Chap 18 Solns-6E - CHAPTER 18 ELECTRICAL PROPERTIES PROBLEM...

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CHAPTER 18 ELECTRICAL PROPERTIES PROBLEM SOLUTIONS 18.1 This problem calls for us to compute the electrical conductivity and resistance of a silicon specimen. (a) We use Equations (18.3) and (18.4) for the conductivity, as σ = 1 ρ = I l VA = I l V π d 2 2 = (0.1A) 38 x 10 3 m ( ) (12.5 V )( π ) 5.1 x 10 3 m 2 2 = 14.9 ( -m) -1 (b) The resistance, R , may be computed using Equations (18.2) and (18.4), as R= l σ A = 51 x 10 3 m 14.9 ( Ω− m) 1 [] ( π ) 3 m 2 2 = 166.9 18.2 For this problem, given that an aluminum wire 10 m long must experience a voltage drop of less than 1 V when a current of 5 A passes through it, we are to compute the minimum diameter of the wire. Combining Equations (18.3) and (18.4) and solving for the cross-sectional area A leads to A= I l V σ From Table 18.1, for aluminum σ = 3.8 x 10 7 ( -m) -1 . Furthermore, inasmuch as A = π d 2 2 for a cylindrical wire, then 1
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π d 2 2 = I l V σ or d= 4 I l π V σ = (4)(5 A)(10 m) ( π )(1V) 3.8 x 10 7 ( Ω− m) 1 [] = 1.3 x 10 -3 m = 1.3 mm 18.3 This problem asks that we compute, for a plain carbon steel wire 3 mm in diameter, the maximum length such that the resistance will not exceed 20 . From Table 18.1, for a plain carbon steel, σ = 0.6 x 10 7 ( -m) -1 . If d is the diameter then, combining Equations (18.2) and (18.4) leads to l=R σ A=R σπ d 2 2 = (20 )0 .6 x10 7 ( 1 ( π ) 3x10 3 m 2 2 = 848 m 18.4 Let us demonstrate that, by appropriate substitution and algebraic manipulation, Equation (18.5) may be made to take the form of Equation (18.1). Now, Equation (18.5) is just J = σ E But, by definition, J is just the current density, the current per unit cross-sectional area, or J = I / A . Also, the electric field is defined by E = V / l . And, substituting these expressions into Equation (18.5) leads to I A = σ V l But, from Equations (18.2) and (18.4) 2
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σ = l RA and I A = l RA V l Solving for V from this expression gives V = I R , which is just Equation (18.1). 18.5 (a) In order to compute the resistance of this copper wire it is necessary to employ Equations (18.2) and (18.4). Solving for the resistance in terms of the conductivity, R= ρ l A = l σ A From Table 18.1, the conductivity of copper is 6.0 x 10 7 ( -m) -1 , and l σ A = 2m 6.0 x 10 7 ( Ω− m) 1 [] ( π ) 3x10 3 m 2 2 = 4.7 x 10 -3 (b) If V = 0.05 V then, from Equation (18.1) I = V R = 0.05 V 4.7 x 10 3 = 10.6 A (c) The current density is just J= I A = I π d 2 2 = 10.6 A π 3 m 2 2 = 1.5 x 10 6 A/m 2 (d) The electric field is just E= V l = 0.05 V = 2.5 x 10 -2 V/m 3
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18.6 When a current arises from a flow of electrons, the conduction is termed electronic ; for ionic conduction , the current results from the net motion of charged ions.
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Chap 18 Solns-6E - CHAPTER 18 ELECTRICAL PROPERTIES PROBLEM...

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