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bulbsfuseswire.jpg Bulb, Fuse, & Wire Current Specs

For bulb resistance, use Ohm's law: V (Volts) = R (Ohms) x i (Amps), so:
R (Ohms) = V (Volts) / i (Amps)
A 194 bulb has 14 / .72 = 19.4444444 Ohms resistance
An 1815 bulb has 14.4 / 0.2 = 72 Ohms resistance
For an incandescent bulb labeled in volts & watts: V x V / W = R
A 15W @ 12V bulb has 12 x 12 / 15 = 9.6 Ohms

- When the diameter of a wire is doubled, the AWG will decrease by 6. (e.g., No. 18 AWG is about twice the diameter of No. 24 AWG.) 
- When the cross-sectional area of a wire is doubled, the AWG will decrease by 3. (e.g., two No. 18 AWG wires have about the same cross-sectional area as a single No. 15 AWG wire.)
- When gauge is decreased by ten gauge numbers (from No. 18 to 8 ), the area and weight increase, and the resistance decreases, each by a factor of approximately 10.

See also:

[url=http://www.supermotors.net/registry/media/1007183][img]http://www.supermotors.net/getfile/1007183/thumbnail/wiregaugecurrentlengtht.gif[/img][/url] . [url=http://www.supermotors.net/registry/media/849725][img]http://www.supermotors.net/getfile/849725/thumbnail/fusesblades.jpg[/img][/url] . [url=http://www.supermotors.net/registry/media/830776][img]http://www.supermotors.net/getfile/830776/thumbnail/fusiblelinkrepair.jpg[/img][/url] . [url=http://www.supermotors.net/registry/media/748790][img]http://www.supermotors.net/getfile/748790/thumbnail/tsb951411inoplamps.jpg[/img][/url]

https://www.fleet.ford.com/truckbbas/non-html/2002/281.pdf 
https://www.fleet.ford.com/truckbbas/non-html/1997/c37_39_p.pdf
bulbsfuseswire.jpg | Hits: 7310 | Posted on: 10/25/10 | View original size (316.18 KB)

Bulb, Fuse, & Wire Current Specs

For bulb resistance, use Ohm's law: V (Volts) = R (Ohms) x i (Amps), so:
R (Ohms) = V (Volts) / i (Amps)
A 194 bulb has 14 / .72 = 19.4444444 Ohms resistance
An 1815 bulb has 14.4 / 0.2 = 72 Ohms resistance
For an incandescent bulb labeled in volts & watts: V x V / W = R
A 15W @ 12V bulb has 12 x 12 / 15 = 9.6 Ohms

- When the diameter of a wire is doubled, the AWG will decrease by 6. (e.g., No. 18 AWG is about twice the diameter of No. 24 AWG.)
- When the cross-sectional area of a wire is doubled, the AWG will decrease by 3. (e.g., two No. 18 AWG wires have about the same cross-sectional area as a single No. 15 AWG wire.)
- When gauge is decreased by ten gauge numbers (from No. 18 to 8 ), the area and weight increase, and the resistance decreases, each by a factor of approximately 10.

See also:

. . .

https://www.fleet.ford.com/truckbbas/non-html/2002/281.pdf
https://www.fleet.ford.com/truckbbas/non-html/1997/c37_39_p.pdf
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