Voltage drop and circuit design..
I know that there is voltage drop in wire as a result of its length and gauge. I’ve noticed in the houses I’ve worked on and diagrams that I’ve seen lighting and small appliance wires usually go from the circuit breaker to a series of consecutive boxes.
Is there any reason to not wire off in two directions (with two separate cables) from a given box? In other words, if it results in the use of lesser total length of wire in a circuit, is there any reason you should not branch off in two directions and have more than one box in which the wires end on the same circuit?
Also, is the first box in a wiring circuit subject to less of a voltage drop than the last box because there is a shorter length of wire between the first box and the breaker box? Or does the total length of the wire in the circuit determine the voltage drop which is the same at each box?
Replies
Do not use a junction box if will become unaccessable later. By this I mean if you have to cut drywall to get to it.
The distance traveled adds to the voltage drop. It isn't a factor of the total length of the wire.
Energy in motion, the more motion the less energy at the other end.
Thanks for the response. Just to make sure I have it right:
Let's say there is 75' run of 12/2 wire on a 15 amp circuit. There are two boxes on the the circuit, one 15' from the breaker box, and the other 75'. The only thing that will be plugged into the circuit is a toaster.
Given that added wire length increases both resistance and voltage drop, and what you've said, are the following statements true:
1. Running the toaster will use more electricity, and it will be operating at a slightly lower voltage, if it is plugged into the 75' outlet than if it was plugged into the 15' outlet.
2. Running the toaster on the 15' outlet will use the same amount of electricity, at the same voltage, regardless of whether the remaining 60' of cable is connected.
In other words, if you later decided you were only ever going to use the toaster at the 15' outlet (and never use the 75' outlet for anything), there would be no advantage to disconnecting the 60' run of wire that goes to the 75' outlet because it is not causing either added resistance or voltage drop.
(This is just a hypothetical I'm using to try to understand voltage drop and resistance.)
since it is only a toaster you would use less power at the 75' plug. a toaster is only resistive and will run at what ever voltage is available, it just won't make any toast as the voltage goes downbut a motor would draw more amperage, because it is designed to operate at a constant speed and torque, when the voltage gets too low, the smoke will then be let out.
.
., wer ist jetzt der Idiot ?
Great point. So the toaster would use less energy when plugged into the 75' run, but is that because it would be operating at a lower voltage?
And wouldn't you be wasting some amount of electricity due to the resistance in the extra 60' of wire?
Basically, if you took the same questions as I posed above, but you substituted a small air compressor, what would be the result? Would that extra 60' of wire be wasting electricity due to resistance or dropping voltage even if the compressor was plugged into the 15' outlet?
From previous responses, I take it the answer is "no." Is the answer that the extra 60' would not cause any voltage drop or loss to resistance if the compressor was plugged into the 15' outlet.
In other words, if you decided that the compressor is the only device you will ever plug into the circuit, and it would only ever be plugged into the 15' outlet, would it make sense to disconnect the extra 60' of cable solely to lessen voltage drop and loss to resistance?
No you are concerned with voltage drop at the device you are using. no need to disconnect anything beyond that point. BUTyour compressor : plugged into an outlet where the circuit has VD problems will affect everything BEYOND the compressor.
and any wiring beyond the compressor will not affect the compressor.
.
., wer ist jetzt der Idiot ?
No you are concerned with voltage drop at the device you are using. no need to disconnect anything beyond that point. BUT your compressor : plugged into an outlet where the circuit has VD problems will affect everything BEYOND the compressor. And any wiring beyond the compressor will not affect the compressor
So, you're saying that with the compressor plugged in and running at the 15' outlet, anything plugged in at the 75' outlet will experience even more voltage drop than it would when the compressor is off?
All of this is a bit surprising to me. I would have thought that the whole wire would be energized when anything in the circuit is running, and that that would cause resistance along the entire wire. But I guess what it boils down to is that due to electricity following the path of less resistance, it isn't coursing through the entire length of the cable, but only through the cable up to the device being used and then back to the breaker box?
Thanks again
the wires are energized as long as the breaker is OnThere is VD in any circuit that uses wire ! 15' 75' 450' they all will have some VDIf you have TOO MUCH VD at the end of your circuit AND you want toast at that 75' outlet AND you have a compressor RUNNING at the 15' outlet
the toaster will have even more voltage drop to contend with.also the voltage being used by the compressor will not "travel" to the end of the circuit either.. it's already there, it will just be a bit lower .. especially when starting up !!remember that Voltage is Pressure........ it does not flow anywhere.
.
., wer ist jetzt der Idiot ?
the wires are energized as long as the breaker is On
that much I know, but always it's always a good reminder :)
There is VD in any circuit that uses wire ! 15' 75' 450' they all will have some VD
maybe I've phrased my questions wrong, but I do know that there is going to be some miniscule voltage drop even 1' from the breaker; I'm concerned about the amount of voltage drop
If you have TOO MUCH VD at the end of your circuit AND you want toast at that 75' outlet AND you have a compressor RUNNING at the 15' outletthe toaster will have even more voltage drop to contend with.
and while that probably wouldn't harm the toaster (I'm guessing the toaster wouldn't get as hot, it wouldn't be great for the compressor motor). In fact, on a 15 amp circuit, even my small compressor would trip the breaker if used with something like a toaster
also the voltage being used by the compressor will not "travel" to the end of the circuit either.. it's already there, it will just be a bit lower .. especially when starting up !!
remember that Voltage is Pressure........ it does not flow anywhere
Yes, I know I don't quite understand amps, volts and watts. I find their definitions difficult to understand. I know there is an analogy to water in a pipe, but that analogy is limited.
you can measure Volts, and Amps and until a few years ago you could only calculate Watts which is this formula Volts X Amps = Watts, which is what we all pay for per thousand Watts or Kilowatt
so using Algebra
your hypothetical toaster consumes 1200 Watts and you measured 120 Voltsyour Amps will be 1200 W / 120 Volts = 10 Amps .
.
., wer ist jetzt der Idiot ?
Let me further addle this confusion.
Watts = volts times Amps.
On the other hand, voltage drop is expressed by Ohm's law. It is a misdemeanor to disobey it. It's simple: Volts = Amps times Resistance.
Resistance per foot for various wire gages#14 .00307#12 .00193#10 .00121#8 .00076#6 .00049
You have to know the Amps for each device. A 60 watt bulb would be 60 = 120 volts X ___ amps. A little algebra gives you 60/120 = .5 amp. Suppose the compressor uses 7 amps and is plugged in 15' from the full blooded panel. [#14 wire] 30' [the current or amps goes a total of 2 X 15' = 30'.] X .00307 = .00921 ohms resistance. 7 amps X .00921 = 0.6447 volts which is your voltage drop. Subtract this from your 120 volts and you have 119.35 volts available for your air compressor.
Now for your 60 watt bulb 60' additional feet away. 120' X .00307 ohms/foot = 0.3684 ohms. .5 amp X .3684 ohms = 0.1842 volts. Your box @15' has 119.35 volts available, so 119.35 - .1842 = 119.1658 volts available for your light bulb.It will be theoretically a weensy bit dimmer but not to notice.
~Peter
If you replaced the Sulfer in "cosine" with Calcium, what do you get?
Check your arithmetic in that first calculation. I think you slipped a digit.
If your view never changes you're following the wrong leader
Okay, thanks to your actual numbers for various wire gauges and yours and DANH's examples, I think I understand the voltage drop at least enough to understand that it isn't enough to be concerned about in a moderately-sized home.
But what about the electricity lost to resistance. Is it similarly negligible?
It's my understanding that like voltage drop, resistance increases with length. If you have a fairly heavy draw on a 15 amp circuit (a bunch of lights, ceiling fans and a big tv, etc.) wired with 14 gauge wire. The resistance causes the wire to warm up, and that warmth is the lost electricity. If that same circuit with the exact same load was wired with 12 gauge wire (and all the same devices on), would their be much savings in electricity due to the lower resistance? Or is it too small to ever be concerned with.
In a residential setting the savings of larger wire (in terms of power actually consumed by the wiring) is too small to measure. But voltage drop affects the efficiency of devices such as AC compressors, and a little drop can be enough to seriously affect a compressor's efficiency. So observing mfgr's recommendations is always important.
If your view never changes you're following the wrong leader
Think about it this way, Assume you have a 15 foot cable with a compressor on it, and the voltage drop on that cable is, just as an example, 5 volts. So instead of 120V at the compressor you have 115V.Now you connect a 60 foot cable (of the same diameter) to the junction box where the compressor is. Because the compressor is running the voltage in that box is 115V, so the MOST you can ever see at the other end of the 60 feet is 115V (until the compressor stops, at least). Any load (like your toaster) on that 60 foot cable will contribute ADDITIONAL voltage drop, adding, for example, 1V of additional drop for each 15 feet. So the voltage at the compressor becomes 114V and the voltage at the toaster is 110V.(Note again that the above voltage drops are just for example, and not particularly realistic.)
If your view never changes you're following the wrong leader
Think about it this way, Assume you have a 15 foot cable with a compressor on it, and the voltage drop on that cable is, just as an example, 5 volts. So instead of 120V at the compressor you have 115V.
Now you connect a 60 foot cable (of the same diameter) to the junction box where the compressor is. Because the compressor is running the voltage in that box is 115V, so the MOST you can ever see at the other end of the 60 feet is 115V (until the compressor stops, at least). Any load (like your toaster) on that 60 foot cable will contribute ADDITIONAL voltage drop, adding, for example, 1V of additional drop for each 15 feet. So the voltage at the compressor becomes 114V and the voltage at the toaster is 110V.
I realize that this is all hypothetical, including the voltage drops, but in your above example, if the voltage drop at 15' was indeed 5v, wouldn't the drop be a total of 25v after another 60' of cable (because there are 4 15' lengths in 60', and the drop in your example is 5v every 15')? (Again, I know these are crazy numbers, I'm just speaking in terms of the example.)
Or are you saying something else about the amount of voltage drop not just being related to the length of the cable, but also what's plugged into it? Because, if that's the case, then that's something else I don't quite have a grip on.
No, the original 5V drop at 15 feet is due to the current flowing through that 15 feet to the compressor. If there's nothing attached to the end of the other 60 foot section then there's no current flowing and hence no additional voltage drop -- voltage drop is current times resistance, but it's only the resistance of that part of the cable that current is flowing through.Now, if you turn off the compressor and attach the toaster at the far end, you get 1V of drop per 15 feet for the entire 75 feet, meaning you measure 119V at the compressor connection and 115V at the toaster. Now turn the compressor back on and you can simply add the two sets of voltage drops -- 1+5V drop at the compressor, and 1+4+5V at the toaster.Like I said, this stuff is not always intuitive, unless you've got a pretty good background in electricity.
If your view never changes you're following the wrong leader
Now, if you turn off the compressor and attach the toaster at the far end, you get 1V of drop per 15 feet for the entire 75 feet, meaning you measure 119V at the compressor connection and 115V at the toaster.
I appreciate your explanations as this is something that I've always wanted a better understanding of. I'm curious, though, about one thing in this example. Why would there be 5v of drop with the compressor plugged in (and running) at 15', but only a total of 6v drop when you add the toaster at the 75' outlet?
Am I getting lost in this hypothetical? I thought voltage drop was a function of the length of the wires. I mean, it sounds like you are saying that plugging the toaster in the 75' outlet is going to raise the voltage at the compressor (plugged in at 15') from 115v to 119v.
Let's say nothing is plugged in at 15'. Are you saying that the voltage drop with only the toaster plugged in at 75' would be different than the voltage drop with only the compressor plugged in at 75'? If so, why?
Edited 1/27/2008 7:30 pm ET by JDLee
Let's keep it simple. Assume that the wire we're using has one tenth of an ohm of resistance per 15 feet. The compressor draws 50 amps. The toaster draws 10 amps.
The compressor at 15 feet therefore sees a voltage drop of 0.1 ohm x 50A = 5V. The compressor at 75 feet would see a voltage drop of 0.5 ohm x 50A = 25V. The toaster at 15 feet would see 0.1 ohm x 10A = 1V. The toaster at 75 feet would see 0.5 ohm x 10A = 5V.
Now, if you have the compressor at 15 feet plus the toaster at 75 feet then the current through the first 15 feet is 50 + 10 amps, or 60A. So the voltage drop on that section is 0.1 ohm x 60A = 6V. The current in the remaining 60 feet is only 10A (the toaster), so the voltage drop through that segment is 0.4 ohm X 10A = 4V. But this is just the last 60 feet, and you must add in the voltage drop on the first 15 to get the total voltage drop (6V + 4V) to the end of the cable.
If your view never changes you're following the wrong leader
Thanks to you, also, Dan. While I'm no good at math, that example really helps me understand the precise answer to my question.
On a related note, I live in southern California. It's my understanding that the utility companies here have been reducing the voltage of electricity they provide. I guess they do this to lessen the overall demand on the on the grid (which you likely know has been a problem). Kind of like continually reducing the gallons per flush on toilets.
Of course, they claim it's not a problem. But how low can the voltage go before it becomes a problem for various devices like motors and TVs, etc.
Standard line voltage is 120V in most of the US. In general stuff should work to reasonable efficiency down to about 110V, and should function reasonably well down to around 105V. The first to suffer is motors -- they'll begin to slow down, overload, and overheat around 105V. Most computers built these days should be able to function down to about 95V, TVs maybe 100V. Incandescent light bulbs don't give a rip -- they just keep getting dimmer, and around 90V the light output will be about 50% of normal. Fluorescent lamps will probably run down to 90V, but refuse to start from cold below about 105V.
If your view never changes you're following the wrong leader
When I was in grad school we were doing some navigation experiments. We had a DC3 with a mini-computer. It was one of the first ones uses IC's in a low cost computer. It had mag core so we could preloaded it.But I had an ASR33 without legs in the back of my hatch back that we would carry onboard with the paper tape program.And you had to enter about a dozen instrcutions by hand by flippig switches first. So it was always a fun time.Well we got the computer running and all working. Then taxied out and just as we where turning around the computer would crash.We did that several times in a row.The 120 on the plane came from a mo-gen set running off the ships' 24v DC.We set the computer up in the lab on a variac.The computer run fine until the voltage dropped to about 95. Then a low voltage shut down circuit did what was suppose to do and shut the commputer down safely.But the circuit pulled IN a relay to do this. And at about 80 the relay released and the computer started running again. But the operating voltge was way to low for it to run correctly and it wiped out the memory.Changed the proceedure to turn off the computer after see that it worked on the ground. Then as soon as the throttles where advanced to full power for takeoff, turn it back on again..
.
A-holes. Hey every group has to have one. And I have been elected to be the one. I should make that my tagline.
I'm not sure what all those acronyms are, but I followed the story!
I always have APC BackUPS units on my computers, and now I have one on my HD LCD rear projection TV. It's really important on that because if the power were to go out while I'm watching TV, and I didn't have the backups, the cooling fan that always continues to run after the set is off would not run, and this would cause the bulb to die immediately, or at least shorten its life. I had the same problem in high school with the spotlight in our auditorium. When someone on stage crew would shut it down without leaving the fan on (it was VERY old), the bulb would go and it cost close to $200 back in the mid 80's.
I'm going to get an LCD flat panel soon. I think I'm going to put it on a BackUPS, too. Not because I'm concerned about any fan (I don't think they have them), but rather because I'd like a steady power flow to the set, and the BACKUPS provide that, too. Of course, maybe I should just get a power filter for that?
(I know I have to keep an eye on the watts drawn by the TV. You can't use the BackUPS on a CRT TV because I guess there's a huge electrical draw when you first turn a glass-tube TV on.)
"(I know I have to keep an eye on the watts drawn by the TV. You can't use the BackUPS on a CRT TV because I guess there's a huge electrical draw when you first turn a glass-tube TV on.)"Sure you can.When I had CRT monitors they and the computer where on the UPS..
.
A-holes. Hey every group has to have one. And I have been elected to be the one. I should make that my tagline.
Now that you mention it, it seems like I did that, too, back when I had CRT monitors. Maybe it was just the size of the TV that was the problem. It was a 40" CRT standard (not wide) screen. I was trying to figure out whether I could (or should) hook it up, and I remember the guy at APC telling me that I couldn't go by the manufacturer's information on how much wattage the set used. He said that initial "pop" you hear when you turn that kind of TV on is it doing something that takes more electricity than the TV uses the rest of the time it's on. Then again, maybe he didn't know what he was talking about. I figured it was in his interest to tell me his product would work, but that doesn't mean he knew what he was talking about.
There is an antigausing coil that gets hit with a big jolt at the start and with each osillation of the AC decrease slighly.Don't have any idea of how much current that draws..
.
A-holes. Hey every group has to have one. And I have been elected to be the one. I should make that my tagline.
"Of course, they claim it's not a problem. But how low can the voltage go before it becomes a problem for various devices like motors and TVs, etc."The standard supply is 120 +/- 5%, except in CA. I found that spec a few years ago. Don't remember what the lower spec if for CA, but it was done just for the reason that you say.Now for things like lights the current and wattage is reduced porportionally and the light also is reduced. Aslo an heading (stoves, space heaters, toasters).Most good electronic equipment will work over a very wide range. A number of devices use switching power supplies. Look on the label for some computer monitor and I have seen 100 - 250 volts listed for the operating range. That is without any switches or adjustments, although different cordset are needed for other countries.Motors are a different situation.First they are speced to run at 115 so they should be should be OK at the lower limits. But they are more of a constant power device, so as the voltage drop the current increases.I am not sure, but I would think that at something between 110 and 105 tht they would start getting flaky..
.
A-holes. Hey every group has to have one. And I have been elected to be the one. I should make that my tagline.
Hi Bill--
I just checked my voltage and it is 120v, which made me remember that I think I heard they only lower it when power is scarce. It figures that the rest of the time they'd want it back up at 120v so that they can sell more.
I'll have to check in the hot summer days when the TV anchors ask us to wash our clothes at night and turn our air conditioners up to 92 degrees. It will probably be low then, when it suits them.
Oddly, running the toaster on the end of the longer wire will use less electricity per minute. But it will take longer to toast your bread.
If your view never changes you're following the wrong leader
"Oddly, running the toaster on the end of the longer wire will use less electricity per minute."That 1200W toaster uses less electricity due to VD, but back at the meter isn't the electricity consumption the full 1200W because line loss is consuming the difference?BruceT
Assuming that the toaster resistance is constant (all material have some tempature coeficent, but should be small enough to ignore in this example).Sp now you have the toaster resistance in series with the wiring resistance so the current draw will be slightly less.And if the toaster used a sensor of the color or surface temp of the toast it would have to run longer and in turn use more total power. The amount needed to actually toast the toast, but also the lost power in the wiring..
.
A-holes. Hey every group has to have one. And I have been elected to be the one. I should make that my tagline.
No, because the voltage drop causes the toaster to consume less CURRENT.Remember, current is voltage divided by resistance. If the toaster consumes 10A at 120V, it has a resistance of 12 ohms. If it only receives 115V due to voltage drop, that 12 ohms will only allow 9.58 amps of current. (Of course, the reduced current means that the voltage drop won't be quite the 5V, which means that the current will be higher, etc, but you have to get into higher algebra to solve that exactly, and rarely do you need to be that exact.)Without the voltage drop the toaster consumes 120V * 10A = 1200W. With the voltage drop the toaster + wire consumes 120V * 9.58A = 1149.6W. The toaster alone consumes 115V * 9.58A = 1101.7W, while the wiring consumes the remaining 5V * 9.58A = 47.9W.Of course, since the toaster is receiving less power it takes longer to make toast, so you do end up paying more due to the voltage drop.
If your view never changes you're following the wrong leader
You said, "Motors are a different situation. ...they are more of a constant power device, so as the voltage drops the current increases."So my 1200 watt toaster is really a 12 OHM resistive device rated for service up to 1200W, therefor it uses less power as voltage drops due to line losses, resulting in less electricity consumed at the meter? A 10amp/120V rated motor, on the other hand, would try to deliver 1200 watts and so would pull more amperage to make up for voltage drop in a long wire run, thereby consuming more electricity at the meter?
BruceT
You got it.
If your view never changes you're following the wrong leader
Thanks, Bill & Dan, for taking the extra time and patience to explain the why-and-how of electricity to those of us who know just enough to otherwise be dangerous. I know that a lot of guys here [even lurkers] learn from your responses.
BruceT
It's always a pleasure to tutor someone who's making an honest effort to learn.
If your view never changes you're following the wrong leader
"Also, is the first box in a wiring circuit subject to less of a voltage drop than the last box because there is a shorter length of wire between the first box and the breaker box?"
That is correct. But it also depends on the loads.
If the first one is say a 10 amp heater and the last one is 60 watt lamp. Most of the voltage drop would be at the heater and the voltage at the light would be about the same.
"Is there any reason to not wire off in two directions (with two separate cables) from a given box? "
You can do that. And it is sometimes done when there is a reason logical reason. But doing it too much can get confusing to someone that works on it later.
But in a house it is general does not gain you anything. First you need to be in the 75-100 ft range until the voltage drop is a concern. There are a number of online voltage drop calculators. A google will find them.
Typically you have a long run from the panel to the first box. From their is typically a short run to say all of boxes in a living room.
if you do have a large house with long runs then you should consider installng sub-panel(s).
If you only have a couple of long runs where voltage drop would be a problem then upsizing the long run can be done.
.
A-holes. Hey every group has to have one. And I have been elected to be the one. I should make that my tagline.
Thanks for the response, Bill. Could you read the reply I justed posted to McPlumb to see if I'm understanding things?
On the home I'm working on now, I don't think cable length is much of an issue. It's a typical 1700 square foot, single-story slab home in a subdivision that was built in 1979 or 1980. I'm just trying to get a better understanding of how voltage drop and resistance work.
There's only one circuit that concerns me in terms of length. There's a run of 6/3 NM cable, with a 10 gauge ground, that is about 80'. It originally connected to a free-standing range, and I want to connect it to a full-size double wall oven instead (and use a gas cooktop).
From what I've read, this existing 6/3 cable would be large enough to supply the double wall oven, and it would be wasteful to replace it. The only difference with a newer 6/3 cable would be that it would have better insulation and maybe a larger ground.
If that's true, the only reason I would replace it would be to upgrade to 4/3. Would you think that is a worthwhile upgrade? Right now I can only see replacing the existing 6/3 cable if I find it is damaged. The original owner did all kinds of crazy things to the wiring, so I may find a problem with it.
Your response to Mc Plumb is correct.Now I want to warn you one one things. For cable the standard is not not use the ground wire in designating cable.So a 15 amp/ 120 circuit is wired with 14-2 romex. If it was very old it would would be only the 2 wires. New work would have a ground wire, but still only called 14-2 and have a ground wire.Now stove and ovens use 120/240 use 120/240 and require 2 hots and neutal. Until 1996 the NEC allowed using ONLy 3 wires and the neutral was also allowed to be used as the ground. New work requires 4 wires, with a separate ground wire. But many local codes required 4 wire long before that.Now ranges where often wired with SE cable. That will have 2 hot wires and the neutral is bare wire that is wrapped around hots and not insulated except fo rthe other sheath.You can reuse an existing circuit for replacing equipment. However, if you "change" the circuit then it needs to be upgraded to 4 wire circuits. It is really up the the local inspect about what "change is".As to your question. The #6 should be more than plenty for the ovens, but what are the installation instructions about the size of the circuit..
.
A-holes. Hey every group has to have one. And I have been elected to be the one. I should make that my tagline.
Hello Bill, thanks again for the response.
I do know that, for instance, modern 12/2 cable actually contains 3 wires (white, black, bare). But I only recently became aware of the issue of whether grounds are present in all generations of NM cable.
They are in everything in this house, including the 6/3 wire for the range. I just checked it and it is a 4 wire cable... 1 white, 1 black, 1 red, and one bare neutral. The neutral is only 10 gauge, and it's my understanding that that wouldn't be the case with currently available 6/3 cable. I'll check the instructions for the double wall oven, but I believe they state 40 amp.
Presuming that 6 gauge is big enough, would you be at all concerned with the existing 6/3 having the undersized ground, or with the quality of the insulation on the wire given its age?
The smaller ground wire is still grandfathered in.The number #16 in 14-2 is a little iffy.But the purpose of the ground wire it to provide a path so tht if the the applicance shorted to ground that it would draw enough current to trip the breaker. And #10 can handle that..
.
A-holes. Hey every group has to have one. And I have been elected to be the one. I should make that my tagline.
So, if it was in your own house, would that undersized #10 ground concern you? Or the age of the cable in terms of the insulation?
It is all a balacing act of condition and difficulty of replacing it.If it was in very good condition and 75 ft through a 12" high crawlspace it would stay.If it 5 ft from the panel and above an unfinished basement, and the insulation was crumbly I would replace it.Now inbetween???.
.
A-holes. Hey every group has to have one. And I have been elected to be the one. I should make that my tagline.
Very tall attic, not too difficult to replace. So, if it was easy enough, you'd replace it. Why would you replace it? I mean, what benefits would you be seeking?
I would not replace it just because of the small ground wire.More likely if the insulation showed deteriation, but again it depends on how bad..
.
A-holes. Hey every group has to have one. And I have been elected to be the one. I should make that my tagline.
Okay. I guess I'll let the condition of the cable's jacket (and anything that might be exposed underneath) be the determining factor. This cable, and some of the others in the house, are black. And I'm finding that the moron that owned the house before me cut into wires at different points (intentionally mostly, so it shouldn't be a problem with this one) and did things like splice wires together and tape it all up with black electrical tape. Due to the jacket and tape being black, it's hard to always see what he's done.
Keep in mind that a ground wire, under normal conditions, carries no current and contributes no voltage drop. The wire need not be large enough to carry the circuit load without unreasonable voltage drop but only needs to be large enough to not burn up when subjected to a short. There are probably all sorts of footnotes on the NEC regarding this, but in general a wire one size smaller than the current-carrying conductors is fully sufficient for this task. You have a wire that's two sizes smaller -- probably sufficient but with a bit less margin of safety.But consider that many homes are still wired with 3-wire range plugs (no separate ground at all), a far less safe arrangement. And the wire you have was code when installed and should be "grandfathered" from an inspection standpoint. I'd frankly leave it be unless you find some other reason why the cable ought to be replaced.
If your view never changes you're following the wrong leader
Your instincts are correct (which isn't a given when it comes to voltage drop). Voltage drop is a function of the distance traveled from the breaker panel (and of course also a function of the diameter of the wire). Y-type circuits will reduce voltage drop but aren't commonly done simply because 3 wires are harder to tie together than two, and it's easier for the electrician to plan things (and keep the plan straight) as a daisy-chain.
Plus of course you want to avoid junction boxes with no "device", since they must remain accessible, and that may mean an unsightly blank cover.
You will often see, in commercial property with suspended ceilings, the main "trunk" of a circuit run through the ceiling, with junction boxes every so often to feed down to lights or outlets. Simpler (& cheaper) than daisy-chaining if conduit is being used.
Keep in mind, though, that voltage drop is rarely a problem in residential construction, since the wire distances are so short and loads are relatively small in most cases.
At first wiring off in two directions would make some sense, but what you will run into is that very quickly you will have too many wires to connect with a standard wire nut . This will cause problems with space & wire fill requirements
Sorry for chiming in late. I am not a Sparky. I noticed on the American Copper Association web sight a few years ago they did some studies on the most efficient wire sizes to use in certain situations. Their conclusion was to go to a one size larger wire only in situations where a constant load is on the circuit. Such as a light left on all the time, refrigerator circuits and furnace circuits etc.. It was a waste to go oversize on circuits that saw intermediate use. They probably still have the info their. Jay
A waste in terms of the added cost of the heavier-gauge wire?