- #1
fulmenatrum
- 10
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Recently, I made a laptop cooler using an old CPU fan (rated 12V, 0.37A) I removed from a desktop. The cooler was comprised of the cooling fan, a base for the laptop to go on, and a DC power adapter (rated 12V, 1A)...nothing too complicated. Now from what I know, the current being drawn by the circuit should be equivalent to what the device, in this case the old CPU fan calls for, or 0.37A. However, when I connect the multimeter in series with the circuit, the rating i get is 0.28A.
The following calculations are all based on what I thought to be correct which may be totally wrong so please correct me by all means. Using V=IR, I found what I thought the resistance of the fan should be using its ratings (so, R=V/I...R=12V/0.37A...R=32.42Ohms). Next, I found the resistance of the fan using the measurements I got from the whole circuit which included the power adapter, cheap wire, and fan (R = 12V/0.28A = 42.85Ohms).
My question is if I could credit the 10.43Ohms (42.85Ohms - 32.42Ohms) to the wiring (and thus, the lower reading of current) since it is in series with the fan or whether it would be totally incorrect to assume this as the fan itself is composed of other electrical components I have no way of including in my calculations.
Any help would be much appreciated. Thank you!
The following calculations are all based on what I thought to be correct which may be totally wrong so please correct me by all means. Using V=IR, I found what I thought the resistance of the fan should be using its ratings (so, R=V/I...R=12V/0.37A...R=32.42Ohms). Next, I found the resistance of the fan using the measurements I got from the whole circuit which included the power adapter, cheap wire, and fan (R = 12V/0.28A = 42.85Ohms).
My question is if I could credit the 10.43Ohms (42.85Ohms - 32.42Ohms) to the wiring (and thus, the lower reading of current) since it is in series with the fan or whether it would be totally incorrect to assume this as the fan itself is composed of other electrical components I have no way of including in my calculations.
Any help would be much appreciated. Thank you!