# How to check light bulb power?

#### rambomhtri

Joined Nov 9, 2015
551
Hi, I have a very simple question. I have this little 230V (it's marked) light bulb, and I want to know how much power it needs. If I check with my multi the bulb resistance, I get 450 ohm.

First, since we are in alternating current, the lightbulb is (or has) an impedance, which is:
Z = R +jX (I guess a light bulb is an ideal resistor in series with a very small coil)

I don't know if these 450 ohm refer to R or |Z|. That's the first question.

Second, if I divide 230V/450ohm I get that it needs 0.5A aprox. So, then, the power must be V•I, which is 230x0.5 = 115W. Second question, I don't know if these are the S 115VA, or the P 115W. Like in the power formula S [VA]= P [W] +jQ [VAr]

Third, that's a very high power for a little light bulb, so I checked the amperage with my multi, setting it in series with the light bulb. I read 0.045A, so it's actually 10W, which makes more sense.
So the third question is why am I getting theoretically 115W and in practice 10W.

Thank you

#### OBW0549

Joined Mar 2, 2015
3,566
So the third question is why am I getting theoretically 115W and in practice 10W.
The resistance of a tungsten light bulb when cool (i.e., when checking it with an ohmmeter) is very low. When operated at its rated voltage it is around 12-15X higher due to the heating of the filament and tungsten's large temperature coefficient of resistance. So you can't predict the wattage of the bulb by making a room-temperature resistance measurement.

For example, here's a graph of the resistance of a 120V 100W bulb versus applied voltage:

#### rambomhtri

Joined Nov 9, 2015
551
Thanks!
I was thinking about that, that the temperature affected a lot the resistance. OK, so my first assumption was wrong. Can you also please answer the 3 questions?

Thank you very much. By the way, if the bulb at 25°C has 450 ohm, that means that when you plug it to the wall, in the very first milliseconds you get a huge amperage peak, right? Of exactly 0.5A as I calculated previously?

#### OBW0549

Joined Mar 2, 2015
3,566
Sure:

First, since we are in alternating current, the lightbulb is (or has) an impedance, which is:
Z = R +jX (I guess a light bulb is an ideal resistor in series with a very small coil)

I don't know if these 450 ohm refer to R or |Z|. That's the first question.
At mains frequency, any inductive reactance in the filament is insignificant to the point of being almost unmeasurable. So you're dealing with a pure resistance for all practical purposes.

Second, if I divide 230V/450ohm I get that it needs 0.5A aprox. So, then, the power must be V•I, which is 230x0.5 = 115W. Second question, I don't know if these are the S 115VA, or the P 115W. Like in the power formula S [VA]= P [W] +jQ [VAr]
That power (which occurs only the first few milliseconds until the filament heats up) is almost entirely real power (W) with negligible reactive power (Volt-amps).

Third, that's a very high power for a little light bulb, so I checked the amperage with my multi, setting it in series with the light bulb. I read 0.045A, so it's actually 10W, which makes more sense.
So the third question is why am I getting theoretically 115W and in practice 10W.

By the way, if the bulb at 25°C has 450 ohm, that means that when you plug it to the wall, in the very first milliseconds you get a huge amperage peak, right?
Yup.

Of exactly 0.5A as I calculated previously?
I don't know about the "exactly" part because I haven't calculated it, but it sounds about right.

Incidentally, that huge current surge at the moment power is first applied is the main reason tungsten bulbs usually fail at turn-on; over time, as the tungsten evaporates from the filament, it develops "hot spots" where the filament is thinnest and the turn-on current surge eventually becomes too much for the filament to bear. !POP!

#### rambomhtri

Joined Nov 9, 2015
551
Wow, that information was really interesting. Thank you so much, I really loved the bulb failing explanation, it makes all sense in the world. Now another question comes to my mind...
Why this light bulb design (which has been around for 150 years) doesn't incorporate "something" to increase the current slowly until the working temperature? Wouldn't that increase a lot the life-time of a light bulb? May be it was not incorporated because they wanted light bulbs to fail overtime so we must buy a replacement?

I'm thinking about a dipole or something that at first it has a lot of resistance, but over time (ms) the resistance decreases as the current flows, which gets bigger and bigger. Is there a component that behaves like that? At first open circuit but later a short?

#### OBW0549

Joined Mar 2, 2015
3,566
Why this light bulb design (which has been around for 150 years) doesn't incorporate "something" to increase the current slowly until the working temperature? Wouldn't that increase a lot the life-time of a light bulb? May be it was not incorporated because they wanted light bulbs to fail overtime so we must buy a replacement?
Anything I can think of that would accomplish that would make the light bulb MUCH more expensive. No corporate perfidy or skullduggery involved, just simple economics.

I'm thinking about a dipole or something that at first it has a lot of resistance, but over time (ms) the resistance decreases as the current flows, which gets bigger and bigger. Is there a component that behaves like that? At first open circuit but later a short?
The only component that comes to mind is a negative temperature coefficient (NTC) thermistor; this technique was used in TVs of yore, the old kind with vacuum tubes, to limit inrush current which would otherwise put huge stress on the tube filaments. Doing so increased the cost of the TV but slightly; adding that same component to a lowly light bulb would probably double the cost, or more.

#### wayneh

Joined Sep 9, 2010
17,493
There were slow-start gizmos for light bulbs. Not worth it to protect most bulbs which are nearly free.

“Bulb miser”