text: # 5 ix iy 25 vab 20 ix 125 a iy 125 a 4 12114 example 1
  
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  Unformatted text preview: = ix + iy + 2.5 ix + iy = 2.5 vab 20 ix = 1.25 A iy = 1.25 A 4 1/21/14 Example 1Ω 4Ω v1 7Ω 10 A 3Ω 6Ω v2 v1 = ? v2 = ? Find : P7 Ω = ? P1Ω = ? Hint: Use circuit reduction technique. Solution: 1Ω a a b 4Ω v1 3Ω / /6Ω = 7Ω 10 A c 3Ω d d 1Ω b 4Ω 7Ω 10 A v1 c 2Ω d v2 d d a a 6Ω ( 3) ( 6 ) = 2Ω ( 3) + ( 6 ) d v2 d 5 1/21/14 1Ω a a b 4Ω 7Ω 10 A v1 c v2 2Ω d d 1Ω + 4 Ω + 2Ω = 7Ω d a a 7Ω 10 A d 7Ω d d a a 7Ω 10 A 7Ω 7Ω / /7Ω = 3.5Ω d d d a a vad 10 A d 3.5Ω vad = (10 ) ( 3.5 ) = 35V d 6 1/21/14 1Ω a a 7Ω 10 A b i1 4 Ω i d i= Ohm’s Law: d d 35 7 i = 5A P7 Ω = ( i ) ( 7 ) i1 = 5 A P1Ω = ( i1 ) (1) i2 = i1 Current division: 3Ω 3Ω + 6Ω 2 i2 = v2 = ( i2 ) ( 6 ) Ohm’s Law: 2 v1 = 20V v1 = ( 5 ) ( 4 ) Ohm’s Law: 6Ω i2 10 = 5 + i1 KCL: v1 c 3Ω d vad = 35V 5 A 3 v2 = 10V Review Problem (1) a Head light(1) • a • + • − 12V b b • • a Head light(2) Given:head light draws 3A P12V = ? 7 1/21/14 SoluCon: a Head light(1) • a • + • − 12V b b • • a Head light(2) KCL node (a) a I = 3 + 3 = 6A 3A I P12V = − (12 ) ( 6 ) = −72W 12V P12V = 72W (delivered ) b• Problem (2) 40Ω Ix 12 mA Iy Ix = ? Iy = ? Iz = ? 3A 20Ω 30Ω 10Ω 2 mA 4 mA Iz P12 mA = ? P4 mA = ? 8 1/21/14 SoluCon Ix KCL node C: I x + 12 m = 4 m a Iy 12 mA b c I x = − 8 mA 4 mA I x + I y = 2m KCL node a: 2 mA −8 m + I y = 2 m d I y = 10 mA KCL node b: Iz d I z = I y − 12 m I y = I z + 12 m d I z = 10 m − 12 m I z = −2 mA Assume the voltages : vbc vcd Ix KVL a- c- b- a () − ( 40Ω ) ( I x ) + vbc + ( 20Ω ) I y = 0 a 40Ω Iy b + vbc 20Ω 30Ω − ( 40Ω ) ( −8 m )...
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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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