There was a thread a while back about the usefulness of voltage drops....
Last week I found a bad connection on a motor starter where the voltage drops correlated with the heat very well, which isn't always the case. Most of the time I have found that the difference in Voltage drop between the "Bad" Phase, and the "Good" phases are not much, to the point of being statistically insignificant. But in the case of the attached thermograph, it correlated rather nicely. The load was 27 amps. Wire gauge was 8 or 10. Temp was 120 C on B phase, and voltage drops were 94mv 294mv 49mv. Perhaps Voltage drops do not become significant until the severity of the anomaly is at this stage.

Voltage_drop.doc (92 Kb, 49 downloads)

Voltage drop confirms the location of an anomaly. It is usually used to confirm there is an internal problem. If all you have is a simple loose connection on a cable, it wont create a voltage drop.

Right, because the current will simply go through whatever remains of the good connection. Depending on load, you may have plenty of good conductor left to carry the load. You essentially have a parallel resistance circuit. But this begs the question: If there is heat, there must be resistance, if resistance, than a voltage drop. In this case, a load of 27.1 amp, VD of .294mv equals 8 watts of power used and resistance of only .01 ohms. I have never seen an anomaly with a voltage drop greater than 1.

quote:

Originally posted by Bob Berry:
If all you have is a simple loose connection on a cable, it wont create a voltage drop.

The above contradicts basic rules of electricity.

Assuming that a loose connection manifested itself by temperature increase with current not affected, addtional power deltaP generated at loose connection will be:

deltaP = I^2 * deltaR
deltaR - connection resistance increase

or additional voltage drop
deltaU = I * deltaR

It is a separate question as to whether or not temperature increase deltaT is directly proportional to deltaU. This relationship may not be linear - the situation Martin runs into.

David, if you assume a serial connection you are correct, but a degrading connection becomes essentially a parallel connection, where the load goes through the path of least resistance.

Martin,

I assume that connection resistance is in series with the rest of the power circuit. Of course, you can view connection resistance itself as made out of several parallel branches. But once one of them opens up, overall connection resinstance goes up and my previous conclusion stands. Resistance goes up at poor connection - this is the key.

David

Yes, but if you read my post, it depends on the load. If you run 10 amps through a 0000 stranded cable (rated 300 amps+) even if 50% of the strands have burned parts which have very high resistance, the amount of contact area required to conduct 10 amps will certainly remain in the 50% that still has good contact. There will be no voltage drop other than the normal voltage drop associated with the normal resistance of copper wire because the load is less than the current carrying capacity of the remaining good contact areas. It is when the remaining good contact area becomes smaller than the load needs that we really begin to encounter meaningfull resistance to the current.

Connection resistivity vs. load is practically a flat straight line except for very-very low or high loads.