How would you measure the ohms of a light bulb?

[alsetalokin4017]

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, heavy-duty 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.       )

[Kalvin]

^ ^ That was nice one. 

[Gregg]

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

[StillTrying]

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/20w-halogen-bulb-viewed-by-a-photodiode/

[CD4007UB]

Our first-year students use an approach similar to IanB above, measuring the I-V 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.

[Shock]

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.

[MrAl]

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.

[brucehoult]

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.

[Mr. Scram]

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

[innkeeper]

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!

[Seekonk]

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.

[Belrmar]

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

[rstofer]

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

[innkeeper]

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?

[Mr. Scram]

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.

[Damianos]

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!!!

[Dave]

Y'all need more Jes- err... Pease in your lives!

A Seeley box 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 non-ideal transformer.

Now, whether you consider shifting decimal points and reading millivolts as ohms a mathematical operation, is up to you.

[Mjolinor]

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?

[technogeeky]

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 (band-pass 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.

[Brumby]

...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.

[RoGeorge]

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.

[Brumby]

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.

[Mjolinor]

The question is flawed.

[MK14]

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.

[kalel]

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.