# 52v 12 ma d 4 ma d kvl b c d d b vbc vcd

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Unformatted text preview: + vbc + ( 20Ω ) (10 m ) = 0 10Ω d c vcd Iz 2 mA vbc = − 0.52V 12 mA − + d 4 mA − d KVL b- c- d- d- b − vbc − vcd + ( 30Ω ) ( I z ) = 0 +0.52 − vcd + ( 30Ω ) ( −2 m ) = 0 vcd = 0.46V P12 mA = + ( vbc ) (12 m ) P12 mA = + ( −0.52 ) (12 m ) P12 mA = −6.24 mW (deli ) P4 mA = + ( vcd ) ( 4 m ) P4 mA = + ( 0.46 ) ( 4 m ) P4 mA = +1.84 mW (abs ) 9 1/21/14 Ix Problem(3) b a Ix = ? d Ix = ? KCL node a : b a i1 + 4 m = 1m i1 = −3mA d d Ix Assume i1 as shown KCL node b: c 4 mA 2Ix Problem(3) 1mA 1mA c i1 4 mA 2Ix 2 I x = i1 + I x d d d I x = i1 I x = −3mA 10 1/21/14 2 kΩ Problem (4) • + Vx 5 kΩ 24V − Vx = ? • 6V SoluCon 2 kΩ 2 kΩ a b i Assume the current i as shown 5 kΩ 24V Ohm’s law: Vx = i ( 5 k ) e KVL (a- b- c- d- e- a) 6V − i ( 2 k ) − i ( 5 k ) − i ( 2 k ) − 6 + 24 = 0 + Vx − c 2 kΩ d d i ( 9 k ) = 18 i = 2 mA Vx = ( 2 m ) ( 5 k ) Vx = 10V 11 1/21/14 SoluCon 2 kΩ a b Using voltage division rule: 5 kΩ 24V vx = ( 24 − 6 )V ( 2 kΩ + 5 kΩ + 2 kΩ ) ( 5 kΩ ) − c e 6V 18 vx = ( 5 kΩ ) 9 kΩ + Vx 2 kΩ d d Vx = 10V Problem (5) 20 kΩ 40 kΩ + V1 = ? Vx 2 + − V1 + Vx − 5V − 12 1/21/14 V1 = ? Problem (5) Assume the current i as shown 20 kΩ a i KVL (d- a- b- c- d) Vx − i ( 20 k ) − i ( 40 k ) − 5 = 0 2 Vx 2 40 kΩ b + + − + Vx c − V1 5V − Ohm’s law: Vx = i ( 40 k ) d d SubsCtute to get: i ( 40 k ) − i ( 20 k ) − i ( 40 k ) − 5 = 0 2 KVL (d- b- c- d)...
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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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