Heat margin of resistor

 I measured the temperature of resistor of 3.2L DME. It is over 90deg.(C). I think it's too hot.


  So I calculated the power consumption of the resistor.
R=0.1/2=0.05ohm
V=RI=0.05*10=0.5V
P=area of left graph=1/2*0.5*10*6/10=1.5W.
P(one resistor)=1.5/2=0.75W
Engine is running at 2700 rpm when above calculation. If I increase the revolution of engine, power consumption of resistor is increase too.
  BOSCH chose 1W resistor. I replased it 3W.





  The temperature decleased to 69deg.(C). I satisfied.



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