Electronics experts, 2 questions

[bsoplinger]

If I stick 2 resistors in parallel I know I end up with half the resistance, that is if I put 2 400 Ohm resistors in parallel I end up with 200 Ohm of resistance. But I have a question about the power. If I need say a half W of power handling, will using 2 1/4W 400 Ohm resistors in parallel do the trick? Ie, does putting 2 of the same resistors in parallel get me half the resistance but twice the power handling?

Is there a good source of LEDs for model railroading, vs just 'where can I buy LEDs?' I can go to any of the standard online places that sell resistors and capacitors and the like and find pages of info on LEDs they offer but I have no idea what to look for to get ones that are like the Miniatronics I get via Walthers:
http://www.walthers.com/exec/productinfo/475-1231010
Yeloglo White LEDs pkg(10) 3mm for 15 bucks.

[engineshop]

I get my LEDs from this place.

http://cgi.ebay.com/ebaymotors/50X-3mm-SUPER-BRIGHT-WHITE-LED-LAMP-10K-mcd-50-Resistor_W0QQcmdZViewItemQQcategoryZ33713QQihZ002QQitemZ120123730904QQrdZ1

It even comes with resisitors for 12V operation.

[up1950s]

"does putting 2 of the same resistors in parallel get me half the resistance but twice the power handling?"

Yes . The watt capacity is the addition of the 2 individual resistor watt ratings . If you had 2 resistors of unequal watts , such as 1/4 and 1/2 watt , and they were put in parallel with each other , its capacity would be 3/4 watt , though it is better to use 2 resisters if equal watts in such a ckt . If put in series the watt capacity would be the smallest value of 1/4 . The 1/4 watt resistor would be a watt bottle neck in this series example .

I am showing the un-equal watt resistor's to make the math example clearer only , and I am not an expert , but this is what I beleive to be the facts .

[Iain]

If I stick 2 resistors in parallel I know I end up with half the resistance, that is if I put 2 400 Ohm resistors in parallel I end up with 200 Ohm of resistance. But I have a question about the power. If I need say a half W of power handling, will using 2 1/4W 400 Ohm resistors in parallel do the trick? Ie, does putting 2 of the same resistors in parallel get me half the resistance but twice the power handling?

1  = 1  . 1
Rt  R1+R2

(Ignore the period over the +, I had to put it there to preserve formatting)
Rt = total resistance of both resistors
R1 and R2 are the two resistors in question.  Although it comes out being half the resistance when you have resistors of equal value, at other times this does not work.

[wm3798]

While the experts are standing around... will this work?  I'm getting close to wiring the lighting on my turntable.  I want to try to sacrifice as few lights as possible to the Learning Curve Gods...

Lee

[up1950s]

While the experts are standing around... will this work?  I'm getting close to wiring the lighting on my turntable.  I want to try to sacrifice as few lights as possible to the Learning Curve Gods...

Lee

Yes it will . What I do on things like this is put a resistor to limit the voltage to the highest brightness I want , then in series with that add a variable resister to control the voltage lower for effect . This does 2 things , it protects the lights , and gives you control . Many times I use a variable resistor to make it easy to decide on the value of the resistor . I start with the highest setting , then lower it to what I want or need by sight and measurements . Then I remove that variable resistor and measure its value to determine what resister to solder into the ckt . I have made a resistor box , and cap box to make this easier for me , but a spliced in variable resistor works just as well for saving from doing the math .

[Nelson]

While the experts are standing around... will this work?  I'm getting close to wiring the lighting on my turntable.  I want to try to sacrifice as few lights as possible to the Learning Curve Gods...

Lee

NO! do not do this. If one bulb were to burn out, the voltage would increase to the remaining bulbs and quckly burn them out. The 1.5v g.o.r. bulbs are very sensitive to over voltage. You either need a resistor for each bulb, or wire the bulbs in series through one resistor, but it is much better to use a voltage regulator rather than a resistor for these small bulbs.

[TrainCat2]

A good place for electronics online is Mouser
http://www.mouser.com

[mmyers]

There's a handy caculator here: http://www.1728.com/resistrs.htm

[DKS]

I believe the simplest and best way to power 1.5 v GOR bulbs is by using a pair of silicon rectifier diodes and a ballast resistor. The voltage drop across the diodes will always be ~.75 v each, regardless of the supply voltage. This way, there is no chance of over-powering the bulbs, and no issue with calculating the correct resistor value for the first circuit shown in this thread, which will change depending on the supply voltage as well as number of bulbs being powered. It's also simpler and much cheaper than a voltage regulator circuit.

[DKS]

Forgot to add that, for an AC power supply, wire two pairs of diodes back-to-back; otherwise, the lamps will be dim. You can use a full-wave rectifier in place of the four diodes for a very simple circuit.