# Risk in limiting current with light bulbs in parallel

I'd like to make a current limiting device on the cheap using some light bulbs. I've seen somewhere the idea of connecting light bulbs in series to limit the current.

For example, I'd wire a 250w light bulb in series to limit the current used by the circuit to 250w. Which has a resistance of 193.6 Ohms according to this formula:

P = VI
V = IR

250 = 220 * I
250 / 220 = I
1.136 = I

V = IR
220 = 1.136 * R
220 / 1.136 = R
193.6 = R


simulate this circuit – Schematic created using CircuitLab

Here, the lamp is limiting the current going into the main winding of the transformer to 250w.

In theory, if I wanted to increase the maximum current going into the transformer, I'd have to modify the circuit to have more light bulbs in parallel like this.

simulate this circuit

Since the resistance in parallel can be calculated this way:

1 / Rtotal = 1 / R1 + 1 / R2 + ... + 1 / Rn
1 / Rtotal = 1 / 193.6 + 1 / 193.6
1 / Rtotal = 2 / 193.6
1 / Rtotal = 1 / 96.8


Which reduce the resistance to 96.8 ohms and then the maximum current flowing in the transformer should be:

V = IR
220 = I * 96.8
220 / 96.8 = I
2.272727273 = I
I*220 = 500


In theory adding any light bulb with a certain wattage rating in parallel to other light bulbs but in series with the transformer should increase the maximum current flowing to the transformer.

That said, I was wondering what happens to the sine wave coming out of the light bulbs? Is it possible that the sine waves goes out of sync even though all light bulbs are rated for 50hz or something else that I'm not aware of?

• The thing that you should be most aware of is that the "rated" resistance is for full on bulbs with glowing filament. When the filament is cooler, the resistance is lower. And unless you have really really long differences in wires, how should the sine waves go out of sync? Jul 26 '17 at 11:17
• Don't try it with LED lights. :)
– pipe
Jul 26 '17 at 11:55
• It may technically work, but burning off all of that power is pretty wasteful.. Jul 26 '17 at 13:55

Yeah, you can do that.

Also, incandescent lightbulbs (not CFL/LEDs of course) have a desirable property here: when the filament is cold, resistance is much lower. This means much less voltage sag at low currents than a resistor would offer.

Check out this answer, everything you need is there.

Now, for AC, the thermal mass of the filament is small, but it is enough to smooth out temperature variations across mains cycles, so the resistor will stay roughly the same during a cycle, and your sine should look good.

If there are rectifiers after your transformer, your current won't be a sine anyway, but a series of spikes. Your filament will average them over time, allowing max current to be exceeded during the rectifier pulses, but not on average (which is useful).

• I have successfully used a light bulb to protect an industrial single-phase input servo drive. That last point is very good and one I hadn't considered fully before: the voltage will be high during the portions of the mains cycle while the current is low. You could expand on this. +1 Jul 26 '17 at 11:29
• Right, so as I understand, the lightbulb would light bright only if the transformer was trying to use more than 250w as the resistance would be lower, it wouldn't be enough to turn the light on completely. And yes, I'm talking about incandescent lights. Jul 26 '17 at 11:42
• D'uh, my link was bad. Fixed. Click on it, it has a nice explanation. Jul 26 '17 at 11:58

You may not need to.

The resistance when cold will be much lower than the calculated resistance ... measure it and add that to the question.

So unless the load takes a lot of power (e.g. due to a fault), the current (I) will be relatively low, meaning there is relatively little power wasted in the lamp, relatively little resistance increase due to heating, and therefore relatively little voltage drop across the bulb.

But there are no obvious safety risks to paralleling bulbs if you think you have to.