Input and output powe

were you able to find out? Even i'm wondering the same

He just meant that in practical situations, people analyse circuits in incremental sense, so the Power Gain calculated practically or mentioned practically makes sense only in incremental sense If you have a small signal (suppose 10 mV sin (wt)) and you try to amplify, then you might probably add some constant voltage to make it something like 3V + 10 mV sin (wt) [by connecting to some power source], and then use the 3V as the operating point. Now, since you need to amplify the 10 mV part, the Power Gain is used to refer to how much the 10 mV part is getting amplified. For example, for the 3V + 10 mV sin(wt) input, the output could be 5V + 20 mV sin(wt) The 3V part is subject to a non linear gain, so the ratio 5/3 and 20/10 are not equal But the power gain would be related to the 20mV part because that is what's important to us

here Vgs is remaining constant so in ID formula w/l will be equal to 2w/l ...that's why current will get doubled !!...

BInod