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IRWIN 9e 3_97

# IRWIN 9e 3_97 - Irwin Basic Engineering Circuit Analysis...

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Unformatted text preview: Irwin. Basic Engineering Circuit Analysis, 9/E 1 3.9? Using 1011]: analysis, ﬁnd L, in the: circuit in Fig. P’F-Q‘II‘. Figure P343? SOLUTION: " KCL‘. I“, I+ 21x = I I 3 121’ 21-1 3 Jed-2:5" IZ+ 13-21]. L: IA’IL'Z‘C‘H' If) 101‘11’21‘111’ 2'13"" 16 ____________________.____———-———————- Chapter 3: Nodal and Loop Analysis Techniques Problem 3.97 2 Irwin, Basic Engineering Circuit Analysis, 9/E KVL'. 1k II + 12+ 1kL—-I'):o 1k. 1” lk CIH‘ 1'): "" ,2— l2k1.— 11:11, :42] KVL: ik1+ ikL—Io): 1'): {21:11— in; ma, eff-11:1; = l 2.] KW: 2vx+iic£2r 11:11:, VX: iklé ZULU‘) ”(Ia )*ik(1w1§)= 0 MM»: Wm... Wm N .w 1L1 + 2k L‘“ 1k 15+ 21:14:: l nwwmw w... Wmmw «a «MW .9»- 1-,...- Ww~ 4». WWW... .( KVL: 2V rikl +iklbrik1” Hilde: o 4153‘) wwill; 1)+ikll 22v+21y*14)+i'<(I4-13)*’“14 l - 1:31,“: ﬁlial MWm—y,x‘u:' w . w Chapter 3: Nodal and Loop Analysis Techniques Problem 397 Irwin. Basic Engineering Circuit Analysis, 9/E 3 // ZkIprOI; ikIHJrOIs-nvolﬁ ~12 OI.+ 2H,,r ZkIw—leg‘iklé=é _1k1, +01L+2quvik 13.“. MIUO 01'1" 011‘? 9.1% -15-"16: -1LII- iklz— 1L1? + 2L1? +5LIé:-é if: —I,_— 0217+ «IsmIé, 10: ”3-22,“ «2 (’0'96m)+2(0‘29m)—3d6m 10 7. —' IR' 88mA Problem 3.97 Chapter 3: Nodal and Loop Analysis Techniques ...
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IRWIN 9e 3_97 - Irwin Basic Engineering Circuit Analysis...

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