ECE-320 CH02.pdf - Unix.eng.ua.edu

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G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2Therefore,1 1 11= +R 90eq( ) ( Ω −) R+ += 4750 14.2 34.3Ωeq.b) The current through and the power absorbed by the 90-Ω resistor are:IV110=90S90 Ω= =R1( V )1.22 A2110 2S90 Ω= = =R190P134.4 W________________________________________________________________________Solution:Problem 2.43Known quantities:Schematic of the circuit shown in Figure P2.43.Find:The equivalent resistance at terminals a,b in the case that terminals c,d are a) open b) shorted; the same forterminals c,d with respect to terminals a,b.Analysis:With terminals c-d open, Req= ( 360 + 540) (180 + 540) Ω = 400Ω,with terminals c-d shorted, Req= ( 360 180) + (540 540) Ω = 390Ω,with terminals a-b open, Req= ( 540 + 540) (360 + 180) Ω = 360Ω,with terminals a-b shorted, Req= ( 360 540) + ( 180 540) Ω = 351Ω.________________________________________________________________________Solution:Problem 2.44Known quantities:Layout of the site shown in Figure P2.44; characteristics of the cables; rated voltage of the generator; rangeof voltages and currents absorbed by the engine at full load.Find:The minimum AWG gauge conductors which must be used in a rubber insulated cable.2.34
G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2Analysis:The cable must meet two requirements:1. The conductor current rating must be greater than the rated current of the motor at full load. Thisrequires AWG #14.2. The voltage drop due to the cable resistance must not reduce the motor voltage below its minimumrated voltage at full load.KVL :R−VR−V+ I+ R110 V −105V=7.103 AMaxGC1RC=dGC 2+ VM −FL1RV=RC1C1GRC=d2−VI+ V+ VM −MinM −FLM −MinM −Min+ V+ I= 703.9 mΩ=12=[ 703.9 mΩ]150 mRC 2M −FL= 0RC 2= 0Ω= 2.346 mmTherefore, AWG #8 or larger wire must be used.________________________________________________________________________Solution:Problem 2.45Known quantities:Layout of the building shown in Figure P2.45; characteristics of the cables; rated voltage of the generator;total electrical load in the building.Find:The minimum AWG gauge conductors which must be used in a rubber insulated cable.Analysis:The cable must meet two requirements:1. The conductor current rating must be greater than the rated current of the motor at full load. Thisrequires AWG #4.2. The voltage drop due to the cable resistance must not reduce the motor voltage below its minimumrated voltage at full load.KVL :R−VR−V+ I+ R450 V − 446 V=51.57 AMaxSC1RC=dSL−FLC 21+ VRC1V=RC1SRC=d2+ V−VI+ VL−FL=L−MinL−MinL−Min+ V+ I== 77.6 mΩ12[ 77.6 mΩ]85 mRC 2L−FLR= 0C 2= 0Ω= 0.4565 mmTherefore, AWG #0 or larger wire must be used.________________________________________________________________________2.35
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ECE-320 CH02.pdf - Unix.eng.ua.edu

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