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# Answer to Question #170588 in Electrical Engineering for Douglas Carter

Question #170588

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.

1
2021-03-11T06:56:06-0500

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-\u03a9\/ft * \\frac{1\\space \u03a9.m }{601530493.4\\space CM-\u03a9\/ft } = 1.33 * 10^{-6} \u03a9.m"

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

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 = \u03c1 \\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 "= \u03a0 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 = \u03c1 \\frac{L}{A} = 1.33 * 10^{-6} \u03a9.m * \\frac{0.376m}{5.00*10^{-9} m^2 }"

### Resistance, R at 20 0C = 100.00 Ω.

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-Tref )]

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]

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

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