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Unformatted text preview: P3.1 [a1 Thea island 12 Miresistcrssicin seriesm acctheir island? its}
resistors. The simpliﬁed circuit is she'sn below:
13:} 101: El [Is] The 3 1d}, 5 its}, and 7 k0 resistors are in series. The simpliﬁed circuit is
shown belm:
4kg 3111A ﬂ lﬂikﬂ [c] The song: 4005’}: and sent} Hiﬁtﬂl'ﬁ are in ssria. The simpliﬁed circuit is shown. below:
1200:]
ZWnNEi Iiﬂﬂﬂ
P313 [:1] 60f20= 1200f80=15il 1224=238/Sﬁ = 352
15+8 +i'= 3M1 30lﬂﬂ=36ﬂﬂflﬁﬂ=24§2 H.s= 1542445 = 54:71 [h] m+4n=7sn 75sn=375m125=3cn
m+2n=scn scum =3?5[if125=3[iﬂ
ans10:40:: dﬂllﬁﬂ+9lﬁ=24+ﬁ=3iiﬂ
3030=15r2 R¢=m+1s+5=3on [C] iii#302809 BUIIZO:IEQ
lﬁ+ld=3llﬂ 3ﬂ+24=54ﬂ
5427 =1sn us +12 = sun 3030=15i1 Rah=3+15+2=20ﬂ 133.11 [a]
an R... = {111 + 211}[12 +1311111}] = 31115 = 11111
1.524“ = 1D(2.4} = 24 141".
21'.) 11., = 1131;: = 10+ 2ﬂﬂ24} = 16 V
um = %{241 = 1—1124} = 14.4 V
1... = % = 11.15 A
[b] pg]: M = 0411—144}: =15.35 w [c] p2” = —(2.4}{24} = —5?.ﬁ W
Thus the pan:21 {11211911113911 by the Currant 311111119 is E1716 W. 2BR; 133.14 4:112411 so 1122:1111)
21111e 1211
3‘1111+11e 5“ RFFQ Thus. E: IDRL an 111:2411 P 3.13 Begin by using the relationships among the branch currents to express all
branch currents in terms of 1.: i1 = 21': = Lima} = seen)
i: = if: = 2(214}
i3 = 21.1 Now use KCL at the top node to relate the branch currents to the current
supplied by the source. f1+iz+i3+i4=lﬂ1A
Exprew the branch Currents in terms of is and solve for ii: 1not:sitculls.Hittn:15:",l so i4=%a Since the resistors are in parallel, the same voltage, 1 V appears sum each of
them. We loiow the current and the voltage for R; an we can use Uhm‘s law to calculate R4: R¢=i 1" :4 ={1f15)1nA =15 ki’l Calculate 1'3 from i; and use Dhm‘s law as above to ﬁnd R3; QED? o 1 1t"
i = J = — =
15 A . . R: is {2115} l 7.5 k9 i3= 2i4= Calculate a: from at and use Ohio’s lat».r as above to ﬁnd R2: [1.004 1; 1 v
:3 3“ 15 3 a (4/15) M Calculate 11 from a; and use Ohio’s lat».r as above to ﬁnd R1: 0.003 1.: l V
“I “‘4 15 l a (3/15) M The resulting circuit is shown below: 1 Ski] F" 3.22 [a] At .10 lﬂﬁLL' Fa : kn, = “2 R1+R21:,.
At Fulllaﬁd: 1}u=::m,= Re 1}”. where)?“
R1+Re I
_ Hg _ {1H}
Therefere L: — 111—112 and R1_ L RE
Re {le}
= 1:1 R =
ﬂ RI—Re Em ' e R“
rl—ﬂ' HERO (1—H
Th — R
1.15 [k t] ) Ro+R2: k 2
1;
Selﬁngfer R3 jn'elde R3 = {1: 1 ER"
A1501 Rl={1'k}eg R1=[k_ﬂ}Ra
.i: :13:
[LEE g
[13] R1 _ (Eyed—2mm
LLDE
R2 = (ﬁﬁﬁlelsr m
[C ]
+ E
R1
BEI'U'
R2 Iaﬁmum diaeziipetiﬂn in R2 menus at 1m lﬂﬁd, theIefere, HEUHUEEZ'F
.F' = — = 133.5 w
3‘4”} 14.15? “1
lfaximum diﬁﬁipatiﬂn in R1 mum at full 143321.
[50' — 0301:5012 Pnltm} = = 57.33 Inﬁnr 250D shun {an}? FRI — 2500 — 1.55 w — 1555 um:
_ {'3}? _ 1
FR? _ 131152—333 P3.25511251’2=51‘1: 512—51121311: 15{15—12—13}=52; Therefore. 19 = % =1:.5 5 5 20
:5n=ﬁ{135}=1CA' 31:: ll —{lﬂ]=2 A P 3.32 [a] 11m... = {51] >1 1113115 55{5555 — 25}] =15512 V
[5] 55.5 ={5551 15 3;{1555}_ — r: 5525 V [1.5512
n 5.1525 52 51151 = f — 1) 5: 1:15 = 4122525 NUT ASSIGNED. DHIJIFTD BE USED TO ASSIST 1WITH PRGB LEM P 3.33 P 3.33 The measured value ii EUIIEUJ = 130551333. 51:1
 = .—= 1.555525 .5; _ 1555 _ 1455155 .5
15' 115.55515—15} ' ' 511 760': } T1115 true value is EJ2EI = [5571 . Ell] , EC
19—m—2 151... ImuE—E(2}—l.5 151 1.494758 3151512201" =[ l: — I] H 100 = —C.343733'E1 m—Uﬂﬁﬁ P 3.53 Begin by transforming the ﬁ—oonnected resistors {it} 51, all) [1, ﬁﬂﬂ} to
Y—ccnneeted resistors. Both the Y—connected and lit—connected resistors are
shown below to assist in using Eqs. 3.44 3.45: 150 Now use Eqs. 3.4! 3.46 to calculate the values of the Y—mnncctod renters: C40) [10] R = = [JﬂHtDj
‘ m+m+m _ _ , storm}
2—m5”* R “a; Fm: EDD. The transformed circuit is shown below:
159 3:15.}
a h The equivalent resistance seen by the 24 V source can be calculated by rushing
series and parallel combinations of the resistors to the right of the 24 V source: aq = (15+5}(1+4) +ec=2u15+2e=4+sn zine Therefore: the current i: in the 24: V source is given by __sdv_ t——24ﬂ _1A Use current division to calculate the currents i1 and :13. Note that the current
211 ﬂows in the branch cont aiming the 15 ﬂ and 5 ﬂ series corniected resistera while the current is ﬂows in the parallel branch that contains the series
connection of the 1 ﬂ and 4H resistors: 4
15+5 4
1:1: ejzﬁumzasa, and czis—oeazust Now use. Kill. snrl Uhmls law to naimrlate m. Note that 1'}: is the sum of the
voltage shop across the 4H resistor, =ls'g, and the voltage drop across the QUE—E!
resistor, 2th: is — £1422 + 201i — 41311.8 A} +2J(1 A} — 3.2 1— 2D — 33.2 V
Finally, use KVL and Uhru’s law to calculate s2. Note that s? is the sum of the voltage drop across the 517! resistor, 51;], and the voltage drop across the
235] resistor, QUIZ: w=s1+2ni=5m2ay+aos A} = 1+au=21o F 356 [a] Calculate the 1values of the Y—eonnee'terl resistors that are equivalent to the
It] £1,413 [1. and 505] ﬁ—eoruaeetod rosistcazs: {lﬂﬂtﬂ} (ammo;
I m+m+m ’ m+s+m
_ {lﬂ}[dﬂ} _
R3 ‘11:: +413 +51} ‘4“ Replacing the He—Ra—Ra delta with its equivalent 1r gives
13 u New calculate the equivalent resistance Rab by making series and parallel
combinations of the resistors: Rah:13+5+[[s+4}[2o+4)]+r=3so [b] Calculate the values of the ﬂ—oonnected resistors that are equivalent to the lﬂﬂ, 351, and 41:”? Y—eouneeteel resistors:
[lﬁHS] ‘1' {311040} +[1UJE4tll = sou Rx: _____§___—— Eﬁnmo
R, = (lﬁ){8)+{8}$ﬂ}+(1mt4ﬂl = % = m,
R2 2 use) + {sen} + onus} _ ﬂ = ,,,,, 4U _ 4U Replacing the R3, R41 H5 “rye with its equivalent at gives
13 3 Male: eeriea and parallel eembinatieaa of the reeietcnre to ﬁnd the
equivalent reeiatance Rah: emanate = 33.3351; 333nm = 3.313
33(33.33+331} 2 13a
ea =13+13+T=33ﬂ [1:] Convert the delta connection R4—R5—Rs to its equivalent wye.
Convert the we eenneetien RQ—Rd—Rg ta its equivalent delta. P 3.61 159 25lﬁ.25 : 5 ﬂ GUIISD : 20 ﬂ ._(5}{15}_ , _ _ r
11 _ (1“) _ 2.25 A. 1:1. _ 2011 _ 4.; V v3 = 2521 = 56.25 V
136.25 = ivy — 1:1 211.25 V 11.252 + £3 + 55.259
6.25 30 15 Filmme = = 293ﬂ3’f5 1W ...
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