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

For example i have 48vdc 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 dont understand how the coltage drop can exist without a current.

I believe my multimeter is eqivalent of 10Mohm when reading voltage. Does this mean the joint is probably 2.3 Mohm, and its current thought the joint and meter of microamps 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 wire with no load? does this mean there must be leakage somewhere? If the meter is 10Mohm then this implies 160kohm through wire??

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?

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

For example i have 48vdc 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 dont understand how the coltage drop can exist without a current.

I believe my multimeter is eqivalent of 10Mohm when reading voltage. Does this mean the joint is probably 2.3 Mohm, and its current thought the joint and meter of microamps giving the reading of 37v?

Also when i make the connection good, at the other end of the long 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 wire with no load? does this mean there must be leakage somewhere? If the meter is 10Mohm then this implies 160kohm through wire??

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

For example i have 48vdc 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 dont understand how the coltage drop can exist without a current.

I believe my multimeter is eqivalent of 10Mohm when reading voltage. Does this mean the joint is probably 2.3 Mohm, and its current thought the joint and meter of microamps 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 wire with no load? does this mean there must be leakage somewhere? If the meter is 10Mohm then this implies 160kohm through wire??

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

For example i have 48vdc 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 dont understand how the coltage drop can exist without a current.

I believe my multimeter is eqivalent of 10Mohm when reading voltage. Does this mean the joint is probably 2.3 Mohm, and its current thought the joint and meter of microamps giving the reading of 37v?

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

For example i have 48vdc 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 dont understand how the coltage drop can exist without a current.

I believe my multimeter is eqivalent of 10Mohm when reading voltage. Does this mean the joint is probably 2.3 Mohm, and its current thought the joint and meter of microamps giving the reading of 37v?

Also when i make the connection good, at the other end of the long 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 wire with no load? does this mean there must be leakage somewhere? If the meter is 10Mohm then this implies 160kohm through wire??