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

A. 20

Step-by-step explanation:

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

`d_(i)` = 0.005 x
`92^{((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^{((36 - s))/(39)} }`

`(0.032)/(0.005)` =
`92^{((36 - s))/(39)} }`

6.4 =
`92^{((36 - s))/(39)} }`

Find the log of both sides to have,

log 6.4 = log
`92^{((36 - s))/(39)} }`

log 6.4 =
`((36 -s)/(39))` log 92

`(log 6.4)/(log92)` =
`((36 -s)/(39))`

0.410523 =
`((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.