I would not expect to find leads to be supplied with the bench supply.

Do you mean it would be unusual for the output leads to be included?

 

Do you mean it would be unusual for the output leads to be included?

Yes. I have never had leads included with a bench power supply.

 

So the supply voltage at the source is 12.000 volts, while at the load the voltage is 11.7 volts, with 1.82 amps flowing. So we can calculate the total loop resistance with Volts= current x resistance, and then resistance = voltage/current -> R=0.30/ 1.82=0.164 ohms, which includes the resistance of the wire, the connections , and any splices.And that includes the assumption that the source supply regulation is perfect.

 

Simple.
Measure the voltage at the load resistor and compare with the measured voltage at the terminals of the PSU.

 

At the risk of being redundant, I explained this in post #2 and reinforced it several times after some misleading, irrelevant information was posted.

 

In this case, it is entirely possible that the voltage drop is caused by resistance in the connection points of the heavy jumper wire connections. That source of resistance is difficult to duplicate and a challenge to measure accurately. In addition, consider that the millivolts of voltage drop in a loop add up, and accurately determining the location and magnitude of such voltage drops is not simple nor easy..