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Old 01-21-2003, 04:07 PM   #1
avk
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Default Ohm testing of light circuits

Got a set of fog lamps and decided to bench test the components. The resistance of both bulbs measured at about 0.5 Ohm, although a bulb rated at 55 W should have about 2.6 Ohm. So this is how much the resistance changes as the filament heats up. Might be worth noticing. In fact a while ago, someone posted about troubleshooting a fog light circuit, believing there was a short somewhere because the resistance between the power and ground wires was 0.5 Ohm.
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Old 01-21-2003, 06:38 PM   #2
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what formula are you using to calculate the ohms.

R = E^2 / W <----E is sq. then divide by W

R=.436 or .50 wich is close enough.

you cant ohm test a ckt that is powered, you must first isolate the ckt from power. and ground, if you disconnect from power and gound, and measure from power wire to the the end of the ground wire, the only resistance you should have is that of the light bulb. 0.50, unless you have a corroded wire, wich would creat a high resistance in the ckt, wich you do not have.

so sounds good to me.

Jay

Last edited by munkis; 01-21-2003 at 09:55 PM.
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Old 01-21-2003, 09:53 PM   #3
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Jay,
I used R=(V^2)/P because P=(V^2)/R: V is voltage, R resistance and P power (Watts).
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Old 01-21-2003, 10:01 PM   #4
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Resistance = Voltage squared 24 divded by Watts, 55...

comes up with .50, unless my math is wrong and it could be.

and your ohm test of each ckt seems right also, you should only have the resistance of the bulb wich we have established is .50 ohms.

Jay

Last edited by munkis; 01-21-2003 at 10:07 PM.
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Old 01-21-2003, 10:29 PM   #5
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Jay,
Sorry, somehow at first I only read the question in your post and not the rest of it. Voltage squared is 144, so 144/55=2.62
Ohms. I only tested the individual parts using a meter, so it wasn't a powered circuit. Hope this makes it more clear.
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Old 01-21-2003, 10:46 PM   #6
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ok...
P=55watts
E=12volts
R=?

R=(E^2)/P... 12x12=144, 144/55 =2.6181ohms ??ok??
check answer...
P=(E^2)/R... 12x12=144, 144/2.6181 = 55.001 watts...

looks good on paper...

The main thing I can think of is that a heated element has a higher restance...
The hotter something usually gets the worse it conducts...
I don't have the equipment to test ohms under load...
usually if an element is bad or shorted it will have a short life.
The element should burn out, but I have seen where a fuse will smoke in the process.
That is the safety of OCP over current protection or a "fuse"
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Old 01-21-2003, 10:58 PM   #7
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Quote:
Originally posted by McDade
Resistance = Voltage squared 24 divded by Watts, 55...
Jay
yup, your math is a little off

It seems if I rember that the low ohms reading is probibly right...
Checking light bulbs isn't like checking plain old resistors...
Plain ole resistors don't see 1000*+ tempatures
I have never checked lamps for an absolute reading, only that there was any reading at all.
Gabe
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Old 01-21-2003, 11:13 PM   #8
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It sure is known how the resistance changes with temperature, I just didn't realize it shows in this particular way.
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Old 01-21-2003, 11:26 PM   #9
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I'm not saying it is absolute...
I'm only speculating.

"dam* it jim, i'm a electrican not a scientist"
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Old 01-22-2003, 07:29 AM   #10
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haha ooops, was never good at math, thanks

now I know, and knowing is half the battle.

Jay
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Old 01-22-2003, 07:39 AM   #11
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You can do a power on test to determine R using ohm's law. If you power the circiut(sp) with a multimeter in series measuring current then use R=V/I. Where R=resistance V=volage and I=current. R=12/reading from current meter. That should give the resistance of the bulb when powered. If you are interrested.
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Old 01-22-2003, 08:12 AM   #12
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To get an accurate value, one'd need to do what you say (and measure the voltage directly). To see the difference between "cold" and "hot" resistances, the calculated "hot" figure would be enough because the change is about 5-fold.

Last edited by avk; 01-22-2003 at 09:20 AM.
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Old 01-22-2003, 03:34 PM   #13
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at this point, you guys are going to need the derating curve from the manufacturer of the bulb......resistance will increase(exponentially, depending on device) w/heat.......

not sure why u'd wanna find the resistance of the bulb, a straight continuity check should be fine to see if the bulb is still functioning.

also, if u assume the car was on, then the potential would be roughly 14v, instead of 12v....so that 55w rating is a bit mis-leading as it does not tell u the condition in which it was measured(i'm sure there is a standard)

also, i think jay got 24v cuz he did this: 12*2=24v, as oppose to 12^2=144v(i make these algebraic mistakes all the time)

so y r u trying to find the static resistance of the bulb?
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Old 01-22-2003, 04:12 PM   #14
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Quote:
Originally posted by newbie sewbie
<...>not sure why u'd wanna find the resistance of the bulb, a straight continuity check should be fine to see if the bulb is still functioning<...>.

True. My comment was about what you see if you do measure it.

<...>
so y r u trying to find the static resistance of the bulb?
You probably mean that since the bulb resistance depends on the temperature, it varies with voltage. That's also true.
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Old 01-29-2003, 01:48 PM   #15
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it's not valid to measure resistance while a current is going through a resistor...also you don't know if the bulb is only a resistor, it may be treated like one, but those who've designed and built them really only know. So measuring resistance (usually referred to as input impedance) of an unknown unit shouldn't be done by measuring the resistance of the unit with an ohm-meter.

do you guys only have voltmeters? Does anyone have multimeters? Try reading the voltage and current drawn by the bulb when it's on.

Measure the voltage by:

R = resistance(ohms), V = voltage(volts), I = current(amps), P = power( watts)

R = V/I
and P = I * V
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Old 01-29-2003, 03:27 PM   #16
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cRayZee: It's a DC circuit, so it's only the bulb's resistance which matters, although it should have some inductance if that's what you imply. As already mentioned, the resistances were measured not under any power, other than suppplied by the meter which calculates the resistance pretty much the way you describe. The filament heating was negligible, but if the ohmmeter supplied 12V to the load, it'd read the "hot" value.

Last edited by avk; 01-29-2003 at 03:41 PM.
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Old 01-29-2003, 04:32 PM   #17
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But if you don't trust what it says on the specs, then you wouldn't want to use the 50W assumption. Then you make your own measurements by measuring the voltage and current under power and get the actual wattage and resistance values of your bulb.
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Old 01-30-2003, 11:03 AM   #18
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The low initial resistance of a tungsten filament bulb is well known. This is what produces the inrush current (as much as 5-10x the nominal current) that occurs when you first turn on the bulb. The resistance of the filament increases very rapidly (about 0.1 seconds) as it comes up to temperature.

You'll have to measure the current into the bulb and voltage across the bulb if you want to know the running resistance of the bulb.
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Old 01-30-2003, 11:34 AM   #19
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Yes it's well known and using nominal specs would be enough to point a several-fold increase.
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