# Wiring electric baseboard

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• posted on May 26, 2016, 4:59 am
Hi, I'm wiring four new electric baseboards in my basement. All four are 220Volt units: 500W, 500W, 1000W, 1500W, for a total of 3500W. From the main panel to the farthest unit is less than 100ft. If I understood the rules, putting all of this on one circuit will require a 20Amp breaker (1.25*3500W/220V), and the wire can be #12AWG. Did I get that right, and am I missing any other design factor?
Many thanks, especially if you can reference the NEC so I know I'm doing it right. Theodore
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<%-name%>
• posted on May 26, 2016, 5:50 am
On Wed, 25 May 2016 21:59:39 -0700 (PDT), snipped-for-privacy@yahoo.com wrote:

OK first things first, heaters are rated at 240v not 220 so your total load is 14.6a.
OK lets look at the code (Fixed electric space heating art 424) ************** 424.3 Branch Circuits. (A) Branch-Circuit Requirements. Individual branch circuits shall be permitted to supply any size fixed electric space-heating equipment. Branch circuits supplying two or more outlets for fixed electric space-heating equipment shall be rated 15, 20, 25, or 30 amperes. ************* OK there, 20a is a listed size
******************* 424.3(B) Branch-Circuit Sizing. Fixed electric space-heating equipment shall be considered continuous load. *******************
"Continuous load" means you size the circuit to 125% of total load (14.6*1.25.25a) so 18.25a is OK on a 12 ga wire with a 20a breaker
Modern space heating equipment should meet the "disconnecting means" requirement with a switch marked "off" that disconnects both hot wires. (424.19(C))
I think you are good to go.
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<%-name%>
• posted on May 26, 2016, 12:09 pm
Fantastic. Thank you soooo much for the detailed reply. Very much appreciated.
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<%-name%>
• posted on May 28, 2016, 8:04 pm
snipped-for-privacy@aol.com Thu, 26 May 2016 05:50:22 GMT in alt.home.repair, wrote:

You're going over the 80% trade standard load (16amps on a 20amp circuit) at that point, though...If you're going over 50ft or so, I'd probably use 10/2wg for the main run and come off that with a 12/2wg tail for each of the heater units. As each tail is only going to be asked to carry whatever heater it's connected to will ask for, not the combined load for all of them.
While I wouldn't expect to see an overheat fire risk situation using 12/2wg for the main run (If it's not for a long distance run...), I'd feel better knowing the main line I ran to feed them was more than upto the task and wasn't near full capacity if I ran one or all of the heaters at the same time.
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MID: <nb7u27\$crn\$ snipped-for-privacy@boaterdave.dont-email.me>
Hmmm. I most certainly don't understand how I can access a copy of a
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<%-name%>
• posted on May 28, 2016, 8:19 pm
On Saturday, May 28, 2016 at 4:08:17 PM UTC-4, Diesel wrote:

Where is the over 16 amps coming from? All I see are 14.6 and it meets NEC.
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<%-name%>
• posted on May 28, 2016, 11:17 pm
wrote:

Excuse me but the load is 14.6a. The 18.25a IS 125% of the load so the 80% has been accounted for.
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<%-name%>
• posted on May 29, 2016, 3:47 am
snipped-for-privacy@aol.com

The load is 14.6A under perfect conditions of a wire 100ft or less in total length, yes. However, if the last heater happens to be a bit further away from the panel than 100ft as the OP suggested, there's a real risk of voltage drop to the last heater with the others running because the 12/2 wire just can't provide the full load that far out to the last device. A voltage drop will increase the amount of amps required by that heater to do it's job. The 12/2 isn't going to appreciate that and neither will the breaker. It will tolerate being over 20amps until it heats up enough to trip.
It's why I suggested 10/2. So, the OP won't even have to worry about problems down the road if their estimates are off. It certainly won't hurt the situation.
100feet is the normal usual maximum spec'd The code does not rule on lenghts and voltage drop.. --only suggests!
The end of run outlet is where the starvation is the worst and any inductive load will suffer the worst at that point so you must consider this as you overall design limit at 100feet.
A 20amp circuit fully loaded at this outlet will drop 5.6% where 5% is the RECOMMENDED maximum and we must install 10gage wiring if we still insist on a 20amp rated circuit. The other choice is to derate this long run to a 15 amp circuit on this 12g wire and you could possibly expect to be at 5% drop at 140-150feet. Tthe full 15 amp dribbles off at around 120 feet!
20amp circuit loaded at 18amp is 5%
16amp will result in 4.5% missing at that far outlet.
The loss will be accumulative along the run as other 'stuff' is plugged in.
Most equipment that uses electricity is designed to function well at plus or minus 10%,whetner it is a resitive load like light bulb or an inductive load like a lighting ballast or motor.
This all becomes even more crucial when the power company is experiencing voltage sags in the summer high loads and the volatage that arrives at the pole down the street is already considerably LOW!!!---and if you designed your circuit lengths at the extra long end ---you are causing motor loads to labor,and lights to be dim!
Go with the far outlet at 100feet MAX...and you will be with the majority!!!
from the OPS original post: From the main panel to the farthest unit is less than 100ft.
So, the last heater on the circuit, depending on how far away it actually is from the panel and if the others are all running or not, could exceed the rated capacity for the wire and/or the breaker. IE: it's too far and the wire is too skinny to get the needed amps to the device.
I understand there is a cost difference going with 10/2 and some individuals consider that to be more important than the total amount of amps the circuit might pull when fully energized (all heaters on - high) So, for short distance, and if the OP is very close with his figure on distance to last device on the circuit, your suggestion will save him money and work fine. If he's wrong or a bit off and it's a bit further, he might have to break up the heaters into seperate fed circuits to keep the amps low enough. Which will use additional wire and another breaker or two in his panel, assuming there's room for that.
At this point, whatever cost savings the OP had by not running 10/2 have disappeared.
Why even have to be concerned with it? Use the thicker wire, you KNOW each device will get all the amps it needs, without any possibility of an overload condition.
Like I said though, whatever choice the OP makes, it looks fine from here.
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MID: <nb7u27\$crn\$ snipped-for-privacy@boaterdave.dont-email.me>
Hmmm. I most certainly don't understand how I can access a copy of a
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<%-name%>
• posted on May 29, 2016, 7:40 am
wrote:

The heater is not going to decrease it's resistance and draw more current to compensate for the voltage drop. It will just be a slightly smaller heater. Heaters are rated at 120, 208 or 240v. If the voltage is lower, the heater will just have a lower output. An example is an oven element. At 240v it is rated 3600w, at 208 it is only 2700w
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<%-name%>
• posted on May 29, 2016, 2:45 pm
On Saturday, May 28, 2016 at 11:51:10 PM UTC-4, Diesel wrote:

Like Gfre said, a resistance heater isn't going to increase it's current draw because there is some additional resistance in the circuit. It's like saying a light bulb will burn just as bright with a rheostat in the circuit to dim it, because it's going to pull additional current to compensate for the lower voltage. Put additional resistance in the circuit and the total current goes down, not up. And the difference here, the resistance in that wire is negligible.

There won't be problems if he follows the code. I can see using the heavier gauge wire if he thinks he might need to go to a larger load in the future, ie adding a heater or using larger ones.

A random internet posting is your cited source of information?

You don't show the calculations that is based on, and IDK how you came up with it. The resistance of 12 gauge wire is .0016 ohms per foot. 100 ft, you have .16 ohms. At 20A, that produces a voltage drop of 3.2V . There are two conductors so double it, 6.4V. It's a 240V circuit, 6.4V drop from 240V is just 2.7%, not 5.6%.

IKD where any of that is coming from, the relevant calcs are above.

What plugged in stuff? He's wiring a dedicated circuit for baseboard heaters, there are no receptacles.

And he's within 2.7%.

There is no motor load. If there were, we'd use the appropriate NEC to size for that.

The worse case is with all heaters on and that is what we sized for, 12 gauge is sufficient.

Good grief. We have used the rated capacity of the heaters. And there is already margin in the NEC, they don't say do the calcs, then upsize it more. You can if you want to, but there is no operational or safety issue here.
So, for short distance, and if the OP is very close with his

Nonsense as the NEC and above calcs show.

That's right, just use 12 gauge which NEC says is perfectly fine. You're the one throwing all kinds of FUD in and using incorrect calculations.
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<%-name%>
• posted on May 30, 2016, 2:23 am
Sun, 29 May 2016 14:45:19 GMT in alt.home.repair, wrote:

That isn't what I'm saying at all. I'm saying that the further out you go with 12/2 (or any wire really, it just depends on the thickness of the wire, the power you started with, and how far you're actually going), the less available voltage you'll have at the end of the run.
The less volts you have, the more amps you'll need to make up the difference to get the desired wattage.
Also, the 10/2 isn't going to be near capacity at any time, either. even if he has them all going on high at the same time. The 12/2 however, will be depending on the amount of heaters going and their setting(s).
going by Ugly:
7.9 volts lost 12/2 at 100ft 4.97 volts lost on 10/2 at 100ft
240-7.9#2.1 on the 12ga 240-4.97#5 on the 10ga
1500/232.1=6.5 amps *rounded up* 1500/235=6.39 amps *rounded up*
And that's due entirely to distance with no loads present on the line yet.
100ft on 12/2 according to my 2014 edition of Ugly's electrical reference is a net loss of 7.9 volts at the end of the run, with no load present, yet. At 125 ft out, the loss increases to 9.8 volts. And, obviously gets worse from there.
OTH, the 10/2 wire loses 4.97 volts at 100ft and 6.21 volts at 125ft out. I'd rather get as many volts to the device (heaters) as is realistically possible. The closer I can get to their expected input voltage, the less amps they'll require to do their jobs. The more heat I'll get (which is obviously the point here) and the less power I'll use doing it. A win win win.
To do anything less is only costing me more money and time in the future. The 12/2 is going to heat up a bit more under various conditions than the 10/2 ever thought about doing, even if all heaters were on at the same time, on HIGH. Over time, the 12/2 wire will degrade due to heating/cooling cycles that the 10/2 won't have experienced. As it degrades, it's own resistance will increase.
Another side effect of running the wire warm/possibly hot to the touch at near full load over a period of time is that the connection points and terminals in the panel and the heaters will also become a little 'warmer' than they would if they'd been powered from the 10/2 line and pigtailed to it with a short 12/2 run.

Uhh, no. I remembered it was 100ft or so off hand as the general rule of thumb, but didn't remember how much voltage was lost as a result. I've also got various NEC books and my 2014 edition of the yellow Ugly electricians reference book. It's where the voltage drop figures I used today came from, actually.

Just to clarify, the calculations I provided in this post aren't FUD, unless you're able to dismiss the yellow Ugly electrician reference book. I haven't seen any 240volt baseboard heater wired with a 12/2 in sometime, actually. I think the last time I actually observed that was with a trailer. Alas, they're built with heavy consideration on cost. IE: as cheap as you can get away with. I don't work like that, I don't offer advice with that in mind first and foremost.
If it were me, writing only for myself, I'd spend the extra money for the heavier gauge wire and use the 12/2 wire for pigtailing off of it into the heater units. I've already explained why I'd run it in this manner. No real point in doing so again.
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MID: <nb7u27\$crn\$ snipped-for-privacy@boaterdave.dont-email.me>
Hmmm. I most certainly don't understand how I can access a copy of a
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<%-name%>
• posted on May 30, 2016, 2:46 am
wrote:

With no load, there is no voltage drop at all.
If you are actually dropping 7.9v you do not have a 1500w heater anymore. The element is going to be still around 38.4 ohms (the element did not change, just the voltage) so at 232.1 v heater draws 6.04a and becomes a 1402.8w heater.

This means nothing at this point unless you are changing the element to get 1500w at 232v.
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<%-name%>
• posted on May 30, 2016, 3:05 am
On 05/29/2016 09:46 PM, snipped-for-privacy@aol.com wrote:
[snip]

Heaters are supposed to be intelligent enough to figure out when they need to shorten their elements :-)
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<%-name%>
• posted on May 30, 2016, 5:57 am
snipped-for-privacy@aol.com

The voltage drop is due to the wire's own resistance, and the distance the voltage must travel. The wire we're using isn't a super conductor. It has a certain amount of resistance to it. As a result, some volts are no longer available to us, we spent them getting the rest down the line. Nothing for free, you know.
Watts is watts, man. Available voltage determines how many amps it's going to take to get them, though. Lower voltage=more amps to do the same job.
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MID: <nb7u27\$crn\$ snipped-for-privacy@boaterdave.dont-email.me>
Hmmm. I most certainly don't understand how I can access a copy of a
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<%-name%>
• posted on May 30, 2016, 6:09 am
wrote:

Voltage drop still depends on the load. If there is no load, there is no voltage drop. If you have a wire with no load at all, there will be full circuit voltage at both ends.

If you have a lower voltage, the watts will be lower. In our example, that 1500 watt heater (at 240v) will be 1407 watts at 232v. Current will actually be less, not more.
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<%-name%>
• posted on May 30, 2016, 6:53 am
snipped-for-privacy@aol.com Mon, 30 May 2016 06:09:42 GMT in alt.home.repair, wrote:

No, there won't. The wire isn't a super conductor. It takes a little to push the electrons on it. The thinner and longer the length, the more is lost in transit. If it were a super conductor, what you've written would be absolutely true. As long as the wire has resistance of it's own, we're subject to voltage drop. We have various ways in which to minimize the voltage drop, though. Short of using a super conductor however, we can't outright prevent it.

Not exactly. You can have high voltage and next to no current behind it (think of a taser) or low voltage and a considerable amount of current behind it (think of your car battery) it's no chump and your starter motor isn't exactly a low drain device...You can also lose a few volts in our example, and, still be able to pull enough current to make up the difference.
The easiest way I know of to explain the relationship is this:
It may be useful to consider the image of water in a hose. Voltage is equivalent to pressure, water flow is equivalent to current and the diameter of the pipe is equivalent ot the thickness of the wire - or resistance.
--
MID: <nb7u27\$crn\$ snipped-for-privacy@boaterdave.dont-email.me>
Hmmm. I most certainly don't understand how I can access a copy of a
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<%-name%>
• posted on May 30, 2016, 11:13 am
On 05/30/2016 12:53 AM, Diesel wrote:

Voltage drop is represented by the formula E=I*R Seems to me if the current flow is zero, then the voltage would be zero as well.
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<%-name%>
• posted on May 30, 2016, 12:17 pm

You are correct. If there is no current flowing, there is no voltage drop. The same is true for water pipes. If no water is flowing, the pressure is the same at both ends.
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<%-name%>
• posted on May 30, 2016, 2:37 pm
on 5/30/2016, Al Gebra supposed :

That's a good theory, but IMO it is wrong.
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<%-name%>
• posted on May 30, 2016, 3:14 pm
On Monday, May 30, 2016 at 10:37:10 AM UTC-4, FromTheRafters wrote:

No theory, it's correct from electricity 101. With an open circuit, no current flow, the voltage across the wires is the same at both ends.
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<%-name%>
• posted on May 30, 2016, 3:45 pm

That I can agree with. There is no "voltage drop" because there is no circuit in which current can flow. Voltage drop is defined in terms of a circuit with current flowing from the source through a load and back to the source and voltage drop is across all of the resistances in the circuit including the internal resistance of the source.
Think of it this way:
Distance equals rate times time (D=RT) and you have two trains on a railroad track. Detroit to Chicago on one end and Chicago to Detroit on the other. Neither train is moving. Does the distance between them drop to zero?
No, it doesn't.
Voltage drop has no meaning in an 'open circuit', which isn't actually a circuit at all, just as 'rate' has no meaning for objects which aren't moving.
From Wikipedia:
"An electronic circuit is composed of individual electronic components, such as resistors, transistors, capacitors, inductors and diodes, connected by conductive wires or traces through which electric current can flow."
If current can't flow, it stands to reason that it is not a circuit, and 'voltage drop' has no meaning.
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<%-name%>
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