If Al is still following this thread I am sure he will have a better technical answer.
Not necessarily, Jim :-)
Your finding that having the hairdryer plugged into the outlet where the measurement was taken resulted in a DC offset of around 730 mv, while measuring at the same outlet but with the hairdryer plugged into various other circuit branches on the same leg resulted in around 115 mv, is indeed a bit of a headscratcher.
But not being familiar with the design of hairdryers I did a little research and found this interesting paper:
http://www.idc-online.com/technical_references/pdfs/electrical_engineering/MEASURING_ACOUSTIC_NOISE_EMITTED_BY_POWER.pdf
As stated on pages 4 and 5 and as depicted in Figure 13, at least when hairdryers are operated "at lower power" they place a half-wave rectifier diode in series with the load they place on the AC, or at least a substantial part of the load. So DC offset results from the difference between the amount of current that is drawn during the positive half-cycle of the AC waveform and the amount of current that is drawn during the negative half-cycle, and the differing voltage drops that occur in the resistance of the AC wiring between the two half-cycles as a result of that current difference.
In tests 2, 3, and 4 essentially zero current was being drawn through the dedicated wiring between the service panel and the outlet where the measurement was taken, so no voltage drop would have been occurring in that wiring. And the DC offset that was measured would have resulted essentially from the voltage drop differential between half-cycles that was occurring in the panel and in the outside wiring, since the additional voltage drop in the wiring between the panel and the outlets where the hairdryer was plugged in would not have been in the path to the outlet where the measurement was being taken. And presumably the outside wiring is considerably heavier gauge than the 10 gauge Romex used for the dedicated line, and therefore it would present a lower source resistance (per unit length, at least) for DC offset to develop across as a result of the asymmetrical current draw.
Whereas in test 1 that asymmetrical current was being drawn through an additional resistance in the path to where the measurement was taken, corresponding to the sum of the resistances of the dedicated line’s two 75 foot 10 gauge conductors, which amounts to about 0.15 ohms.
Looking at it quantitatively, if my theory is correct the difference in the amount of current drawn by the hairdryer between the positive and negative half-cycles would be:
(730 mv - 115 mv)/0.15 ohms = 4.1 amps.
In the context of the large current draw of an 1875 watt hairdryer, and one that is placing a diode in series with much of the load it presents, I suppose that is consistent with your findings.
Best,
-- Al