quick question?

[blanik]

question is:

i want to put 20k resistors on high-brightness LEDs (give me sufficient glow) but i only have 1/8 Watts 20k resistors, would this fry the 1/8 W or should i join 2X 10K 1/4 W...

thx

[Faber]

how many volts are you trying to drop?

P=(V^2)/R

(power equals voltage squared divided by resistance)  if the answer is less than 1/8 watt, then you should be fine.

*edit* or you could use current... P=(I^2)R

[blanik]

it's 9V, so according to your trick, it would be:

9X9=81   /   20000ohm =    0.004 W    is that right?

[Faber]

Almost.  Let's just say the LED wants 2.0V to run.  If you drop all 9V of the battery through the resistor, then the LED has nothing to run on.  So you want to drop V(in)-V(led) through the resistor.

If the LED needs 2.0V, then that's 9-2=7V and that's what you use in the formula.

So... 7x7=49/20k is .00245 watts.

[blanik]

so obviously the 1/8W resistor would have done the job...  too bad, i already put the 2X10k @ 1/4W... i had place anyway... 

[Faber]

Yeah, either one will do the trick.  I probably just should have said yes... hmmm...