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Fall2010_Exam1

# Fall2010_Exam1 - ECE 430 Exam#1 Fall 2010 Name 30 LUT 1 0 M...

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Unformatted text preview: ECE 430 Exam #1, Fall 2010 Name: 30 LUT 1 0 M S 90 Minutes Section (Check One) MWF 10am MWF 2pm 1. / 25 2. / 25 3. /25 4. /25 Total / 100 Useful information sin(x)=cos(x—90°) 17:2}- §=V—I* 52¢ =J§VLIL46 0<6<180° (lag) I, =\/§I¢ (delta) —- _ ‘ Zy=Za/3 ,u0=4ﬂ-10’7H/m ~180° <0<0 (lead) VL =\/§V;, (wye) ch.dl=[SJ-nda ICE-d1=—3 SB-nda in=i MMF=Ni=¢iR ’ 6t ,uA ¢ 2 BA 2=Li=N¢ k: M 1hp=746Watts 1'le / m '\ , . . H 105m. ‘2 L L ' _N . - z‘ 1 L2 2 + . . + N10,: NZOL +v — V' VZ' VI N: I + VL ’ q’ _ : ,ATL ~ Au 4~ N Na V _ L __ _ 0; I ' ‘ar Mg; Problem 1. (25 points) A single phase source with voltage 120Volts (RMS) is serving a passive load through two wires with total impedance (0.5 + j1.0)Q . The load has a power factor of 0.85 lag and it draws a current with magnitude 8 Amps (RMS). What is the magnitude of the voltage across the load? £2 F5}: 0.?S‘lqc3 L'DQV—CQH: 6’ [email protected]:ZL9° [email protected] ©;:-—§l. ° 3 we 8 \LVL \ﬁ \ZOZC9 : (0s— +3lD_\(€A:Z).Q°I4\) + VAQ‘ lZOmsa +3taocfvxc9 ; <8.%[_3LQK° v + v RQm—l Ford-1‘0!“ W IN? PM” W lZOcoHQ .: 37.63 +V \ZOSIHCQ : H. .70 \20ms(2.2w°3:;.(73 ~\r\} \i: Z.zb/° V: llZ-ZS’V Problem 2. (25 pts) A three-phase power system consists of a wye-connected generator connected to a delta connected load through a transmission line having a per-phase impedance magnitude of 1.5 Ohms and angle 75 degrees. The delta-connected load consumes a total three-phase apparent power of 450 kVA at 0.8 power factor lag when the generator voltage is set so that the voltage at the load is 4,160 Volts (line to line). - a) Find the magnitude of the generator line current for this system. b) Find the magnitude of the line-line voltage of the generator. 0) Find the total three-phase complex power supplied by the generator. PCP- we... Ezpwdm'l— IA I Lem—c I __ \7 __ —> °"‘ 9 van — Q ._ a 3 3H,: 3..” : Vb, I}: \50 Qﬁogx 10114 If: 150434» o _....____b Loud. 2-K Vol‘l‘ag‘, Mu“ V? [‘0‘ as fol-emu. emu i i m i — l x I. l 53W Van : «r \ZV, : (\.S‘L§'9°3(cz.er-3e.9°j + 'dJVgM‘ = ?3.7a+ 55?.9 + Diem; ’ Van : Zane; Au" i \744, '2 C) €739} '-’ 3 \TQ" fa = 3 (24%; [.1330 v3 (tows/36.9%) Eng : H619 439.25 Kw. Problem 3. (25 points) An iron core with two coils is shown below. The permeability of the iron core is inﬁnitely large, hence the magnetic reluctance of the metal materials can be ignored. The dimensions of the air gaps are summarized in the ﬁgure. Ignore fringing effects. Gap 1: Distance: 5 mm; cross—sectional area: 1 cm2 Gap 2: Distance: 6 mm; cross-sectional area: 0.5 cm Gap 3: Distance: 5 mm; cross-sectional area: 1 cm2 2 a) Identify the dot marking of the two coils; b) Draw the magnetic equivalent circuit; c) Find the ﬂux density in air gap 1 if i1=10 Amps (DC)and i2=20 Amps (DC) : 3.970] X109 F”: wk More space on next page Blank page for more space (:3 gm“ = ? KVL N 5 ~ \ooo I44: NM 2 R5““" ¢' + REM)?— ¢C Nziz': Nu} : \000 At Nziz : 1'3sz (lc. + fqug ¢L a? ¢c 7— g! +¢L # NH.‘ :NL;2 2: N‘ Nxf; : ‘ * <5: “asap. +£ﬁqh) + 0% (quhA NZ L2 2 I t ¢' + C6; #BuL-ha ofﬁcohi- gwwr?‘ ;+— mu m {Mod-4 th r¢z 2¢ Ni 3 ¢ (ngpz ‘F 125%! + RSOLPKX (a: lace We a (3.6%x109%3 + (an “09%) 2 (Azgﬁz : 14.33 my" wk .4 T B3043. :: (ﬁl'lqskﬂ :: (LIPS xlO—GWM ~ O'OH33 ELL/7 (Mo-Lima) “ n42. Problem 4. (25 points) Two identical coils (each with zero resistance) are located near each other. Coil #2 is open circuited. When a 60Hz sinusoidal voltage of 120 Volts (RMS) is applied to coil #1, the coil #1 current is 6 Amps (RMS) and the voltage measured on the open—circuited coil #2 is 70 Volts (RMS). a) What are the self inductances of coil #1 and #2 in Henries? b) What is the magnitude of the mutual inductance between coil #1 and coil #2 in Henries? c) What are the current magnitudes in coil #1 and #2 if a short circuit is placed across coil #2 while the given voltage is applied across coil #1? CtB'\//4\ 5133 0)) KV L OV\ RH 1002 '1'" ' 0 ‘l’ d; Vic?) L- L- Vzﬁl = - M 1% Vl : ; _ ~ V1L+\:\ZC>JE COS (now-{A { {LA ' @ A’Z 0 gm: 70JECosLt’z01r-h) W: LLH’l — . _ tzcm Ind 495 \c.t+3\=edZ= 120 02/ 12017 L “\$36 «9 it it {Vze’ld't : deQ ~ (it 7o~lismgizon£3 2 Liz‘éﬂ» __ ML. m \lmr “'6‘” W w> M= to IZmrM Bow-6 More space on next page Blank page for more space C3 1—: /’*\ 5: \(l/L O . + . . - - O : Ldia , Vx&:‘ZOECoS (Izmvf) F v—s a W: Mid—L Lclu‘g dt E M 2 5293=Ata.c+) I, In € :5 V.L+\:L_d;u_ _M<§;‘) G/F ﬂaw“ t 35 V‘Lﬂ: (L__ Z d“ {IZOECOQ(‘ZOJW£XJ£ :ﬂL_Al/Z~)Olh \ZOJE SivaLOrrH : - L Row (L a} L+3 5.1» ILtU’)‘: HE... , r“ \zou(L-ﬁ‘:z) 2 \ 0% -(Cmsm HSZ Yr( 3“ 0.033“ ) létGHms a: CLOCHA L” ERR \LVL ZLA'ib-l ...
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Fall2010_Exam1 - ECE 430 Exam#1 Fall 2010 Name 30 LUT 1 0 M...

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