If you choose to use a wire gauge with too small a Constant, or attempt to push more power across the wire run than calculations indicate, the result could be dangerous or expensive. If the wire over heats, it can start a fire. If certain bulbs do not receive enough power (due to excessive voltage drop) the bulbs will burn out prematurely.
Wire Constant “C” - Resistance “R” and Wattage “E/R” formulas
* gauge of wire has these values of Resistance ”R” and Constant “C”
| 10 GA -> |
R = .001020 -> |
E/R = 11,764 |
C = 5,800 |
| 12 GA -> |
R = .001588 -> |
E/R = 7,547 |
C = 3,700 |
| 14 GA -> |
R = .002525 -> |
E/R = 4,752 |
C = 2,300 |
| 16 GA -> |
R = .004016 -> |
E/R = 2988 |
C = 1,500 |
Watts (on cable) x Length of wire run |
= Voltage Drop on wire |
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Constant of wire |
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Voltage drop should not exceed 1volt on a 12 v system; Use V = 1 as a standard.
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If you know the Wattage (W) and Constant (C) of the wire,
and desire the Length of run (L):
L = C ÷ W
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If you know the Wattage (W) and Length of run (L),
And desire to know what Gauge wire; use wire gauge with constant greater than:
C = L x W
Ex: 50 watt lamps, with 100 foot run
C = 50 x100 = 5,000 = 10GA (C= 5,800; greater than 5,000)
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If you know the Constant (C) of the wire and the Length of run,
But desire to know how many watts you can supply:
W = C ¸ L
Ex: 12 GA wire (C = 3,700) with 50 foot run
W = 3,700 ÷ 50 = 74 watts
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