2 - 900w Tansformers On A 15A Circuit? - Electrical - DIY Chatroom Home Improvement Forum

 DIY Chatroom Home Improvement Forum 2 - 900w tansformers on a 15A circuit?
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07-19-2008, 02:11 PM   #1
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## 2 - 900w tansformers on a 15A circuit?

I am installing 2 900 watt Malibu LV transformers and have been told 1 15 A circuit is OK. According to my calcs, the 2 will pull about 17A, so I see the breaker a poppin all the time............. So who is the idiot here, me or "Goober" (my goofy brother)

07-19-2008, 02:17 PM   #2
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I'm not sure how you came up with 17 amperes for total load since 900 watts plus 900 watts equals 1800 watts and 1800 watts divided by 120 volts equals 15 amperes.

Your brother may be a "Goober" (mine sure is) but as long as nothing else is on the 15 ampere circuit he is also correct this time. Furthermore, unless you install the maximum number of lights on each transformer the actual load will be less than 900 watts each. So install one less lamp than the maximum on each transformer and you should be fine.

07-19-2008, 02:26 PM   #3
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Quote:
 Originally Posted by Furd I'm not sure how you came up with 17 amperes for total load since 900 watts plus 900 watts equals 1800 watts and 1800 watts divided by 120 volts equals 15 amperes. Your brother may be a "Goober" (mine sure is) but as long as nothing else is on the 15 ampere circuit he is also correct this time. Furthermore, unless you install the maximum number of lights on each transformer the actual load will be less than 900 watts each. So install one less lamp than the maximum on each transformer and you should be fine.
You are rite on the math...... LOL, never was my best subject. Question: Even if I leave out a 25w bulb, wont the rest of the bulbs receive that 25w evenly dispersed?? If so, the load will still be 900w!! "?

07-19-2008, 02:38 PM   #4
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Quote:
 Originally Posted by tonyBagadonuts You are rite on the math...... LOL, never was my best subject. Question: Even if I leave out a 25w bulb, wont the rest of the bulbs receive that 25w evenly dispersed?? If so, the load will still be 900w!! "?
Tony-

No. The transformers are rated for 900W which expresses their maximum capacity. Add up all the bulb wattages on the low voltage side and divide by the low voltage number (probably 12V DC). This will tell you what that particular string will draw. Basically, P(in) = P(out) (not taking into consideration losses due to efficiency/losses of/in the transformer). The actual load will dictate what the current draw is.

Jimmy
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07-19-2008, 03:02 PM   #5
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Quote:
 Originally Posted by BigJimmy Tony- Add up all the bulb wattages on the low voltage side and divide by the low voltage number (probably 12V DC). This will tell you what that particular string will draw. Jimmy
Not sure that I am getting this. Assume I max at 900w (36 lites @25w lv bulb). Assume all other things equal, forgetting voltae drops, etc. By the above formula (900w/12v = 75) So your saying I am only drawing 75w?? I must be in that lower 50% of dumber people cause that does not seem at all rite to me. What am I not getting here?

 07-19-2008, 03:04 PM #6 Member     Join Date: Mar 2008 Posts: 43 Rewards Points: 29 If my above post is correct, I can put 20 900w LV transformer on a 15 A circuit
07-19-2008, 04:22 PM   #7
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Quote:
 Originally Posted by tonyBagadonuts Assume I max at 900w (36 lites @25w lv bulb). Assume all other things equal, forgetting voltae drops, etc. By the above formula (900w/12v = 75) So your saying I am only drawing 75w??
No you're not dumb, I am an engineer and it's my job to confuse people!

The basic equation for power is P(watts)=I(current) x V (voltage). If your bulbs are 25W and the LV level is 12V, then

P=I x V,

I=P/V,

I=25W/12V=2.1A (approx) on the LV side for each bulb connected. 36 of them is therefore 2.1A/bulb x 36 bulbs = 75A. Your post indicated that if you removed one bulb that the load would remain the same which is untrue. In that case, the output current would drop by 2.1A/per 25W bulb (again this is approx) and the actual output power would drop accordingly.

Read the litterature for your transformer. Some will dictate the maximum load that they want to see connected to take into account transformer loss, output voltage, etc., and this may be less than the rated power figure. On the other hand, some take this into account and simply publish a lower power rating.

Quote:
 Originally Posted by tonyBagadonuts If my above post is correct, I can put 20 900w LV transformer on a 15 A circuit
OK, lets get back to the transformer rating of 900W. In a perfect world, there are no losses, no friction, etc, and P(in)=P(out). Remember how I mentioned that P=IV? Well,

P(in)=P(out)
and P=IV, then by substitution,

IV(in)=P(in)=P(out)=IV(out)

for our discussion, "in" will be the primary or 120V AC side and "out" will be the secondary or 12V DC/low voltage side.

So now, assume that your transformer is loaded to its full rating, i.e. 900W. As I pointed out above, on the low voltage side,

900W=P(out)=12VDC x I
so 900W/12V=75A

But you are concerned about the line or primary current (i.e. what you'll be drawing at the plug). Again,

900W=P(in)=V x I
900W=120VAC x I
and re-arranging, 900W/120VAC=7.5A.

So, two of them, run at full capacity, will reprent a primary current of 15A. I could have expressed this another way by saying that the connected load would be 2x900W or 1800W. With our trusty formula, P=IV, I could have said 1800W/120VAC=15A. But I figured that I'd toss in some stuff about how the transformer works from one side to the other.

You should be OK with both on one 15A circuit but you'll want to make the circuit dedicated for this purpose. If they are truly loaded to 1800W, throw any other load on the same circuit and you should trip.
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 07-19-2008, 04:49 PM #8 Member     Join Date: Mar 2008 Posts: 43 Rewards Points: 29 OK BigJimmy, I got it. I hadnt used Ohms Law in years, silly me...... Thanks Now I got one for you.......... When we buy electricity, are we buying watts or amps? I say amps and if we have voltage drop, wont the amps increase? So if the power supplier drops violtage, say 5%, wont that increase billable usage ?? PS. I dont like voltage drop, so I will use #12 for above circuit but a 15A breaker.
 07-19-2008, 06:11 PM #9 Member     Join Date: Mar 2008 Posts: 43 Rewards Points: 29 Well, not Ohms Law, but I am sure its somebodys law !!
07-19-2008, 06:14 PM   #10
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Quote:
 Originally Posted by tonyBagadonuts OK BigJimmy, I got it. I hadnt used Ohms Law in years, silly me...... Thanks Now I got one for you.......... When we buy electricity, are we buying watts or amps? I say amps and if we have voltage drop, wont the amps increase? So if the power supplier drops violtage, say 5%, wont that increase billable usage ?? PS. I dont like voltage drop, so I will use #12 for above circuit but a 15A breaker.
Ok, your power company (i.e. often quoted in this site as "POCo" for short) bills you some cost per kilo-watt hour or KWh. Again, this is really just P=IV again. They don't charge you cost/amp because they have not only residential customers but comercial and industrial customers as well. That being said, you are likely connected via a split-phase 3-wire service whereas a non-residential customer may have a 3-phase 480V or higher service, depending on their load. That being said, 5A at 240V is much less instantaneous power consumption than 5A @ 4160V. So, a more standard way of measuring is by watt or kilowatt hours. As an example, you have a 100W bulb that is on for 10 hours. This is simply 1000Whrs or 1 KWh. Remember, the P=IV formula represents the power consumed at any instant in time whereas energy consumption is the power measured over time (and instant. power is generally variable at any given moment depending on what you're using in your house). So, if all service voltages/phases were the same, I guess we could talk about being billed for current consumption but this is not practical as I mentioned above.

Now voltage drop is interesting. In the case of a resistive (or nearly resistive load) like an incandescent bulb, the energy consumed is again based on P=IV. Using ohm's law for a moment, V=IR, if the voltage decreases and the load resistance remains the same, the current will decrease. From the perspective of P=IV, the power consumed will decrease as well. So why don't we just all install variacs at our service entrance and adjust the voltage to say 90V and save a bundle? Well, devices are designed to operate at a nominal voltage which is assumed to be 120VAC. If we turn down the voltage, the lights get dimmer, it takes longer to toast our bread, etc. Then you'd be putting in higher wattage bulbs to increase the light output, re-toasting your bread, etc. In the end, you'd probably be consuming just as much power. However, there is a bad side to low voltage. The most common motor type is induction. They are designed to consume a certain amount of current at their design voltage. When they are subjected to low voltage, they compensate by consuming more current. The problem is, I^2R losses dominate and this leads to increased heating which can shorten their lives and decrease their efficiency (which of course increases the cost to run them).

Most utilities guarantee the end-users' voltage to be some nominal value plus/minus say 10%. To you and I, this means that if I measure the 1ph voltage at a receptacle, it should be 120V +/-10% (108-132V). If you were to install a strip chart recorder in your home, you'd likely see the voltage changes over the course of the year, especially if you live in an area that is subjected to wide temperature fluctuations over the year/seasons. The amount that you'll see can be affected by several things including where you are physically connected to the POCo's distribution system (i.e. are you near or far from a dist. transformer), the degree to which it is loaded, the size of your service, the extent that you are loading the service, etc. Highly inductive loads from industrial customers do a great job of depressing voltages simply due to the nature and size of their loads (but if they are a big consumer, they are typically required to maintain their power factor above some agreed upon value or face fines for failure to do so).

When we concern ourselves with voltage drop at the residential level, most people shoot to keep the terminal voltage at the end of the circuit w/in 5% of the line voltage at the service panel (the NEC does not require us to comply with this. They only recommend). Since the branch circuit conductors have some finite resistance, voltage will always be less (excluding capacitive loads) at the far end of the circuit for non-zero current. Longer conductors and higher currents both lead to more voltage drop. In your case, there is nothing wrong with using #12 wire on a branch circuit protected at 15A and given that it sounds as if you are pretty much expecting that load to be 15A, it's probably not a bad idea. But if your receptacle is 15ft. from your panel, at the end of the day (w/o running a calc), it probably won't make much difference.
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 07-19-2008, 06:53 PM #11 Member     Join Date: Mar 2008 Posts: 43 Rewards Points: 29 Very fasinating stuff............ I have a small grasp of wiring, but theory is great............. Thanks for taking the time to expalin it to me...... In my first read thru I got most of it.......... Couple of re-reads and I will own it !! lol Thanks again
 07-19-2008, 06:59 PM #12 Power Gen/RS Engineer   Join Date: Jun 2006 Location: Oak Park, Illinois Posts: 785 Rewards Points: 566 No problem! If you have other questions that are OT w.r.t. this thread, you are welcome to PM me (although you may not be able to with only 14 posts). If so, let me know and I'll see if I can PM you with a real addy. Then again, I doubt that the moderator would look down on technical talk that is somehow related to your original question. Plus, if we keep it public, then I have the eyes of other professionals to help me keep my facts straight. Jimmy __________________ Well, now, there's what's right and what's right and never the twain shall meet.
 07-19-2008, 07:24 PM #13 Member     Join Date: Mar 2008 Posts: 43 Rewards Points: 29 Really appreciate your help and I may be back with more ?? soon..........
 07-20-2008, 05:21 AM #14 " Euro " electrician     Join Date: Apr 2006 Location: WI & France { in France for now } Posts: 5,369 Rewards Points: 2,000 Even thru Big Jimmy explain very good on this one but however there one thing ya guys overlook most lowvoltage luminaires circuit useally limited to 300 watt per circuit on secondary side so watchout when you make the plans for LV lighting secondary circuits { they are very senstive with voltage drop } IIRC per NEC code the 300 watt = 25 amp at 12 volts on low voltage circuits. So.,, Therefore on 900W transfomer you will see three secondary { LV } circuits so they are divied up even each. for connection it will varies depending on LV transfomer set up some do have multitap to adjust the secondary voltage to boost up a little to comperise the distance. Merci,Marc
07-20-2008, 11:03 AM   #15
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Quote:
 Originally Posted by frenchelectrican Even thru Big Jimmy explain very good on this one but however there one thing ya guys overlook most lowvoltage luminaires circuit useally limited to 300 watt per circuit on secondary side so watchout when you make the plans for LV lighting secondary circuits { they are very senstive with voltage drop }
Bon point, mon ami! I even forgot that mine (Malibu I think) has two outputs A&B. Also, if I remember correctly, the manufacturer wants you to load the circuit pretty well since if you do not, the secondary voltage will be elevated which will lead to premature bulb failure.

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