Problem 3-25

(a)

(b) For the prlmary (or secondary) ph.se (or llne) cufientl

., =+ f: *4,e, de = !Ln^+""'!!

4 =! f:''naeve - !L"n)'nicn=ltsin) ana +^ = tan-r (aJb") = t'nJ(cot n'y'2)

(c) Cr = 2l3un and er - 0L = crl',12 = '12 IJr, and I' = l"l(2/3)

f--;-pF = t/t , = 3/r = 0.95a9 .nd HF-Jotnt)z'r - 0 3108I

Pmblem 3-19

(a) Wlth q = 6, Eq, (3.39) gtves the output vottage as

" rut =o.esoe r-1, * t,*1ea,ry-fr "os1rz,x1..o]rhe lo.d tmpecfance, 2 =xn11*q=tFlSrlzen

and 0n=wr-l1naq1X1

and the load current ts

tL(J - Idc-

where t. -U-09549Y.,R

(b) Vn = 170 V, f - 60 Hz, R = 2OO at o = 2n t = 377 rad/sThe rms value of the rtpple current Is

': = #'.-r!ffi (*)'. #ffi,,i (#l.'Conslderlng only the lowest oder harmontc (n = 6) and neglecting othe6,

| _ 0.9s49r- (2\,.-J'FffiIxJuslng the value of Ic. and after stmp flcaHon, the rtpple factor ts

IProblem 3-2j (a)

$JY (tFor the prlii-ifilie cfrreni'--",

= ) t' h**,eve - t ff" fr cos6eve=-4-"n4

Problem 3-21

(a)

( . )\9J

9d-1

(b) For the primary (or srrpply) current: Fromcu'ent rs

Eq (3'23), the prlmary

, ,,., _

rr, J'i"",r * 'i" ta,r *1!l9+.o1t " t L I 3 5 I

rr = 4u(n./2)The rms current fs If = L PF = lr/1. = 2'l2ln = o'9 and HF =

0.4834.(c) For the rectlfler InPut (or secondary) current:

" =!( t .a,eve=o

. lbn=+fi I asin(nqde = -3-(1-ccr, ,e)

9" = tan{ (a,,/b") = oc,, = ./(a"'? + b^'?) and It = ctl"lz = 'l2rhand I3 = I!/Jz

PF = r1ll. = 2la = 0.6366 and ItF - ,fi;'J- = '.rtt

Problem 3-17

' '' glng t|me of capacltor' For a slngle'! Let t1 rnd t2 be the charglng and 06cndrphase full-wave rectlfler, the period of output voltage

ls T/2' where T ls the

perlod of the Input voltage and the supPly frequency ls f = 1/"T'

t1 + t, = T/2 If h>>tr whlch ls norm'lly the case' t! 'T/2

Durlng dlscharglng of the caprcttor, the c6pacltor dlscharges exponntlally

.

and the outpul (or capacltor) voltage lsk\=Y e-t l RC

.

where Vm ls the peak value ofsupply voltage,The peak-to-peak rlpple voltage ls

v =v t t=t . \ -v t t -r^\=v -y "- '21RC

=y t t ' "nzlRC

1

slnce, .". 1- x, vr = V6 (1 - 1+ t/Rc) = Vm txlRc = Vh/(2fRC)Thus, the rms value of tfie outprit voltage harmonlcs ls

rYy ---+=-D-

ac 2,t2 4.12 | RC

R E 20 a. L = 5 rhH, f = 60 Hz,.n = 2nf = 377 radlsTaklng a ratlo of 1011, the value of the capacltor ls glven by

l0= L92.4 ttF6"37i.1R2 .{6"ni,f

From Eq. (3.39), the rms v.lue of the 6th harmonlc lsV. = -2-"s.s5a9Yo 35.J2From Eq. (3-64), the rms vale of the rlpple voltage ls

f. q O.9549Yy = -----=9-= -:-*----=----4-' l@), rac -t 3s,lz t6a)' \c -l

The rlpple f6ctor ls

R'zr'* =#