Question:
I want to reduce fan noise on my PSU. I'm quite confident that All these different designs (pabst...etc) are pretty samiliar so I'll try to do it myself without buying another fan.
the current fan moves about 8.3 cfm, so I figure if I reduce it about 50%, the Decibels will also be reduced a samiliar amount. Seeing that decibels is a log function, i'm guessing a 10dba from original 28 reduction will result in 50% noise decrease, making it virtually inaudible.
So given that, i'm setting out to put a resistor on the 12V to make it around 7V or so. I'm pretty sure i'll trip to start up. Question is:
(i know i can use the 12-5 trick, but I want to use a resistor because i'm a nerd)
what kind of resistor do I need, i haven't had physics since high school but this is what I remember:
I'm guessing because V=IR, and if I want (7/12) of the current, I need 12/7 of the resistance...right...right.
to find the current resistance, I take P=IV, 2.4W=12V*I, and find that the current is 2.4/12 A.
So I take that and plug into V=IR, to get R=144/2.4.
So thats the resistance of the wire Right...?
So I just take 12/7 of that and subtract 60 and get my new resistor right?
it comes out to 42 ohm. So a 40ohm resistor would do the trick right?
(these are all questions in cause you don't know)
But my question is...if the wire itself is a resistor, it should dissapates heat itself right?
So my initial calculation based on the power consumption of the fan is somwhat faulty...my question is...how negligable is this number.
my intital equation should have been (input power)=heat disappation+transferred energy to the fan.
Heat dissapation of a resistor is P=I^2*R and the transferred energy should be p=IV
so the sum should equal 2.4W
Using this function, I get the R is only 30 ohm, so my new resistance should only be 15 ohm???
Can someone verify this for me?