Question

Figure shows thecircuit of a flashing lamp, like those attached to barrels at highway construction sites. The fluorescent lamp L (of negligible capacitance) is connected inparallel across the capacitor C of an RC circuit. There is a current through the lamp only when the potential difference across it reaches the breakdown voltageVL; then the capacitor discharges completely through the lamp and the lamp flashes briefly. For a lamp with breakdown voltage VL = 63.0 V, wired to a 94.6 V idealbattery and a 0.198 mF capacitor, what resistance R is needed for 3 flashes per second?

Answer 1

The break down voltage, V_L = 63.0 VThe emf of the battery, ξ = 94.6 VThe capacitance of the capacitor isC = 0.198 μF=(0.198 μF)(10^-6 F/1 μF)=0.19810^-6 FThe required time, t = (1/3) s= 0.3333 s______________________________________________When the emf ξis applied to a resistor R and capacitance C,the charge on the capacitor increses according toq = Cξ(1-e^-t/RC)q/C= ξ(1-e^-t/RC) ...... (1)The relation between charge q and voltage V_L is given byq = CV_Lq/C= V_LThen, the equation (1) becomesV_L=ξ(1-e^-t/RC)V_L/ξ=(1-e^-t/RC)e^-t/RC= [1-V_L/ξ]e^-t/RC = [(ξ-V_L)/ξ]Apply the natural logarithm on both sides we get-t/RC = ln[(ξ-V_L)/ξ]_____________________________________________________Therefore, the resistance of the resistor isR = -{t/C[ln[(ξ-V_L)/ξ]]}=-{(0.3333 s)/(0.19810^-6 F)[ln[(94.6 V-63.0V)/94.6 V]]}= 1.53510^-6 Ω