Wire
Wire
WIRE CHART
12 Vol
t
acceptable cable size (mm²)
Amps
0.5
1
1.5
2
3
4
5
7.5
10
15
20
25
30
40
60
80
100
125
1
0.4
0.4
1.84
1.84
1.84
1.84
1.84
1.84
1.84
1.84
2.9
4.6
4.6
7.9
13.6
25.7
32.2
49.2
2
0.4
0.4
1.84
1.84
1.84
1.84
1.84
1.84
1.84
1.84
2.9
4.6
4.6
7.9
13.6
25.7
32.2
49.2
Cabl
e Length (metres)
5 10 15 20 25
0.4 0.4 1.84 1.84 1.84
0.4 1.84 1.84 1.84 1.84
1.84 1.84 1.84 1.84 2.9
1.84 1.84 1.84 2.9 4.6
1.84 1.84 2.9 4.6 4.6
1.84 2.9 4.6 7.9 7.9
1.84 4.6 4.6 7.9 7.9
2.9 4.6 7.9 13.6 13.6
4.6 7.9 13.6 13.6 13.6
4.6 13.6 25.7 25.7 25.7
7.9 13.6 25.7 25.7 32.2
7.9 25.7 25.7 32.2 49.2
13.6 25.7 32.2 49.2 49.2
13.6 25.7 49.2 49.2
25.7 49.2
25.7 49.2
32.2
49.2
30
1.84
1.84
2.9
4.6
7.9
7.9
13.6
25.7
25.7
32.2
49.2
49.2
1
0.4
1.84
1.84
1.84
1.84
1.84
1.84
1.84
2.9
4.6
4.6
7.9
13.6
25.7
32.2
49.2
2
0.4
1.84
1.84
1.84
1.84
1.84
1.84
1.84
2.9
4.6
4.6
7.9
13.6
25.7
32.2
49.2
All the methods ofdeterminingvoltage drop on this
page are for DC only. AC electricitybehaves quite
differently.
M etriccables are specifiedbythe copper area (in square
millimetres),the number ofstrands ofwire and the
number ofconductors or cores in each sheath. The
voltage drop is the same regardless ofvoltage,assuming
that amps,distance and cross sectional areas are the
same. Ifthe wattage remains the same for different
voltages,the amps can be calculatedbydividingwatts
byvolts.
TheFormul
a
Ifyou needto calculate the voltage drop under a given
set ofcircumstances,there is a formula bywhich it can
be determined.
Let:
A = cross sectional area ofcable in (mm²
)
L = route length in metres
I= current measuredin amps
R = resistance ofcable ( )
aluminium = 0.028
resistance ofcopper = 0.017
steel = 0.18
Example:
You have a power point connectedto a power source.
The route length is 8 metres. Ifthe wire is 4.6 mm²
multi-strandedcopper cable andthe expectedcurrent is
expectedto be 10amps,we have:
acceptable cable size (mm²)
Amps
1
2
3
4
5
7.5
10
15
20
25
30
40
60
80
100
125
Copper Conductor
Voltage Drop = 2x L x Ix R ÷A
24 Vol
t
Cabl
e Length (metres)
5 10 15 20 25
0.4 1.84 1.84 1.84 1.84
1.84 1.84 1.84 1.84 1.84
1.84 1.84 1.84 1.84 2.9
1.84 1.84 1.84 2.9 4.6
1.84 1.84 2.9 4.6 4.6
1.84 2.9 4.6 4.6 7.9
1.84 4.6 4.6 7.9 7.9
2.9 4.6 7.9 13.6 13.6
4.6 7.9 13.6 13.6 25.7
4.6 7.9 13.6 25.7 25.7
4.6 13.6 25.7 25.7 25.7
7.9 13.6 25.7 25.7 32.3
13.6 25.7 32.2 49.2 49.2
25.7 25.7 49.2 49.2
32.2 32.2 49.2
49.2 49.2
Pl
asti
c Insul
ati
on
Wh at:
Low voltage power systems often operate at rather high
current levels. Ifthe interconnectingcables are too small,
a large proportion ofthe power available will be wastedin
the cable itself. This loss can be reducedbyusinga larger
cable,but this increases costs. The acceptable maximum
voltage drop for DC loads is 5% of nominal battery
voltage. The chart and the formula on this page are
providedto help you in selectingthe best cost /power loss
compromise.
A =5
30
1.84
1.84
2.9
4.6
4.6
7.9
13.6
25.7
25.7
25.7
32.2
49.2
NOTE: The Cable Length in the above tables are route
length which is halfthe total conductor length. Ifthe
positive and negative leads are different lengths an
average must be taken.
L =9
I= 10
R = 0.017
Voltage drop can then be calculatedto be 0.58volts. If
this figure is consideredto be acceptable it wouldavoid
spendingmore moneyon larger wire.
mm²
per metre
30 m rol
l ampaci
ty
1.84
twi
n sheathed
W IR-M02 W IR-302
15 amps
2.9
twi
n sheathed
W IR-M03 W IR-303
20 amps
4.6
twi
n sheathed
W IR-M05 W IR-305
25 amps
7.9
si
ngl
e (bl
ackor red)
W IR-M08 W IR-308
45 amps
13.6
si
ngl
e (bl
ackor red)
W IR-M14 W IR-314
70 amps
25.7
si
ngl
e (bl
ackor red)
W IR-M25 W IR-325
90 amps
32.
si
ngl
e (bl
ackor red)
W IR-M32 W IR-332 110 amps
49
si
ngl
e (bl
ackor red)
W IR-M49 W IR-349 150 amps
or extra low voltage
NOTE: The above cables are ratedf
133