The circuit shown below uses a neon-filled tube in the setup below. The neon lam

Question

The circuit shown below uses a neon-filled tube in the setup below. The neon lamp has a threshold voltage V0 for conduction, because no current flows until the neon gas in the tube is ionized by a sufficiently strong electric field. Once the threshold voltage is exceeded, the lamp has negligible resistance.
The capacitor stores electrical energy, which can be released to flash the lamp. Assume that C = 6.10

The circuit shown below uses a neon-filled tube in the setup below. The neon lamp has a threshold voltage V0 for conduction, because no current flows until the neon gas in the tube is ionized by a sufficiently strong electric field. Once the threshold voltage is exceeded, the lamp has negligible resistance. The capacitor stores electrical energy, which can be released to flash the lamp. Assume that C = 6.10 times 10 -2 ?F, R = 2.7700 times 10 6 ?, V0 = 84.00 V and ? = 110.0 V. Assuming that the circuit is hooked up to the emf at time t = 0, at what time will the light first flash? What happens after the first flash? The capacitor charges again and nothing happens The capacitor charges for the same time again and repeats The capacitor charges for half the time it charged initially and repeats Nothing occurs because the capacitor has discharged all its energy How much energy is released from the flash? (You may assume that the neon lamp shorts the circuit when it flashes and ignore the effect of the battery for this short time.) A typical light bulb emits at 100W. What is the ratio of energy emitted by the neon flash to the light bulb (Assuming the light bulb is turned on for the same amount of time as the neon bulb)?

Explanation / Answer

a)

You will have to show or describe the circuit to be certain. I will assume that 110 V is connected to R and C in series and that the neon bulb is in parallel with C.

The voltage across a discharged capacitor is 0 at t=0 and grows to the applied voltage over time. (In engineering, 5*RC is usually considered charged.)

Vc=emf(e^(-t/RC))

84 = 110 * e^-t/RC = 110 e^-t/(2.77*10^6*6.1*10^-8)

t = 0.4556 s

(iii) The capacitor charges for half the time it charged initially and repeats

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