The figure shows the circuit of a flashing lamp, like those attached to barrels

Question

The figure shows the circuit of a flashing lamp, like those attached to barrels at highway construction sites. The fluorescent lamp L (of negligible capacitance) is connected in parallel 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 voltage VL; then the capacitor discharges completely through the lamp and the lamp flashes briefly. For a lamp with breakdown voltage VL = 71.0 V, wired to a 95.0 V ideal battery and a 0.210 F capacitor, what resistance R is needed for two flashes per second?

Explanation / Answer

t = 0.5
Vc = 95(1-e^-t/RC) = 71
(1-e^-t/RC)=71/95 =0.7473
1-0.7473 = e^-t/RC=0.2526
-RC*ln(0.2526) = t = 0.5
R = -.5/ln(0.2526) * 1/C = -3.63/0.021 = 173.04 Ohms

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