 reply to dandeman
Re: How to calculate the amperage of paralleled conductors??? said by dandeman: If one length of cable turns out to be for example 400 microohms, and another cable 100 microohms, then ohms law says the 100 microohm cable will carry the by ratio the brunt of the current.. That makes a sense the current will always prefer the path of least resistance, but what might be the actual difference in the current carried by the other conductors? As an example what I am doing, impedance aside, let’s take 60 amps divided by 4 wires equaling 15 amps per number 10 conductor and lets say I think those conductors are 100 feet long but unknown to me two of those conductors are 110 feet in length. That stated at 120 volts how much of an additional burden might be placed upon the shorter conductors due to the additional resistance of the longer conductors? 

leiboldPremium,MVM join:20020709 Sunnyvale, CA kudos:10 Reviews: ·SONIC.NET
 said by 54067323:let's take 60 amps divided by 4 wires equaling 15 amps per number 10 conductor Sorry, but that is exactly the kind of thinking that will get you into trouble. The current will only divide equally over the 4 wires if their electrical characteristics are identical (and in real life they are never truly identical). In your example you will have 2 wires with 100 * R and 2 wires with 110 * R (where R is the resistance in Ohm per feet, about 0.001 for 10awg copper) in a parallel circuit. The voltage across that circuit is equal but due to the difference in resistance the current distributes unevenly. This is all theoretical since you are ignoring the resistance of any splicing/bonding along the way. Combined resistance of the bonded circuit is 0.0262 Ohm resulting in a 1.571V drop at 60A current (the 120V supply is irrelevant, the result is the same at 12V or 240V). The current through the 2 shorter wires is 15.71A each while the current through the 2 longer wires is 14.29A each. This small difference in current is to be expected since there is only a 10% difference in wire length in your example (resulting in an identical 10% difference in currents: 14.29 + 10% = 15.71).  Got some spare cpu cycles ? Join Team Helix or Team Starfire! 


2 edits  said by leibold:The current through the 2 shorter wires is 15.71A each while the current through the 2 longer wires is 14.29A each. This small difference in current is to be expected since there is only a 10% difference in wire length in your example (resulting in an identical 10% difference in currents: 14.29 + 10% = 15.71). Based on that do you think a tinned number 10 in good condition measuring out at nearly zero ohms end to end can or cannot safely carry the differental amperage? 

mackeyPremium join:20070820 kudos:14  reply to 54067323
said by 54067323:That makes a sense the current will always prefer the path of least resistance, but what might be the actual difference in the current carried by the other conductors? When they ran the building I work in off a generator for a week while repairing the transformer vault, they paralleled 2 4/0 cables for each phase. Most of the pairs were fairly balanced, but one pair had about 325A on one cable but only about 100A on the other. The one that was carrying 325A was quite warm.... /M 

garys_2kPremium join:20040507 Farmington, MI Reviews: ·Callcentric
 reply to 54067323
It sure sounds like you're going to go this way, so if you do please check the amperage on each conductor while drawing a lower, test load (say, 100 watts). Any imbalance there should scale up (assuming the connections don't change) with the larger load and you'd at least know where to look for possible problems. 

leiboldPremium,MVM join:20020709 Sunnyvale, CA kudos:10 Reviews: ·SONIC.NET
 reply to 54067323
I'm sure that what you are planning to do can be done safely if it is done properly but comments like "nearly zero ohms" don't inspire confidence in me that you appreciate how different "nearly zero ohm" values can be. The 100 micro Ohm and 400 micro Ohm example from dandeman is a great example: both values are "nearly zero Ohm" (less then 1/1000th of an Ohm) but the 2nd wire has 300% more resistance which means that the first wire will carry 300% more current! That is the kind of imbalance you should try to detect and avoid. I would not use a simple multimeter in Ohm setting to measure the conductivity of wires (the resistance is too low and the precision insufficient). The test clips may have a higher resistance then the entire length of the wire being tested. There are special meters for conductivity (inverse of resistance) but I would probably determine the resistance by applying a constant current source (lab power supply; alternatively use constant voltage and fixed resistor) and measuring the voltage. This is a much better way to get accurate results (especially when using a high test current) since voltage is what multimeters measure best. The test suggested by garys_2k is another way to achieve confidence that the wires will share the load somewhat equally (less precise but perhaps easier to perform with what you have available).  Got some spare cpu cycles ? Join Team Helix or Team Starfire! 

Bob4Account deleted join:20120722 New Jersey  reply to leibold
Another way to calculate it is:
100/420 * 60 = 14.29 amps
110/420 * 60 = 15.71 amps
(420 is the total length of the conductors) 

2 edits  reply to leibold
said by leibold:I'm sure that what you are planning to do can be done safely if it is done properly but comments like "nearly zero ohms" don't inspire confidence in me that you appreciate how different "nearly zero ohm" values can be. Oh I do but at the amperage levels versus the gauge of the conductors I am working with an ohm one way or the other is not a deal breaker. As for the work I just got in from the job and it's done and tested and worked without a hitch with a maximum voltage drop on the longest run of 3.7 vac under full load. Thankfully this weekend didn't count on the calendar so we actually completed the work on time keeping my fanny intact and unchewed. FWIW the reason I was using a vom (an old fluke 77) was to ensure continuity, I've had it in the past where the tone from the fox would pass through a bad connection on a conductor that should ohm out at nearly zero yet it wouldn't pass any serious amount of current due to a burnt up splice, by ohming out the conductors I can be reasonably assured I have good copper to work with. There are days though when I feel like this guy.
But for now all I need is a six pack on ice and some sleep. 

 reply to Bob4
said by Bob4:Another way to calculate it is:
100/420 * 60 = 14.29 amps
110/420 * 60 = 15.71 amps
(420 is the total length of the conductors) Either way it's within a safe amperage margin for number 10 copper. 
