20095ee10_1_HW 3_edition8

# 20095ee10_1_HW 3_edition8 - i 4 = 15 A i 5 = 9.77 A i 6 =...

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UCLA E LECTRICAL  E NGINEERING  D EPARTMENT :  EE 10: C IRCUIT  A NALYSIS  1   HOMEWORK 3 The following are the assigned problems from the textbook with numerical solutions: 2.29)      a)  i 0  = 0,  b)  i 1  =  -60mA   , c)  i 2  =  -240mA   Please solve this problem using the KCL, KVL, and Ohm’s Law methods of the first two lectures 3.56)   i o  =  2.4 A and P ABS (140  ) =   72.576 W 4.7)  (a) P 3A   (developed) = -30 W (b) P 60V  (developed) = 300W 4.11) i 1  =  23.76 A,  i 2  = 5.33A,  i 3  = 18.43 A,
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Unformatted text preview: i 4 = 15 A, i 5 = 9.77 A, i 6 = 8.66 A 4.17) (a) v o = 50 V , (b) P ABS = 31.875 W, (c) P ABS (3 A) + P ABS (80 V) = -270 W; P Developed (3 A) + P Developed (80 V) = 270 W 4.29) P ABS (20 V) = -602.5 W (As a check on intermediate results, the node voltages you should find in your problem solving sequence are: v 1 = -20.25, v 2 = 10V, v 3 = -29V, i φ = 0.25 A, v ∆ = 10V 1...
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