# Basic resistor question

I am getting to install resistors for the bulbs on a first generation Genesis and have some questions.
1. Can the wires on the resistor be bent for installation?
2. Can the wires be cut? 3. Why do you need a resistor for each bulb?
Thanks Pat
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In a message on Sun, 10 Jul 2005 14:45:25 GMT, wrote :
"C> I am getting to install resistors for the bulbs on a first generation "C> Genesis and have some questions. "C> "C> 1. Can the wires on the resistor be bent for installation?
Yes. You need to keep the wires (leads) from touching each other or any random metal. Electrical tape, heat shrink tubing, discarded insulation from wire stripping, etc. can be used.
"C> 2. Can the wires be cut?
Yes. You need to leave enough for soldering! :-) Actually they generally should be trimmed as short as you can and still reach the connections you need to reach.
"C> 3. Why do you need a resistor for each bulb?
Current and/or voltage limiting. Incandescent bulbs have a certain voltage rating. Grain-of-wheat (grain-of-rice) bulbs are often 2V bulbs. Putting a full 12volts across them will turn them into extremely bright and very short-lived bulbs! Often modern headlights bulbs are LEDs which are all about 2Volts as well and will also 'burn up' if given a higher voltage. The series resistor drops the voltage by turning the excess voltage into heat. The resistor value is determined by Ohm's Law:
V = I*R     (R = V/I)     (I = V/R)
* == multiplication, / == division V == Voltage in Volts I == Current in Amps R == Resistance in Ohms.
If the track voltage is 12 volts (typical analog H0 scale track voltage, DCC is sometimes higher, but you are probably NOT wiring the bulb to the track in that case, but to the DCC decoder's 'headlight' function output, which would typically be 12 volts) and the bulb (or LED) is rated at 2 volts, the resistor needs to drop 10 volts. Typically a LED uses 20miliamps (.020 Amps). You'll need to check what your bulb actually draws, but lets assume it draws 100miliamps (.100 Amps). This works out to:
20MiliAmp 2V LED:
10 / .020 == 500
100MiliAmp 2V Grain-Of-Wheat:
10 / .100 = 100
You also need to figure the power rating of the resistor. This is:
P = V * I
P == Power in Watts V == Voltage in Volts I == Current in Amps
For the 500 Ohm resistor for the LED:
10 * .020 = .2
A 1/4 Watt (.250W) resistor will do.
For the 100 Ohm for the grain of wheat bulb:
10 * .100 = 1
You'll need a 1 Watt resistor for this.
I have a resistor calculator program as part of my Model Railroad System (http://www.deepsoft.com/MRRSystem /)
"C> "C> Thanks "C> Pat "C> "C> "C>
\/ Robert Heller ||InterNet: snipped-for-privacy@cs.umass.edu http://vis-www.cs.umass.edu/~heller || snipped-for-privacy@deepsoft.com http://www.deepsoft.com /\FidoNet: 1:321/153
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Patrick Carcirieri wrote:

Yes. Best way is to use needle nose plier and hold the wire straight right at the body of the resistor and to bend out past the needle nose. Less chance for breaking the wire at the body of the resistor this way.

Yes. I would suggest using a pair of flush cutters and have the flush side of the cutters towards the resistor. Less mechanical shock sent down the wire to the body of the resistor or other component and less chance for failure of the component.

If the lamp is a low voltage lamp, let's say 6 volts as an example. It will be rated at some specific current flow through the lamp at 6 volts, let's use 0.02 amps, also called 20 milliamps. Using Ohms law we can calculate out that the resistance of the lamp when it is operating is 300 ohms. The lamp will last a good long time at 6 volts, won't get too hot and will not be real bright. Now hook that same 6 volt lamp up to 12 volts and with all other factors being the same you have doubled the current going through the lamp. The original lamp was dissipating 0.120 watts. Now lets see what the same lamp on 12 volts does for dissipation. To get power I am using the formula of I^2 * R where I is the current in amps and R is the resistance in ohms of the lamp. After doing the math the result for this lamp on 12 volts is 0.48 watts. That is four times the dissipation that the lamp is designed for. It will be Real Bright for a Short amount of time.
With the above items known it is easy to see that to hook a 6 volt lamp to a 12 volt power supply there has to be a way to make the lamp see only 6 volts. That is what the resistor does. Ohms law would be used to calculate out that you need a resistor of 300 ohms to keep the current in the circuit down at 0.02 amps. Put this resistor in series with the lamp and the lamp will only see the 6 volts it was designed for. The resistor will have 6 volts dropped across itself. In this instance I would recommend a 1/4 watt resistor to keep the temperature down on the resistor body. a 1/8 watt resistor, while still within the limits of it's ratings, is going to get real hot. The 1/4 watt resistor while dissipating the same amount of power will have significantly more surface area to radiate the heat away. It will stay quite a bit cooler than the 1/8 watt device.
For multiple bulbs the calculations get a little bit more complicated and I don't plan on going into them here. Suffice it to say that as long as ALL the lamps are the same and they are all working, yes you could use just one resistor and put all your lamps on that one resistor. BUT, if one uses different rated lamps on the circuit they all have to be the same voltage. And then one would have to calculate the equivalent resistance of the lamps when they are in parallel. When one of those lamps open up then the rest of the lamps see an increase in voltage. As that happens the power dissipation in the remaining lamps goes up and soon you have to replace all the lamps at the same time.
Sorry I got so long winded.
If interested the Ohms law formula is E = I * R Where E is voltage measured in Volts, I is current measured in Amps and R is resistance measured in Ohms.
Al
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MREG's Resistance Chart might be of some use:
http://www.merg.org.uk/resistor/index.htm
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