Question

CUP 6. A wire has a diameter of 0.032 inches. The AWG rating of this wire is most likely to be O A. 20 O B. 14. O C. 12. O D. 18.​

Answer 1

Answer:

A. 20

Explanation:

The AWG rating of the wire can be determined by applying the formula;

$d_{i}$ = 0.005 x $92^{\frac{(36 - s)}{39}} }$

where $d_{i}$ is the diameter of wire in inches, and s is the diameter of wire in AWG.

Given that $d_{i}$ = 0.032 inches, then;

0.032 = 0.005 x $92^{\frac{(36 - s)}{39}} }$

$\frac{0.032}{0.005}$ =  $92^{\frac{(36 - s)}{39}} }$

6.4 = $92^{\frac{(36 - s)}{39}} }$

Find the log of both sides to have,

log 6.4 = log $92^{\frac{(36 - s)}{39}} }$

log 6.4 = $(\frac{36 -s}{39})$ log 92

$\frac{log 6.4}{log92}$ = $(\frac{36 -s}{39})$

0.410523 = $(\frac{36 -s}{39})$

36 - s = 0.410523 x 39

         = 16.0104

⇒ s = 36 - 16.0104

      = 19.9896

s = 20

Therefore, the AWG of the wire is 20.