the question asks :- briefly explain why the drain voltage is specified as being 5volts? (or VDD/2) it refers to a jfet transistor amplifier circuit of a constant current source. it has a resistor in both the drain and source paths of the circuit, as well as the gate. the supply voltage (VDD) is 10 volts +ve. the question also refers to a transfer curve and a GATE/Source Voltage = 0. My answer is (and i am hoping you can either add or comment on): A JFET transistor amp needs to be biased using resistors that: give a drain voltage equal to half the supply voltage. A gate source reverse bias voltage to give a drain current of IDSS/2.This is the most linear part of the transfer curve, which is the operating point of the circuit, if distortion is to be avioded. Question 2. as on the attachment for the same circuit. They ask if Rs was reduced, describe the effect this would have on Vd and Vs of the circuit. My answer: By the equation: Vs=Id*Rs Id=.005v Rs=500ohms Vs=.005*500=2.5v If Rs=250ohms Vs=.005*250=1.25v therefore Vd=Id*Rd If Rd=1k Vd=.005*1000=5 v Vd would remain unchanged if Rs reduced If this is not correct, could u help please? the circuit i refer to is on the right hand side of the page attached.