Solution 2 - ECE 205 Section A, B, and C. Homework #2...

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Unformatted text preview: ECE 205 Section A, B, and C. Homework #2 Solutions 1.18 Calculate the power absorbed or supplied by each element in Fig. 1.29 – 4A + 3V – Io = 3A + 10 V – 1 24V R1 4A + 6V 1 9V R1 + – _ 40 60 2400 /150 16 40 60 50 R1 40 60 2400 /150 16 40 60 50 2 + – _ 40 60 2400 /150 16 40 60 50 R1 40 60 2400 /150 16 40 60 50 3Io 2 3A – 5V + (b) + – R1 40 60 2400 /150 16 40 60 50 (a) R1 40 60 2400 /150 16 40 60 50 Figure 1.29 For Prob. 1.18 R1 40 60 2400 /150 16 40 60 50 R1 40 60 2400 /150 16 40 60 50 Solution: Note, we will express absorbed power as positive terms and delivered power as negative power absorbed. (a) For the 9-V voltage source, p = -4 (9) = –36 W For element 1, p = 4 (6) = 24W For element 2, p = 4 (3) = 12W (b) For the 24-V voltage source, p = 24 (-3) = –72W For the current-controlled voltage source, p 3 I0 (3) 27W = 27 W For element 1, p = 10 (3) = 30 W For element 2, p = 3 (5) = 15 W 1.20 Since p = 0 -306 + 612 + 3V0 + 28 + 282 - 310 = 0 72 + 84 + 3V0 = 210 or 3V0 = 54 V0 = 18 V 2.4 (a) (b) i = 15/100 = 150 mA i = 15/150 = 100 mA 2.7 Find the number of branches and nodes in each of the circuits of Fig. 2.71. 12V 1 2 -+ 1A 4V + _ 10 5 3 1 5 4A 2 3 (a) Figure 2.71 For Prob. 2.7 Solution (a) 1 2 (b) 4V + _ 10 5 3 6 branches and 5 nodes. (b) 12V 2 -+ 1A 4A 1 5 3 7 branches and 5 nodes. ...
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