# Question: Is it possible to connect in one wire only?

1. Jan 29, 2010

### xiv_wolf

Hello guys!

I would just like to ask if this connection is possible.

2. Jan 29, 2010

### MATLABdude

Hi Xiv_wolf,

This connection is possible. It just won't do anything as you probably won't have any voltage across the load (unless the connection points are really, really, far apart, and even then you won't have much)

3. Jan 29, 2010

### Carl Pugh

Worked for company that built welders using this circuit. (80 volt instead of 220 volt)
Used copper or aluminum bus bars between power source and (main) load.

4. Jan 29, 2010

### zgozvrm

No voltage drop across leads to bottom load = no circuit = no work.
The bottom load in your picture has been short-circuited, and therefore, bypassed.

On the other hand this will work:

... but each load will only utilize a portion of the available 220V AC, such that the voltage drops across the load will add up to 220V.

If the company you worked for actually connected ammeters this way, I hope they're not in business any longer! Without removing the wire between the 2 leads to the "bottom load," there is a short-circuit which bypasses whatever type of load you place there. Those ammeters would have always read zero!

5. Jan 29, 2010

### Averagesupernova

Speaks a voice with little real-world practical experience when it comes to high current ammeters.
-
A wire that is not a superconductor will have a small resistance. This can be enough to cause a small yet easily measurable voltage across what appears to be a short piece of wire. This old trick is certainly nothing new. The same thing has been accomplished using a metal plate and drilling holes in it for calibration.
-
I will admit that it may be poor practice to draw a circuit that way without noting that the 'wire' across the device that measures/indicates the current is doing more than just being a wire.

6. Jan 29, 2010

### zgozvrm

I deal with voltages as high as 12,000V AC, 3-phase @ up to 200A, as well as 480V, 3-phase up to 2,000A. I don't know about you, but I consider this to be "high current." At those levels, I have only seen current measured by the use of current coils (know as current transformers) which divide the current down to a safer level (while also providing electrical isolation).

In the case of an inline ammeter like what you're speaking of, you need to place a shunt in parallel with the ammeter in order for the meter to be able to read currents above its normal capability. This shunt is more than a piece of wire, it is another load (albeit a small one) that has a very precise, known resistance as well as the ability to allow the full current of the circuit to pass through it without burning up. This allows for the current to be split so that the meter only handles a known maximum amount of current, while the shunt passes the remaining current.

The schematic shown by the OP shows no device of any type at this location. A straight line in a schematic represents a wire. For a shunt, there should be a resistor symbol at that location (between the 2 leads to the lower load).

So ... the answer is "No, you cannot make connection like 'this' and expect it to work."

7. Jan 29, 2010

### Averagesupernova

That straight line in the schematic represents a wire with a non-zero resistance which will develope a voltage across it when current flows through it. A 100 micro-amp meter movement with a 600 ohm coil needs about .06 volts across it to get full deflection. You don't think it is possible to run roughly 100 amps through a foot of #8 copper wire with .000739 ohms per foot and develop more than enough voltage to make the meter movement go full scale when it's leads are hooked on a piece of said wire a foot apart? Granted, the #8 wire needs to be cooled with a fan or something to prevent overheating.
-
I already mentioned in my other post that a schematic drawn like this is poor practice. But to say that the welder in carl pugh's case could not have possible shown any current on the meter is quite simply wrong.

8. Jan 29, 2010

### zgozvrm

It's not poor practice ... it's simply incorrect!

If I owe you $1,000.00 and I write you a check with one less zero on it than there should be, then you will only be getting$100.00 from your bank when you deposit the check. Is that "poor practice?" No! It is a mistake and it is incorrect.

The original circuit as shown gives no wire size or distance, so it can't be assumed that there is a shunt in this circuit. Without such qualifiers, there is nothing preventing a person who is building a circuit from the diagram from placing the leads to the bottom load at whatever distance they choose (even to the same exact point).

If a meter is to be connected in this fashion, then there needs to be a shunt (which has a specific, known resistance) and it should be indicated on the schematic. As shown, there are no restrictions and therefore the ammeter on the welder would show little or no current (at the very least, it would vary from one welder to the other).

My point is that I seriously doubt (I hope) that the welders where wired like that. Rather, they were likely wired as such:

9. Jan 29, 2010

### Averagesupernova

Not EVERYTHING can be represented on a schematic. Work in electronics for a while (not as an industrial electrician) where positioning components physically is just as important as connection and you will know what I mean. It is quite possible and acceptable to use a piece of #8 wire as I described.
-
The choice is yours. Would you represent it on a schematic as a resistor? I wouldn't, but I would make a note to indicate that it is there for more than carrying current. BTW, you haven't answered my question:

10. Jan 29, 2010

### Bob S

Use of calibrated 50-millivolt shunts for measuring very high dc currents is standard practice. The voltage signal is read using a very high impedance voltmeter (insulated), so there is negligible voltage drop in the voltmeter connection. See

http://www.emproshunts.biz/product.aspx

Bob S

11. Jan 29, 2010

### Averagesupernova

The argument isn't what is standard, BobS. My main point is that Carl Pugh gave an example that zgozvrm has said is not possible.

12. Jan 29, 2010

### Averagesupernova

Ya know, something has just ocurred to me. The OP in this thread has not been back. So we don't know the motivation behind his post. It is possible the OP has seen this sort of thing on a schematic or has been into a machine that was wired as Carl Pugh has suggested and is wondering what the he11 is up.

13. Jan 29, 2010

### zgozvrm

Yes, but then there will be more information, such as a layout diagram. The point being, with the information given, you can't assume there is a shunt in the location in question.

Yes, I would represent it as a resistor, but there would be a notation as to the fact that it is a shunt of some specification. I would also accept a rectangle with the word "shunt" (or the resistance value) inside with other pertinent information outside. The point here is that if you require a shunt in a circuit, you need to show it somehow.

And to answer your question (if it makes you happy), of course there could be enough of a voltage drop in your scenario. In your example, there would be a 73.9 mV drop across 1 foot of 8 gauge wire, so the meter would have to have a full-scale deflection of 73.9 mV (possible, but unlikely, as this is an odd value for full-scale).

Last edited: Jan 29, 2010
14. Jan 29, 2010

### zgozvrm

Not possible as drawn. The existence of a shunt (a known resistance between to points) is not indicated.

15. Jan 29, 2010

### Averagesupernova

I can't agree with you here. This is why I said not everything can be represented on a schematic. There is nothing that says how long the distance is between the generator and the load on the right in the OP's schematic. There is also nothing that states the distance between the leads of what the OP calls the lower load. They could be attached with several feet of wire between them easily, or they could be attached right next to each other. Schematically each one would be correct, but electrically in the real world they would behave differently. It goes back to MATLABdude's first post in this thread. The schematic cannot show this without extra notation. I don't know of any machine/product that could (or should) be assembled from just a schematic. I have done technical writing for assembly procedures in an electronics manufacturing plant. You shouldn't count on a schematic only for assembly.
-
Of course it is unlikely. That's the whole point of the schematic being unreliable. The assembler has instructions on how many inches apart to connect the leads. The unit is tested and if the meter is within accuracy spec all is well. If not, the leads are moved or a trim-pot is adjusted. Now BobS gave an example that would solve the problem, but for less-than-precise measurements, the example I give is a good alternative.

16. Jan 29, 2010

### zgozvrm

What you're saying here is that you are agreeing with me: As shown, this circuit is no good. More information is needed.

17. Jan 29, 2010

### Averagesupernova

LMAO Wow. Twist it how you like. The main goal in my posting was to try to set you straight in the example of Carl Pugh's post. For instance: If someone gave me a schematic like that and I built it on a 4 x 8 sheet of plywood with the lower load 100 micro amp meter movement as I described with a lead on each end of the plywood and sent 100 amps out of the generator through the load on the right, I would get a meter reading assuming I used appropriately sized wire as previously discussed. No one could say I built the circuit incorrectly. The meter reads, end of story. What I'm really curious about is what motivated the OP to post this.
-
And BTW, I was the first to say something like this:
-
Your argument will be: "But you could build it with the leads right next to each other and the meter won't read." You would be correct, but you can't say it won't read either if I put the leads where I want them.

18. Jan 29, 2010

### zgozvrm

First of all, the original schematic never stated that this load was an ammeter (after all, an ammeter is typically such a minor load that it generally is considered to be no load at all). Usually, meters are indicated in diagrams as such.

Second, you'd be a fool to try to build this circuit without more information.

I never doubted that you could build a circuit in which the lower load was actually an ammeter, just that you couldn't assume that there was a known resistance between the 2 points. Even if you did only use wire (rather than an actual shunt), it would likely be of different wire gauge than the rest of the circuit and would be easily identified visually as "something other than just wire" and the OP most likely would have mentioned it.

Drawn as is, it would be assumed by most people that there is no resistance (negligible at most, definitely not a precise known amount) between those 2 points. If a resistance is part of the circuit, than it needs to be displayed (or referred to) in some way. If that was the intention of the diagram, then there is missing info.

So ... good for you for explicitly say that first. (I said it implicitly).

Lastly, taking the diagram at face-value, this connection would be considered a short-circuit. If I posted a schematic of a circuit that was fed by an AC source and flashed an LED at a rate of 2 flashes per second with several components on it and then later told you that one of the wires represented by line was actually a shunt resistor, you would ask, "How come the schematic doesn't show it?" That fact would likely change the circuit in some way. Also, if someone wanted to add an additional load (another LED) to this circuit, it would not suffice to connect it in the way that the OP showed in the original pic.

Granted, it is difficult to show everything in a schematic, but not when it comes to parts and their inter-connectivity (that's what schematics are for). Proximity of parts is whole different issue ... there are different diagrams for that (plus proximity wasn't part of the original question).

So you're right, there can be a circuit like the OP's if you have a known resistance between those two points and the bottom load is an ammeter. And, you can build one. Big deal. Get off your high horse! You are obviously one of those people that is always right (or you just like to create arguments with no real purpose ... just stir the pot). I can make this circuit do just about anything, if I start adding parts that aren't indicated on the schematic.

That's surely not what the OP was asking.

Taking the length of wire and its resistance into account is a special case, one that deals with very sensitive circuits. For the standard schematic diagram (which, without additional information, is what this is), the circuit is not valid; the lower load is short-circuited.

"End of story!"