Determining Circuit Length Through Resistance Testing

[infinity]

I have a situation where I need to find the distance of 7 runs of MC cable from a low voltage transformer to 7 LV lighting tracks. The tracks are about 10' apart so each set of LV conductors from the transformer are about 10' longer than the next. This makes the longest run about 60' longer than the shortest. The engineer thinks that this will create a problem because the lighting level at the farthest track will be dim compared to the level of the closest due to voltage drop. By making all of the circuits the same length the varying light output from track to track will be eliminated.

Question is what is the simplest way to test each individual cable run to find their respective lengths? Someone suggested splicing the two conductors on one end of the cable together and measuring the resistance on the other end with a DMM. Then you could compare the readings and find each cable length. Will that actually work? Or is there a better method? The goal here is to add cable to the shortest runs so that they're all approximately the same length. The cables are already in place. Thoughts?

 

[iwire]

A DMM does not have the accuracy needed for this. Conductor length testers are available and expensive.

FWIW I agree with the EE

 

[gar]

080811-0739 EST

infinity:

Have these cables already been installed? The implication is yes, and you have seven pairs of wires.

If already installed, then there are at least two methods two methods.

First method: Apply power to one pair of wires at a time. This will identify what pair belongs to what strip. From the physical placement of the strips you know the length relative to the longest run. Just add the corresponding amount of wire.

Second method: Loop resistance test. Make a very low resistance connection of the two wires at the remote end, use a good screw compression method.

As an example #12 copper wire has a resistance of 1.588 ohms per 1000 ft. Assume your longest run is 60 + 20 = 80 ft. The loop length is 160 ft or 160 * 1.588 / 1000 = 0.254 ohms. The change of resistance for each 10 ft is 20 * 1.588 / 1000 = 0.032 ohms. Apply a known current of about 10 A to your loop, a reasonable amount of current for #12. You need to measure both current and voltage simultaneously, if possible.

Current measurement is just a matter of getting the ammeter in series with the loop. The voltage measurement method is more critical. Even if you use low resistance connections to the loop from the current source, then the voltmeter probe contact points should be on the loop wire rather than on the current source side of the connections. Generically this is referred to as a four terminal resistance measurement.

You would expect at 10 A a voltage across the loop of V = I * R = 10 * 0.254 = 2.54 V. If the current was 11.7 A, then the voltage would be 2.97 V.

The incremental voltage for each 10 ft of distance (20 ft of loop) is I * 0.032 and for 10 A = 0.32 V. My value of resistance per ft of #12 is from a table and you actual resistance may differ somewhat from table values.

How to supply this kind of current. If one strip of lamps draws about 10 A, then put a strip in series with the transformer as the current limiter. Tungsten incandescent lamps are somewhat of a constant current load and thus are better for this experiment than a normal resistor.

Somebody else check that what I said makes sense and has no errors, and maybe to clarify my comments. I have to leave at the moment.

.

 

[derek22r]

Can someone please explain this to me.

I am understanding that there are 7 lights each wired with a homerun back to the transformer, Correct? How will adding wire to the closest light affect the voltage drop on the furthest light? It will still have the same load and resistance.

 

[gar]

080811-1335 EST

derek22r:

There are 7 different strips of lights. Each strip is fed with its own pair of wires. The furthest away strip will have the lowest voltage at the lamps. There is no desire to change the voltage at the furthest strip since this has the lowest voltage.

What is desired is to make the light intensity at each strip the same. This means that the voltage at each strip needs to be the same. Since there are different resistances to each strip because of the different cable lengths it is desired to make each path have the same resistance. One way to increase resistance is by increasing cable length. Basically make the cable length to each light strip the same.

.

 

[rook81108]

Why is it not possible to locate the LV transformer at the end of each LV strip? The best way to match distances would be to eliminate the low voltage transmission. If that is not possible, then your extra wire to match voltage drop on each is expensive and bulky.

Finaly, the appearence of light will be altered by many other factors (surroundings, surface colors, objects, etc..) so that often identical lights swapped in their positions may appear different brightnesses without any merrit to the claim.. Too often these quirky judges are halucinating.

 

[infinity]

This is a job we inherited from someone else. Although we know can find both ends of the cable there is no guarantee that each individual run is in a straight line from point A to point B. So it seems that by determining the distance of each run, by it's finding its resistance, would allow someone to make each run length the same. I've seen expensive equipment that will tell you the length of a run by testing it and measuring it's resistance. Someone suggested the DMM method but I've been skeptical of using one for this type of testing due to the limited accuracy of that type of test.

 

[brian john]

If the job is completed NOTHING beats the eyeball test. Fire it up and take a look, Simple.

If the fixtures are not installed, do a voltage drop calculation based upon estimated lengths and know load current and use a variac.

 

[nakulak]

I have little knowledge of low voltage lighting systems, but I recall seeing something called an "equalizer hub" which I assume was an array of pots or var resistors of some kind to balance low voltage drop just like in the scenario you have mentioned ?

 

[gar]

080811-1501 EST

infinity:

With a Fluke 87 meter and the approximate voltages I mentioned above you will be able to measure resistance within 1% using the 4 terminal resistance measurement method. In reality you do not need accuracy of 1%.

What is the load current for each strip at what voltage?


rook81108:

The probable reason there is not a transformer at each strip is to eliminate running 120 V to each strip.

Long ago, before 1879, Edison determined that a low voltage distribution system was not practical. Obviously this is a function of current, distance and voltage.


nakulak:

The practicality of variable resistors is a function of current, and resistance required. Suppose my 10 A current from above is in range, then for 0.032 ohms per 20 loop feet the power dissipation is 10*10*0.032 = 3.2 W. For the closest strip this is 18 W and 0.192 ohms. You won't find these as commercial resistors, but you can wind them with Nichrome wire. You have to have a safe place to put these high power things. However, the same is true of all the extra cable that needs to go somewhere to generate these resistors with cable.


A more practical and lower power loss method would be to use phase shift modulation to adjust the current, and no additional wire.

Another possibility is to break the seven into three groups of 2 strips, and 1 of 1. Feed each of these at the mid point to each group, then use equal lines to the center of each group. Maybe move the transformer to a more central point relative to the groups.

.

 

[infinity]

Another possibility is to break the seven into three groups of 2 strips, and 1 of 1. Feed each of these at the mid point to each group, then use equal lines to the center of each group. Maybe move the transformer to a more central point relative to the groups.

Unfortunately the only real option is to extend some of the transformer load conductors to make the lengths equal. The remote transformers are already installed in a room chosen by the architect and all of the wiring is in place. It would be possible to install additional cable within the ceiling to lengthen the real short runs. The only question is how much cable is needed, hence this thread.

 

[gar]

080811-1904 EST

infinity:

I have told you how to determine how much resistance and/or additional wire to use. Have you made any of the measurements?

.

 

[infinity]

080811-1904 EST

infinity:

I have told you how to determine how much resistance and/or additional wire to use. Have you made any of the measurements?

.

Yes you did, thank you. The real question now is will a simple DMM give an accurate enough reading to make the test worth the effort or should just an approximation of the different lengths be used? The powers that be were looking for a scientific approach.

 

[gar]

080811-2059 EST

infinity:

What wire size is being used? What is the load current for one light strip? What is a simple DMM? What is the minimum resolution in millivolts, smallest displayable increment? Is the wire copper or aluminum? What is the transformer output voltage? Do you have some of the same wire?

.

 

[infinity]

080811-2059 EST

infinity:

What wire size is being used? What is the load current for one light strip? What is a simple DMM? What is the minimum resolution in millivolts, smallest displayable increment? Is the wire copper or aluminum? What is the transformer output voltage? Do you have some of the same wire?

.

1800 watt Q-Tran transformer w/ 7-10 amp secondary CB's
7-Runs 50'-125' in length
24 VAC
225 watts/track
#10 MC cable

 

[gar]

080812-0601 EST

infinity:

24 V at 225 W is 9.37 A, thus one track of lights will provide a good test current. If this is not enough use two tracks in parallel to provide double the current. You can use one or two existing installed tracks including their cables. An exact value of current is not required. You just need to measure and know the current when the resistance measurement is made.

125 ft * 2 is the loop length for the longest run. Assuming copper for the conductor the calculated resistance is 250 * 0.9989 / 1000 = 0.250 ohms. Using two tracks for the current source and about 18 A means:
4.5 V for the 125 ft run
and 4.5*50/125 = 1.8 V for the 50 ft run.

The difference between 125 and 50 is 75. Divided by 6 is 12.5 ft between tracks and an incremental loop of 25 ft. The incremental resistance is 0.25*25/250 = 0.025 and the incremental voltage between tracks at 18 A is 0.45 V.

Another way to get approximate information follows:
Measure the current to a particular track, the voltage between that track's wires at the transformer, and the voltage difference between the wires at the track. Subtract these two voltages and divide by the current to get the loop resistance. This will be somewhat less accurate because of a lower current and normal line voltage fluctuations into the transformer. I am assuming all three measurements in this case can not be simultaneous.

The ratio of aluminum to copper is approximately 2.655/1.678 = 1.582 .

From Wikipedia
http://en.wikipedia.org/wiki/Incandescent_light_bulb

[edit] Voltage, light output, and lifetime
See also: Lamp rerating
Incandescent lamps are very sensitive to changes in the supply voltage. These characteristics are of great practical and economic importance.

For a supply voltage V,

Light output is approximately proportional to V^3.4
Power consumption is approximately proportional to V^1.6
Lifetime is approximately inversely proportional to V^16
Color temperature is approximately proportional to V^0.42 [40]

This means that a 5% reduction in operating voltage will more than double the life of the bulb, at the expense of reducing its light output by about 20%. This may be a very acceptable trade off for a light bulb that is in a difficult-to-access location (for example, traffic lights or fixtures hung from high ceilings). So-called "long-life" bulbs are simply bulbs that take advantage of this trade off. Since the value of the electric power they consume is much more than the value of the lamp, general service lamps for illumination usually emphasize efficiency over long operating life; the objective is to minimize the cost of light, not the cost of lamps. [41]

The relationships above are valid for only a few percent change of voltage around rated conditions, but they do indicate that a lamp operated at much lower than rated voltage could last for hundreds of times longer than at rated conditions, albeit with greatly reduced light output. The Centennial Light is a light bulb which is accepted by the Guinness Book of World Records as having been burning almost continuously at a fire station in Livermore, California since 1901. However, the bulb is powered by only 4 watts. A similar story can be told of a 40-watt bulb in Texas which has been illuminated since September 21, 1908. It once resided in an opera house where notable celebrities stopped to take in its glow, but is now in an area museum.[42]

The approximate voltage difference between your 125 ft track and the 50 ft track for copper wire is about 1.35 v . This is a (1.35/24)*100 = 5.6% difference. Thus, the light output at 125 ft vs 50 ft is about 0.944^3.4 = 0.822 or 82%.

Do as brian john said and do an eyeball test.

.

 

[rook81108]

Calculate, eyeball and estimate all you want but if you run low voltage in MC (I have) it should be labeled very well. The inspectors around here won't allow coils or bunches of MC above a cieling. Idiot proofing and cat hearding are their speciality.

The service hack who stumbles upon it in the future needs to "Know" not to "power it up". Smoking in the building is not allowed with tobacco let alone the lights above the building tennants. :wink:

 

[gar]

080812-1054 EST

infinity:

rook81108 has made a good point on unexpected problems you might encounter. Also cost is a factor. So I looked at some resistor sources. Ohmite might have some useful units.

www.ohmite.com

Track 7 requires no additional resistance. Track 6 might require about 0.025 ohms and would dissipate about 2.5 W. Track 5 would need 0.05 ohms, and so on based on my above calculations. Track 1 requires 6 times 0.025 = 0.15 ohms and the power dissipation is about 15 W.

As an example see
http://www.ohmite.com/cgi-bin/showpage.cgi?product=270_series

Note: that high precision is not required. If you are within +/-10% on resistance that is probably more than adequate. Also it may not be necessary to use any resistor on track 6. You need visual evaluation of the relative brightnesses.

Correct resistance values would need to be determined for you actual wires. My above figures are based on limited information. Out of time for now and I haven't checked anything above.

.

 

[infinity]

Calculate, eyeball and estimate all you want but if you run low voltage in MC (I have) it should be labeled very well. The inspectors around here won't allow coils or bunches of MC above a ceiling. Idiot proofing and cat hearding are their speciality.

The service hack who stumbles upon it in the future needs to "Know" not to "power it up". Smoking in the building is not allowed with tobacco let alone the lights above the building tennants. :wink:

I would ask the inspector where in the code book does it say that you can't leave extra MC cable within a ceiling?


I did some quick voltage measurements today with the tracks energized. The end of track 1 (the closest track to the transformer) had a VAC reading of 21.4 volts. The end of track 7 (the farthest track from the transformer) had a voltage of 20.2 VAC.

IMO these values are unacceptable because the amount of light output is far less than anticipated by the lighting designer who bought a system that runs at 24 VAC. However visually, given the distance apart between tracks 1 & 7 the 1.2 VAC difference isn't really noticeable.

 

[gar]

080812-1550 EST

infinity:

The differential resistance of 0.15 ohms I calculated from track 1 to 7 and a current of 9.37 produces 1.41 V and that is in good agreement with your 1.2 V difference measurement. Current may be somewhat less because of the lower lamp voltage.

Change the transformer so that you get the desired voltage at track 7. You might consider a 28 V unit with somewhat higher current rating. Thus, proably a lower internal impedance. Or use a seires boost transformer. Just on a ratio basis a 28 V transformer will raise track 7 to 23.4 V. But this is quite dependent upon internal impedance.

Is the transformer a nominal 120 input and is line voltage at that level?

It now appears that you have the additional problem of insufficient voltage, and maybe differential intensity is not a major factor.

.