gunrunnerjohn posted:You're right, of course, I just divided by two. I should have thought about it. In any case, I'd limit my planned power dissipation to around half the rated capacity.
Got it. Good advice! It's hurting my head thinking about it.
I actually confirmed it using Ohm's Law: V=I * R and W=I * V
So substituting: W= I*(I * R), or W = (I)^2 * R
Re-arranging: I^2 = W/R or I = (W/R)^.5
Now if you use W = 0.25 watt and R = 0.7 ohm; I works out to 597 ma, or call it 600 ma for simplicity.
Whew, time for a gin and tonic!
Rod