hw5a - EE302 Homework #5 Chapter 3, Solution 36. Applying...

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Unformatted text preview: EE302 Homework #5 Chapter 3, Solution 36. Applying mesh analysis gives, 12=1011~612 -10 =-61;+ 8b 6 5 —3 11 Ol' : —5 —3 4 I2 5 —3 6 ~3 5 6 A: :1], 1: :9, A2: :47 —3 4 —5 4 —3 #5 A} 9' A: _7 :-——:-— [7-34—— I' A 11’ 2 A 11 i1=-i1=-9:’ll =-{}.818l A, i2=11—Tz=10m=1-4545 A- vo = 6i2 = 6X1.4545 = 8.727 V. 321—5 permissibi‘i 75m» 1431-1 matrix +23 be wri‘i‘fem f/LS ‘{ 50mg 0’1,"V (Lariat/Cider 7 fl 4440 “(0 Ii {'L 4-!- .4— _.(0 (9+1 1 4C) Chapter 3, Problem 39. Determine the mesh currents i1 and i2 in the circuit shown in Fig. 3.85. mv |2 V Figure 3.85 Chapter 3, Solution 39. _ For mesh 1, 40—21r +1011—613 =0 But 1‘ :1E 42. Hence, 10=—211+212+l011-612 —) 5:411—212 (1) For mesh 2, 12+312—6Ilzfl ———> 6:3II—th’2 (2) Solving (1) and (2) leads to 11:0.8A, I2 =—0.9A Babb +6 m manLrikJ‘lcvrm was? aql‘ov 50[L4+}DN (714/ #93er 3.40 )1 +10 .40 3 E ‘3 47. Chapter 3, Problem 40. F or the bridge network in Fig. 3.86, find 1,, using mesh analysis. 5» 2m Figure 3.86 Note — since the bridge is balanced, you can check your answer using parallel and series R combinations to find Req and then use ohm’s law to find in. But you must first work it Using three loops. Assume all currents arefimhfid apply mesh enaiysis for mesh 1. 30=12i1—6i2—4i3 —p 15=6i1w3i2*2i3 (1) for mesh 2, 02-6i1+i4ig-2i3 —' 0=-3i1+?i2—i3 (2) for mesh 3, 0=—4ile2i2+10i3 0=-2i.-i2+5i3 (3) Solving (l), (2), and (3), we obtain, i0 = i1 = 4.286 mA. ' Wm} iiaib i0 er'c WMWfly pit/Vi FEE/:‘i’rfm" & 504 V‘CA 03¢ 0U :MJHP ‘(0 “Li ,0, ’50 I O ,(o Low—*2. #2— A51 3 mo wLi) ’2— Lid—Lh- :13 Chapter 3, Solution 44. Loop 1 and 2 form a supermesh. For the supermesh, 6i1+4i2-5i3+12=0 (1) For loop 3, -i1—4i2+7i3+6 = 0 (2) Also, i2=3+i1 .7‘47‘123 (3) Solving(1)to(3), i] =-3.067,13=—1.3333; io=iI —i3=-l.7333A firs permissibfib +0 5%, ML“ 5 i/L mch'W’YK 7%Y’m M5 Sajwa, jaw/M To 4 “5 A *VL @ «I l, o X123@)+J «I 94/ fl 3;,» fig L9 no-re. {M Li) Mal 2, (New) raw/meg SO fg‘b diagcmagg {it} {ft-73"— O‘ R Chapter 3, Solution 50. For loop 1, 16i1—10ig —2i3 = 0 which leads to 8i1— 5i; — 13 = 0 (1) For the supermesh, -60 + 10i2 - 10h + 10i3 +2i] = 0 or -6i1+ 5i; + Sig = 30 (2) Also, 3io=i3—i2 and io=iI which leads to 3i1=i3—i2 ’j‘jfim "#8) Solving (1), (2), and (3), we obtain i] = 1.?31 and i0 = i1 = 1.731 A 134m germisnéfo 1T7 puff“ H394 334401501/‘5 rim wmwa’r‘fx (TM/zap SC/{VCJ or» 0L/ 0&1 GRLC’Y“ lb #10 *2 11" 0 0) 40-1 IO ((3 it 7/ 60 (1) ’5 I ~\ 33 O ('9 Chapter 3, Solution 52. For mesh 1, — i2) +401 — 13) — 12 = 0 leads to 3i1—i2 —2i3 : 6 m 403 — 51) ._ F J _ - . . . . .‘ *A '1 But V0 = 201 — 12) whlch leads to -1; + 312 213 = 0 (2) x '\ For the independent current source, i3 = 3 + i2 .90 1+ 1,3 ¢ 3 (3) Solving (1), (2), and (3), we obtain, il=3.5A, i2=-0.5A, i3= 25A. C) "'“( f 3 ' 1%» mad/Ix firm find/5 mi‘o/ WW5 ...
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This note was uploaded on 05/01/2008 for the course EE 302 taught by Professor Mccann during the Fall '06 term at University of Texas at Austin.

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hw5a - EE302 Homework #5 Chapter 3, Solution 36. Applying...

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