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HW3-Solutions-enme350

# HW3-Solutions-enme350 - ENME 350 Electronics&...

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Unformatted text preview: ENME 350 Electronics & Ins’rr'umenfa’rion I Homework #3, Due 9/30 Problem 1 For the circuit shown in Figure 1 131 Figure 1 a) Find the equivalent resistance seen by the source. b) How much power is delivered by the source? Page 1 of 8 ENME 350 Electronics & Instrumentation I Homework #3, Due 9/30 Problem 2 Find the equivalent resistance between terminals a and b in the Circuit of Figure 2. 3 Q Figure 2 3 O‘EF‘ W ’T‘Wu. a'r'cwﬂ‘ Can be. rcdfaum a: ‘x 351» ............ _ a—W‘” ca, 2 1» (4!" 1.3-— 5 a’_ __ _____________ .................................... 2 , ' , 4 {I Q , “J” :1 L/,2J1 _.L,~--——~ " 2: +! 4;}; 6 z r:— 12% ’" -_L- +__L. ,L ””73: {HI HQ 5 6 12,41.» Z in Sew drU’x tr» :5 ask 3““ ,_ JD WI, 3"” -.._—~— «or-m «a 6-5" " qﬁw ’ a" la &‘,...—" """"""" Page 2 of 8 ENME 350 Electronics & Instrumentation I Homework #3, Due 9/30 Problem 3 Determine the power delivered by the dependent ‘ ‘ source 1n the crrcurt of Frgure 3. > } Figure3 -EO‘UELT Tm wage an L»- New” “ USN] Ohm’s \ou ”VMFQZK“ 3'3"“— : vg US\$34 C jf UL" Vx ﬂ: QoﬂthA) + {0V 1:an ”T‘KL Tr) ”TM. Niger'- HW— pawutek intrigue} ”0"" Sta!“ . U) Mean Poi-dd {-5 + 2- O ‘ '1 5 Agligtrd Page 3 of 8 ENME 350 Electronics & Insfr‘umenfa’rion I Homework #3, Due 9/30 Problem 4 Determine the voltage between nodes B and A , VBA, in the circuit shown in Figure 4. Figure 4. VS=5 ¥R1=2.2k§2, R2 = 18 k9, R3 =4.7k§2, R4=3.3 k9. \‘vs Serfts CMA {L3 4 {2:4 cm: m 2mm. 5;":10‘ \lu;Vz 1V3&VH Solution KT+¢L Q-ttﬁ". [gICJ'l-v g g V ,_ (J. \I’L; Eff/Vs :: 13"“ 7‘5— “ Q ' l Sun/“‘0“. / 15V jﬁs Vs ; iii“; 2&5“ f 2 q; V3: 5237;: 9.71441 +33'H' é¢§V {244 v; ~ 53:9: '6’" 3 1’0 Vu 1. 6:71” " 4.:UCJL +3.3)CJL Page 4 of 8 ENME 350 Electronics & Ins’rrumenfm‘ion I Homework #3, Due 9/30 Problem 5 (P2.22 from text) *P2.22. F ind the voltages v1 and 122 for the ‘ circuit shown in Figure P222 by combining I osi'stauces in sexies and parallel 12 L2 25 Q JWv—w -— T- aw»?g 4. + u-«12V6 lUQ . w, 30\$). L2 59 t 6 s: i ‘‘‘‘‘‘‘‘‘‘‘‘‘‘‘‘‘‘‘‘‘‘‘‘ ‘WV‘ Figure P232 Solution We combine resisfonces in series and parallel uni-ii fhe cincui‘l‘ becomes an ecnii‘voleniL resisf'once across the vol‘i'oge source. Then, we solve ihe simplified circuit and ironsfer information back along line chain of equivoien‘is un'i'il we have found “line desired resul'l's.‘ Page 5 of 8 ENME 350 Electronics & Instrumentation I Homework #3, Due 9/30 Problem 6 (P228 from text) P228. Consider the citcuit Shown in Figure P228 Find the values of vi, 2);, and ya}, we 39 4 o. 27 ........... WWW OJWVMVV min + l m\$v2 ’1», 9 V ‘‘‘‘‘‘‘‘‘‘‘‘‘‘‘‘‘‘‘‘ Mi 6 £2 Figure P223 Solution The currents through the 3-0 resistance and the 4-9 resistance are zero because they are series with an open circuit. Similiarly, the 6-9 resistance is also in series with the open circuit, and its current is zero. Thus, we can consider the 10-{2 and the 5-552 resistances to be in series. The current circulating clockwise in the left-hand loop is given by _ 9 _ 10 + 5 ' counterclockwise in the right hand loop is 1 A. By Ohm's law, we have v2 = 2 V. Then, using KVL we have vim : 1/1 —v2 =1 V. ’1 and we have i/I : 5/; : 3 V. The current circulating Page 6 of 8 ENME 350 Electronics & Instrumentation I Homework #3, Due 9/30 Problem 7 Using node voltage analysis in the circuit of Figure 7, find the current 1' through the voltage source. Let R1 = 100 9; R2 = 5 £2; R3 = 200 9; R4 = 50 (2; V 2 50 V; I = 0.2 A. — V : o azi/+ Y}.— VL-W :0 3% VE'V...‘ 4— Z}— + Viv;— +— I g + -———--"'-P:Ll ﬂ‘ 8'“ KL 5 P5 ngV\ 4- av; + %LV*’L-Vi) 4— I“? T 5 ° -—lfb cw.) : ~20 C?) ;> --zt\/\ +7-0V1 +3V3 : ~l6‘b’i. 2 g. Q—clqéﬁ V7, &\/3 [:0 MM. mum? swrCe V using KVL -Va. - V+V3 : o :9) lVB-‘ﬂ 6r V5’Vz :: ‘50 f'"“ 3 in Matrix {2ch @[email protected] 5Q) 3N“ . Q3 440 «L V\ __ O] sch/M3 usihj MAJ“? YWM‘ ~24 2° 3 V=— ‘ So \,l\$,t,.§.'zg_3é;.\f,\/,,;.L4 3.67I\$/\132I.3099V o —l | V3 ' WW teCLCeD Melt-1 =5 [gl‘t—I4‘C , ‘ \/"Vz. , __ o4 * 0.49233; +u8'67lé) .. aim/6 A a 5- ‘9 .v———-——-‘- " ___________————-——-———‘——" Q, ' Page7of8 '— .a z A 1, .... o 9 3'0 3. ENME 350 ElecTr‘onics & InsTr‘umenTaTion I Homework #3, Due 9/30 Problem 8 Using mesh current analysis, find the voltage, 12, across the source in the circuit of Figure 8. o. . ’--~r——., 29 r"1£2 39‘” . \I ~¢V+ ’25, +3Q/,—CL) =5 S‘o‘,——3c',,:2. ._ ,,,,@ mglﬂl &wsh3, ’vaskca all non“ ‘1 V2»; year; PW”? 6‘; "mﬁk 4 (Jim) + 36'; 4-2 ('1 + 3 (Cz’bﬂ :§ .‘3L\ +L1-U'L tlfu. 2A (anb smw‘C/k Relate U2, (st/3 ta ' ‘ C5) ,O'L-f-LZ 5 2 («‘3 — 0'2 .. ‘ S _ ' d trix g rm / (D, 6;) (E @ cam be wmﬁen in Ma 0 ' 2— ' ' ‘ (61c 15 ’3 Q U" ’ O K Sc‘W'] uSMj MAE ‘3 L4 5 0.1 v 2. ﬁkﬂs 03 ‘ P. o —l 1 (A 3 -o.37>3‘\$ b deg-1‘27?" P ICVL at N Sk 3 T‘rcS; L3 3 0.7.77 8 D _. V 1 O r 3 , , . ‘ _, V r. g (/3 ’— _W v>> Page8of‘8 j“ V i: ”5&8ng ...
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HW3-Solutions-enme350 - ENME 350 Electronics&...

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