Answer to Question #170588 in Electrical Engineering for Douglas Carter

Question

The resistivity and temperature coefficient of resistance of Evanohm are 800 CM-Ω/ft and

20 x 10-6 /C, respectively, at 20C. A #40 round wire has a diameter of 0.00314 in. Find the resistance of a 14.79 in long sample of a #40 round Evanohm wire at (a) 20C, and

(b) 520C.

Answer 1

Q170588

The resistivity and temperature coefficient of resistance of Evanohm are 800 CM-Ω/ft and

20 x 10-6 /C, respectively, at 20C. A #40 round wire has a diameter of 0.00314 in. Find the resistance of a 14.79 in long sample of a #40 round Evanohm wire at

(a) 20 ⁰C, and

(b) 52 ⁰C.

Solution:

We are given resistivity, ρ = 800 CM-Ω/ft.

Convert this to Ω.m by using the conversion factor, 1 Ω.m = 60153.49.3 circular mil Ω/ft.

resisitivity in Ω.m = $800\space CM-Ω/ft * \frac{1\space Ω.m }{601530493.4\space CM-Ω/ft } = 1.33 * 10^{-6} Ω.m$

Temperature coefficient of resistance, α  = 20 * 10^-6 / ⁰C.

Answer a:

To find the resistance of the 14.79-inch long sample of a # 40 Evanohm wire at 20 ⁰C.

Resistance R is related to the length and cross-sectional area of the wire by the relation:

$R = ρ \frac{L}{A}$

Length of the wire, L $= 14.79 \space inch * \frac{1 \space m}{39.37 \space inch } = 0.376 m$

diameter of wire = 0.00314 inch

radius of wire, r = 0.00314 inch/2 = 0.00157 inch.

radius of wire in meter, r $= 0.00157inch * \frac{1m}{39.37inch} = 3.988 * 10^{-5} meters.$

cross sectional Area, A $= Π r^2 = 3.1447 * ( 3.988 * 10^{-5})^2 = 5.00 * 10^{-9} m^2.$

plug A= 5.00 * 10 ^-9 m²,

resisitivity, ρ 1.33 * 10^{-6} Ω.m

and lenght, L = 0.376 meter in the formula, we have

$R = ρ \frac{L}{A} = 1.33 * 10^{-6} Ω.m * \frac{0.376m}{5.00*10^{-9} m^2 }$

Resistance, R at 20 ⁰C = 100.00 Ω.

Answer b :

To find the resistance of the 14.79-inch long sample of a # 40 Evanohm wire at 520 ⁰C.

We will use the formula

R = R _ref [ 1 + α (T-T_ref )]

resistance at 20 ⁰C will be R _ref.

T _ref = 20 ⁰C.

T = 520 ⁰C.

Temperature coefficient of resistance, α  = 20 * 10^-6 / ⁰C.

R = 100.00 Ω  [ 1 + 20 * 10^-6 / ⁰C.  (520 ⁰C - 20 ⁰C )]

= 100.00 Ω  [ 1 + 20 * 10^-6 / ⁰C. *  500 ⁰C ]

= 100.00 Ω  [ 1 + 0.01]

R = 101 Ω.Resistance, R at 520 ⁰C = 101 Ω.

Answer a : Resistance, R at 20 ⁰C = 100.00 Ω. Answer b : Resistance, R at 520 ⁰C = 101.00 Ω.