Elgar SmartWave SW1750A, Operation Manual

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SW 5250A•SW 3500A•SW 1750A Installation
Operation Manual 2-17
The following notes apply to Table 2–9 and to the power cable definition:
1. The above figures are based upon insulated copper conductors at 25°C (77°F),
two current carrying conductors in the cable plus a safety (chassis) ground.
Columns 3 and 4 refer to “one way” ohms and IR drop of current carrying
conductors (e.g., a 50-foot cable contains 100 feet of current carrying conductor).
2. Determine which wire gauge for the application by knowing the expected peak
load current (I
peak
), the maximum tolerated voltage loss (V
loss
) within the cable,
and the one way cable length.
The formula below determines which ohms/100 feet entry is required from
Column 3. Read the corresponding wire gauge from Column 1.
(Column 3 value) =
V
loss
/[I
peak
x 0.02 x (cable length)]
Where:
Column 3 value =
Entry of the table above.
Cable length =
One way cable length in feet.
V
loss
=
Maximum loss, in volts, permitted within cable.
Special case: Should the V
loss
requirement be very loose, I
peak
may exceed the
maximum amperes (Column 2). In this case, the correct wire gauge is selected
directly from the first two columns of the table.
Example: A 20 ampere (I
peak
) circuit which may have a maximum 0.5 volt
drop (V
loss
) along its 15-foot cable (one way cable length) requires (by formula)
a Column 3 resistance value of 0.083. This corresponds to wire gauge size 8
AWG.
If the cable length was 10 feet, the Column 3 value would be 0.125 and the
corresponding wire gauge would be 10 AWG.