I don't know if this is the right place to ask this, but I'll give it a try.

I have a PC with a rather noisy fan which I'd like to replace so I got a new one with lower RPM and of better quality. However, there is one little problem: The PC won't boot with the new fan. I contacted the manufacturer of the motherboard and they told me that there is a "safety check" done on cold boot on the CPU fan connector to determine whether there is a fan connected or not.

Apperantly the new fan doesn't use enough current (on boot) so I need to increase the load.
(The fan works perfectly fine when connected while the PC is running. The only problem is a cold boot)

The original fan is: 12v 0.5A
The new fan is: 12v 0.18A

If I connect 2 of the new fans in paralell the PC boots just fine. So I believe I need to add something that increases the load from 0.18A to about the double. (0.18 x 2 = 0.36A)

I would assume adding a resistor in paralell to the fan is all that's needed. I've been looking at Ohm's law but I haven't figured out how it's done yet. Once I know what resistor to use, does the direction matter when connecting it?

I'd be very thankful if anyone could help me.

 

So you want 0.18A to flow through the resistor. R=V/I, so R=12/0.18=66.6ohm
power dissipated is P=V.I, which is 2.16W, so you would want to use at least 5W rated resistor.

 

So you want 0.18A to flow through the resistor. R=V/I, so R=12/0.18=66.6ohm
power dissipated is P=V.I, which is 2.16W, so you would want to use at least 5W rated resistor.

Thank you!
Would this be a safe "fix and forget" solution? Or would I have to worry about heat being generated and possible resistor explotions?

 

if you use the rule of having the resistor rated for twice the acual dissipated power you should be ok. Just keep the resistor somewhere where it gets enough airflow.