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		1 -	V_GS		…(a)
			V_P

given that, I_DS = 6.4 × 10^{– 3} amp
V_P = – 5 V
I_DSS = 10 × 10^{– 3} amp. 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 = – (– 1 V)

R_S =	1	= 156
	6.4 × 10^–3

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