I’m not a complete dunce with electricity (but I’m close)[D)] and I do find it fun. Am working on doing some turnout indicator lights on my control panel. I’m stumped on a few things:
Resistors can be used to reduce voltage to a light bulb, right?
But how do I know the current-draw of a light bulb? (I even have a little multi-purpose type meter I bought decades ago and never learned to use… will it help?)
Then how do I know what resistor to use? I see them rated as watts and as ohms. What’s up with that?
In experimenting (and while being too cheap to buy resistors when I’m obviously lost as to what to buy) I’ve had success using a Christmas light in series with an LED (use the Christmas bulb like a resistor – the LED lights, the bulb doesn’t which is what I want) Is this dumb?
I know. You’re thinking, “This guy really shouldn’t be playing with wires.”[:)] But I will figure it out!! (With your help).
Thanks for taking time to help!
Hi jxtrrx,
"Resistors can be used to reduce voltage to a light bulb, right? " Well, sort of. It has the same effect. Actually the voltage stays the same, the resistor adds to the total load keeping the current in check so the light doesn’t blow out. I wouldn’t try to find out much with the meter, myself.
“Then how do I know what resistor to use? I see them rated as watts and as ohms. What’s up with that?” The watt rating is how much current the resistor can take before it burns up. Generally resistors come in .5 watt, 1 watt, etc. up to about 5 watts. For the small little LEDs, GOW, GOR, etc. bulbs used on the layout, a .5 watt would be fine for up to 2 or 3 bulbs in a circuit. Ohms is the measure of resistance that the resistor puts into the circuit. I have a little box from Loys Toys that has several different resistors wired into it for testing what ohm resistor works best in an application. It starts at 1000 ohms and goes down to 120 ohms I think. If it were me, I would spend a couple of dollars (they are very cheap even at RS) and get an assortment of different resistors and try them out to see which one works best. Start with a high rated resistor and work down to where you get the brightness you want. When you install the resistor, make sure there is airspace around it as they will get quite hot in use.
I’d suggest that you go to the library and get a basic electricity/electronics book, or possibly buy one of the “wire your layout” books on the market if they cover some basic Ohm’s Law stuff. There’s probably too much to teach all at once here in the forum. But I’ll try to answer your questions as best I can:
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Yes, resistors can be used to reduce voltage and also to limit current.
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If you didn’t get any specs with the bulbs, then yes, learning to use the meter will allow you to measure their current draw.
3a) Ohm’s Law is your friend here. Lots of good Web sites with info on how it’s applied. Here’s one example: http://www.wiringfordcc.com/gorhlite.htm#a16
3b) Watts are how much energy the resistor is rated to dissipate without overheating. Ohms is the measure of how much resistance to current flow a resistor has. See the Ohm’s Law discussion for more on how the two interact.
- No, it’s not dumb, but it might not be the best solution.
P.S. You mentioned turnout indicator lights on your control panel. If you’re using Tortoise or other slow-motion turnout motors, you can usually wire LED’s in series with them without using a resistor. Check the Tortoise instruction sheet for details.
HTH ,
Steve
In DC volts times amps equals watts. If a motor or other device runs on twelve volts and draws one forth of an amp it is using three watts. Howmus is incorrect in one statement. Cutting a resistor into a circuit DOES reduce the voltage. That’s why those locomotives with 1.5 volt headlights need a resistor in series with a DCC decoder’s headlight output to drop the 12 volt output to 1.5 volts. Of course if it’s a P2K loco you get a Digitrax DH163LO decoder and elimiate the need for the resistors. Don’t get too hung up on all this electrical stuff. Just do what makes sense and you should be fine.
Cheers
OK. Try this. Place a meter in series in a circuit with a resistor in line. If your power source is 12 volt, even with the resistor, your circuit is still 12 v. The resistor adds resistance (ohms) to the circuit limiting the current that flows. It has the same effect that lowering the voltage would have since V=IR. Here is a good page that will show this: http://www.physics.uoguelph.ca/phyjlh/Fendt/phe/ohmslaw.htm If you increase the resistance, the voltage stays the same but the current is lowered. If you keep the same resistance and lower the voltage which is the potential across the circuit, you will also reduce the current draw. At least that is how I was taught Ohm’s law way back when…
That would depend on the impedence of the meter you’re using. Older style analog meters will introduce more of a load to a circuit (lower ohms per volt rating) than most of the newer digitals. Placing a meter in series with a resistor of the same value as the meter’s impedence will show about half the circuit’s voltage. However, this is the normal way for connecting an Amp meter, in series with the load. Connecting a voltmeter like this, in many cases, will fry it.
At the risk of getting more complicated, voltage across a series circuit will be divided depending on the resistance or impedance of the components in the circuit, but the total voltage will be the output of the power supply (unless it’s overloaded). Example: Two 100 ohm resistors in series across a 12 volt power supply will drop 6 volts each, measured by connecting the voltmeter leads across each resistor
Sorry, to get back to your original question, if it’s a standard incandescent bulb, you can measure the resistance with an ohm meter. It should be one of the functions on the multimeter you were talking about.
P.S. You mentioned turnout indicator lights on your control panel. If you’re using Tortoise or other slow-motion turnout motors, you can usually wire LED’s in series with them without using a resistor. Check the Tortoise instruction sheet for details.
Radio Shack sells LED panel lights with resistors already installed. They’re around a dollar or so each, come in 3 colors (red, green, and orange) and work great with the Tortoise.
You are right! I knew that… Had a brain F@&t when I wrote it.
“At the risk of getting more complicated, voltage across a series circuit will be divided depending on the resistance or impedance of the components in the circuit, but the total voltage will be the output of the power supply (unless it’s overloaded). Example: Two 100 ohm resistors in series across a 12 volt power supply will drop 6 volts each, measured by connecting the voltmeter leads across each resistor (parallel), the normal way you measure voltage.”
Exactly! The circuit is still a 12v circuit. The current across the lamp (hence the “effective” voltage) is lowered. But it is still a12v circuit. That is why locos with 1.5 volt lamps need a resistor in the circuit. The voltage is still 12 volts but It is shared with both the lamp and the resistor. The total potential of the circuit is 12 volts. Right?
Measuring the resistance of a bulb when it is cool is not very useful. The bulb’s resistance changes as it heats. You can estimate the resistance by dividing the voltage rating squared by the power rating. For example, a .5 Watt, 12 Volt bulb would have a resitance (in use) of around 144/.5=288. Now were going to make an assumption just to help us along. We’re going to assume that the bulb has about that resistance whenever it is lit, no matter what the voltage. I have no idea how reasonable that is, but it should be “good enough” to do some quick calculation. Let’s say we want to only drop 9 volts across the bulb. That’s 3/4 of the total, so we need the bulb to be 3/4 of the resistance. So, a total resistance of around 400 Ohms, leaving a resitor of about 100 Ohms. What wattage? Well, we’ve got 12 volts across 400 Ohms total, which gives 144/400 Watts, or about .36W. a quarter of that or a little less that a tenth of a Watt is disappated by the resistor, so a 1/4 W resistor should work fine. (Please check my math before the fire, rather than after).
The equations you need:
V=Voltage in Volts
I=Current in Amps
P=Power in Watts
R=Resistance in Ohms
V=IR
P=VI
You susbstitute some things to get the other handy things:
P=IIR
P=(V*V)/R
There will be a test Tuesday.
V=IR
Time to brush up on your algebra! (who’d a thunk model railroading was the place where you’d actually have to use it?[:D]
In summary:
Voltage: E(or V) = I times R
Current(Amps): I = E divided by R
Resistance (Ohms): R = E divided by I
[quote]
QUOTE: Quoted from:
http://www.electronics-tutorials.com/basics/ohms-law.htm
What are the ohms law formulas?
To make it much easier for you I have put all the relevent formulas together for you here complete with worked examples of ohms law. You will notice the formulas share a common algebraic relationship with one another.
For the worked examples voltage is E and we have assigned a value of 12V, Current is I and is 2 amperes while resistance is R of 6 ohms. Note that “*” means multiply by, while “/” means divide by.
For voltage [E = I * R] E (volts) = I (current) * R (resistance) OR 12 volts = 2 amperes * 6 ohms
For current [I = E / R] I (current) = E (volts) / R (resistance) OR 2 amperes = 12 volts / 6 ohms
For resistance [R = E / I] R (resistance) = E (volts) / I (current) OR 6 ohms = 12 volts / 2 amperes
Notice how simple it is?
Now let’s calculate power using the same examples.
For power P = E2 / R OR Power = 24 watts = 122 volts / 6 ohms
Also P = I2 * R OR Power = 24 watts = 22 amperes * 6 ohms
Also P = E * I OR Power = 24 watts = 12 volts * 2 amperes
That’s all you need for ohms law - remember just two formulas:
for voltage E = I * R and;
for power P = E2 / R
You can always determine the other formulas with elementary algebra.
Ohms law is the very foundation stone of electronics!
Knowing two quantities in ohms law will always reveal the third value. I suggest you print these formulas out and paste them onto scrap cardb
They may work okay, but if you’re using a Tortoise with the recommended 12v supply, the resistor is unnecessary and will only slow down the tortoise. If you buy your LED’s and mounting clips at an electronics store, instead of a cell phone store like RS, you’ll get them for probably half that dollar for each indicator. Plus, you can get red/green bipolar LED’s (not bi-color with three leads) that will change from red to green depending on which way your turnout is thrown. Not to knock RS, they can be a lifesaver sometimes. But for stuff like this, they’re expensive.
Steve
Please remember that diodes do not follow Ohms Law. They have their own math, but its way simplier than anything else.
Diodes only allow current to flow in one direction…we use to them to recitfy AC into DC. So when current is flowing, the diode is said to be ‘on’ or forwarding.
When forwarding current, a regular diode will have 0.7 volts across it. So if you had a 12V (DC) source, and one diode, and one light bulb, and the diode is in the forward direction the light will come on and receive 11.3 volts. If the diode is in reverse the light will not come on, the light will have 0 volts across it, and the diode will have the full 12.
If you have a LED the same rule applies only it is 1.2 volts I think. Using the secenario above the light bulb would have 10.8 volts across it.
ZENAR diodes are different. The are specifcally designed to have different types of voltage drops. So you can buy a 5 volt zenar diode. Using the secenario above the light bulb would have 7volts across it.
Diodes have varying abilities to handle current. Some are rated at millamps, so if you use to small of a diode it will ‘pop’ so fast you would not even see the light come on.
Since diodes do not obey Ohms Law, and since Ohms Law is present in every circuit, then you must use a resistor with a diode, or LED, always. If not the diode will live a very short life.
OK. I started this thread by saying I wasn’t a complete dunce about electricity. I now realize I am a complete dunce. Eeeek. [:D][D)]
By the way, not using tortoise (yet), using Atlas with their #200 relay for the lights. I did take Ray’s advice and I picked up a .5 watt assotment of resistors, and am experimenting with those.
Thanks All.
any dunce that wants to live long, prosper, and remember things should keep one hand in his pocket while touching anything electrical, and don’t stand in water.
BUY an MultiMeter (can be as cheap as $8.
To read VOLTAGE: connect in PARALLEL with the whatever is being tested.
To read AMPERAGE: connect in SERIES (between Power Pak & whatever)
Turn UP the power pak until lthe desired brightness is obtained - or bulb burns out - whichever comes first. You now have a point of reference.
G.O.Wheat bulb’s are generally 12v.
G.O.Rice bulb’s are generally 1.4v.
BAYONET bulb’s generally have the number on the side.
LED’s require a resistor on one side to keep it from burning up. (TRY 1000 ohms 1/4 watt.) @ 12v. Amperage should be around 2ma. (resistor’s are plus or minus 20% - cheap). Higher voltage reqires higher resistane and vice versa. Trick is selecting resistance value to keep ma. below LED’s stated rating (example 2.2ma)…