Ac current limiting per light bulb?

Dear all..
Please help me to know: how wattage of light bulb(please see photo) is needed for limiting my circuit 120ac 60 watt maximum
Thank you so much.

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I think have seen this trick before.

A filament style light bulb
http://en.wikipedia.org/wiki/Incandescent_light_bulb
can act as a overcurrent protector in low-current region of its I-V curve,
http://en.wikipedia.org/wiki/Current%E2%80%93voltage_characteristic
where the filament is cold and has low resistance. 

And I think for this trick to work, the normal power draw for the device (at the mains voltage) has to be much less than that of the rated power for the light bulb (at mains voltage).

For example, if you have a radio that normally draws around 50 mA, at 120 VAC, or 0.05A*120V = 6W, and you put it in series with a 60 W filament light bulb, and then both share the same current, since they're in series. 

However 50mA is not enough current to make the light bulb filament hot.  The filament gets hot at its rated current, which is 60W/120V= 0.5 A=500 mA.  At that temperature, white hot, its effective resistance, is 120V/0.5A = 240 ohm. 

In contrast, when the light bulb filament is cold, its resistance might only be like 48 ohm.  So with a current of 50 mA, it only sees a voltage drop of like 0.05A*48 ohm = 2.4 VAC.  The rest of the voltage, 117.6 VAC, goes across the load.

The protection occurs, when, for some reason, the load draws too much current.

Supposing the load turns into a short.  Then the current rises to 500 mA, the normal current for the light bulb.  In that case, power dissipation in the light bulb is 120V*0.5A = 60 W, and power dissipation in the load is 0V*0.5A .

At current of 50 mA, when everything was cool and froody: Power dissipation in the light bulb was 2.4V*0.05A = 0.12 W, and power dissipation in the load was 117.6V*0.05A = 5.88 W.

I'm not sure about the worst case power dissipation in the load.  I think the exact number depends on the shape of the light bulb's IV curve.  However I am almost certain this number is somewhere in the range:

5.88 W  < Pmax < 60 W 

lam (author)  Jack A Lopez3 years ago
Dear Jack...
Very very helpful Thank you for your concern explain and your time advise me.
iceng3 years ago
I use this all the time and start with a high resistor ie a low wattage bulb
if things look to be working try a lower series resistance ie higher wattage

If you really need more power put two 100W lamps in parallel.

The worst that can happen is the lamps light at full brightness .
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lam (author)  iceng3 years ago
Dear Iceng.
Thank you for your understand my question
and helpful with your photo,Thank you again for your time.
iceng lam3 years ago
BTW
Keep in mind that the initial cold resistance of an incandescent light
is five times lower then when hot and lighted.
This may explain why some strange effects can and do happen.
So you want to use the bulb like a fuse? Not a very efficient option since the bulb will pull an additional 60W load on the power source. You also run the risk of your circuit failing before the bulb does. The wattage rating on the bulb is how much power it pulls. Not it's maximum rated power. It might not fail till it's got over 100W flowing through it.
lam (author)  mpilchfamily3 years ago
Dear mpilchfamily.
Thank you for your concern and advise me.
I don't see what you have. What's the resistance of the "circuit" ?
lam (author)  steveastrouk3 years ago
Dear...
I am very sorry for my question is not clear (my English not well ).
I want protect (over current) my circuit ( 120ac - 0.5 amp = 60watts) by connect an light bulb en serial with my circuit , so I don't know how wattage of light bulb must to used is right for protect my circuit.Please help me
Thank you for your time
If your circuit takes 60W, it takes 60W, putting a resistor in series will make it take less power, and the resistor will heat up. To protect your circuit, use a fuse !
lam (author)  steveastrouk3 years ago
Dear steveastrouk. Great helpful .Thank you a lot for your time help me.