IRWIN 9e 3_95 - Irwin, Basic Engineering Circuit Analysis,...

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Unformatted text preview: Irwin, Basic Engineering Circuit Analysis, 9/E 1 3,95 Using loop analysis. tin-d 1Hum the circuit in Fig. P3135. Figu re F335 SOLUTION: : 1.41 KCL I}: I"+13 IU— 11:13 I” I : KVL “ + "’ 'L’ O Chapter 3: Nodal and Loop Analysis Techniques Problem 3.95 2 Irwin. Basic Engineering Circuit Analysis, 9/E v1: Jun) = (Is--Ié)2k 2(2k)<1$.* 14) + lkIl +ikC1‘-I,_)= o 4klg " (“£16 +2kIl" .‘ULIZ:O ‘1 1F? 4?: is“ “W16 ‘ O] m .__. “m “m... We..— mu. 12+ik<—1')+ikllo *ik(1.'lz)+ ik (11—13): , l2. d.le akllv—iklagdq. Kw. '. 11%. '1")+ ikl3+ 21%. 1‘“): o ‘ “(1539*#53131: Ia>= o KVL', 19.: 2kQ1)+2ka ‘ZKCIH'I;) "l’ Qkklg’ 16): '2- IL‘; 4MA \(VL‘. 2k1"'+ iujé + 214—10“; .‘r (Is'flI-‘JL? 0 @3'5». 1L. Jélx = o 2k I,— ikxgog ~l Misrfirlclézo ~ik1I ,r 2U; iklb 1.013% 016:4). OI. ’ ill; ‘T4k13‘i015— 2mg : o 01, + 011+ 013+ 4U: * 2k 16:10 Oil—+012, - 2k15—2l613-+'5k16=o Problem 3.95 Chapter 3: Nodal and Loop Analysis Techniques INvin, Basic Engineering Circuit Analysis, 9/E 3 2,: 1:3 : '- 4MA 13- = SK'ZSMA Chapter 3: Nodal and Loop Analysis Techniques Problem 395 ...
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