Use simple arithmatic:
The resistance of the fan is about 5V/0.38A= 13.2 ohms.
The current with an 8V supply is 8V/13.2 ohms= 0.61A. The resistor value is (19V - 8V)/0.61A= 18 ohms. It dissipates (11V squared)/18 ohms= 6.7W so use a huge 10W resistor.
Use simple arithmatic:
The resistance of the fan is about 5V/0.38A= 13.2 ohms.
The current with an 8V supply is 8V/13.2 ohms= 0.61A. The resistor value is (19V - 8V)/0.61A= 18 ohms. It dissipates (11V squared)/18 ohms= 6.7W so use a huge 10W resistor.
That's awesome. I guess what I was worried about was, how do we know this is correct, if we don't know the Amperage of the 19V line? Is that irrelevant? I think you only need one (V I or R) to determine the other two.
I broke this down to make sure I understand the concepts behind it. I think I'm starting to understand.
Fan volt rating is 5v / .38A
5 divided by .38 is 13.157. Round it to 13.2
Goal is to run on 8 volts
8 (voltage you want) divided by 13.2 (ohms of device) is .60606 - round to .61 A
Tapping into a 19V rail. We don't want 19 volts. We only want 8 .
19 (voltage you have) minus 8 (voltage you want) is 11.
11 (voltage you want) divided by .61 = 18 Ohms (This is the resistor size needed)
To determine the wattage - square your new voltage, and divide it by the number of ohms
11 volts squared is 121. 121 divided by 18 (Ohms) is 6.72
So, we need a resistor capable of dissipating more than 6.72 watts. A 10 watt resistor is sufficient.
I read that there is another way to determine wattage. Current times the voltage at a given point.
The current here is
.61 amps. Multiply that by
11 (the amount of voltage we plan to dissipate.)
.61 times
11 is
6.71! Everything checks out!
I'm also going to keep reading Ohm's law definitions and tutorials until it really clicks.