Sizing Cables

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Very high currents can flow in cables connecting inverter to the batteries and between the batteries. Cables must be adequately sized to suit the maximum current drawn by the inverter. Low resistance links and tight connections are essential to maintain even use of a battery bank.

In order to minimise the voltage drop between battery and inverter, and so maintain inverter efficiency, cables should be kept as short as possible and be sufficiently sized.

Recommended sizes (2m long) are:Conversion of American Wire Gauge to Metric

Conversion of American Wire Gauge to Metric
Size AWGAmp Rating per single conductorArea mm²
20 7 0.52
18 10 0.82
16 15 1.31
14 20 2
12 25 3.31
10 40 6.68
8 65 8.37
6 95 13.3
4 125 21.15
2 170 33.62
1 195 42.41
1/0 230 53.50
2/0 265 67.43
3/0 310 85
4/0 360 107.20
250 405 126.70
300 445 152
500 620 253.40
750 785 300.66
1000 935 506.70

 

Battery to Inverter Cable Size (2m long)
Inverter WattageBattery Voltage
12 V24 V48V
150 W 10 mm² 6 mm² -
250 W 16 mm² 6 mm² -
500 W 35 mm² 10 mm² -
1000 W 50 mm² 25 mm² -
1500 W 50 mm² 35 mm² -
2000 W 70 mm² 50 mm² -
2500 W 95 mm² 70 mm² 50 mm²
3000 W - 95 mm² 50 mm²
3500 W - 95 mm² 70 mm²
4500 W - - 70 mm²


To calculate an acceptable size of wire to do a particular job, it is necessary to know the wattage or amperage of the appliance, it's distance from the battery, and the system's voltage

In general most 12 V lighting and appliances will work on voltages as low as 11-11.5 V. But if possible, wiring should be sized so that losses are less than 2% ie a 0.25 V drop.
(See Table 2.)

The following equation is useful in choosing the required wire:

R = E / I x L

R = Resistance of wire in Ohm/m
E = Maximum allowable voltage drop in the wiring in volts
I = Current drawn by appliance in amps = Wattage/Voltage
L = Length of wire of the complete circuit in metres (count twice - both in & out).

For Example:
If a 60W (60/12 = 5A) appliance needs to be situated 10m from a 12V battery and the maximum allowable voltage drop is 2% (0.25 V):

R = 0.25 Volt / [(5 Amps)x(20m)]  =  0.0025 Ohm/m.

Therefore resistance of the wire should be less than 0.0025 Ohm/m. Looking at Table 1 this means that 6 mm² wire or thicker must be used. The thicker the wire the better the efficiency.

Voltage drop becomes less of a problem with higher voltage systems.

Table 1

Cross Sectional AreaMaximum Current Rating (Amps)Resistance
Enclosed Cable In Conduit Or TrunkingCable Clipped Direct To Open SurfaceOhm / metre
mm² Twin 3-Core Twin 3-Core Per Core
1.0 11 9 12 10 0.018
1.5 13 12 15 13 0.012
2.5 18 16 21 18 0.0074
4.0 24 22 27 24 0.0046
6.0 30 27 35 30 0.0031
10.0 40 37 48 41 0.0018
16.0 53 47 64 54 0.0012
25.0 60 53 71 62 0.00073
35.0 74 65 87 72 0.00049

Table 2

Maximum length (in m) of cable from power source to load for less than 2% voltage drop.  (For a 12 V system.)   (For 24 V systems cables can be twice as long; 240 V wiring can be 20 times as long)

Cable Size (mm² cross-sectional area)
Amps 1 1.5 2.5 4 6 10 16 25 35 50 75 100
1 7 10.91 17.65 28.57 42.86 70.6 109.1 176.5 244.9 - - -
2 3.53 5.45 8.82 14.29 21.4 35.3 54.5 88.2 122.4 171.4 - -
4 1.76 2.73 4.41 7.14 10.7 17.6 27.3 44.1 61.2 85.7 130.4 -
6 1.18 1.82 2.94 4.76 7.1 11.7 18.2 29.4 40.8 57.1 87 117.6
8 0.88 1.36 2.2 3.57 5.4 8.8 13.6 22 30.6 42.9 65.25 88.2
10 0.71 1 1.76 2.86 4.3 7.1 10.9 17.7 24.5 34.3 52.2 70.6
15 - 0.73 1.18 1.9 2.9 4.7 7.3 11.8 16.3 22.9 34.8 47.1
20 - - 0.88 1.43 2.1 3.5 5.5 8.8 12.2 17.1 26.1 35.3
25 - - - 1.14 1.7 2.8 4.4 7.1 9.8 13.7 20.9 28.2
30 - - - - 1.4 2.4 3.6 5.9 8.2 11.4 17.4 23.5
40 - - - - - 1.8 2.7 4.4 6.1 8.5 13 17.6
50 - - - - - - 2.2 3.5 4.9 6.9 10.4 14.1
100 - - - - - - - 1.7 2.4 3.4 5.2 7.1
150 - - - - - - - - - 2.3 3.5 4.7
200 - - - - - - - - - - 2.6 3.5