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# Homework4Key - Irwin Basic Engineering Circuit Analysis 9/E...

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Unformatted text preview: Irwin. Basic Engineering Circuit Analysis. 9/E 1 3.11 Find V0 in the network in Fig. P3.ll using nodal analysis. Figure P341 SOLUTION: V0 : V, —- 6 V0 3 O V Chapter 3: Nodal and Loop Analysis Techniques Problem 3.11 Irwin, Basic Engineering Circuit Analysis, 9/E 3.35 Find In. in the circuit in Fig. P335 using nodal analysis. 12 kl! Figure P335 SOLUTION: CD 1 12 ks! KLL M j I, : 41% + Io 1% : __\CL IZK I, :— AI 'f' 1‘5 lLK VI’VQ— , __\_/L —l— V2 l2_K ’ 3K /9_K 2V!" V2. 3 I414 3V, + 2V2 :- O Chapter 3: Nodal and Loop Analysis Techniques Problem 3.35 2 Inn/in, Basic Engineering Circuit Analysis, 9/E Problem 3.35 Chapter 3: Nodal and Loop Analysis Techniques Irwin, Basic Engineering Circuit Analysis. 9/E 1 3.36 Find 1’; in the circuit in Fig. P136 using nuda] aimhr'gis. 12".Ir 1 m e + 1:}? , T 2 m 1 in L0 Figure P336 SOLUTION: W,» ~— \ \ Chapter 3: Nodal and Loop Analysis Techniques Problem 3.36 2 Irwin, Basic Engineering Circuit Analysis, 9/E Problem 3.36 Chapter 3: Nodal and Loop Analysis Techniques lnNin. Basic Engineering Circuit Analysis, 9/E 1 351 Find It. in the circuit in Fig. P35 l. 5"! Figure P351 SOLUTION: ,, /~~ » N Chapter 3: Nodal and Loop Analysis Techniques Problem 3.51 Irwin, Basic Engineering Circuit Analysis, 9/E 1 3+9:- Using 1001:: analysis. ﬁnd if; 1'11 111:: circuit in Fig. P190. Figure P3490 SOLUTION: KCL ‘- 1.: I+I2 I: 1"12. KCL‘. '15; I‘TI. I’: I34:I KVL‘. 2k1,+ ik l+2k(-I')= o 2klniéifj2,:35(I-’VI‘>=O “351' - .455 :r?‘:f§75) KVL: 2LI': 12+ 2V1, 2%} Q 1%" 2— w 1L) [2311‘— Zklikipiklj) : :2. \KVL: 2H. + in): meagre} Chapter 3: Nodal and Loop Analysis Techniques Problem 3.90 2 Irwin, Basic Engineering Circuit Analysis, 91E “L " “‘5 *%rt:-:.3+ Skip ikIL’2kI3+OIL,=0 v2k 1\«2k11—r2k13+ 01,:12 2k1‘+ LkIl+ 013+ ikTV:ll :01!“ I” 013* I912“ ——_—_———_—_——————-———— Problem 3.90 Chapter 3: Nodal and Loop Analysis Techniques Irwin, Basic Engineering Circuit Analysis, 9/E 1 3.94 Usng 1041}: analysis. ﬁnd 1’}. in the network in Fig. Plﬂri. 1kﬂ + 2 Kﬂ in Figure P3435. SOLUTION: 1 m + 2m U0 Chapter 3: Nodal and Loop Analysis Techniques Problem 3.94 2 Irwin. Basic Engineering Circuit Analysis. 9/E KVL '. ik1+ ikI': 2V0 v0: 2m, 1k(IyI.)+ 1“ (I343 )= 2(2k 1.1) ‘_ ikl‘ +2k15 _ SkIEiz o? KVL'. 2V0 = 1kI|+ ikIla- 2kIH \lo: 1k 11+ 1k122kL1= on I\‘11.‘. 3’ 4m ' 01H 01,: ik13+3>k11;i7_ -ik1|+ 01,4 Zing—511.1?!) 1L1” 11:1ﬂ 0:5 -2L1\1;o Jn 4N,- mod'Mx MW (4% combo AL/DTMQMQDL/ ’ a" “idem, cu '. Gib/3 ~43) (Meow-ac, W yam—«a Mo mozwcmy, : VD; Act 0Uth {9pm CLO/youif) Problem 3.94 Chapter 3: Nodal and Loop Analysis Techniques ...
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