Solution 3

# Solution 3 - v = 10 V For loop 2, –2 + 8 + 3i x =0, i x =...

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ECE 205 Section A, B, and C. Homework #3 Solutions 2.12 For loop 1, -20 -25 +10 + v 1 = 0 v 1 = 35v For loop 2, -10 +15 -v 2 = 0 v 2 = 5v For loop 3, -v 1 +v 2 +v 3 = 0 v 3 = 30v 2.13 I 2 7A I 4 1 2 3 4 4A I 1 3A I 3 At node 2, 3 7 0 10 2 2     I I A At node 1, I I I I A 1 2 1 2 2 2 12   At node 4, 2 4 2 4 2 4 4     I I A At node 3, 7 7 2 5 4 3 3   I I I A Hence, I A I A I A I A 1 2 3 4 12 10 5 2     , , , + v 1 - + v 2 - + v 3 - 25v + + 10v - + 15v - + 20v - loop 1 loop 2 loop 3 2A

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2.15 For loop 1, –12 + v +2 = 0,
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Unformatted text preview: v = 10 V For loop 2, –2 + 8 + 3i x =0, i x = –2 A 2.18 Applying KVL, -30 -10 +8 + I(3+5) = 0 8I = 32 I = 4A-V ab + 5I + 8 = 0 V ab = 28V 2.19 Applying KVL around the loop, we obtain –20 + 10 + 3i –(–4) = 0 i = 2A Power dissipated by the resistor: p 3 = i 2 R = 4(3) = 12W Power absorbed by the sources: p 20V = 20 ((–2)) = –40 W p 10V = 10 (2) = 20W p 4V = (–4)(–2) = 8W...
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## This note was uploaded on 02/03/2010 for the course ECE 201 taught by Professor C during the Spring '10 term at Miami University.

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Solution 3 - v = 10 V For loop 2, –2 + 8 + 3i x =0, i x =...

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