lol...overkill
Let me throw some variables into your experiment.
I think the stock wiring is 20awg.
Resistance of 20awg wire is appx. .01 ohms per foot
We will say that there is 20ft of 20awg wire.........20*.01=.2 ohms
Let's say an amp will deliver 100w into 4 ohms...
P=(I^2)*R
I=sqrt(P/R)
=sqrt(100/4.2)
=4.879
So your wire will carry 4.879 amps of current at max draw...keep in mind this is DC current and the AC current that your amp produces isn't likely to reach this amount of current regularly.
Let me throw some variables into your experiment.
I think the stock wiring is 20awg.
Resistance of 20awg wire is appx. .01 ohms per foot
We will say that there is 20ft of 20awg wire.........20*.01=.2 ohms
Let's say an amp will deliver 100w into 4 ohms...
P=(I^2)*R
I=sqrt(P/R)
=sqrt(100/4.2)
=4.879
So your wire will carry 4.879 amps of current at max draw...keep in mind this is DC current and the AC current that your amp produces isn't likely to reach this amount of current regularly.
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