# Tag Info

## Hot answers tagged lamp

59

Leaving it on would use more energy, absolutely. Sometimes, people try to convince themselves that turning a light on and off uses more energy because there is some high inrush current, or some such thing. Firstly, incandescent lights hardly have any inrush current, because they don't have any capacitors to charge, and they need not strike an arc in the ...

40

Okay, let's set up a simple simulation: According to the Wiki page on incandescent bulbs, for a 100W, 120V bulb, the cold resistance is ~9.5Ω, and the hot resistance ~144Ω. It takes around 100ms for the bulb to reach the hot resistance on turning on. So armed with this info, we can simulate and prove the initial surge would be absolutely ...

19

According to a Mythbusters episode summary on Wikipedia: " The MythBusters calculated that the power surge from turning on a light would only consume as much power as leaving it on for a fraction of a second (except for fluorescent tube lights; the startup consumed about 23 seconds' worth of power)". So in fact it is possible that on/off would consume more ...

13

This is a common phenomenon: neon pilot lights have a limited lifetime, and after many years of use, they begin to flicker, then they finally go dark. They no longer can operate at line voltage, but instead require a higher voltage for stable operation. Also, neon pilot lights can act as photosensors. Try this with a flickering neon bulb: shine a red ...

12

It depends on the type of lightbulb! Halogen, incandescent, florescent, and vapor lights all use tungsten filaments that heat up and emit electrons via thermionic emission. In that sense, they are similar. However, the method to "turn on" the lights varies. Incandescent bulbs are simply turned on once and left on. The inrush current is on the order of 12 ...

10

I'll presume your bulbs are ordinary resistors first. The 60 W bulb is R1, the other one R2. Resistance can be calculated as $R = \dfrac{V^2}{W}$ For R1 and R2 that's $R1 = \dfrac{(110 V)^2}{60 W} = 202 \Omega$ $R2 = \dfrac{(110 V)^2}{110 W} = 110 \Omega$ Then Rp || R1 = R2, or $\dfrac{Rp \cdot R1}{Rp + R1} = R2$ Filling in ...

10

The constantly on setting would consume more energy powering the bulb. A possible counter-argument would be that the turn-on/turn-off cycling would shorten the bulb life, and thus the energy cost of manufacturing, transporting, and disposing of it would be amortized over fewer service hours. But without digging up actual numbers, my gut feeling is that ...

9

Nothing was said about the lamp types, or that they need to be identical. So, assuming that L1 and L2 are both 120V incandescents, and L1 is a 1 watt bulb and L2 is a 100-watt bulb, simulate this circuit – Schematic created using CircuitLab when the switch is open, L1 will be lit, and the current through L2 will be so small as to produce ...

9

To quote from the great wiki - https://en.wikipedia.org/wiki/Neon_lamp "When the current through the lamp is lower than the current for the highest-current discharge path, the glow discharge may become unstable and not cover the entire surface of the electrodes.[6] This may be a sign of aging of the indicator bulb, and is exploited in the decorative "...

7

There are two things to consider in trying to match the light output of multiple LEDs Controlling the current thru each LED. Compensating for light output variances between LEDs even when driven with the same current. The first is not that hard to guarantee electrically. The brute force way to ensure the same current thru all LEDs is to put them in ...

7

Look at the listed amp rating of your bulb You need to look at your bulb's documentation/data sheet for its listed draw in amps. If it provides a VA rating, you can compute amps from VA/volts. If it provides actual watts and power factor, you can compute Watts/Volts/PF. LED bulbs are often marketed as "Same brightness as a 60 watt incandescent bulb", ...

6

No. It seems very unlikely there is any need for a fuse. The LIFX gadgets are presumed to be rated for direct connection to a mains branch circuit without any additional fusing. The fuse was probably there to protect the switch as incandescent lamps do not require fusing. They act as their own fuse. The "dimming" feature where it connected two bulbs in ...

5

I see that you're referring to the light fixture itself. In this case, the rating is the maximum that the fixture construction can withstand. Remember that the fixture is really only wires, switches, and connectors. It only serves to pass the current from the wall to the light bulb. In your case, the fixture can handle any voltage up to 250V. You can use a ...

5

As Steven states, this is only true when the bulbs act like ordinary resistors. The solution is easy. The voltage across the 'divider' will be evenly distributed when power at the top half and power at the bottom half are equal. Power at the top half is 60W. Power at the lower half is 110W. To have equal power both at top and at bottom halves, you have to ...

5

You'd need to provide brand and model of projector and bulb plus web links. What you need to know, at least, is ANSI lumen rating of projector and present technology and rating. All that said, this is getting more viable but is not something attempted lightly. The fact (as you recount it) that available solutions are from unknown makers and that they don'...

5

All of the energy that goes into an incandescent bulb will get converted into heat, which must then be dissipated somehow. Some of that heat will then be radiated off in the form of light, but the energy must start out as heat. Therefore, the only way an incandescent bulb can use more power is for it to dissipate more heat. A bulb which is cold consumes ...

5

For full brightness they can be driven either with (about) 12 VDC or 12 VAC RMS. Use of half wave rectified AC will operate them at somewhat less than half power but they may flicker and will have a lower color temperature. I've never seen them operated with half wave AC. If this works for some application and meets some purpose then that's fine - but I ...

5

According to the U.S. Dept. of Energy: It is best to turn off incandescent and halogen bulbs whenever they are not needed, due to their high consumption of electricity. For a compact fluorescent bulb, a rule of thumb is to leave it on if you leave a room for 15 minutes or less (depending on several factors). For LED lighting the operating life is unaffected ...

5

You are overloading your current supply, it is shutting down to prevent damage. 2 x 10W = 20W Which is more than 15W. It seems the only options are buying 5 of this "converters" (one for each lamp) or buying another model which can handle more power (more lamps per converter).

5

The (now obsolescent) conventional tungsten-filament consumer incandescent bulbs have a trade-off between life and efficiency. The longer you make the bulb last for (by running the filament cooler) the less efficient the bulb is, so the higher the lifetime cost of the electricity required to illuminate it. The light also becomes redder the cooler the ...

5

It's more than that -- polarized plug make cheap single-pole switches much safer. For example, look at the extension strip: With a polarized plug, it is perfectly safe to have a single-pole switch there-- one just makes sure it interrupts the "live" contact With a non-polarized plug, one either has to put a more expensive double-pole switch, or to accept ...

5

TL;DR The electric arc may conduct much higher current than the filament, this is not normal mode of bulb operation so you can see a flash and maybe even have the bulb exploded. Typical incandescent bulb failure develops while current is flowing through the filament. The filament is a piece of wire about half a meter long coiled such it forms a filament ...

4

See http://hackaday.com/2011/11/21/simple-touch-sensors-with-the-arduino-capsense-library/ Ever thought of using touch sensors on your projects but didn’t because it would be too much work? [Paul Stoffregen] proves that it can be pretty easy if you use the CapSense library for Arduino. Here he’s created three touch sensors, connecting them to the Teensy ...

4

How safe is this? Relatively safe, this is how relays are used. Is the only thing separating wandering hands from live wire just the screw terminal? Yes. Of course, touching only one side of a wire, with only one hand, is not terribly dangerous, compared to completing a circuit through your body. Of course, the only thing separating a live outlet and ...

4

Responding to old post but it comes up at top of google search: Neon lights can't work in complete darkness. This is the "neon lamp dark effect". I know it sounds fake, but that's why neon lamp bulb makers sometimes add a little ionizing radiation inside the bulbs. Dark Effect: All ILT Neon lamps are subject to a condition called dark effect. Dark ...

4

Just rotate the plastic from the back (CCW to "unscrew", it's sort of a bayonet base-like action). I have the same general kind of lamp in the instrument cluster of an E28 chassis automobile. The metal contacts work against pads on the single-sided PCB. It's a 24V 1.2W bulb (photo from linked site). As @Jack mentions some such bulbs have removable bulbs ...

3

Couple of comments about the optical portion of your project: Both the electrical and optical portion of the project are important but you won't be able to achieve a homogenized optical output even with perfect drive electronics unless you worry about the optical portion of the design which in turn may constrain how you plan to build/drive the device. ...

3

No, there is no such thing as standard insulation for a particular wire size. The wire gauge only tells you the physical size of the wire. From that, and knowing the material (usually copper), you can determine its resistance per length. That in turn tells you how much voltage it will drop for a given current, and how much power it will dissipate. These ...

3

Flouro lamps are actually more efficient at say 30KHz than 50Hz .The efficiency curve doesnt keep increasing ,it kind of limits out .This is why you tend to not see stuff in the MHz range despite it being technicaly feasible .The electronic ballast makes the tube itself more efficient .Even if you use rubbish for the electronics your overall efficiency ...

3

You've drawn an NPN transistor as a high-side switch. This won't work since you need to drive the base at least 0.7V higher than the emitter, and your RPi GPIO outputs only push 3.3 volts. You should move the transistor into the ground path to operate as a low-side switch, and add a current limiting resistor to the base. Even then, you will probably run ...

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