Final answer:
The maximum drain current for the JFET is the same as IDSS, which is 20 mA. The gate-source cutoff voltage is -4 V. The value of the drain-source resistance (RDS) cannot be determined with the given data.
Step-by-step explanation:
The question revolves around finding various parameters of a Junction Field Effect Transistor (JFET) given some specific values. The maximum drain current (IDSS) is provided as 20 mA, which is the current flowing through the device when the gate-source voltage (VGS) is zero, and the pinch-off voltage (VP) is given as 4 V, which is the gate-source voltage where the channel starts to pinch off and the current begins to decrease.
The maximum drain current for a JFET, which is equal to IDSS, is the highest current that can flow through the device. Since the value for IDSS is given as 20 mA, we can say that the maximum drain current is also 20 mA. The gate-source cutoff voltage (VGS(off)) is another name for VP; hence, the gate-source cutoff voltage is -4 V, indicating that this is the voltage at which the drain current will be completely pinched off and reduce to zero.
The value of RDS, which refers to the drain-source resistance in the ohmic region, cannot be determined from the given information alone without additional data such as the drain-source voltage (VDS) and specific device characteristics beyond IDSS and VP.