The JFET in the circuit shown in figure has an IDSS = 10

The JFET in the circuit shown in figure has an I_DSS = 10 mA and V_P = 5 V. The value of the resistance R_S for a drain current I_DS = 6.4 mA is (select the nearest value)—

We know that I_DS = I_DSS …(a)
Given that, I_DS = 6.4 10_–3 A
V_P = 5V
I_DSS = 10 10_–3 A
V_GS =?

6.4 10^–3 = 10 10^–3		1 -	V_GS		²
			- 5

1	V_GS	= √.64= 8
	- 5

or V_GS = 1V
Now, I_DS R_S = V_GS = (1V)

R_S =	1	= 156Ω
	6·4 × 10^–3

Since 156 Ω is very close to 150 Ω.
Hence alternative (A) is correct answer.