
Are the following four steps correct in determining the power rating of the 38Ω resistor from the above pic?
1) find max current draw of gate from datasheet of triac

2) Calculate voltage drop across 38Ω resistor
v=IR
=(max gate current from triac)x(Resistance at pin 6 of optocoupler)
=0.05 X 38
=1.9V
3) calculate power rating (wattage=currentXvolts)
=0.05 X 1.9
=0.095W
4) Derate max power rating by 2 to yield minimum power rating of 38 ohm resistor
=.095 X 2
=0.2W power rating of resistor