supplementary notes - ac power.pdf

7/29/2019 Supplementary Notes - AC Power.pdf

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F4Ezl +\lMEzt+3e Syu-^-"+-1 Nat- AC n.3--{n .-i (""'-",C"G(UT CfiEG FdlSG

A 3*-it^so.zfJ u'AFae-: vCt2= Vonc^C*+e)Ar*.**'( :tLJ O ,i .I .[-2*r,-) t** = -* k T.

v&): V,^ oa Cuv +e)Vr.*^ c,,t (wt + e ) - V,^

^* tn^-n* , * c*+ e) a too1 t'*o* t e s a.+ =-e t tra.F an

'r- LtT5,'...


2/7

Vn-. ), J* F--^.r, d. S.^- -r4.h; v--L* . U.{^ g^. (wv^!i,.,*- b.^n.^r. , + t -."1 t" cJcrJ*J. St*-+t-F 4 v"l*.-g* t Yl"*I t. a\ }^e4 *-'+a.^-44

VCt ) = V* cz's C uL + g )

(-) 2-

V n^,

vi*o(*? +e) ,{en^^s:J+ I!_t Oea


3/7

ME2143/]VIE2143E AC Sienals and Power: ExamplesExample: Power Delivered to a Resistance by a Sinusoidal SourceSuppose thnt a voltage given by v(/) - 100 ecs(100nr) V is applied to a 50-Q resis-tance. Sketch v(t) to scale versus time. Find the rms value of the voltage and theaverage power delivered to the resisJance. Find the power as a function of time and

T:llf:20 ms.p(r) (V)

100

The peak value of the volt*ge is Vn * 1{ifi Y. Thus. the rrns value is Y'*'Vr,r/&:7A.77V. Then, the average pcwer isp-.* : vil' - {7a'7D2 : 1oo wavs R 50

The power as a lunction oJ time is given by

sketch to scale.From the given v(t) equation,ar : 1002u, andf: atl2n : 50 }Jz,

P.,,,- = 100

v(t): Vu,.t'E(*t+e)

t(rns)

-r 00

7t(t1 :+ :IIb#Sg, = 2oocos2(1oozr) w(La- f*-fua-*'*. k-t{t"-*'*'CI ***\ luo VItk

,;(ms)A.G UC.ll- Ttuw'a-tt'

2t) i,tg '---7/lmp(r) (w)

{oo Q,


4/7

Ll..t*1 f L*"-= +, A-SJ S.;t'*Y" c&s

Vr :- V. -t Ut"+ (t> {-}": zo Lqs"

= zo co.c-vs*)'rj Lo *;^('-ts "J

Lo Z-Y5 "

X Lozs (rs" ) ?-Vu- ( c'4 o

z (+" tct",o (-Y 5') "

V* cea (tc 'oiU,r.^

- (q"1yJVv*c-t e -J Uu'^ s^l $

r@


5/7

C i.,,,, '* \ o-,- u; ,f .o-r"'u'Vr_/>+'LLc )ut =\ U =71 c L _L

C*.-*. ,t. .L!1 ,n..t..#t'.te+- fl-- /'r . Lr; = T,* s^l

C".t + Sj{'1"-..-I( . .t tect )L- i*d{ \-

r T co:('.--t* + e')t; L- btutr.- c't )V-Ct) '=

? [,*o,, ,k e.* '*t*T-, = 7'^i/, : '' LT'- ze :

#.*;.-.-*.* (n{, 1l'- ""G'-Y)hu.,e- .,J-a +=-u t6.t. .J

*^J " -,{"(12,/p -.?. "


6/7

{t',-/ { 11 L\r///\/'

S(i- c,*E-.

sctll*-( =.5 .^"-{*

4,. {:1..-ua1-Laaor t.. '*}*r-**tal qv(+) = \r* c*. t(.rf .-F f )

flv-s"wP->+L- z-q0

@ \,&) - \/ u- c,- (.rL + e )1,\/v t, rX\9.A l, ,- 'i lou ,; yo[,a-t."'

l-


7/7

Example: Analysis of a Wye-Wye system with 0.2H inductance and 50 ohmsresistanceA balanced positive-sequence wye-connected 60 -Hz three-phase source hasline-to-line voltage of Vr : 1000 V. This source is connected to a balancedwye-connected load. Each phase of the load consists of a0.2-H inductance inseries with a 100-ohms resistance. Find the line-to-neural voltages, the linecurrents and the power delivered to the load. Assume that the phase of Vun iszero' .--"VL = fT Vyntt'?u.t''& d! r'"-- {-l* \ - ,.,.!..-t, " !rr*r{-$e- ' -- !a"r i*.]t" -,*;,.J q 1 t{aqd?&s line-t*-n*xfr*l v*l**g* is 1S00 / r,6 = 577,4 V, trll* phuxo arg*e w*s*px*1{iu,d in *hs p:.*b.**n'r sf*lemsrt" $* r** wiJl *ssxxs th*? the pk*se ofVsruisa&r*. ?hexwe**v*, 'ld**u Vu,n ,ve* i \i^*-to*'u-"-"-[ *t'"g*J .^*.f*t--rc.,*^ce tY**SIT,Sc0' U*,*3??"4x.-1t0" {n*S7?.'{El?il" L["^"b. L )?ha niruuit {ur th* s phuse is Eh*wn hel*ry, {}ilo *xT **nsidur * x**tnsl*p*rx***i*N ** axisl i:t * **lqnqqd Y-Y *qnnuuti'sn &ve* if *xu is **tplryui *ul ly pru*e,nt.)

fun * j#S--n*+iuL= +i7f.faJLgdM-.-v, , I

I : t*r .-q;. {.f??.sg

Yh* **phas*, 1ix* +xrroxt {*? Von 577.4t".Qre.{ * 1* 100;J?5AOr!* 4.610/* -37.A?' !u"cr a f :q z.(

j?qry*rL{- z-tst'Tli:e *urr*::?x f ur ph*sar * snd s dra. 1f'rs s*mo e.usp? fur pi'r*se.fr* * 4,*rfid" #jS' 3*r * 4.$1"0dq4j$-ar-:J.r)--r2oo t-)-i-1 .b)o+12.ron - *1&.*${#} - ,f,8#@{33t$r.#4"} * 3.Iss k}rY

supplementary notes - ac power.pdf

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