=i i in = is = v v v L L LN = L L LN v Leq = v L eq L L LN

13 3 Capaciors in Series Apply KVL The equivalen capaciance vs=v v vn = v s= i i i C C CN = C C CN i C eq = i C eq C C CN

14 4 Capaciors in Parallel Apply KCL is=i i in dv dv dv C C N dv = C C C N =C The equivalen capaciance is=c eq dv C eq =C C C N

15 5 Two-Elemen Shorcus Two resisors in parallel: R eq = R R R R Two inducors in parallel: Leq = L L L L Two capaciors in parallel: C eq =C C Two resisors in series: R eq =R R Two inducors in series: Leq =L L Two capaciors in series: C C C eq = C C

16 Simplify he shown circui using series and parallel combinaions. The capaciors 6 μf and 3 μf are in series C eq= = F Then, Ceq is in parallel wih μf C eq=c eq F=3 F The inducors H and 3 H are in parallel Leq= 3 =. H 3 Then, Leq is in series wih 0.8 H Leq=Leq 0.8= H 6

17 7.4 Consequences of Lineariy Now, we will modify he previous definiion of a linear circui o include he capaciors and inducors Kirchhoff s laws, nodal analysis, and mesh analysis can be applied o a circui wih resisors, inducors, and capaciors Generally speaking, he mesh and nodal analysis of RL and RC circuis lead o consan coefficien linear inegrodifferenial equaions A he middle node, apply KCL il ir ic =0 v v dv v v d il 0 C =0 L s R A he righ-hand node 0 ir is=ic v v d v vs is=c =0 R 7

18 8 7.5 Simple Op Amp circuis wih Capaciors The op-amp Inegraor Performing nodal analysis a he invering inpu, vs 0 ic = R f d v C vs = R Inegraing and solving for vou, we obain Cf f v d vc = R C f vs vc vc 0 = v R C f s vc f Cf f 0 f 0 f 0 Since vc =0 v ou, we have f vou = v vc 0 R C f s 0 f

19 The op-amp differeniaor Performing nodal analysis a he invering inpu, 0 vou =ic Rf vou d vs = C Rf Rearrange and solving for vou, we obain vou d vs = Rf C R Homework Assignmen 6 P7., P7., P7.3, P7.0, P7.3, P7.4, P7.5, P7., P7.8, P7.35, P7.36, P7.4, P7.53, and P7.54 9