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In this drain to gain feedback resistance biasing circuit, it is said that the feedback resistance Rg is large and usually in \$M\Omega\$ range. But no current flow through this resistance.
So why it is needed to be large?
Any value of this resistance should serve the purpose as there is no current or power absorption.
At first order approximation, you are right in that Rg could be replace by a short and you'd get the same thing, at least in the steady state. However, the dynamics in response to Vdd changes would be different. MOSFETs have a fairly significant effective gate to drain capacitance, which forms a low pass filter together with Rg.
However, I suspect the real reason for high Rg is that this is just a small piece of a larger circuit. The intent is to bias M1 at some reasonable operating point, then feed in small changes to that bias point onto the gate. In the wider circuit, there is probably a capacitor connected to the gate, with the other side of the capacitor being driven by a signal that this transistor is supposed to amplify. Rg is therefore set high enough to be large relative to the impedance of the capacitor over the frequency range of interest. That means Rg supplies the DC bias value, and the AC component comes from the signal thru the capacitor.
In most cases, such a circuit is used as an amplifier vor signal voltages. Thus, it is desirable that the input resistance of the amplifier stage is large in comparison to the source resistance of the signal source. The total input resistance is mainly determined by Rg - thus, Rg should be large. But keep in mind that it is not the real value of Rg that is to be considered. Due to the Miller the effective value of Rg is reduced.
