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hw2_sol - P3.1[a1 Thea island 12 Miresistcrssicin seriesm...

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Unformatted text preview: P3.1 [a1 Thea island 12 Miresistcrssicin seriesm acct-heir island? its} resistors. The simplified circuit is she's-n 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] 60|f20= 1200f80=15il 12||24=238/Sfi = 352 15+8 +i'= 3M1 30||lflfl=36flflflfifl=24§2 H.s= 154-2445 = 54:71 [h] m+4n=7sn 75||sn=375m125=3cn m+2n=scn scum =3?5[if125=3[ifl ans-10:40:: dflllfifl+9||lfi=24+fi=3iifl 30||30=15r2 R¢=m+1s+5=3on [C] iii-#302809 BUIIZO:IEQ lfi+ld=3llfl 3fl+24=54fl 54||27 =1sn us +12 = sun 30||30=15i1 Rah=3+15+2=20fl 133.11 [a] an R... = {111 + 211}||[12 +1-311||111}] = 311||15 = 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 = 04-11—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 cur-rents 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:sit-culls.Hit-tn: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 lflfi-LL' Fa : kn, = “-2 R1+R21:,. At Fulllafid: 1}u=::m,= Re 1}”. where)?“ R1+Re I _ Hg _ {1-H} Therefere L: — 111—112 and R1_ L RE Re {l-e} = 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: [LE-E 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 l-faximum dififiipatifln in R1 mum at full 1433-21. [50' — 0301:5012 Pnltm} = = 57.33 Infin-r 250D shun {an}? FRI — 2500 — 1.55 w — 1555 um: _ {'3}? _ 1 FR? _ 131152—333 P3.25511||251’2=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] >1 111-3115 |55|{5555 — 25}] =15512 V [5] 5-5.5 ={5551 15- 3;{15||55}_ — 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 _ 1455-155 .5 15' 115.55515—15} ' ' 511 760': } T1115 true value is EJ||2EI = [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-! 3.4-6 to calculate the values of the Y—mnncctod renters: C40) [10] R = = [JflHtDj ‘ m+m+m _ _ , storm} 2—m-5”* 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=2u|15+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 resist-era 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} +2|J(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 3-56 [a] Calculate the 1values of the Y—eonnee'terl resistors that are equivalent to the It] £1,413 [1. and 5-05] 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+3-31} 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 25l|fi.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 Film-me = = 293fl3’f5 1W ...
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