I have an air-cooled resistor which lies in series with a 55W lamp, and it's used to dip the lamp on a car when driving during the day. The car runs 12V DC.

It's resistance is 1.7 ohms, and the coil had burned out due to dampness and such.

My plan is to replace the wire wound unit with a standard high wattage resistor of a similar resistance.

My maths goes as follows:

The resistance of the 55W lamp is:

E^2/W

144/55 = 2.61818 ohms.

The combined resistance of the two components is 2.61818 + 1.7 = 4.3818 ohms.

This will dissapate the following watts in total:

E^2/R = W

144/4.3818 = 32.863 W.

The wattage per component is the ratio of the resistance:

(W/R)*R1 = R1W

(32.863/4.3818)*1.7 = 12.465W for the 1.7 ohm resistor.

(W/R)*R2 = R2W

(32.863/4.3818)*2.61818 = 19.1979W for the 2.61818 ohm light.

So that means I need a minimum wattage of a ~16W resistor.

Is my calculation of the wattage correct or am I making a mistake?