Mismatch between measured current and calculated current - what am I missing?

[mindcrime]

All these years, I thought I understood something as basic as Ohm's law, but tonight I was disabused of that notion, apparently.

To wit...

I have one of those cheapo 12V automotive "circuit tester" light pen things. I wanted to repurpose it to work with a different voltage by cutting the ground lead and adding in a resistor. Simple stuff, right? So to calculate what resistor to add in at the new nominal voltage, I decided to start by measuring the resistance through the device as it is now (just the bulb, an old incandescent flashlight bulb jobbie). So I grabbed a multimeter and measured it at around 9.5 ohms.

OK. V=IR right? So I divide both sides by R and get I = 12V / 9.5 ohms, = 1.26A. OK, looking good. Then I dial up 12V on my Rigol DP832 bench supply, and clip the light in and see it light up. And the display reads 0.132A. WTF? At first I assumed the power supply had some setting for the units to display and that it was somehow off, but there's that big ole 'A' right there, which seems to leave no room for uncertainty. Then I started thinking my DP832 had a bug or that my multimeter was wrong.

So... grabbed my real multimeter, an EEVBlog BM786 and measured the resistance again. Same thing - right around 9.5 ohms. Then I decided to do one more "sanity check" and so I wired the multimeter into the circuit in current mode, with the range set to Amps.  Powered everything up, and the multimeter and the DP832 both give the same current reading, 0.132A or so.

So two meters confirmed the current draw, and two different meters confirm the resistance of the circuit. And I've read and re-read the Wikipedia page on Ohm's Law to confirm that Ohm's Law works in units of Volts, Amps, and Ohms. But yet there's a discrepancy of basically a factor of 10 between what I'm measuring and what I'm calculating. I know this is something obvious and trivial that I'm missing, but somehow it's escaping me at the moment (I blame lack of coffee). Can anybody point out what I'm missing in this?

I haven't added any other resistors or anything yet, so what I'm measuring is literally just the circuit tester pen thing with it's bulb and a little bit of wire. I'm defeated here. 

[ataradov]

Resistance of the filament goes way up as it heats up.

Strangely enough, there is a whole Wired article on that - https://www.wired.com/story/need-an-ohms-law-party-trick-take-a-lightbulbs-temperature/

This effect was used in practice for automatic gain control in the Wien bridge oscillators - https://en.wikipedia.org/wiki/Wien_bridge_oscillator . I remember when I was very young my dad had a signal generator with a light bulb deep inside it. The purpose of the light bulb always puzzled me.

[basinstreetdesign]

First of all, the use of a 15 Watt light bulb as a penlight tester would be a bit on the heavy side.  More like 1 - 2 Watts sound more reasonable.
What you are seeing is the cold resistance of the bulbs' filament and you are trying to reconcile that with the power draw of it when hot.  The resistance of an incandescent bulb when cold is a fraction of what it is when hot and 1/10 sounds about right.

[mindcrime]

OK, fair enough. I knew, of course, that the resistance of the filament goes up. What I guess I misunderstood, was that it happens as quickly as it does, as well as the magnitude of the increase. I guess I thought it would creep up slowly as the filament heated and that the increase would be something more like 2x or 3x. Shows you that I don't mess with incandescent bulbs much.   

In practice, if the current draw ever registers more than the 0.132 range, it happens to fast for me to notice it on the meter(s) with my naked eye.

Anyway, that helps. At least I know I'm not going (totally) insane and that my test gear didn't all suddenly get infected with the DaVinci Virus or something. 

Follow-on question - if I want to add enough resistance to where I can use this thing with a nominal 48V circuit, and avoid blowing the bulb all the time, should I use the cold resistance or the hot resistance to do the subsequent calculations?

[mindcrime]

The more I think about it, the more I think "I'm a dumbass". Of course the resistance change is quick... the bulb starts to glow more or less immediately when you apply power, and it's glowing because it's hot! Derp, derp. I don't know if I can get by with my "lack of coffee" excuse on this one or not. 

Strangely enough, there is a whole Wired article on that - https://www.wired.com/story/need-an-ohms-law-party-trick-take-a-lightbulbs-temperature/

Great article, thanks for sharing!

[ataradov]

Do you actually need it to be a light bulb? It will be way-way easier to re-implement that tester with an LED.

Even setting aside resistance selection issues, you will need to use a pretty beefy resistor to drop that voltage. It will be somewhat expensive and very warm after prolonged use.

[mindcrime]

Do you actually need it to be a light bulb? It will be way-way easier to re-implement that tester with an LED.

No, I just happened to have this thing lying around unused, and it's a fairly close match to what I need, other than the voltage range difference. But yeah, I could definitely build a better "purpose built" device, I just can't be bothered. 

Even setting aside resistance selection issues, you will need to use a pretty beefy resistor to drop that voltage. It will be somewhat expensive and very warm after prolonged use.

There won't be any prolonged use. I just need to touch it to something and see if it lights up for the briefest period detectable by the human eye. I just need a binary "yes this is hot, no this isn't hot" indicator.

[ataradov]

Then calculate based on the cold resistance. It will under-power the warm bulb, but it would still light up somewhat. May be lower the resistance from there to get it brighter at the risk of .burning the bulb more often.

It is still way easier to find a couple LEDs. And you will need to look for a resistor anyway.

[Ian.M]

*WRONG*
If you calculate the series resistance based on the cold resistance, as the current drops as the filament heats, it will grossly overvoltage the bulb.

Use the calculated hot resistance, or simply calculate what resistor will drop 36V with 132mA through it.  270R would give full brightness at 48V.   Note that the bulb will be significantly slower to heat up so there may be a slight but perceptible lag from contact to indicator visibility.
Also note that the 270R resistor will dissipate close to 5W steady state so it *MUST* be wirewound, cermet or carbon composition >=1W to survive repetitive overloads <1 second duration.  If you need to keep the test lamp on while you fiddle with something else use a 5W resistor!

[mindcrime]

*WRONG*
If you calculate the series resistance based on the cold resistance, as the current drops as the filament heats, it will grossly overvoltage the bulb.

Use the calculated hot resistance, or simply calculate what resistor will drop 36V with 132mA through it.  270R would give full brightness at 48V.   Note that the bulb will be significantly slower to heat up so there may be a slight but perceptible lag from contact to indicator visibility.
Also note that the 270R resistor will dissipate close to 5W steady state so it *MUST* be wirewound, cermet or carbon composition >=1W to survive repetitive overloads <1 second duration.  If you need to keep the test lamp on while you fiddle with something else use a 5W resistor!

Yeah, 270R @ 5W is what I came up with as well. I just ordered a box of 10 of 'em. It's probably overkill since I need the light to be on for about 250-500 milliseconds or so at most. But you never know... I like to leave some "head room" on stuff like this.

[tunk]

An advantage of using an LED is that it will work over a range of voltages.
E.g. if you set it to use 2mA@12V, it will be around 5mA@24V and 10mA@48V.

[Ian.M]

That's not always an advantage.  The LED will start to light with only a couple of volts across it, so will indicate on 'phantom' voltages, and its very hard to distinguish a LED passing 5mA from the same LED passing 10mA so will also give misleading results on bad connections with excessive resistance.   An incandescent test lamp is low enough impedance not to respond to phantom voltages, and has a notable variation in brightness with moderate voltage change, so its quite clear if the voltage tested is notably low e.g. due to excessive contact resistance, which its far more sensitive to anyway due to the higher test current.   LED test lamps do have their place, and if a bicolour LED is used, have the additional utility of indication polarity, or if AC is present, but in no way are they equivalent to an incandescent one with a bulb voltage appropriate tothe circuit.