I would like to know how to reduce the brightness of the running lamp in the tail lights of my tow car. It is a MINI Cooper and each tail light contains one single filament running lamp. These lamps also act as the brake lights in normal(not towed) driving. I believe the car's electronics change the voltage from say 5.5v as a running light to 12v as a brake light although I haven't confirmed this. Since I will tow this car with my RV, the running light feed from the RV will range 12v-14.6v. This high of a voltage will appear that the brake lights are illuminated and also create unnecessary heat within the tail light. I don't know how to calculate the proper size resister to safely reduce the brightness to say maybe half. There are two lamps, 21w each that are fed from one line. I have diodes in the circuit so there won't be any type of back feed into the car's electronics. Thank you for your help. Howard
I don't know how to calculate the proper size resister to safely reduce the brightness to say maybe half. There are two lamps, 21w each that are fed from one line.
The relevant formulas are \$R = V / I\$ and \$P = V * I\$ (where V = voltage in volts, I = current in amps, R = resistance in ohms, and P = power in watts).
Calculating the required resistance for an incandescent lamp is a little tricky though, because the filament resistance has a positive temperature coefficient so it reduces at lower voltage (causing higher than expected power draw), while the light output spectrum shifts more towards infrared which is less visible. So you might have to experiment with different resistor values to get the exact effect you want.
Another option might be to simply wire an identical lamp in series. Each lamp should then drop half the total voltage, and you could use the 'ballast' lamp as an indicator that the tail light is on.
So how much current and power would each lamp draw, and what would be the equivalent resistance to do the same job? I tested a 12 V 15 W festoon lamp on a range of voltages from 0 to 12 V, and this is what I got:-
At 12 V the lamp drew 1.22 A, so its resistance was 12/1.22 = 9.84 Ω. At 6V it drew 0.86 A, ~70% of the current at 12 V (a fixed resistor would draw 50%). Its resistance at 6 V was 6/0.86 = 6.98 Ω, ~70% of the value at 12 V.
We can expect that a 12 V 21 W lamp would act similarly. At 12 V it should draw 21/12 = 1.75 A, and the resistance should be 12/1.75 = 6.86 Ω. 70% of that is 1.23 A and 4.8 Ω. So a 4.7 Ω (standard value) power resistor should drop the voltage from 12 V to ~6 V. It would dissipate ~6*1.23 = ~7.4 watts, so should be rated for 15 W or higher (derated 50% to keep surface temperature down), and kept away from anything that doesn't like high temperatures.
So you could put either a 4.7 Ω 15 W resistor or a 12 V 21 W lamp in series to reduce the tail light voltage to ~6 V.
Would it be safer to wire across the Mini's brake light switch, using a relay controlled by your RV's brake light? Then you don't have to break into the Mini's wiring loom (always a bad idea when dealing with a BMW product.) The brake light circuit should be working regardless of the car's key switch position.
Diodes might be more effective in reducing the voltage and current (and therefore power) through your bulbs.
Each diode would drop about 0.7-0.9V, so you would need about 5-8 pieces.
You might need to use 3A diodes, especially if using the same diodes for both bulbs.
A better solution would be to get lower wattage bulbs or replace the bulbs with LEDs, which would use a lot less energy (about 10 times less) and would be easier and less wasteful to control.