what wattage of resistor to use

[Lestat]

i have pretty much my choice of how many ohm resistor i can use,, here at work we have tons., but i am limited to 1/8th watt 1% and 5% tolerance and 1/4% @ 5%.

now logic would dictate 1/8th watt @ 1% as that would limited any bleed off or voltage passthru to 1.8th of a watt and 1% leway.

as that the correct way to view it.

i pretty much can go anywhere from 100 ohm to 1,000,000 ohm.

sorry this is for a vdroop on a mobo

[Lestat]

ok i have a 50k 60k and 100k

they are 1/8th watt @ 1%

now my electrical is horrible i have always shyed away from this type of stuff but i am happy to learn.
i have not seen people talk about anything other than "go grab a 50k resistor and use that" never what the wattage and % rating is on the resistor.

these are resistors we use here at work for PCB boards. ut i dont know what the input rating is on them, ie 12v 5v or 3v

sorry this sounds so stupid of me, but like i said electrical has never been something i have been into.

so i would rather do this right.

[sluggo]

You don't select a 1/8 watt resistor to "bleed off" power. The wattage rating of the resistor you use is determined by how much current is passing through it. How much current passes through it is determined by the value of the resistor and the voltage drop across it.

In your case, the voltage drop is 2V or so. The value of the resistor is (say) 50k. So the current is 2V divided by 50k, or 40 uA (40 millionths of an amp). the power is the current squared times the resistance, so (doing some math) you are dissipating 80 microwatts. The 1/4 watt resistor is plenty.

A faster way to figure it out is just calculate the voltage squared divided by the resistance (4/50000) and you get the same answer, but the current calculation helps explain what really going on.

For basic DC circuits:
V=IxR (Voltage drop (in volts) equals the resistance (in Ohms) times the current (in Amps))
I=V/R
R=V/I
P=VI (Power (in Watts) equals voltage times current)
P=Isquared x R
P=Vsquared/R

PS: The "drop" is merely the voltage change across the resistor. If one side of the resistor has 2V on it and the other side has 0, that's a 2V drop. If one side has 500V and the other side has 498V, it's still a 2V drop.

[Lestat]

oh god i just lost ten years off my life trying to understand that

did i forget to mention math scrambles my brain.. has since high school. i suck at it.

so the quater at 5% should be fine.

i was actually talking to the owner of our company today as he is an engineer and has been for decades and he said the same thing.. in general the 1/4 watt @ 5% is sufficient.

its for vcore droop. incase that changes anything.
just something i want to try on an old crap board i have laying around.

[septim]

Could you use this product for basis with resistors etc.
been trying to brush up on my volts watts amps studies
but its around 8years ago.

How to wire this product for use with cold cathode lights?
Illuminated Bulgin Vandal Switch -Ring Illumination
The LED runs on 3.3V DC. was hoping to wire the LED from either PSU 12V or 5V, suggested resistors are
* 12 V DC source: 490 ohms
* 5 V DC source: 140 ohms

so i tried running 12V thru said 490 ohm but it still came out 12v
any help and corrections would be appreciated