EEVblog Electronics Community Forum
Electronics => Beginners => Topic started by: Beamin on October 29, 2017, 10:45:05 am

To see it go from zero while is cold to high impedance when bright? Use a meter that can put out high voltage like VTVM? How would you do that with modern equipment?
And the question is MEASURE not calculate.

Measure the current while you apply varying voltage, and then use a simple application of ohms law to get the resistance from the known voltage and current. Multimeter just does the calculations for you by using a known current and displaying the resistance after doing the math internally.

The ohms increase by the PTC of the filament,
A light bulb that expects to run of 240V AC can also be fed off 240V DC to get the same equivilent power, e.g. heat.
So use a variac or step down transformer to get about 170V AC, which when rectified gives 240V DC, then simply measure the current at the switch on vs the stead state to know your cold / hot resistance.
Of if the AC current on your multimeter is good enough resolution, simply measure in series.

Measure the current while you apply varying voltage, and then use a simple application of ohms law to get the resistance from the known voltage and current. Multimeter just does the calculations for you by using a known current and displaying the resistance after doing the math internally.
For this challenge you are the average American and you don't know any math; ANY. "math is hard/sucks" and your iPhone doesn't have a calculator app, not that you would know how to use it anyways. So you have to take measurements.

Since you don't mention what equipment you have available to you, an oscilloscope with a voltage and current probe would suffice. Measure the voltage at the terminals of the
bulb and the current in the wires. Use the builtin math function to calculate a division, resulting in ohms. The time aspect can be observed on the oscilloscope screen.

Ok, so how to directly measure resistance, measure it cold, stick it in a socket, heat it up, and then use a relay to switch both contacts of the light bulb to a meter measuring resistance, and disconnecting the mains, it will begin dropping quickly, but it will measure for the first fraction of a second.

Buy an expensive SMU that can deliver enough voltage and power for your light bulb.
It should provide you with an builtin Ohms measurement function that measures the resistance at any voltage you apply.
Check the manual before buying, I don't know if an average American is able to read a manual ...

Aside from all of the snide remarks about Americans, what does the OP mean in differentiating measurement from calculation. I can't think of a single "measurement" of electrical quantities that doesn't have some implicit calculation. Old school d'Arsenval meters are based on some rather sophisticated relationships between current, voltage, magnetic field strength and Hooke's law. The newer digital voltmeters are based on integrating current over time before the final mathematical step of a ratio.
I am guessing that one of two things is the objective. A plot of resistance over time as the bulb heats up or a measure of the resistance at a particular operating condition in a circuit (possibly a small range of values as occurs in the classic Wien bridge oscillator). The test set ups to get the best answer for these conditions is slightly different, so maybe it is time for feedback on the objective.

Heat the filament to the same temperature as the filament would have when the bulb was lit normally, and measure the resistance.

Heat the filament to the same temperature as the filament would have when the bulb was lit normally, and measure the resistance.
That is the best answer for providing a direct resistance measurement.
Of course, there are two other questions ....
1. Knowing what the operating temperature is
and
2. How to heat the filament to that temperature
I take it these are left to the experimenter....

Another method is to use an analog panel meter with two needles, use the other needle for voltage measurement and the other needle for the current measurement, print a new background for the panel so that you can read the resistance from the needles' crossing, and now you have a measurement instrument for measuring the bulb resistance without calculations.
https://i.ebayimg.com/thumbs/images/g/vWcAAOxyVaBSpVKv/sl225.jpg

Use a DC lab power supply and set the current limit to 1.000A and measure the voltage across the bulb when it is lit. Now the resistance is equal to the voltage across the bulb. No need to calculate anything.
Edit: Won't work.

Use a DC lab power supply and set the current limit to 1.000A and measure the voltage across the bulb when it is lit. Now the resistance is equal to the voltage across the bulb. No need to calculate anything.
Who's to say that 1A is the correct operating current?

Another method is to use an analog panel meter with two needles, use the other needle for voltage measurement and the other needle for the current measurement, print a new background for the panel so that you can read the resistance from the needles' crossing, and now you have a measurement instrument for measuring the bulb resistance without calculations.
https://i.ebayimg.com/thumbs/images/g/vWcAAOxyVaBSpVKv/sl225.jpg
A thought. Somebody just needs to design the appropriate scale.

Use a DC lab power supply and set the current limit to 1.000A and measure the voltage across the bulb when it is lit. Now the resistance is equal to the voltage across the bulb. No need to calculate anything.
Who's to say that 1A is the correct operating current?
P=U*I, thus varying the voltage one can determine the power flowing through the bulb. Of course, if the bulb won't work with 1A current, then one has to calculate something and that situation is beyond the scope of the requirements ;)
Edit: As a second though, this method may not work as Brumby noted. I will overstrike my previous message, too.

Another method is to use an analog panel meter with two needles, use the other needle for voltage measurement and the other needle for the current measurement, print a new background for the panel so that you can read the resistance from the needles' crossing, and now you have a measurement instrument for measuring the bulb resistance without calculations.
https://i.ebayimg.com/thumbs/images/g/vWcAAOxyVaBSpVKv/sl225.jpg
A thought. Somebody just needs to design the appropriate scale.
Sure, but that wasn't excluded from the specifications, only requirement was to be able to measure bulb's resistance directly without any further calculations. So, once a person has an instrument one can measure the resistance directly without any further calculations. ;)

Aside from all of the snide remarks about Americans, what does the OP mean in differentiating measurement from calculation. I can't think of a single "measurement" of electrical quantities that doesn't have some implicit calculation. Old school d'Arsenval meters are based on some rather sophisticated relationships between current, voltage, magnetic field strength and Hooke's law. The newer digital voltmeters are based on integrating current over time before the final mathematical step of a ratio.
I am guessing that one of two things is the objective. A plot of resistance over time as the bulb heats up or a measure of the resistance at a particular operating condition in a circuit (possibly a small range of values as occurs in the classic Wien bridge oscillator). The test set ups to get the best answer for these conditions is slightly different, so maybe it is time for feedback on the objective.
If rescaling a meter is legitimate (and I don't know how you do that without math), just scale the ammeter in Ohms assuming a constant voltage. It's as legitimate as the rest of this thread.
You can't do electronics without math. You simply can't avoid Ohm's Law at any level  hobby or professional. The EEs around here have taken a boat load of math or subjects that don't have math in the title but are really math in drag. Field Theory comes to mind as does Control Systems. You can't avoid curl and divergence nor can you avoid Laplace Transforms. It just comes with the paper.
Heck, you can't even calculate the ballast resistor for an LED without math. Sure, you can copy somebody else's work but what if your LED has vastly different characteristics? How in the world could you bias a transistor? Calculate the gain of an op amp? I suggest that the premise of this thread is wrong: Everybody doing electronics can do some math. Otherwise, they do something else.
Not a realistic proposition, doing electronics without math!

My suggestion:
1) Remove the glass of the bulb and measure the resistance of the filament.
2) Then take a flame and heat up the filament.
3) Measure the resistance of the filament.
The filament may not survive long, but maybe long enough to make the measurement.

Since those power meters you plug into a wall socket and the load into the power meter can measure voltage and current, could there be a model that also shows the resistance of the load?
There's something for USB:
(https://i.imgur.com/sEOf5G0.png)
USB Charger Mobile Power Current Voltage tester meter capacity 330V QC 2.0 + load resistance 2A/1A discharge With a switch
The above is DC 2A 05A 330V so it would need to be some low power light bulb, but the same option might be available on an AC power meter.

This is an interesting experiment that anyone who is curious should perform.
I attach the results that I obtained when I did the experiment myself some time ago. One particular point to note is that there are two "resistances" for the light bulb filament at any point on the graph. One "resistance" is the value of V/I at that point. The other "resistance" is the slope of the graph, the value of dV/dI. Since the slope of the graph is the value that remains nearly constant around any local region of operation, this is IMHO the value that is most useful where something called "resistance" is needed.
(https://www.eevblog.com/forum/beginners/howwouldyoumeasuretheohmsofalightbulb/?action=dlattach;attach=365259;image)

Resistance is a derived unit. It is not possible to have a resistance value without calculation, it can only be "derived" using volts and amps.

Resistance is a derived unit. It is not possible to have a resistance value without calculation, it can only be "derived" using volts and amps.
Definitely, but unless I misunderstood, the discussed task is to have it calculated by an instrument, rather than a calculator, so that the user doesn't need to know the formula.

Sense resistor to an A/D and a micro implementing ohms law, job done though why bother is for sure a lot more difficult to answer. :)

Definitely, but unless I misunderstood, the discussed task is to have it calculated by an instrument, rather than a calculator, so that the user doesn't need to know the formula.
But if you don't know the formula how is the value useful to you, other than as a trophy to stick on the wall?

Having it done for you is easy, just use 2 meters that have remote measurement measuring ability, and connect to a computer, with the software there ( probably Mathlab to use a sledgehammer to open a peanut, but simpler is just Python and use a little scripting to do the simpler graphical display of instantaneous resistance) doing the scaling of the 2 measuements and the multiplication, and then giving a real time instantaneous resistance value.
If you have 2 HP GPIB meters you can get increased resolution using some nice 6 digit meters, and have the result accurate to 6 digits, but in most cases with any arbitrary lamp 2 decimal points in scientific notation will be more than enough for most uses. you will see from that how external reflectors that focus the radiation back on the filament does have an effect on the resistance, and how forced and convection cooling of the envelope also does change it.

Really? OK, I'll play.
Step 1: Connect DC power supply, ammeter, and bulb in series. Run up DC voltage until bulb is at desired brightness. Read current from ammeter.
Step 2: Remove bulb from circuit and replace with variable resistor (rheostat, heavyduty potentiometer, etc.) and adjust resistor until you see same current as in Step 1.
Step 3: Measure resistance of variable resistor.
Done.
(Just make sure your variable resistor can handle the power. A 1/4 watt pot probably won't be suitable. :DD )

^ ^ That was nice one. \$\Omega\$

1. Hire an artsy type to draw up fanciful pictures of your new product idea
2. Start overly optimistic Kickstarter that will find resistance of said lamp using Arduino and/or Raspberry Pi; a goal of excess of $250K should do it.
3. Wait for funds to roll in
4. Hire “expert” to actually design product(s)
5. Update Kickstarter with new information and new $ goal
6. Wait for more funding
7. Find cheap Chinese company to actually make the product
8. Congratulations, you are now officially in Marketing

When I was doing the halogen and 60W bulb tests, I tried a few times to see if I could detect any change in the bulbs' resistance with its 100Hz ~10% change in light level using the scope's math on V/I, but never managed to get any meaningful math waveform, just noise and large spikes when V and I pass through 0.
https://www.eevblog.com/forum/chat/20whalogenbulbviewedbyaphotodiode/ (https://www.eevblog.com/forum/chat/20whalogenbulbviewedbyaphotodiode/)

Our firstyear students use an approach similar to IanB above, measuring the IV characteristic for a filament lamp (using a DMM and a series resistor). Since the tungsten filament has a known temperature coefficient, they can find the temperature of the filament as a function of power. When it's glowing, most of the heat is lost as radiation, and the students can then verify the blackbody T^4 law.
By monitoring the lamp output with a green LED acting as a photodiode, it's also possible to determine Planck's constant. Hence, one can demonstrate some fundamental quantum physics using just a humble torch bulb.

Someone could characterize the light bulb for you, then you would just need to cross reference the voltage to get the resistance and whatever brightness is required, no math needed. Or if you were able to measure a combination of voltage, current or power, a slide ruler/chart could be used instead of math. That is the historical method of not using your brain.
A multimeter would need to perform two measurements simultaneously (and do the math for you) based on the rules you have provided. A power meter or oscilloscope that had the right math could automatically calculate it, or modified firmware on any device that could perform the two measurements. The easiest way of doing the math is obviously an ohms law calculator.
The easiest device to use is a power meter and work it out with ohms law. For example my toasters elements are 19 ohms each, hot water jug is 24 ohms, takes seconds to setup and calculate.

Hi,
The resistance of a bulb is not zero with zero voltage. It is finite and non zero.
For example, a 100 watt 120vac bulb is around 10 ohms cold (zero current).
For a 1.5v bulb i have it is around 6 ohms cold (zero current)
When you measure you apply a voltage and measure the current and use Ohm's Law:
R=V/I
however with zero current you cant do that. You really have to use the slope, but you can get a good idea by just using a very very low current like 1ua, but even 100ua will work in most cases and even 1ma in larger wattage bulbs.
So all filament bulbs will have a curve similar to:
R=Ro+R(v)
where Ro is the zero current resistance and R(v) is a function of the applied voltage.
There are equations that approximate the character of filament bulbs and they even can predict life expectancy in the absence of vibration.

I attach the results that I obtained when I did the experiment myself some time ago. One particular point to note is that there are two "resistances" for the light bulb filament at any point on the graph. One "resistance" is the value of V/I at that point. The other "resistance" is the slope of the graph, the value of dV/dI. Since the slope of the graph is the value that remains nearly constant around any local region of operation, this is IMHO the value that is most useful where something called "resistance" is needed.
About 20 ohms for the latter, by the looks, fairly constant right down to one or two Volts or so, but about 1 ohm initially at close to zero Volts.

Why wouldn't you want to calculate anything? You don't even have to do it manually, you can use an actual calculator.

just a note for anyone who wants to plot the resistance ... this is a classic example of needing two meters...
use two meters to measure voltage and current at the same time make sure your measuring the voltage right at the bulb so its accurate as there will be voltage drop both across the current meter and on the leads.
have fun with this, it is a classic experiment!

I have a Hypatia ground bond tester. It can do 100A and can probably get up to 12V. The nice thing is it will constantly give a resistance reading for any current dialed up. I find it a really handy instrument.
Every moron wants to use a lamp as a load resistance.

unless you are looking for a ultra precise measurement i would recommend to use a watt meter , basically you should use that and a multimiter to measure the lines voltage and power. I=power/voltage. and then with ohms law r =v/I. if you want to trust the bulb's wattage ratting is even easyer

But the 'game' is to do it without calculations. It needs to be a direct measurement.

i do know a couple solutions that would give you a direct resistance measurement but, it really depends on the rage of what your measuring
so are you measuring AC? DC?
what is the operating voltage of the bulb?
what is the expected current range of the bulb?

Sure, but that wasn't excluded from the specifications, only requirement was to be able to measure bulb's resistance directly without any further calculations. So, once a person has an instrument one can measure the resistance directly without any further calculations. ;)
Without knowing why calculations can't be present, it's a problem that's both trivial and impossible to solve. What's asked for isn't the actual problem, it's a perceived set of requirements to solve the actual problem. There might very well be an easy short cut to solve the real problem, rather than the proxy question.

Resistance is a derived unit. It is not possible to have a resistance value without calculation, it can only be "derived" using volts and amps.
This ^^^^
Someone has to do the calculation...
Anyway, as it seems like a game, one more "method" is to use the voltage ratio operation of a multimeter. As the voltages on two resistors, connected in series (same current), is proportional to their values so is also the ratio of them.
Connect a reference resistor R_{REF} in series with the lamp R_{lamp}.
Connect a variable voltage source to the series combination.
Connect the reference input (V_{REF}, usually the "sense" input) of the meter to the R_{REF} and the input (V_{IN}) to the R_{lamp} (Ask from someone to read the manual and to tell you how to do this!).
Adjust the voltage to the brightness you like and measure the ratio. If the R_{REF} is equal to one Ohm, the ratio is Ohms of R_{lamp}; if it is one kilo Ohm then the ratio is R_{lamp} in kilo Ohms.
You can stick a piece of paper to the right of the last digit of the display with an omega symbol (or 'k' omega) to avoid confusion ...!
Of course this does not work with zero volt or outside of the instrument capabilities.
In this method the indication is "resistance" but we are measuring voltages, so it is indirect...
If a direct resistance measurement is needed, the only method that I can think is someone to go inside the material and examine the paths of possible electricity routes!!!

Y'all need more Jes err... Pease in your lives!
A Seeley box (https://youtu.be/Sv21wD3FRmY?t=17m55s) would do exactly what is requested. You feed 1mA through the light bulb and measure the resulting voltage with a multimeter. You read the resistance directly in millivolts. Better yet, use 10mA and mentally shift the decimal point one digit over.
Use an integral number of PLC on the multimeter to essentially form a 50/60Hz notch filter, which should take care of the nonideal transformer.
Now, whether you consider shifting decimal points and reading millivolts as ohms a mathematical operation, is up to you. ://

Y'all need more Jes err... Pease in your lives!
Now, whether you consider shifting decimal points and reading millivolts as ohms a mathematical operation, is up to you. ://
What else could you call it? :)

I am not totally sure if this would work and I don't think it's better than several ways already listed, but we should be able to leverage the fact that a resistor is linear with regard to frequency.
Get a SMU with incredible resolution or a very fast DMM with incredible resolution (DMM7510) comes to mind.
Combine using a resistive divider the 120V 60Hz AC with some other smaller, higher frequency signal which doesn't land near any harmonics of 60hz (say, a 2Vpp 1830Hz). Record the highest data rate possible while sweeping the AC 120V 60hz signal up and down. Then take the FFT of this situation both with and without the lightbulb (replaced by a short), and subtract the two amplitudes of the 1830 Hz signal of the high frequency signal component (bandpass filtering it if you want to).
Of course you'll need two very high dynamic range DMMs to do this, one to measure the current of a shunt and the other to measure voltage.
But since the filament is perfectly resistive and since it will affect both signals equally, you can calculate resistance this way.

...you can calculate resistance this way.
FAIL!
Read the criteria....
For this challenge you are the average American and you don't know any math; ANY. "math is hard/sucks" and your iPhone doesn't have a calculator app, not that you would know how to use it anyways. So you have to take measurements.

And the question is MEASURE not calculate.
In physics, to MEASURE means, by definition, to compare with the reference unit size.
That being said,
1. You will need to use some sort of a reference resistor (e.g. a rheostat with resistance gradations, or a potentiometer with gradations, or reference resistor box  the correct reference to pick depends of the size of your light bulb).
2. Some sort of comparator device, like a Weatstone bridge https://en.wikipedia.org/wiki/Wheatstone_bridge, or the procedure given by alsetalokin4017 in steps 1...2, or anything else you can come up with by yourself.
3. In the end you read the gradations from your reference resistor.

And the question is MEASURE not calculate.
In physics, to MEASURE means, by definition, to compare with the reference unit size.
You go too far. Just as (I hope) you don't derive many of your daily calculations from first principles  so do we not need to go this far.
The brief is exceptionally clear. To get the resistance by reading it directly off a device.
Within the brief, it is made very clear that the person taking the reading is not going to have any mathematic skills, which means they are going to be even less likely to have any scientific abilities.
We can go in neverending circles with alternative ways to derive the resistance  but that isn't the question.

The question is flawed.

You can't do electronics without math.
...
Not a realistic proposition, doing electronics without math!
As I always like to say. Electronics needs three important skills:
Electronics and Maths.

A multimeter already does the calculations to get a "resistance" reading for a non powered devices (2 wire measurement).
It's not impossible that there is a device (or that one could be made perhaps) that can do the same with powered devices (e.g. a power meter that displays resistance).
Internally, the device will use calculations. But to read the value, you will not need to do any calculations.
I don't know if this is useful or not in practice, but as a theoretical question, I don't see an issue.

You are right ... theoretically, this is not an issue at all. Building a device that can achieve the stated objective has no fundamental obstacles. It just requires the motivation to do so.
There is no real fun in that, which is why most answers have been offered using offtheshelf equipment and sometimes creative setups.
Some challenges so far have been the "zero calculation" requirement and the fact that resistance of a light bulb is not linear over temperature.

OK, @Beamin, answer this for us: How do you "measure the ohms" of ANYTHING without calculations?
(Hint: You cannot "measure the ohms" of ANYTHING. But you can measure current and voltage and calculate the resistance/impedance.)

The measuring device does the calculation.
The "no calculation" constraint was clearly aimed at the operator...
For this challenge you are the average American and you don't know any math; ANY. "math is hard/sucks" and your iPhone doesn't have a calculator app, not that you would know how to use it anyways. So you have to take measurements.
Mind you, the thing that I find fascinating is trying to figure out what a person with such limited capabilities is going to DO with that resistance measurement.

The measuring device does the calculation.
The "no calculation" constraint was clearly aimed at the operator...
For this challenge you are the average American and you don't know any math; ANY. "math is hard/sucks" and your iPhone doesn't have a calculator app, not that you would know how to use it anyways. So you have to take measurements.
Mind you, the thing that I find fascinating is trying to figure out what a person with such limited capabilities is going to DO with that resistance measurement.
Get a friend to work out the current?

Hi,
The resistance of a bulb is not zero with zero voltage. It is finite and non zero.
For example, a 100 watt 120vac bulb is around 10 ohms cold (zero current).
For a 1.5v bulb i have it is around 6 ohms cold (zero current)
When you measure you apply a voltage and measure the current and use Ohm's Law:
R=V/I
however with zero current you cant do that. You really have to use the slope, but you can get a good idea by just using a very very low current like 1ua, but even 100ua will work in most cases and even 1ma in larger wattage bulbs.
So all filament bulbs will have a curve similar to:
R=Ro+R(v)
where Ro is the zero current resistance and R(v) is a function of the applied voltage.
There are equations that approximate the character of filament bulbs and they even can predict life expectancy in the absence of vibration.
The thread was more of an exercise in creative ways to figure things out using equipment rather than math . Its a very simple problem when using math but not so simple just by making observations. So I wasn't so much asking "I have this light bulb in my house how many ohms is it: I have to figure out what bulbs I can substitute and replace it with". More just a fun exercise where I earned a lot of creative options that I wouldn't have thought up on my own and I can use this knowledge in the future when I say "Hey remember in the light bulb thread the guy said measure the temperature? well I think that might work in this scenario. I try not to do linear single problem single solution thinking. This thread was more of a creative writing /a arts project; two thing you wouldn't think would apply to EE but apparently they can be valuable tools! A well rounded education in action solving real world problems.
For instance Verizion FIOS came to my house to install the internet. A horizontal drill head was placed under ground and the drill progress had to b measured. So that the workers (trabaho as they taught me to call it) would walk out to the site and feel around there hands on the ground for vibration. when they were doing that they weren't working the drilling machine or digging holes because the ha to stop what they were doing every five minutes to check the progress of the bore hole. So I came out side with a bunch of small metal 2" rods with little flags on the tops. I set the flags into the round in a grid pattern around the area of the bore holes. Now instead of stopping what they were doing and getting on hands and knees feeling the ground avery few minutes cutting into work time: NOW they could just take a quick peak at which flags were shaking at the top from a distance and see where the drill was without leaving their equipment. Saving tie and saving backs.
The asked if they could keep the lags so I gave it to them.
EDIT add the following letters to my post: AAADNFFKRNESSSS

The measuring device does the calculation.
The "no calculation" constraint was clearly aimed at the operator...
For this challenge you are the average American and you don't know any math; ANY. "math is hard/sucks" and your iPhone doesn't have a calculator app, not that you would know how to use it anyways. So you have to take measurements.
Mind you, the thing that I find fascinating is trying to figure out what a person with such limited capabilities is going to DO with that resistance measurement.
Get a friend to work out the current?
:DD :DD :DD

Hi,
The resistance of a bulb is not zero with zero voltage. It is finite and non zero.
For example, a 100 watt 120vac bulb is around 10 ohms cold (zero current).
For a 1.5v bulb i have it is around 6 ohms cold (zero current)
When you measure you apply a voltage and measure the current and use Ohm's Law:
R=V/I
however with zero current you cant do that. You really have to use the slope, but you can get a good idea by just using a very very low current like 1ua, but even 100ua will work in most cases and even 1ma in larger wattage bulbs.
So all filament bulbs will have a curve similar to:
R=Ro+R(v)
where Ro is the zero current resistance and R(v) is a function of the applied voltage.
There are equations that approximate the character of filament bulbs and they even can predict life expectancy in the absence of vibration.
The thread was more of an exercise in creative ways to figure things out using equipment rather than math . Its a very simple problem when using math but not so simple just by making observations. So I wasn't so much asking "I have this light bulb in my house how many ohms is it: I have to figure out what bulbs I can substitute and replace it with". More just a fun exercise where I earned a lot of creative options that I wouldn't have thought up on my own and I can use this knowledge in the future when I say "Hey remember in the light bulb thread the guy said measure the temperature? well I think that might work in this scenario. I try not to do linear single problem single solution thinking. This thread was more of a creative writing /a arts project; two thing you wouldn't think would apply to EE but apparently they can be valuable tools! A well rounded education in action solving real world problems.
For instance Verizion FIOS came to my house to install the internet. A horizontal drill head was placed under ground and the drill progress had to b measured. So that the workers (trabaho as they taught me to call it) would walk out to the site and feel around there hands on the ground for vibration. when they were doing that they weren't working the drilling machine or digging holes because the ha to stop what they were doing every five minutes to check the progress of the bore hole. So I came out side with a bunch of small metal 2" rods with little flags on the tops. I set the flags into the round in a grid pattern around the area of the bore holes. Now instead of stopping what they were doing and getting on hands and knees feeling the ground avery few minutes cutting into work time: NOW they could just take a quick peak at which flags were shaking at the top from a distance and see where the drill was without leaving their equipment. Saving tie and saving backs.
The asked if they could keep the lags so I gave it to them.
EDIT add the following letters to my post: AAADNFFKRNESSSS
Hi,
Ok then just glue little flags to the top of the light bulb and when you see them start to shake you know the bulb resistance :)
Seriously though sometimes you need math, at least a little. With this project all you have to do is know how to divide. Divide measured voltage by the measured current.
BTW the temperature of most bulbs will be nearly the same. That's because temperature is not a measurement of power even though it takes power to raise the temperature. The temperature of something is dependent not only on power but also on surface area so you would need to know the surface area and then you might be able to estimate the power. For example a thicker filament would require more power than a thinner filament but the two might reach nearly the same temperature because the envelope of the thicker filament bulb might be larger and thus have more surface area. The power, temperature and surface area are related exponentially and we have other things to consider also such as unforced or forced convection cooling. This means this would be a partly physical measurement, and when looking for results to electrical questions (such as Ohms) it is almost always best to stick to electrical measurements and calculations because they are inherently so accurate.

If someone insisted that I make such a device (OP), I would maybe do the following:
Design and make a custom bench power supply, with a maximum voltage output rating to suit the desired maximum working voltage of your bulbs, and give it an adjustable voltage control knob.
Connect the bulb under test, to the power supplies output. Then the program (if it has an MCU) inside the custom bench power supply, can calculate the resistance across its output, and hence calculate and display both the set voltage and resistance in Ohms.
Maybe call it the "LazyOhm" range of power supplies.

there are already devices that do this