For transfer charc, if

ID = IDSS/4 = 10mA/4 = 2.5mA then VGS

$V_{GS}= V_p/2 = \dfrac{3.5}2V$ The resulting curve representing shockley's equation appears in fig.

The network equation is defunded by

$V_G=\dfrac{R_2V_{DD}}{R_1+R_2}=\dfrac{20\times110}{1020}=2.15\\ V_{GS}=V_G-I_DR_S\\ =2.15-I_D(0.51)\\ When \ I_D=0, V_{GS}=2.15\\ When \ V_{GS}=0, I_D=\dfrac{2.15}{0.51}=4.2$

The point of intersect of these two …