Hi folks
Can anyone help me calculate the output of a transformer when it is rated in “VA” ? I assume a VA is a volt-amp but how does it translate to amperage output?
I have several transformers that I know are rated at, say, 4 amps.
The one in question says the output is 50 VA… it is a 12V transformer. How many amps is that?
THEN how would I express that in watts?
Any Edisons out there?
I had Ohm’s law in physics, and I can do all the theoretical computations. But when it comes to putting in a light or an LED and figuring the resistance of the capacitor I need I just post and figure Randy will just tell me. I hope I’m not hijacking a post, but I figure it is related somehow.
Can someone do a practical explanation of how Ohm’s law applies to model railroading?
If I remember correctly, VA is watts.
I think you just divide the 50 VA by the output voltage (12 V) to get 4.2 amps.
You’re on the right track with VA = Volts x Amps. Therefore, if the transformer puts out 12 volts and is rated 50 VA, it can deliver 50 VA/12V = 4.17 Amps. For DC circuits, Watts = Volts x Amps, so a DC load that was supplied through a transformer and rectifier that pulled 4.17 Amps from the 12 Volt supply would be a 50 Watt load.
For AC circuits, it gets more complicated because the AC voltage and the AC current do not necessarily flow “in phase”. This is because the current not only does real work (measured in Watts) like lighting a light or running a motor, but may also be charging a capacitor or magnetizing the iron core of a motor or relay. This so-called reactive current flows and contributes to the total current, but does not do real work. You might think of it sloshing back and forth but not being used up, and no energy other than minor losses in the course of it flowing through the wires and other components is required for it to exist.
For model railroad work, most everything is DC and you don’t need to worry about all the complications of AC circuits.
As far as Ohm’s Law goes, Voltage (V) = Current (I) x Resistance (R). If you know two of the quantities, you can calculate the third. The problem comes in because many devices we use do not have a constant value of resistance. Lamps, for instance, have much more resistance when the are hot (burning brightly) than cold (just energized). Thus, there is a short surge of current while the lamps come on. Motors develop what is called back electromotive force depending on the load that they are driving. LEDs and other diodes have a voltage drop that is almost constant for any current large enough to start current flowing at all.
If there is something in particular that you want to calculate, describe it and I’ll try to help.
Okay,
Tonight I’m planning on putting a new light in a Proto S1. I have a 12v light a little over an 1/8" long. I also have some 3mm LEDs. In additon, I have some 1K and some 240 ohm resistors.
How do I decide what direction to go?
ahuffman seems to have nailed it.
I’d just add two things:
First, for those of you who flunked algebra, find voltage from current and resistance by V=I x R, find current from voltage and resistance by I=V / R, and find resistance from R = V / I. Ohm’s Law is not actually that useful on a model railroad, as the main time we care is with locos, and they have inductance and BEMF as well as resistance as ahuffman said, ie the current isn’t just a function of the resistance of the motor. It IS handy in working out what resistor to put in series with an LED or a 1.5V bulb
Second, DCC does of course introduce AC and all its complications to the equation. In fact it’s not “pure” sinusoidal AC but square wave AC which is even more complicated. Mostly we don’t care and we just leave it to the manufacturers to worry about. The main times it matters for us, that i can think of, are when we want to measure current usage. If you use an ordinary ammeter you won’t get a true reading of amps. This doesn’t matter much if you just want to compare one loco with another: you can still tell that the Climax draws more than the Pacific without knowing exactly how much it draws. So this is still useful to highlight a loco whose current draw is high.
But it is trickier to predict when your current usage is going to exceed the output of the DCC booster since you need to know the true current. I bought one of Tony’s DCC voltage/current meters so I could get true readings. Does anyone know a rule of thumb for using a DCC ammeter (or voltmeter) with DCC?
Here’s a link to Ohm’s law from a model railroad perspective:
When the page loads, select “Practical Electronics” along the left side.
Tom
Since the blue functon lead is +, the blue lead connects to the ‘arrow’ side of the LED. Current in an LED flows in the direction of the arrow, from the anode (arrow to the cathode( flat bar). Best part is, if it doesn;t work, just flip it around, it won’t blow up, as long as you used the resistor (you want the 1K here, 240 is WAY too low). I’ve been doign this stuff for a long time and I still twist the wires on and test it before soldering - even when I’m sure I have it the right way. If I would just solder away, Murphy would step in and it would be the one witht he super short leads that I stuck in backwards, with no room to cut off and redo.
–Randy
Wow…
I knew I came to the right place!
Thanks for all the help fellows. Although I have many power supplies on my HO layout for DCC, lighting and switch machine power my actual application here was for landscape lighting around my garden railroad. I have a cheap source for the 50 VA transformers and now I know that I need two for my 100 watt lighting load (4.2 A X 12 V = 50 watts)
Thanks so much! ED
Here’s an old, old trick for remembering the relationship between volts, amps and ohms. Draw a triangle. Divide it in half with a horizontal line. Divide the bottom part in half with a vertical line. Write “E” in the top half. Write “I” and “R” in the two bottom sections. E = voltage, I = current (amps) and R = resistance (ohms). You’ve just made an “Ohm’s Law Triangle.” Cover up the quantity you want to solve for and the relationship of the other two will tell you what calculation to make. For instance, to find out how much current is flowing through a resistor, cover up “I”. That leaves E over R so the calculation is voltage divided by resistance. Very simple.
Here’s a site that does some Ohm’s Law calculations. Note the “EIR” circle in the upper right (circle is same as triangle mentioned above by Seamonster).
Interesting, must be something they teach in Canadian electronic schools. I don’t recall any of my college professors ever showing us that, although who knows, if I go dig out my old college notebooks from the intro EE course it could be there.
I always remembered it by thinking about where my Dad worked, Ingersoll-Rand. So the IR go together leaving the E.
Let’s avoid the resistor color code though, this is a family forum [:D]
–Randy