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

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G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2Find:a) The resistance of each coilb) The power dissipation of each of the other two possible arrangements.Analysis:(a) For the parallel connection, P = 2000 W. Therefore,or,2000 =1=1 1+ = R2( 220) ( 220)R12( 220) +R 1R225121R.For the series connection, P = 300 W. Therefore,or,( 220)12300 =R + R1R + RSolving, we find that122+1R2123= .484R = 131. 6Ωand = 29. 7Ω(b) the power dissipated by R 1 alone is:( 220)P R= = 368W1Rand the power dissipated by R 2 alone is( 220)P R= 1631W2R22= .121R .2________________________________________________________________________2.22
G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2Section 2.5, 2.6: Resistance and Ohm’s LawSolution:Problem 2.29Known quantities:Diameter of the cylindrical substrate; length of the substrate; conductivity of the carbon.Find:The thickness of the carbon film required for a resistance R of 33 KΩ.Assumptions:Assume the thickness of the film to be much smaller than the radiusNeglect the end surface of the cylinder.Analysis:d dR = ≅σ ⋅ A σ ⋅ 2πa⋅ ∆td∆t= =R ⋅ 2πa⋅σ33⋅1039 ⋅10Ω ⋅ 2.9 ⋅106−3mS⋅m2π⋅1⋅10________________________________________________________________________Solution:Problem 2.30Known quantities:The constants A and k; the open-circuit resistance.Find:The rated current at which the fuse blows, showing that this happens at:1I = .AkRAssumptions:Here the resistance of the fuse is given by:R = R0[ + A(T − )]01 T0where T0, room temperature, is assumed to be 25°C.We assume that:T T = kP− 0where P is the power dissipated by the resistor (fuse).Analysis:2R = R0 (1 + A⋅∆T) = R0(1 + AkP)= R0(1 + AkI R)2R − R AkI R =0R0−3m2.23
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