Question

The fluorescent lamp, L, has a negligible capacitance and is connected, in parallel, across the capacitor, C, of a RC circuit. There is a current through the lamp only when the potential difference across it reaches the breakdown voltage VL; when this happens, the capacitor discharges completely through the lamp and the lamp flashes briefly. Suppose that two flashes per second are needed. For a lamp with a breakdown voltage of VL = 72.0 V, wired to a 95.0 V, ideal battery and a 0.150 μF capacitor, what should the resistance, R, be? The resistor is in series with the battery and the capacitor.

Answer 1

. The voltage to ignite is 72.0 volts, that is what needs to be across the capacitor C. Time will be less than .5 sec, let time be .5 sec for the solution and select a resistor that is 10 % less than solution. Assume the time to discharge the capacitor to the extinguishing is minimum and not used to obtain the solution.

Use the Formula Vc= E(1 - e^-(t/RC) ) substitute your given info and let t=.5 sec. getting:
\[72.0 = 95(1 - e ^{-(.5/R1.5\times 10^{-6)}}\]