I am running a 12 volt DC 25 watt 6 ohm resistor to excite the alternator in my car but it is extremely hot to the touch, could I reduce the heat output by adding another resistor of the same value, in either series or parallel and still achieve being able to excite the alternator?
most alternators use the small amount of residual magnetism persisting within the iron armature to excite and establish current flow in the field coils, without the need for any external excitation source. Can you explain why you are trying to excite the alternator in this way?
Also note the output of the alternator will flow backwards into and out of this excitation circuit and get the dropping resistor really hot unless you have blocking diodes in the circuit. Have you included these?
And finally- remember that the excitation only needs to be applied momentarily, as the armature is first beginning to spin, and then can be disconnected as the alternator self-excites and comes on-line. Are you doing this?
Your current setup is putting 12 volts through a 6 ohm resistor. Ohm's Law say current equals voltage drop divided by resistance. That is 2 Amps of current. Watts equals voltage (12v) times current (2A). That equals 24 watts, the rating of the resistor. To decrease heat you must reduce the voltage or reduce the current.
You can put another 6 ohm 25 watt resistor in series with the first resistor to get 12 ohms. Here is the math: Current equals voltage (12v) divided by resistance (12 ohms). That equals 1 Ampere of current. And watts equals 12 volts times 1 amp current, that is 12 watts.
You now have half the original heat dissipation, and it is spread out over 2 resistors. Wire-wound resistors can be open-wound around a hollow ceramic tube, or they can be cemented into a finned aluminum body.
If the system voltage is 20% higher, at 14.4 volts, then all the current and watt calculations would be 20% higher. So the orignal setup would draw 2.4 amps and dissipate 34.6 watts, and the dual-series resistor setup would dissipate 17.3 watts.
The Ohm's Law for watts is V-squared over R, so 14.4 volts squared over 12 ohms equals 17.3 watts.