Hi
Your schematic is a voltage controled currend source ("voltage controled resistor")
If you stay within the specs of the opamp and the IRF540 you don't have to think about the Vgs.
The IRF540 in the control loop of the opamp.
The IRF 540 is NOT saturated! at 10Amp's
OK, think with me...
Your powersupply is adjusted so dat it gives 12V @ max 10Amps
Set your electronic load to 10Amps
10Amp and a 0.1 Ohm resistor = 1V
On the "+" input of your opamp = 1V
The opamp ouput wil go up driving the gate until its "-" input wil also be 1V ( its the nature of opams to have the same voltage on both inputs)
The voltage the IRF540 needs for 10Amps you wil find in the data sheets ( about 5.5V) +- 10%
The gate voltage against ground level wil be 6.5V @ 10Amps +-10%
OK, there is 10 amps comming out of your powersupply.
The total load (without your connection wires) = U / I = 12 / 10 = 1,2 Ohm
The "R" from the IRF540 = 1.2 - 0.1 Ohm drain resistor = 1.1 Ohm
The IRF540 is disipating 11V x 10 Amps = 110Watts !!!! (It can boil your thea water
)
A saturated IRF540 wil have a resistance of < 0.05 Ohm @ Vgs of > 10V
I hope my 2 cent helps
Kind regarts,
Bram