I = V/R
I = 10/0.9
I = 11.11
P = VxI
P = 10x11.11
P = 111W
You must be getting quite a temperature! What resistance would you get with 3 in parallel? And how are you insulating the wires from each other?
Graeme
Excellent Graeme!Graeme1858 wrote: ↑Sat Jun 17, 2023 4:30 pm
If your wire is 7Ω and you have 7V across it, then
I = V/R
I = 7/7
I = 1A
And:
P = VxI
P = 7x1
P = 7W
At 12 V it would be:
I = V/R
I = 12/7
I = 1.7A
Then:
P = VxI
P = 12x1.7
P = 20.5W
Your temperature is too high at the lowest voltage, so try adding another 7Ω resistor in series to reduce the current flow.
Then:
I = V/R
I = 7/(7+7)
I = 0.5A
So:
P = VxA
P = 7x0.5
P = 3.5W
On full at 12V:
I = V/R
I = 12/(7+7)
I = 0.86A
And:
P = VxA
P = 12x0.86
P = 10.3W
If that gives you a temperature that's too low then change the 7Ω to a 3Ω resistor in series. I don't know if you can get 7Ω and 3Ω resistors, but you get the picture.
Then on low at 7V
I = V/R
I = 7/(7+3)
I = 0.7A
So:
P = VxA
P = 7x0.7
P = 4.9W
On full at 12V:
I = V/R
I = 12/(7+3)
I = 1.2A
And:
P = VxA
P = 12x1.2
P = 14.4W
Hope that helps a bit more than my last post!
Graeme
oh , i was thinking that running parallel would would lower resistance thus lower heat, i got it, lower resistance = more heat with this ohmic material. it gets very confusing since a normal circuit with copper would be that a higher resistance would produce more heat.Graeme1858 wrote: ↑Sat Jun 17, 2023 7:21 pm
I = V/R
I = 10/0.9
I = 11.11
P = VxI
P = 10x11.11
P = 111W
You must be getting quite a temperature! What resistance would you get with 3 in parallel? And how are you insulating the wires from each other?
Graeme
yobbo89 wrote: ↑Sat Jun 17, 2023 9:28 pm oh , i was thinking that running parallel would would lower resistance thus lower heat, i got it, lower resistance = more heat with this ohmic material. it gets very confusing since a normal circuit with copper would be that a higher resistance would produce more heat.
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