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atsparky:
On my 200 A system tonight with a 1500 W load I get the following results:
With the voltmeter on the opposite phase from the load I got less than 0.1 V change from the load. This was the same for either phase combination. This meant that my neutral impedance was less than 0.1/12 = 0.008 ohms.
The voltage change on the side with the load, not thru the breaker or outlet with the load, was about 0.6 V. This means that the impedance of the transformer (center tap to hot line), hot wire, meter, and main fuse was about 0.6/12 = 0.05 ohms.
Assume a maximum of 5000 W for a dryer. This is 5000/240 = about 20 A.
Your 11 A and 2.1 V was about 2.1/11 = 0.19 ohms if all 11 A was on phase B. At 11 A and 0.6 V, the neutral change, the resistance of the neutral is about 0.6/11 = 0.05 ohms. Therefore the hot side is about 0.19 - 0.05 = 0.14 ohms.
Probably not out of line with what you might expect. If we assume 0.28 ohms for the 240 impedance and 20 A, then total 240 drop would be 5.6 V at dryer full power. 120+119.2 = 239.2 and 118.1+114.5 = 232.6, and 239.2 - 232.6 = 6.6 V. Not grossly out of line with my calculation based on estimated impedances. However, the difference in drop on the two sides is confusing.
The measurements with a specific load shifted from one side to the other may help understand what is going on.
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