1) Can't really answer that without knowing the characteristics of the heating elements. I would think there's a good chance it would be in the ballpark though. The easiest way to find out is to try it and measure the relevant voltages.gspd wrote: ↑Mon Oct 17, 2022 9:56 am
2 questions:
1 -Would this resistor drop the voltage from 14V to around 9 or 10 volts, which is what I need, and would itnegatively affectincrease actual power consumption.
2 - Would it need to be mounted on a heat sink plate or would it cool adequately if it was just hanging in the air on its own.
I'm open to alternate solutions.
Thanks again Rob.
1a) The current would be reduced by putting the resistor in series with the two elements (which are themselves in parallel of course).
2) we can come up with an answer to that if we assume a drop of 4 volts across the resistor...
From Ohms law Amps(i)=volts (v) divided by resistance (r) thus 4 volts/2 Ohms = 2 amps.
From the power equation, Watts(w)=Volts (v) x Amps (i) thus 4 volts x 2 amps = 8 Watts.
The resistor is rated at 50 Watts so it will survive quite well. Looking at the spec for a similar resistor, I find that the 50 Watt rating requires the resistor to be bolted to a heat sink but the unmounted rating is 20 Watts (still well within spec).
That being said I would always mount such a component as a) it is relatively heavy and I would want to prevent fatigue fractures of the wires that might occur if it were suspended and b) even though it's within spec, it will get quite warm so it might damage plastic components if it rests against them.
Finallym I would expect the motor speed control unit you mentioned to work quite well but (a) you'd probably have to turn the variable reg up quite high to get any effect and (b) read the reviews before you buy... they aren't universaly complementary.
Rob