When power if first applied,the capacitor will respond with its short-circuit equivalence since the voltage across the capacitor cannot cahrge instantaneously. The result is that the gate-to-source voltage of the JFET will immediately be set to 0V, the drain current Id position,the capacitor will begin to charge to-9V.Because of the parallel high input impedance of the JFET,it has essentially no effect on the charging time constant of the capacitor. Eventually,when the capacitor reaches the pinch-off level,the JFET and bulb will turn short period of time and then turnoff.It is ow ready to perform its timing function.