Assume the transistor is biased in the saturation region. The dc drain current is then given by
I_{D} = I_{DSS} \left( 1  −  \frac{V_{GS}}{V_{P}} \right)^{2}
Therefore,
5 = 12 \left(1  −  \frac{V_{GS}}{(−3.5)} \right)^{2}
which yields
V_{GS} = −1.24  V
From Figure 3.61(b), the voltage at the source terminal is
V_{S} = I_{D} R_{S}  −  5 = (5)(0.5)  −  5 = −2.5  V
which means that the gate voltage is
V_{G} = V_{GS} + V_{S} = −1.24  −  2.5 = −3.74  V
We can also write the gate voltage as
V_{G} = \left(\frac{R_{2}}{R_{1}  +  R_{2}} \right) (10)  −  5
or
−3.74 = \frac{R_{2}}{100} (10)  −  5

Therefore,
R_{2} = 12.6  k \Omega
and
R_{1} = 87.4  k\Omega
The drain-to-source voltage is
V_{DS} = 5  −  I_{D} R_{D}  −  I_{D} R_{S}  −  (−5)
Therefore,
R_{D} = \frac{10  −  V_{DS}  −  I_{D} R_{S}}{I_{D}} = \frac{10  −  5  −  (5)(0.5)}{5} = 0.5  k \Omega
We also see that
V_{DS} = 5  V\gt V_{GS}  −  V_{P} = −1.24  −  (−3.5) = 2.26  V
which shows that the JFET is indeed biased in the saturation region, as initially assumed.
Comment: The dc analysis of the JFET circuit is essentially the same as that of the MOSFET circuit, since the gate current is assumed to be zero.