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?