Dropping Resister formula | O Gauge Railroading On Line Forum

BaldocOGR Forum Member

I need to drop the voltage for my Ross switchstand lights. The formula for the required resister is [Voltage Drop divided by Amperage of lamp.] My multimeter indicates that the Ross lights are 40 milliamps each.

My required voltage drop is 6 volts. So, 6 volts divided by .040 amps results in a resister of 150 ohms. This size works well for a connection with only one lamp. The switchstand light is nice and bright.

However, I have one place on the layout where it is more convenient to have 3 lights connect to the lighting bus at one terminal. Does this change the formula. Do I triple the amperage in the divisor, resulting in a resister value of only 50?

5/12/158:39 PM

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stan2004OGR Forum Member

Correct.

The power through each single-lamp 150 ohm resistor is Voltage x Current = 6V x 0.04 A = 0.24 Watts so use 1/2 Watt resistors which are of course very common and inexpensive.

As you probably bought a bunch of 150 ohm resistors, put 3 of them in parallel (just like the 3 lamps in parallel) to "make" a 50 ohm, 1.5 Watt resistor. That way you don't have to buy another resistor value with a 3x higher power rating.

Alan MancusOGR Forum Member DIGITAL SUBSCRIBER

i calculate you need a 50 ohm resister.

since you have 3 lamps at .040 amp the I =0.12 amps the E= 6 volt.

you divide 6 volts by 0.12 amps the 3 light currents are added together to = 0.12

. hence 6 divided by 0.12 amps = a 50 ohm resister!

OK

hope this helps

Alan

BaldocOGR Forum Member

Thanks very much for the quick replies. You are right, Stan, I have a ton of 150 ohm resisters. Will try your approach.

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