Volts X Amps = Watts

Hi Paul,

I get 29.16 watts = 40 x 18 x 0.0405 which seems about right for a whole bunch of little bulbs. How did you get 0.81?

Scott, I am just repeating what MTH sent to me. Their numbers were 18volts X .045amps = .81watts. The .045 must be the total for the 40 lamps. aul

Actually, 45ma is probably for one bulb!  32.4 watts for 40 of the bulbs.

One thing you must consider when using postwar and earlier transformers. The stated wattage rating is power consumption not output. For example a postwar ZW is rated at 275 watts but its output is only about 190 watts.

the club uses the Malibu lighting transformers.

You do mean 14 volts, yes? Is that the voltage that gives the desired brightness?

Then the equation would change.

Get a 5 or 7 function digital clamp-on meter from Harbor Freight and get the actual measurements at any point you like and save the math for later.

Save the math for later.  I agree!    I did all that math in college 40 years ago with a slide rule, then calculators started to become affordable.  Now, I don't have to do math at work.  What little we do, we have software that does it for us.  Now math gives me a headache!

This all seems about what I use as a rule of thumb.  If incandescent bulbs, I always figure about one watt per bulb, and like to leave a margin in power supply capacity of about 20% just because . . . .  

That math, well, its like this: MTH's data seems to give you about that one watt/bulb: .045 x 18V = .81, and 40 means 32.4 watts.  Add my 20% margin and you are around 39 watts - a 40 watt at 18 volts (output) supply ought to do it.  

If you reduce to 14 watts, the formula used for light bulbs and other resistive loads is

Power = Voltage squared divided by resistance.  

Since resistance of the bulbs isn't changing this means the power is just the square of the voltage ratio or (14/18) squared =.60,  times 32.4 = about 20 watts of lighting load.  However, current draw has dropped only linearly, to 14/18 x .045 = .32 amps each, times 40 = 1.4 amps.  You will need a power supply rated at at least 1.4 amps at 18 volts or 25 watts to power these.

Actually Lee, the resistance of the bulbs does change when you reduce the voltage.  You won't get a straight-line reduction in power, as the bulbs run cooler, they also are lower in resistance.

Really?  Well, technically yes, practically, no.   I've researched- actually measured it in our lab, and written about it in my books, and yes, it does decrease very slightly due to the lower operating temperatures and even skin effect in the filament and all, but it's a rather small secondary effect that I always recommend one ignore in factory and utility engineering, and it is common practice through the electrical industry to do so, too.  If you plan to depend on the slight change you are just cutting margins and everything too fine, in my opinion - and will likely have problems down the road.  Not to mention power factor, which even with resistive loads like light bulbs can be a factor around power supplies and all, and harmonics around all those choppers, and even my 20% margin starts to look like its maybe not enough.  I would select a supply rated 40 watts at 18 volt  to serve those lights - a CW-40 for example, although like I said, a 25 watt supply might barely do it at 14 volts. 

True, the resistance change is probably not a large effect. 

One effect that is much more pronounced is the loss of light when you lower the voltage.  Reducing the voltage by 30% will cut the light to less than half.  The light wavelength also shifts toward red as well with lower voltage.  That is likely a larger issue in reducing the voltage to the bulbs.

That is absolutely true - that reduction to just 14 volts reduces light (and power of course) by 40%, and like you say really starts to effect the color of the light.  

Incidently, this project is close to home.  I had incandescent store lighting on my Main St that I put in for "nighttime" (only a few overhead lights on, dim) use and ran it at about 13-15 volts at night. Around two dozen bulbs or a few more all fed off the main variable voltage leads of a CW-40 (I seem to have extras laying around everywhere).  It looked good at "night" at around that voltage.  But pretty quickly I found myself wanting to run it during "daylight hours" (with all room lights on) which looked best at full voltage, because with more ambient light the lights had to be dimmer.  The color of the light changes noticeably - not objectionably so, but its different when you look inside some storefronts - from 13V, at which the bulbs are barely on, and 18V.  I'm switching to 12V LED now and I operate them at night at about  8-9 volt, as low as they will go.  They dim too but stay bluer than I really like at any voltage but I can add another thirty bulbs (I want to light up the cathedral and etc. or more this way and still use only one power supply for them (I have more, but I have no place to put a second light supply on my control bench . . .  a further advantage, I hope, is that they last longer.  I've had to replace about a dozen incand now.  

I wonder if mixing yellow LED's in would help with the colors of the lighting?

I probably would, but it's one project that I know I will not get to for a long time. I have not even experimented to see if it would make that much difference.   Every time I turn them on I see the blue and think I should, but  . . . 

Interesting . . .

But I wonder why somewhere along the way, the specified 40.5 milliamps, as provided in the first post, became 45 milliamps.

40.5 milliamps equals 0.0405 amps; not 0.045 amps.

Scott, second post, had it correctly for the 40 bulbs.

Not a big difference, but a difference nevertheless.

Alex

Nice math. I didn't really read all of it.

You may actually have a defective bulb which is shorting when it heats up. I've had that, and in due course it will trigger the circuit breaker.

Gerry

That's interesting...I couldn't put any thing to why his lamps should trip the ZW-C if there were only 40 of them at a max of 36 or so watts. Some bulb or fixture shorting? Odd that it restores itself though.

This is 2nd grade arithmetic by the way...just sayin'.

Lee, I use 'golden white' LEDs, and they give a very pleasant colour light. 

This color graph shows the effects of adding (3) primary light colors. Red, Green, and Blue.
Yellow is a combination of Green and Red light.
Cyan is a combination of Green and Blue light.
Magenta is a combination of Red and Blue light
White is combination of all three lights.  Red, Blue, and Green   
John is correct, adding Yellow to the blue light should give a better white.
Best wishes with your project.
Mike
Wikipedia Reference.

Always thought it was {V X A} - watts = 0

How were the lights wired when they quit after 30 minutes?

Then given the MTH info, how did you then come to choose 14 Volts (I assume 14 watts was a typo?) as the new setting for the lights?

I need to know if I can hook up 2 transformers to the same circuit for powering table accessories?? They would be at about 14 volts each. They aren't the same transformer but both are small, old 75 and 90 watt simple transformers. I would be using a phased common.

Clamp-on (around) meters are capable of measuring amperage without interrupting the circuit.  A plus for model railroading.   This older analog meter simply clamps around one of the circuit wires.  The amperage measured X the voltage  is a very good approximation of the power consumed (watts) in the circuit.


Amprobe, from the Sears website.

It is NOT advisable to parallel transformers, especially variable voltage ones with dissimilar characteristics!  If you're talking about two different circuits, one from each transformer, that's no problem with a common connection on one side.

MiwRd, Since your using a modern ZW-C with 2 180W bricks, you have plenty of power available on the UX outputs unless you are running a heavy load on the main lines.

A & B share one brick and C & D share the other.  You can always connect a separate transformer or brick, even something like a 80Watt in to either B or C would give a dedicated transformer for your lights.   G

   P         This is the FORMULA for: P= Power in Watts

--------                  I= Current         

 I X E          E= Voltage in Volts

You hold your finger over what you are solving for. ie , if you want to know what the Power is in Watts, you multiply I (Current) times E (Voltage). If you want to know the Voltage in Volts, you Divide the P (Power) by the I (Current). If you want to know the Current, you Divide P (Power) by the E (Voltage).

It's simple as that!!

Fred

Assuming a power factor of 1.0, right?

I think you are right ,John.

Fred

Power factor is one reason why I always pad my power estimates with 20% - the other being a healthy regard for harmonics.  PF is never 1.0 even if the load is all resistive, because the source in any real AC system isn't purely resistive and other loads particularly old AC motors, aren't either.   I admit I have never had a problem of any type caused by just assuming PF = 1.0 however.  Harmonics, on the other hand, has bitten me hard a couple of times in the last 25 years.  

Being I am an EE (and one, the old codger added, who just received word he is now an IEEE Life Fellow) I love threads like this, but frankly it boils down to a very simple choice: if you have so many lights that the supply doesn't work/trips the breakers, you either remove a few lights or get a bigger power supply. 

That's an incredible accomplishment Lee. I have been working at Raytheon for over thirty years in the field of heat transfer and thermodynamics. IEEE Life Fellow is as prestigious as it gets.

Where can I find the books you've authored and referred to earlier?

(Average) watts is the integral a time period of the product of instantaneous current and instantaneous voltage.

I think gunrunnerjohn has it right concerning incandescent light bulbs. I remember trying to size an autotransformer (Variac) to buy and reading the sheets on them and coming across the concept of a 'tungsten load', that light bulbs do not act linear. There was also some deal that if a Variac was running 120 in 120 out it had low heat dissipation, but at say 60%, it could get pretty warm, even for a true resistive load, you needed to take that into consideration. Then you needed to put the two of them together for light bulbs. Sorry I can't remember the details.

And a practical Engineer too.  Good advice in that last sentence Lee.  G

Most are on Amazon under H. Lee Willis.  They're published by CRC/Marcel Dekker and are priced pretty high.  It makes more sense if you really want to look at them to find them in college libraries: most universities with EE departments have copies. 

Congrats on becoming an IEEE Life Fellow, that's a great accomplishment!

I like your solution to the lighting issue, sweet and simple.

Congratulations Lee.

On the lamps, we are deciding that the 32 watts of lamps are not the only thing on the two 180 watt bricks? MilwRdPaul is running trains at the same time and after a half hour or so, the transformer trips out?

Thank you.  I was elected a Fellow in '92 and the Life designation is really just a testament to longevity: you have to be at least age 65 and your number of years of  dues paying membership added to your age has to be at least 100).  I found it interesting that it had to be dues paying years: my many, many years as a student member did not count.

As to the solution, there really isn't much else you can do is there.  

As far as I am concerned, a really good power supply with lots of power (low impedance source, ability to run cool at high load), very linear control of voltage, a large lever or knob with long movement so you can adjust that control finely, and good breakers and protection, is the best investment you can make in a toy train layout.  A lot of problems, like the one covered here, just never happen when you have a that, and when they do, you are glad you have the protection and the strength of a unit that can take some abuse.  

Lee, Congratulations! That is a great honor. Now, to the subject at hand. I don't know how this subject got so blown out of proportion but here goes. I have a ZW with two 180W bricks. I have 40 street lights equally distributed between the 2 bricks. That is the only load on the bricks. All I want to know is will the bricks be enough? 

Good grief!  Not only will they be enough, they'll be loafing at about 10% of their capacity!  You could easily power them from one brick and still have capacity for at least 100 more bulbs!