Sure.
The equations are:
P = V * I
and
V = R * I
Where:
P = power (Watts)
V = voltage (volts)
R = resistance (Ohm)
I = amperage (amps)
With a substitution, you can get P=VI to be P=R*I^2.
What we know is:
P=2000 W
and
V=120
so I resolves to 16.7 amps.
then swapping V=RI to read as V / I = R, you get R = 7.2 Ohm.
Now for the target power. What we know is that R=7.2 Ohm, and that P should be 350 W.
P = 350 W
R = 7.2 Ohm
so using the substitution above, where P=R*I^2, and a little swapping around, we get I = sqrt (P / R). I resolves to 6.97 amps.
Then, we go back to V=R*I, and we get V = 50 volts.
So now your task is to find a DC power supply, capable of outputting at least 50 volts DC and at least 7 amps. tip: if the PSU is capable of more, it'll last you longer; there's no sense in taxing a power supply to its limit, all the time. Also, look out for voltage drifts as the unit warms up.
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