dim bulb tester light bulb resistance

[Randall]

I have built, and do use a dim bulb tester pretty often. Dim good, bright bad, easy enough. But I was asked a question today I had never thought much about, and couldn't answer. A friend is building a circuit that involves a meter with a 12v 50mA light bulb in it. He measured the resistance of the lamp at 13 ohms, and is concerned that 12v/13ohms = almost 1 amp. He figured the filament would have to be 240 ohms to pull .05A at 12v, and I can't disagree.

So it brought him to the question, do filament resistances change with current draw, and is this the theory behind how dim bulb testers work? As more current is drawn they get brighter because the filaments are offering more resistance?

[galaxiex]

Short answer... Yes.

The cold resistance can be very low, almost a dead short.

As the filament heats the resistance goes up.

[Enzo]

Yes^^^

Measure the cold resistance of a 100 watt bulb, then calculate the resistance it would need to draw 100 watts at 120v.

[Randall]

I get 12 ohms cold and 144 ohms under load.

[Enzo]

And that should be fairly typical of incandescent bulbs in general.

[the fatch]

An amp in working order draws an amount of current and so will appear as a resistance between the hot and cold wires of the 120V supply. With the 200W lamp in series with the amp, the same current flows in both and the 120V is shared between them in proportion to their resistances. As the filament heats up its resistance increases so that its share of the 120V increases. At the same time the current decreases since it is flowing through a larger total resistance. A point of equilibrium is reached where a further increase of resistance of the filament would increase its share of the 120V, but this increase in voltage is counteracted by the reduction of current due to the increase in total resistance. If an amp draws 1A it appears as a 120V/1A = 120Ω resistance. Assume that the resistance of the filament at equilibrium is 30Ω, for a total resistance of 150Ω. The current is reduced to 120V/150Ω = 0.8A. The voltage across the filament is 120V x 30/150 = 24V with 96V appearing across the amp. The filament is dissipating 24V x 0.8A = 19.2W causing it to glow dimly. A short in the amp means that its resistance appears to be 0Ω so that the whole of the 120V appears across the lamp. The lamp operates as it would normally, glowing brightly, the resistance of its filament has risen to 72Ω and it is dissipating 200W so that even with a dead short in the amp the current through it is restricted to a maximum of 200W/120V = 1.67A.