Power wire dilemma
#11
Absolutely right. Voltage drop must be considered into the equation. To expand in somewhat simplistic terms:
Ohm's law:
V = I x R
I = V / R
R = V / I
Whereas:
Voltage = V
Current = I (amps)
resistance = R (ohms)
To determine voltage drop per 100 feet given load current and wire gauge:
VD = .2 x IL x 1.26 (AWG-10)
Whereas:
VD = Voltage drop per 100 feet (Volts)
IL = Current load (AMPs)
AWG = Wire gauge
To determine wire gauge necessary given paired wire length, load current, and desired voltage drop per 100 feet:
AWG = 10 Log (VD/IL) +17
Ohm's law:
V = I x R
I = V / R
R = V / I
Whereas:
Voltage = V
Current = I (amps)
resistance = R (ohms)
To determine voltage drop per 100 feet given load current and wire gauge:
VD = .2 x IL x 1.26 (AWG-10)
Whereas:
VD = Voltage drop per 100 feet (Volts)
IL = Current load (AMPs)
AWG = Wire gauge
To determine wire gauge necessary given paired wire length, load current, and desired voltage drop per 100 feet:
AWG = 10 Log (VD/IL) +17
#12
Heres another thought..whats the alt rated at..105a? 130a? You can bet GM factured in a bit of overhead to cover spiked loads without harming anything......Use that aS your starting point....like said before, unless your throwing in a couple 1000w nom amps you won't have any problems.....gee, search the threads on what everybody has alreadyt done and you'll see....nobody has ever mention torching a car from overload...most of em don't even use Caps!
#13
true enough, i'm probably just over thinking this thing. I've just never installed a system in a car with the main battery in the back before, so i was a little bit wary. I get the supplies in this week, i'll let you know how it all goes. I plan to make a thread on it when completed.
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