PS9_Solution[2] - Name: Section: Z EE331 Homework PS9 —...

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Unformatted text preview: Name: Section: Z EE331 Homework PS9 — spring 2011 Problems from Alexander & Sadiku: CH 11 — Assume all voltages and currents in the problems are RMS unless otherwise specified — as mentioned on pg. 477 of Alexander and Sadiku 0 11.51 0 11.61 0 11.69 0 11.71 0 11.74 (ans: pf: 0.9 lagging, C = 5.74 mF) Additional Problems (Instructor Option): 0 1 1.85 0 Any as assigned by instructor Page 1 of 1 Chapter 11, Problem 51. For the entire circuit in Fig. 11.70, calculate: (a) the power factor (b) the average power delivered by the source (c) the reactive power (d) the apparent power (6) the complex power Figure 11.70 ForProb.11.51. 2 _ <S.H° or cl.l5.\.o_’.}q3m Qucl -% g: [VrLs -— (DOV) : {SJ-lo W “55.43 2* <z.\1<—5.H° VA PM WA c2 °):O_QQ(> Lasso!“ CELT-‘3 q) pg: Q08 (9}: £03053! L) E: moan C) a —. +1553 VAR A) 34153 VA 6/52! 3L5 ‘3 <5 HO K/A Chapter 11, Problem 61. Given the circuit in Fig. 11.80, find 1,, and the overall complex power supplied. 1.2 kw 0.8 RVMI (up) 4 kw 0.9 pf lagging 200190° V Figure 11.80 A /L_ A N S 2 S + 33 + 35 TOT l 3 , (vogue-(095.8 9405015 _ [HIKQ mm ‘dm’sqkma = $ka _ lH'kvtk 0.9 WAR + QKVA I: 1‘ MW a-s.\\\\s) 33: “ mag-0.82 '1 LHHKUA .- .. .th __ D um +«“(9217)‘ 9'” q (a 0.3% tom 5: MM, 0.1%) = 4.1.: NA cum“: 33.(<—€QH° A 900 (40" o 2 534 <+®H°A Chapter 11, Problem 69. Refer to the circuit shown in Fig. 11.88. (a) What is the power factor? (b) What is the average power dissipated? (c) What is the value of the capacitance that will give a unity power factor when connected to the load? 7.. 2 10+j12§2 Figure 11.88 a) 'Ele‘I-DS: [S.L<So.9°57 9£=9V'91' ‘47 Pf: COS(G%):VO.(9'1 1.433%»3 CELT.) A VQ‘ Va 0 b) tar—=7 s: M 2090) = $93160.) \IA * .L ’ . o ‘ 2 15 <¢soa or #0? M W‘“ 5%.? + \3 Cmbuaak W Anna: 570'? U‘) PAW C) %\\l>\ howd- big'gwkk 3:»; Paybcv W\au6\.eeu/- 00 00 ' 10a VAR ._ 2- Mg 1 ~" 7 W - vw ' (PIG-.0 580-? w NEED 40a VAR +1: (ouud‘er +310?qu (“50) -—> &= Rica VAR: (0'0 ) _, (lD‘OV) (M‘de A”; $494,) =5 Chapter 11, Problem 71. Three loads are connected in parallel to a 120400 V rms source. Load 1 absorbs 60 kVAR at pf =0.85 lagging; load 2 absorbs 90 kW and 50 kVAR leading; and load 3 absorbs 100 kW at pf = 1. (a) Find the equivalent impedance. (b) Calculate the power factor of the parallel combination. (c) Determine the current supplied by the source. \%\// {ASE SUMMQ§40W 0Q me—(Tr‘mflkfi Low '. H I it; 33‘ (60 ’50) : lOde AONM ___‘ A‘ 40“” loot» -- = Q‘WA > M -91, . .<:§:j ‘ «use 0 ' sob“ 92‘:— 0 \———\/—~V ‘ “’5‘ (MST-3‘3 ' («as H0+nool r ' I t. IMAM-NV? ' . _~ -I -50 +(E5LI) i QZ'+“" (f; 98lo~8l<h3 . __ ébdefi 1 : figq. o _ 4 J11 P ' +aq e; — +4“ (3‘63 : Q” “— ° - s : ‘l/osm‘wo‘ ' s - MR 0 WA 9 2 Qibfi‘tKVA Chapter 11, Problem 74. A 120-V rms 60-Hz source supplies two loads connected in parallel, as shown in Fig. 11.89. (a) Find the power factor of the parallel combination. (b) Calculate the value of the capacitance connected in parallel that will raise the power factor to unity. ~s-——\\ $1.0m 1 24 kW 40 k’W pf: 0.8 pf m 0.95 : lagging J lagging E Figure 11.89 1A5 E SMMMAT‘i ow 0 Q (PDQZPFU/ia:\23 729171 1 LoADI 1i 1 £2 Llokw "mama": n .DWAR 31.3 2WD Jam ~. \g WAR .— m E % u- ? “KO Ll 0 K0 M ' 921C06"(o.8)243"fi° ' 5’? cvs“(oxs)—-ts.2° aqko+ How: (9wa 0H d 31.1 ELI —71‘uo‘uu.\l¢ . 2 +4” 1 2L0 : 0.1 LALLINL ol-a.‘ a) pig; C06 9 19> QaMT FIN-4‘ Power Pruittqu ‘5) _____§ 9» 1 QA \/ ((1.80) N *0 3|,QKQMK: if): '2 M :7 MM Gama PEA 3 "“M f an~bo)~c. Chapter 11, Problem 85. A regular household system of a single-phase three-wire allows the operation of both 120-V and 240-V, 60-Hz appliances. The household circuit is modeled as shown in Fig. 11.96. Calculate: (a) the currents I], 12, and In, (b) the total complex power supplied, (0) the overall power factor of the circuit. l0 (1 Lamp qu /\ 30$] Knlchcnramp \L J- c It) (I Rcfrlgcrmor '1’ a , nu [(i‘v' TIL ‘ . ‘ “11045.55 3R : de Figure 11.96 A. . '1— : [go V4090": (00 A ‘ 1057. 0 TN : "QOWDO‘O 2-10.454‘9‘7-50A “‘0 k \D + 5155‘) 31 A O ' TC: [20(3) -‘13040 2 g (o A ____g________. 30 A A’ K D A A ’\ A— x 0 ¥__ 00 + I -§LlO (D .. , 0(0 ‘1‘ I 190< ) L ‘0) g 1 SLAM, \— SRCMWA»; SKI " I4 ‘3 b O : EHLgo<o°+ 128% <a15° + l‘Doéo = anui,\<<z.q° \/A C) Pg: CDSUPT") :O.‘f‘fOS L433.\A fix UgA’xi/f m n 1 _VI* 2 = Vrmslrmsffiu — 6i Apparent Power = S = [S] = Vmslnm = VP2 + Q2 Real Power = P = Re(S) = Scoswu — 0i) 5.— Reactive Power = Q = Im(S) = S sin(l9,, — 6,) Complex Power = S = P + IQ = k3“ H3 ('3) vow/w aw»: (wk) Power Factor = Im +Q (lagging pf) (TchWc) (a) Figure 1 1.21 (a) Power triangle, (b) impedance triangle. (b) Re —Q (leading pf) (annual Figure 1 1.22 Power triangle. A. ’L x §=S,*~5:+“‘5n * ~ J 5 M4 ~=9 ALso cu m E Em “A Mme (Fw MM) ; (PM 45 ‘5’: Lam's) 1- (?3+0,¢)\-" *em («or mam PM“) 2+51+53LHS" E ...
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PS9_Solution[2] - Name: Section: Z EE331 Homework PS9 —...

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