
Landscape Lighting Calculating Voltage Drop across a Wire

StarTip 1021 
Gardening Tips for successful and beautiful Landscapes and Gardens 
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
Constant of wire
Voltage drop should not exceed 1 volt on a 12 v system; Use V = 1 as a standard.

If you know the Wattage (W) and Constant (C) of the wire,
and desire the Length of run (L):
L = C ¸ W

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 x 100 = 5,000 = 10GA (C= 5,800; greater than 5,000)

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

© 2009, Star Nursery, Inc. Copy Provided courtesy of Star Nursery www.StarNursery.com