Light Bulbs

[PeterBlack]

Hi everyone,


I'm a little confused when it comes to light bulbs. Say if I have a 12VDC, 50W bulb, what exactly does that mean? Does it mean my light bulb consumes 50W when there's 12VDC across it?
If I connect the following series circuit are my calculations correct?
A 12VDC battery with a 1Kohm resistor and a 50W DV light bulb.

Current in the circuit: I = V/R = 12V/1Kohm = 12mA (can I ignore the resistance of the light bulb?) R = (V*V)/P = (12Vx12V)/50W = 2.88 ohms.
Basically all my voltage would be dropped across the resistor and not the light bulb, right?

Would my light bulb still light up? Or does it need 12VDC across it to operate?

Thank you very much 🙂

Peter

 

[MrDudeMan]

P=IV

your power is 50w, and voltage is 12V. that makes the current 4.17A. light bulbs dont care if its AC or DC. your equations are all wrong.

edit: and you probably dont have the right kind of resistor to handle 4.17A. little resistors that you can buy from radio shack are typically 1/4 or 1/2 watt, not 50. it will burn up in a microsecond or two in that circuit.

 

[PeterBlack]

I'm sorry I believe I worded my original post wrong. The light bulb I have is rated for 12VDC and 50W? It?s a DC light bulb and not operational with AC. They're used on pleasure boats.

 

[MrDudeMan]

Originally posted by: PeterBlack
I'm sorry I believe I worded my original post wrong. The light bulb I have is rated for 12VDC and 50W? It?s a DC light bulb and not operational with AC. They're used on pleasure boats.

i could be wrong, but i was under the impression that incandescent light bulbs dont care if its AC or DC. we tested a few in lab and it worked fine, but it may not always be the case.

anyway, 50w is 50w, AC or DC. you have 4.17A regardless of the method of delivery of the power.

 

[MrDudeMan]

in the event that im wrong, someone feel free to correct me.

 

[jagec]

Originally posted by: Bigsm00th
in the event that im wrong, someone feel free to correct me.

naw, you're right. You can put any waveform you like through an incandecent bulb and it will light up just fine, as long as you don't go crazy on the voltage.

as to the OPs question, remember that a light bulb is basically a resistor. If you have a 50W, 12V bulb, it has a resistance of ~2.9 Ohms. So normally it takes ~4.2 A of current. If you put a 1Kohm resistor in series with it, the resistance of the bulb will be negligible in comparison, and only ~12mA of current will flow (like you said). The light bulb might light up, but it will be VERY dim, maybe even too dim to see.

The resistor won't "burn up", since the voltage and resistance are fixed you KNOW how much current can pass through it, and it's a pretty small amount.

 

[Calin]

The light bulb will have a power at the level of 40mW. It will be enough for it to be warm, but no light at all should be visible.

 

[Loki726]

Originally posted by: Bigsm00th

Originally posted by: PeterBlack
I'm sorry I believe I worded my original post wrong. The light bulb I have is rated for 12VDC and 50W? It?s a DC light bulb and not operational with AC. They're used on pleasure boats.

i could be wrong, but i was under the impression that incandescent light bulbs dont care if its AC or DC. we tested a few in lab and it worked fine, but it may not always be the case.

anyway, 50w is 50w, AC or DC. you have 4.17A regardless of the method of delivery of the power.

Using AC current directly on a bulb will work fine, the current will oscillate so the bulb will switch between off and on at 60hz. It should be fast enough that you can't notice it though.

As a result, if you are using AC, you will observe voltage (and current) at the average value, not the maximum. The average value is the rms voltage and is .707 times the maximum voltage.

 

[Gibsons]

Originally posted by: Loki726

Originally posted by: Bigsm00th

Originally posted by: PeterBlack
I'm sorry I believe I worded my original post wrong. The light bulb I have is rated for 12VDC and 50W? It?s a DC light bulb and not operational with AC. They're used on pleasure boats.

i could be wrong, but i was under the impression that incandescent light bulbs dont care if its AC or DC. we tested a few in lab and it worked fine, but it may not always be the case.

anyway, 50w is 50w, AC or DC. you have 4.17A regardless of the method of delivery of the power.

Using AC current directly on a bulb will work fine, the current will oscillate so the bulb will switch between off and on at 60hz. It should be fast enough that you can't notice it though.

As a result, if you are using AC, you will observe voltage (and current) at the average value, not the maximum. The average value is the rms voltage and is .707 times the maximum voltage.

Long time ago I read about some AC/DC converters that you can put into your light bulb socket. The idea is that the oscillation of the AC current shortens the lifetime of the filament. Convert it to DC and your lighbulbs will last much longer. I think I read something.. maybe in Consumer Reports? that looked at these pretty hard, they found that the devices really did lengthen bulb life, but also reduced the amount of light they put out so much that you end up spending a lot more energy to get the same amount of light, outstripping any gains you see in lightbulb lifetime. I hope I didn't just dream all that. 😕 🙂 😱

 

[PeterBlack]

Thank you to all that replied 🙂

I connected my circuit today and like you guys said no light was produced with the 1kohm resistor in series with the 12VDC light bulb. It would only light up without adding any resistors in the circuit. Make sense since almost all of the voltage was getting dropped across any resistors I was putting in series with the light bulb.

I think I will try making an LED light bulb. What I mean is take 10 white LED's and solder them on a circuit board along with some resistors to limit the current then I will take an empty light bulb socket and insert my completed circuit. This will be my LED light bulb.
What I just wrote might be a little confusing but here is a similar project someone else completed:
http://www.montage.co.nz/led/theParts.htm

I want to use this LED light bulb to connect to a small 5W solar panel as an experiment.

Peter

 

[Yellow Dog]

The resistance of the bulb is about 2.8 ohms. When placed in series with a 1K resister, there will be only 0.034 volts across the bulb, and the remaining 11.96 volts will be across the resister.

 

[jagec]

Originally posted by: Gibsons
Long time ago I read about some AC/DC converters that you can put into your light bulb socket. The idea is that the oscillation of the AC current shortens the lifetime of the filament. Convert it to DC and your lighbulbs will last much longer. I think I read something.. maybe in Consumer Reports? that looked at these pretty hard, they found that the devices really did lengthen bulb life, but also reduced the amount of light they put out so much that you end up spending a lot more energy to get the same amount of light, outstripping any gains you see in lightbulb lifetime. I hope I didn't just dream all that. 😕 🙂 😱

But unless those were some EXPENSIVE converters, the output wouldn't be "clean" DC, so you really wouldn't prolong the life of the bulbs by that long...

 

[Calin]

Originally posted by: jagec

Originally posted by: Gibsons
Long time ago I read about some AC/DC converters that you can put into your light bulb socket. The idea is that the oscillation of the AC current shortens the lifetime of the filament. Convert it to DC and your lighbulbs will last much longer. I think I read something.. maybe in Consumer Reports? that looked at these pretty hard, they found that the devices really did lengthen bulb life, but also reduced the amount of light they put out so much that you end up spending a lot more energy to get the same amount of light, outstripping any gains you see in lightbulb lifetime. I hope I didn't just dream all that. 😕 🙂 😱

But unless those were some EXPENSIVE converters, the output wouldn't be "clean" DC, so you really wouldn't prolong the life of the bulbs by that long...

4 diodes and a filter to block 50Hz (or 60Hz) frequency - it's not so complicated. And it won't reduce the received power so much...
What you really need is a system to reduce the starting current, as from my experience the starting moment is the moment in which an incandescent light bulb will broke in 90+% of the cases. You would need to feed the bulb slowly increasing power, from 0 to full in some half a second to a second. This system (if made with passive things) will cost, and will consume a fair bit of power while the light was on

 

[AbsolutDealage]

Originally posted by: PeterBlack
Thank you to all that replied 🙂

I connected my circuit today and like you guys said no light was produced with the 1kohm resistor in series with the 12VDC light bulb. It would only light up without adding any resistors in the circuit. Make sense since almost all of the voltage was getting dropped across any resistors I was putting in series with the light bulb.

I think I will try making an LED light bulb. What I mean is take 10 white LED's and solder them on a circuit board along with some resistors to limit the current then I will take an empty light bulb socket and insert my completed circuit. This will be my LED light bulb.
What I just wrote might be a little confusing but here is a similar project someone else completed:
http://www.montage.co.nz/led/theParts.htm

I want to use this LED light bulb to connect to a small 5W solar panel as an experiment.

Peter

As I'm sure you know, if your experiment involves using the light generated by the LEDs to power the solar panel, you will not have much luck.

 

[itachi]

Originally posted by: Calin
4 diodes and a filter to block 50Hz (or 60Hz) frequency - it's not so complicated. And it won't reduce the received power so much...
What you really need is a system to reduce the starting current, as from my experience the starting moment is the moment in which an incandescent light bulb will broke in 90+% of the cases. You would need to feed the bulb slowly increasing power, from 0 to full in some half a second to a second. This system (if made with passive things) will cost, and will consume a fair bit of power while the light was on

why would you want to filter out 50/60hz..? if you applied a filter such as that to the signal, you'd reduce the received power. if a filter is said to be applied directly to 50/60hz.. then hypothetically, at that point the intensity is reduced by 3db.. equivalently 0.707*V is the resultant voltage.

you can also make a full-wave rectifier using a single bjt as the only non-linear component.

 

[MrDudeMan]

Originally posted by: itachi

Originally posted by: Calin
4 diodes and a filter to block 50Hz (or 60Hz) frequency - it's not so complicated. And it won't reduce the received power so much...
What you really need is a system to reduce the starting current, as from my experience the starting moment is the moment in which an incandescent light bulb will broke in 90+% of the cases. You would need to feed the bulb slowly increasing power, from 0 to full in some half a second to a second. This system (if made with passive things) will cost, and will consume a fair bit of power while the light was on

why would you want to filter out 50/60hz..? if you applied a filter such as that to the signal, you'd reduce the received power. if a filter is said to be applied directly to 50/60hz.. then hypothetically, at that point the intensity is reduced by 3db.. equivalently 0.707*V is the resultant voltage.

you can also make a full-wave rectifier using a single bjt as the only non-linear component.

because 4 diodes dont make very clean DC

 

[itachi]

whoops.. nm, i was thinkin that it would filter out the output of the full-wave rectifier.

 

[Calin]

Originally posted by: AbsolutDealage

Originally posted by: PeterBlack
Thank you to all that replied 🙂

I connected my circuit today and like you guys said no light was produced with the 1kohm resistor in series with the 12VDC light bulb. It would only light up without adding any resistors in the circuit. Make sense since almost all of the voltage was getting dropped across any resistors I was putting in series with the light bulb.

I think I will try making an LED light bulb. What I mean is take 10 white LED's and solder them on a circuit board along with some resistors to limit the current then I will take an empty light bulb socket and insert my completed circuit. This will be my LED light bulb.
What I just wrote might be a little confusing but here is a similar project someone else completed:
http://www.montage.co.nz/led/theParts.htm

I want to use this LED light bulb to connect to a small 5W solar panel as an experiment.

Peter

As I'm sure you know, if your experiment involves using the light generated by the LEDs to power the solar panel, you will not have much luck.

Unless you want to make one of those evil machines like in Tom&Jerry, or (at a much smaller scale) in Road Runner

 

[Pulsar]

Using AC current directly on a bulb will work fine, the current will oscillate so the bulb will switch between off and on at 60hz. It should be fast enough that you can't notice it though.

There's a bit of a misconception. The filament in the bulb actually takes time to heat up. The 60hz waveform in AC doesn't give the filament time to cool, so you don't notice a flicker.

At least, that's what I'd always been led to believe. Put an LED next to an incandescent and turn them on and off. There is a visible difference in the time to full light. That is the time it takes the incandescent to heat up.

 

[Loki726]

Originally posted by: LsDPulsar
Using AC current directly on a bulb will work fine, the current will oscillate so the bulb will switch between off and on at 60hz. It should be fast enough that you can't notice it though.

There's a bit of a misconception. The filament in the bulb actually takes time to heat up. The 60hz waveform in AC doesn't give the filament time to cool, so you don't notice a flicker.

At least, that's what I'd always been led to believe. Put an LED next to an incandescent and turn them on and off. There is a visible difference in the time to full light. That is the time it takes the incandescent to heat up.

You're right, at 60hz the filament will not cool down fast enough to produce any noticable flicker. I guess I didn't word that very well. While the voltage to the bulb will alternate between off and on, the bulb will remain more or less constantly lit. You would have to drop the frequency considerably to observe any flicker.

Also, it is worthwhile to note that, because the filament heats up, it will not act as an ohmic material and its resistance will vary with its temperature.