# my question?

Suppose you wish to fabricate a uniform wire from a mass m of a metal with density ρm and resistivity ρ. If the wire is to have a resistance of R and all the metal is to be used, what must be the length and the diameter of this wire? (Use any variable or symbol stated above as necessary.)

(a) l =

(b) d =

### 2 Answers

- billrussell42Lv 710 months ago
Resistance of a wire in Ω

R = ρL/A

ρ is resistivity of the material in Ω-m

L is length in meters

A is cross-sectional area in m²

A = πr², r is radius of wire in m

density ρm ?? density is not resistivity x mass ?? I suspect you are using ρ for both resistivity and density, a bad idea.

let d be density

assume the wire is a cylinder

V = πr²L

d = m/V = m/πr²L

πr²L = m/d

πr² = m/dL

r = √(m/dLπ)

R = ρL/A = ρL/πr²

substitute

R = ρL/(m/dL) = ρdL²/m

L² = Rm/ρd

L = √(Rm/ρd) ⬅

r = √(m/dLπ)

diameter = 2√(m/dLπ) ⬅

(you can plug value for L into the above, but it gets complicated)

- 10 months ago
A young man owns a canister vacuum cleaner marked "530 W [at] 220 V" and a Volkswagen Beetle, which he wishes to clean. He parks the car in his apartment parking lot and uses an inexpensive extension cord 18.0 m long to plug in the vacuum cleaner. You may assume the cleaner has constant resistance.

(a) If the resistance of each of the two conductors in the extension cord is 0.500 Ω what is the actual power delivered to the cleaner?

_ W

(b) If instead the power is to be at least 520 W, what must be the diameter of each of two identical copper conductors in the cord he buys? (Enter the minimum diameter.)

_ mm

(c) Repeat part (b) assuming the power is to be at least 527 W. (Enter the minimum diameter.)

_mm