I am restoring a 1930 Stutz, the headlamps originally used a resistor to dim the single filament bulbs from high to low beam.
The vehicle is now 12 volts, to dim the lamps the correct amount I have determined a 3 volt drop will be fine.
Question- what resistor is required to reduce the bulbs (two 48 watt).
Peter H.
Best Answer
Total full load is 2x 48 watts = 96 watts. On 12V, this is a resistance of 1.5 ohms because power = \$V^2/R\$. With only 9V across the lamps a current of 6 amps flows hence, the dropper resistor has to drop 3V at 6 amps and this means it has a resistance of 0.5 ohms.
However, a word of caution, as the lights dim they don't run as hot and their resistance lowers so you might need a resistor that is a tad under 0.5 ohms.
The resistor's power dissipation is 18 watts (\$I^2R\$) so choose one that can easily handle this power without burning.