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HW_5-soln - mewm mm W 51 i ‘i WWII Wm new 3% gnaw M m...

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Unformatted text preview: mewm ; mm; W 51 i ‘i » ; WWII Wm; new 3%» gnaw M» m @ Mesh #1 (on the left-hand side) 2~211”3(11"12)= 0 If we treat mesh #2 (middle) and mesh #3 (on the right-hand side) as a single loop containing the four resistors (but not the . 77 _)_ current source), we can write L2 “112 —3[3—213 —3(12"11)=0 From the current source: [3 —— I 2 = 2 Solving the system of equations: 11: — 0.333 A 12 = — 1.222 A 13 = 0.778 A MN»: MWWL , _ M ”flush “v j” n , 3g 2:?“ y «a; a»: g g 2’ it a w» M ; WWW.” W 2W” Meshes 1, 2 and 3 go from left to right. , . Stamina @éThe Merw-Hésl Cagnpanim; (32:1. ngmuésxirw ii?5;§l§¥i>(§ m» 21, :ywmmiws a: depict; For mesh #1. 2 £2 ,1. 5} 31'). i1(2+3)+i2(—3)+i3(0):2 For meshes #2 and #3: i1(—3)+i2[1+3)+i3(3 +2) I I 0 For the current source: i1(0)+ £2 (1)+ i3 (—1) = —2 Solving, , m M“. ”$5533 ‘ aw“ fl Mum,m v arfi , MMM 5/ andv=i3(3-l—2)=3.89V 4”“ ...
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