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In a DC circuit, why does a voltage drop appear across a poor connection in an open circuit?

For example, I have 48V DC wires, where a connector is charred from bouncing/bad contact. Under no load, the side of the connector reads 48V (the side attach to the supply), while the other reads 37V.

I don't understand how the voltage drop can exist without a current.

I believe my multimeter is equivalent to 10M Ohm when reading voltage. Does this mean the joint is probably 2.3 M Ohm, and its current of microamps though the joint and meter is giving the reading of 37V?

Also, when I make the connection good, at the other end of the 100m wire, I only get 47.2V. I understand the cable resistance will lower the voltage when a load is attached, but shouldn't it read 48V under no load? How can there be a drop of 0.8V across the wire with no load? Does this mean that there must be a leakage somewhere? If the meter is 10M Ohm, then this implies 160k Ohm through wire?

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    \$\begingroup\$ a diagram would be more helpful than so many words. It is hard to wrap your head around the situation. \$\endgroup\$
    – David Norman
    Commented May 5, 2013 at 23:18
  • \$\begingroup\$ How long is the "long wire" (piece of string)? \$\endgroup\$
    – Andy aka
    Commented May 5, 2013 at 23:18
  • \$\begingroup\$ The situation is simply two wires connect to a supply with no load. The wire is extended and had joins in a few places and is 100m long. \$\endgroup\$
    – MichaelD
    Commented May 6, 2013 at 1:10
  • \$\begingroup\$ You might try testing your beliefs about this meter, by measuring a stable voltage supply and then adding a resistor in series with the meter and measuring a again. \$\endgroup\$
    – Chris Stratton
    Commented May 6, 2013 at 3:59
  • \$\begingroup\$ Circuit and picture of your setup may help. \$\endgroup\$
    – jippie
    Commented May 6, 2013 at 6:38

2 Answers 2

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If...

If the meter is 10Mohm then this implies 160kohm through wire??

If the complete situation is as you described, then that is exactly what it means.

However...

The input impedance of the multimeter is not constant with respect to input voltage and is not exactly 10.000000 Megaohms. Your resistance assessment is sensitive to both voltage measurement accuracy and the (unknown) input impedance of the multimeter.

That's why, wire characterization is done with a significant current and a precision (e.g. known) load.

Why not...

If you have a multimeter, why not just measure the resistance of the wire directly? Disconnect it from the source and use the meter's resistance measurement setting?

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Poor connections dont obey ohms law well .There are things like work functions and contact potentials to consider .If you plotted a V I curve you would see the full story and it wont be a straight line .Normal ohmmeters only measure the resistance at one very low point .A well known example of such nonlinearity is the voltage drop of brushes in a DC motor.

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