Chap 18 Solns-6E

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

This preview shows pages 1–5. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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 = 1 r = I l VA = I l Vp d 2 2 = (0.1 A) 38 x 10-3 ( 29 (12.5 29( 29 5.1 10-3 2 2 = 14.9 ( -29-1 (b) The resistance, R , may be computed using Equations (18.2) and (18.4), as R = l = 51 x 10-3 14.9 ( - 29-1 [ ] (29 5.1 10-3 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 From Table 18.1, for aluminum = 3.8 x 10 7 ( -m)-1 . Furthermore, inasmuch as A = 2 2 for a cylindrical wire, then 201 d 2 2 = I l Vs or d = 4 I l = (4)(5 A)(10 m) ( 29( 1 29 3.8 10 7 ( - 29-1 [ ] = 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, = 0.6 x 10 7 ( -m)-1 . If d is the diameter then, combining Equations (18.2) and (18.4) leads to l = R = 2 2 = (20 29 0.6 10 7 ( - 29-1 [ ] (29 3 10-3 2 2 = 848 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 = 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 = But, from Equations (18.2) and (18.4) 202 = l RA and I A = l RA 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 = = From Table 18.1, the conductivity of copper is 6.0 x 10 7 ( -m)-1 , and R = l = 2 6.0 10 7 ( - 29-1 [ ] (29 3 10-3 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 (c) The current density is just J = I A = I 2 2 = 10.6 3 10-3 2 2 = 1.5 10 6 / 2 (d) The electric field is just E = V l = 0.05 V 2 m = 2.5 x 10-2 V/m 203 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....
View Full Document

## This homework help was uploaded on 03/14/2008 for the course MSE 250 taught by Professor Jabbour during the Spring '08 term at University of Arizona- Tucson.

### Page1 / 43

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

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online