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Photoflash Lamp
Photoflash Lamp
Types of Lamps in Lighting World
Electric-discharge lamps depend on the ionization and the resulting electric discharge in vapors or gases at low pressures if an electric current is passed through them. Representative examples of these types of devices are the mercury-vapor arc lamp, which gives an intense blue-green light and is used for photographic and roadway illumination, and the neon lamp, which is employed for decorative sign and display lighting. In newer electric-discharge lamps, other metals are added to mercury and phosphor on the enclosing bulbs to improve color and efficacy. Glasslike, translucent ceramic tubes have led to high-pressure sodium vapor lamps of unprecedented lighting power.
The fluorescent lamp is another type of electric-discharge device used for general-purpose illumination. It is a low-pressure mercury vapor lamp contained in a glass tube, which is coated on the inside with a fluorescent material known as phosphor. The radiation in the arc of the vapor lamp causes the phosphor to become fluorescent. Much of the radiation from the arc is invisible ultraviolet light, but this radiation is changed to visible light if it excites the phosphor. Fluorescent lamps have several important advantages. By choosing the proper type of phosphor, the light from such lamps can be made to approximate the quality of daylight. In addition, the efficiency of the fluorescent lamp is high. A fluorescent tube taking 40 watts of energy produces as much light as a 150-watt incandescent bulb. Because of this illuminating power, fluorescent lamps produce less heat than incandescent bulbs for comparable light production.
One advance in the field of electric lighting is the use of electroluminescence, known commonly as panel lighting. In panel lighting, particles of phosphor are suspended in a thin layer of nonconducting material such as plastic. This layer is sandwiched between two plate conductors, one of which is a translucent substance, such as glass, coated on the inside with a thin film of tin oxide. With the two conductors acting as electrodes, an alternating current is passed through the phosphor, causing it to luminesce. Luminescent panels may serve a variety of purposes—for example, to illuminate clock and radio dials, to outline the risers in staircases, and to provide luminous walls. The use of panel lighting is restricted, however, because the current requirements for large installations are excessive.
A number of different kinds of electric lamps have been developed for such special purposes as photography and floodlighting. These bulbs are generally shaped to act as reflectors when coated with an aluminum mirror. One such lamp is the photoflood bulb, an incandescent lamp that is operated at a temperature higher than normal to obtain greater light output. The life of these bulbs is limited to 2 or 3 hours, as opposed to that of the ordinary incandescent bulb, which lasts from 750 to 1,000 hours. Photoflash bulbs used for high-speed photography produce a single high-intensity flash of light, lasting a few hundredths of a second, by the ignition of a charge of crumpled aluminum foil or fine aluminum wire inside an oxygen-filled glass bulb. The foil is ignited by the heat of a small filament in the bulb. Increasingly popular among photographers is the high-speed gas-discharge stroboscopic lamp known as an electronic flash.
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Energy stored in a capacitor?
The energy (in Joules) stored in a capacitor with capacitance C (in Farads), having a potential difference V (in volts), is given by W = (1/2)C*V^2 . One application is in the electronic photoflash or strobe light. In such a unit, a capacitor of 9 μF is charged to 327 V. Find the energy stored.
Answer in units of J.
Find the power required to charge the capacitor to this energy level in 23 s.
Answer in units of W.
When this capacitor is discharged through the strobe lamp, it transfers all its energy in 0.00023 s.
Find the power delivered to the lamp.
Answer in units of kW.
Having trouble with a homework problem. First correct answer along with an explanation on how to do it (I want to learn not just have the answer, but I would like the answer to check my work) gets the 10 points for best answer. Thanks!
Power = rate of doing work = Energy/time = W/t = (1/2)C*V^2 /t.
P = 0.5*9e-6*327*327/23 = 0.0209 W
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P = W/t = 0.0209*23/0.00023 = 2090 W = 2.090 kW.
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