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 a RC circuit. There is a current through the lamp only when the potential difference across it reaches the breakdown voltage VL ; in this event, the capacitor discharges completely through the lamp and the lamp flashes briefly. Suppose that two flashes per second are needed. For a lamp with breakdown voltage VL = 69.0 V, a 95.0 V ideal battery, and a 0.110 ?F capacitor, what should be the resistance R of the resistor?

C L

Answer 1

We know that, The voltage difference across the capacitor over time t is given by:

V = V0 ( 1 - e^-t/RC)

Taking natural log both the sides and solving for R, we get

R = t / C x ln [V0 / ( V0 - V )]

In our case, V = VL = 75 ; V0 = $\varepsilon$ = 95 V ; C = 0.11 x 10-6 F ; t = 1/2 = 0.5 s

R = 0.5 / 0.11 x 10^-6 x ln [ 95 / (95 - 75) ] = 2.92 x 10⁶ Ohm

Hence, R =  2.92 x 10⁶ Ohm