600 C-9 Bulbs

| One argument was that this is a limitation on how the system is used | after it is wired and inspected. The code generally stays away from such | limits (and this one is particularly impractical). As you said, the NEC | "has no control over the user". (The CMP response was 'the code is the | way we want it' thus explaining the logic behind the rule.) | Another argument was that some UL listed devices, like hair dryers, with | current draw of 12-15A can have 15A plugs - UL thinks plug-in devices at | 100% are OK. (The CMP response was 'we are right, they are wrong'.) | (Both arguments for change make sense to me.)

That's how I look at it with respect to the first half of 210.6(A)(2) :-)

The code board certainly wants to avoid issues like those outlets you can easily picture with a stack of 2 or 3 triple taps and a mass entanglement of plug-and-cords converging into a fire starter's convention. Of course the code's solution is more outlets, especially in the kitchen, which is why my plan is for one of these:

latter is more likely because I haven't found a plate that has the arrangement of the former.

So why not call a 15A circuit a 12A circuit, and a 20A circuit a 16A circuit?

For a voltage of 130 volts, 100x7 watts or 700 watt load, and a No. 18 copper conductor, the per cent voltage drop is approximately 8.6 volts or 7.2 per cent not 4 volts. The 100 feet is one way circuit length. The 2 multiplier in the below equation is for the 200 foot total length.

VD = 2xKxLxI / cma Per cent VD = (130/VD) x 100

k=12.9 for copper L is one way circuit length of 100 feet I is 700watts/130 volts or 5.4 amperes cma for No. 18 AWG is 1620 circualr mils

This is the way we have been doing this for about 40 years. This voltage drop is distributed along the circuit just like it is in a recreational park lot feeder. At the 50th lamp the drop would be about 3.6 volts or 3.6 per cent.

REF: NEC 551.53(C) FPN: Due to the long circuit lengths typical in most recreational vehicle parks, feeder conductor sizes found in the ampacity tables of Article 310 may be inadequate to maintain the voltage regulation suggested in the fine print note to 210.19. Total circuit voltage drop is a sum of the voltage drops of each serial circuit segment, where the load for each segment is calculated using the load that segment sees and the demand factors of 551.73(A).

| Here is one you guys will love. I was told that a brand "X" breaker, listed | to trip at 20a will actually trip at 16a. I told my coworker he was full of | $#!% and get back to work. I am told it all has to do with the 80% rule in | the code. I assured him that a 20a breaker will trip at 20a.

Breakers are not very accurate devices. Since when operating them right around the rated current only involves the thermal element, this needs to take in other considerations such as manufacturing tolerance, ambient temperature, age, etc. The element in one could get a bit hotter than in another for the same current. The point where the element's thermal bend causes the mechanism to trip can also vary.

I think the 80% rule is well justified on the basis that a wide range of tolerance needs to be taken in to account in the design of permanent installations. So, for example, I don't think there is an issue with a 15 amp appliance plugged into into the upper half of a 15 amp duplex receptacle wired to a 20 amp circuit, other than for very little current capacity remaining on the circuit. As for plugging it into the lower half, that makes it easier to plug something else into the same circuit.

| You're right on the voltage drop. I assumed | the string was like the "drawing" below with 100 | feet total wire and 50 feet total length: | | AC --b---b---b---b---b---+ | | | AC ----b---b---b---b---b-+ | | Your voltage drop is based on 200 feet total wire, 100 | feet total length. That's what is shown on the link | he posted, so my assumption was wrong.

If what you show with a "-" is a _single_ wire, then you have wired the bulbs in series, in which case voltage drop will be uniform and not be an issue. But I believe C-9's are all supply voltage bulbs and as such would be wired in parallel. All that I have ever seen are.

If what you show with a "-" is a a _pair_ of wires, then this looks like one of those British ring circuits.

Maybe you meant:

/-+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+- / | | | | | | | | | | | | | | | | | | | | AC O O O O O O O O O O O O O O O O O O O O \ | | | | | | | | | | | | | | | | | | | | \-+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+-

There would certainly be _some_ voltage drop on the latter bulbs and even more if the strings are wired one following the other into the outlet many have on the far end of the plug.

If you were making a _permanent_ installation, such as a series of lamp posts running down a long driveway or road, where the variation in brightness might be detected or even be annoying, then there is this approach for voltage drop uniformity:

/----------------------------------------------------------+ / | / +--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+ AC | | | | | | | | | | | | | | | | | | | | \ O O O O O O O O O O O O O O O O O O O O \ | | | | | | | | | | | | | | | | | | | | \-+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+

This could be done with "220 cable" by using the red wire to feed the far end's black wire and leaving the black wire capped off at the close end.

Unfortunately this is not an option for Christmas tree light strings after manufacturing. If someone did make a very long string of say 250 bulbs at

250 feet where this approach might help, not only would it cost more for the extra wire, but might not get a UL listing because it's "strange".

I should explain after pulling out my copy of the Neher-McGrath Paper from 1957 that 12.9 is the circular mil ohms per foot for 100 per cent IACS copper at 75 degrees C. Since resistance is inversely proportional to circular mil ohm area the resistance of one foot of No.

18 AWG copper is 12.9 divided by the circular mil area of 1620 circular mils. This the same way the Neher-McGrath paper does this minus the lay factor. For 100 lights over a 100 foot long circuit we actually have 100 parallel circuits. For each light circuit we have the resistance of the conductor to the light and the resistance of the conductor from the light or 2 times the one-way circuit length.

Correction: Per cent VD = (VD/130) x 100 not the other way around.

It makes no difference how I drew it, the point was that my assumption was wrong. I assumed 100 feet total wire and 50 feet total length. Series or parallel doesn't matter for the calculation of the voltage drop in the wire: Drop = Total_Length*Ohms_per_foot*current

Ed

I have redone the voltage drop calculation using a series of voltage drops using a spreadsheet which I think is the correct method. Assuming we have a 100 foot long parallel circuit with a 7 watt lamp every 1 foot for a total of 100 lamps at 130 volts the current is .0583 amperes for each lamp and the voltage drop for the first 1 foot of circuit that carries the entire 5.8 amperes is 0.093 volts and for the last foot that carries only .0583 amperes is 0.00093 volts. If we do the 100 calculations add the series of voltage drops they come to 4.7 volts. The voltage drop comes out at 4.7 volts or 4.7/130 x 100 or 3.6 per cent. This is a very interesting problem.

So the voltage drop is acceptable, but the 80 per cent or 16 ampere limitation would allow only three strings of lights on a 20 ampere circuit. 20 x 130 x 0.8 = 2080 watts - rounding off to 2100 watts. 3 x 700 = 2100 watts.

I should also recalculate the watts. At 130 volts for a 7 watt lamp assuming they have a resistive element and this I do not know, we would have a resistance of 2414 ohms . Then at this resistance the last lamp that sees only 125.3 volts would have an actual power of about 6.5 watts. Using this as a the minimum and 7 as the maximum we could guess the average power at 6.75 watts per light without doing the 100 calculations. The 100x6.75 = 675 watts and 3 x675 = 2025 watts instead of 2100 watts. Now 80 per cent of 130 volts x20 amperes is 2080 watts. So we are well within the 80 per cent using three strings of lights.

On Wed, 22 Nov 2006 20:36:19 GMT ehsjr wrote: | snipped-for-privacy@ipal.net wrote: |> On Wed, 22 Nov 2006 04:43:44 GMT ehsjr wrote: |> |> | You're right on the voltage drop. I assumed |> | the string was like the "drawing" below with 100 |> | feet total wire and 50 feet total length: |> | |> | AC --b---b---b---b---b---+ |> | | |> | AC ----b---b---b---b---b-+ |> | |> | Your voltage drop is based on 200 feet total wire, 100 |> | feet total length. That's what is shown on the link |> | he posted, so my assumption was wrong. |> |> If what you show with a "-" is a _single_ wire, then you have wired the |> bulbs in series, in which case voltage drop will be uniform and not be |> an issue. | | It makes no difference how I drew it, the | point was that my assumption was wrong. | I assumed 100 feet total wire and 50 feet | total length. Series or parallel doesn't | matter for the calculation of the voltage | drop in the wire: | Drop = Total_Length*Ohms_per_foot*current

Yes it does matter.

The reason it matters is because there is load, and thus current, distributed over the circuit. You would not assume the whole 700 watt load (about 6 amps) on the segment at the end. Calculating the voltage drop for this requires calculating 100 segments, even with a different load. The first segment up to the first bulb has the current for 100 bulbs, 200 watts, about

6 amps. The next segment has about 1% less. Each segment has less and less. Additionally, each subsequent bulb getting less voltage draws less current. So the amount of loss steps down over the length. It is not the same as a 700 watt load at the end of 100 feet of 2 conductor AWG #18 cable.

Series puts every foot of the cable at the same current.

On 22 Nov 2006 12:39:06 -0800 snipped-for-privacy@electrician2.com wrote: | | gus wrote: |> We are putting light bulbs on a large outside tree in a city park. We |> have six 100 foot strings of 100 C-9 lights. We have two 20 amp GFI |> circuits available. I plan to put 300 lights on each circuit. My math |>

|> shows 2300 watts available and C-9 bulbs use 7 watts. This comes to |> 2100 watts. I will use 12 guage extension cords. The light strings |> and bulbs are commercial grade. This all looks good on paper, but I am |>

|> worried because there is not much room for real world error. Does |> anybody have experience with this many lights? Wasn't there a movie |> about a guy causing a large scale blackout when he threw the switch on |> his house decorations? I don't want to be that guy. Any help you can |> give me would be appreciated. |> Thanks, |> Gus | | I have redone the voltage drop calculation using a series of voltage | drops using a spreadsheet which I think is the correct method. | Assuming we have a 100 foot long parallel circuit with a 7 watt lamp | every 1 foot for a total of 100 lamps at 130 volts the current is .0583 | amperes for each lamp and the voltage drop for the first 1 foot of | circuit that carries the entire 5.8 amperes is 0.093 volts and for the | last foot that carries only .0583 amperes is 0.00093 volts. If we do | the 100 calculations add the series of voltage drops they come to 4.7 | volts. | The voltage drop comes out at 4.7 volts or 4.7/130 x 100 or 3.6 per | cent. | This is a very interesting problem. | | So the voltage drop is acceptable, but the 80 per cent or 16 ampere | limitation would allow only three strings of lights on a 20 ampere | circuit. 20 x 130 x 0.8 = 2080 watts - rounding off to 2100 watts. 3 | x 700 = 2100 watts.

You should make a version of your online voltage drop calculator that can do multiple segment calculations. The simpler form of it would be the classic Christmas light string where the loads and distances are uniform.

Of course a mix of loads, different reactances, different utilizations, etc, would be an enourmous level of work just to enter and describe the situation, much less calculate it. To make things practical we just have to make some upper level assumptions and figure out our limits. For a house with a 2000 foot service drop, it's a lot easier to do despite not really knowing all the loads at any time because we figure in typical and worst cases and just get an idea what a good wire size is or if it is more economical to go with a higher voltage.

On 22 Nov 2006 14:03:44 -0800 snipped-for-privacy@electrician2.com wrote: | | gus wrote: |> We are putting light bulbs on a large outside tree in a city park. We |> have six 100 foot strings of 100 C-9 lights. We have two 20 amp GFI |> circuits available. I plan to put 300 lights on each circuit. My math |>

|> shows 2300 watts available and C-9 bulbs use 7 watts. This comes to |> 2100 watts. I will use 12 guage extension cords. The light strings |> and bulbs are commercial grade. This all looks good on paper, but I am |>

|> worried because there is not much room for real world error. Does |> anybody have experience with this many lights? Wasn't there a movie |> about a guy causing a large scale blackout when he threw the switch on |> his house decorations? I don't want to be that guy. Any help you can |> give me would be appreciated. |> Thanks, |> Gus | | I should also recalculate the watts. At 130 volts for a 7 watt lamp | assuming they have a resistive element and this I do not know, we would | have a resistance of 2414 ohms . Then at this resistance the last lamp | that sees only 125.3 volts would have an actual power of about 6.5 | watts. Using this as a the minimum and 7 as the maximum we could guess | the average power at 6.75 watts per light without doing the 100 | calculations. The 100x6.75 = 675 watts and 3 x675 = 2025 watts instead | of 2100 watts. Now 80 per cent of 130 volts x20 amperes is 2080 | watts. So we are well within the 80 per cent using three strings of | lights.

Is this assuming 130 volts power coming in? Rating the bulb at 130 might be common, but is that a valid assumption for the actual voltage? At 120 volts, there will be less current. The temperature will go down and the resistance will drop a bit and the current will go up a bit. There will be an equalibrium which I don't have the formula to calculate (but people here have done it before). Now figuring out how all those equalibriums settle out for the 100 bulb distributed-load string could make for some very big calculations if you want to nail the numbers exactly.

For equal loads evenly distributed along the length you can get a pretty close answer by putting all the load at the midpoint.

-- bud--

The 80% limit is for a single plug-in on a circuit of 2 or more receptacles. The "15A" circuit is still rated 15A.

The third argument made for not trying to limit a single plug-in load to

80% was that it is inconsistent with the rest of the code, where 80% only applies to continuous loads.

-- bud--

Thank you - I finally see what you are saying, and you are correct.

Ed

There is a method described in the American Electrician's Handbook that tells how to find the load center for a distributed load and to use that for voltage drop calculations. I might look into this for voltage drop calculators.

snipped-for-privacy@ipal.net wrote: Is this assuming 130 volts power coming in? Rating the bulb at 130 might

I agree, it is an engineering problem to get exact numbers. We electricians work with approximations most of the time.

On 23 Nov 2006 16:50:10 -0800 snipped-for-privacy@electrician2.com wrote: | | snipped-for-privacy@ipal.net wrote: | Is this assuming 130 volts power coming in? Rating the bulb at 130 | might |> be common, but is that a valid assumption for the actual voltage? At 120 |> volts, there will be less current. The temperature will go down and the |> resistance will drop a bit and the current will go up a bit. There will |> be an equalibrium which I don't have the formula to calculate (but people |> here have done it before). Now figuring out how all those equalibriums |> settle out for the 100 bulb distributed-load string could make for some |> very big calculations if you want to nail the numbers exactly. | | I agree, it is an engineering problem to get exact numbers. We | electricians work with approximations most of the time.

Rarely is 1 amp ever a make or break situation. Voltages go up and down. Loads go up and down. The standards give plenty of margin for variations and errors combined.

So why not call a 15A circuit a 12A circuit, and a 20A circuit a 16A circuit?

Because we call them by their rating.

2005 NEC 210.3 Rating. Branch circuits recognized by this article shall be rated in accordance with the maximum permitted ampere rating or setting of the overcurrent device. The rating for other than individual branch circuits shall be 15, 20, 30, 40, and 50 amperes. Where conductors of higher ampacity are used for any reason, the ampere rating or setting of the specified overcurrent device shall determine the circuit rating.