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 V; then the capacitor discharges completely through the lamp and the lamp flashes briefly. For a lamp with breakdown voltage VL = 66.9 v, wired to a 99.0 V ideal battery and a 0.161 mF capacitor, what resistance R is needed for 4 flashes per second? Number 1388889 uniti the tolerance is +/-5%

Explanation / Answer

We know that

V = Vb ( 1 - e-t/RC)

where Vb = battery volatge

solving for R, we get

V/Vb - 1 = -e-t/RC

ln (V/Vb - 1) = t/RC

R = t / C*ln (Vb / Vb - V)

Here, four flashes per second means, t = 1/4 = 0.25 seconds

R = 0.25 / 0.161e-3 * ln ( 99 / 32.1)

R = 0.25 / 1.81328e-4

R = 1378.71 ohms

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