Electricity needs a complete path, from source to load, through load, from load back to source. This involves both the hot and the neutral or grounded conductor.
The current (I) is 2000VA/120V or 16.67A
The resistance of #10 is 1.21/kFT and #8, .764/kFT (NEC table 8)
120FT of #10 is 120 * 1.21 / 1000 = .146ohms
120FT of #8 is 120 * .764 / 1000 = .092ohms
The voltage drop comes from the resistance of the hot wire and the neutral.
The voltage drop for 120FT #10 is 16.67 * .146 = 2.43V
The voltage drop for #8 is 1.53V
#10 neutral and hot is 2.43 + 2.43 = 4.86V
#8 neutral and hot is 1.53 + 1.53 = 3.06V
one #10 and one #8 is 2.43 + 1.53 = 3.96V
3% of 120V is 3.6V, so both #8 is ok, one #8 and one #10 is marginal, and both #10 is poor.
If the load is distributed (or the circuit is 120/240V) the analysis is different.
