# V1 ix il 13 12114 solucon a a i1 6

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Unformatted text preview: 1 i = − mA 8 V1 = i ( 40 k ) + 5 = 0 V1 − i ( 40 k ) − 5 = 0 ⎛1⎞ V1 = ⎜ − m ⎟ ( 40 k ) + 5 ⎝8⎠ V1 = 0 Problem (5) a a I1 + 6 kΩ V1 − 3mA Ix IL 2 kΩ 3kΩ 6 mA b b I1 = ? V1 = ? Ix = ? IL = ? 13 1/21/14 SoluCon a a I1 + 6 kΩ V1 IL 2 kΩ 3kΩ 6 mA − b Ix 3mA b a a I1 + 6 kΩ V1 3mA 1.2 kΩ 2 kΩ a + + 1.5 kΩ 3mA I x 2 kΩ V1 V1 − − b IL 3mA 3kΩ − b b I1 = 3mA (2k ) 4 kΩ I x = 1.5 mA 3mA (1.2 k ) 7.2 kΩ Ix = I1 = 0.5 mA IL = 3mA (1.5 k ) 4.5 kΩ I L = 1mA Problem (5) 2 kΩ 2 kΩ a 3kΩ 6 kΩ 12 kΩ b Rab = ? 14 1/21/14 Problem (5) 2 kΩ a 2 kΩ c 6 kΩ 3kΩ b 2 kΩ b b c 2 kΩ b c a 4 kΩ 2 kΩ 6 kΩ b b b d 4 kΩ b a 3kΩ 2 kΩ 3kΩ b c c a 12 kΩ b a 2 kΩ d b b Rab = 4 kΩ Problem(6) a a c 4 kΩ d 4 kΩ 3kΩ 6 kΩ b 5 kΩ 8 kΩ 3kΩ b b b Rab = ? 15 1/21/14 Problem (6) a 5 kΩ a a a 5 kΩ c 3kΩ 6 kΩ b b b a a 6 kΩ b b a b c 4 kΩ b 5 kΩ b a 12 kΩ b c 6 kΩ 1kΩ b b Rab = 3kΩ a 3kΩ 6 kΩ b c 3kΩ b 1.5 kΩ b 5 kΩ 3kΩ a b a a 6 kΩ 8 kΩ 3kΩ b a d 4 kΩ 3kΩ 6 kΩ b 4 kΩ c b b 16...
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## This document was uploaded on 03/12/2014 for the course EEL 3004C at University of Central Florida.

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