## resistance in series

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togabailey
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### resistance in series

I have three light bulbs in series with 120 volts at the source, two bulbs are 60 watt with 10 volts acrossed each of them and the other is 10 watt with 100 volts acrossed it.
Im having problems finding resistance using the
V-squared / p. Because If I use, It=P/E nothing seems to come out right, what is my dead brain doing?

Joseph
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### Re: resistance in series

I would just use P=V^2/R using V=120V. It gives an answer of 1440 ohms. At 100V, the resistance will be a little less. Those power ratings for the bulbs are only accurate for 120V being across them.

Ron H
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### Re: resistance in series

You can't calculate the total resistance in the circuit unless you can measure the current. That's because the resistance of a light bulb is much higher when hot than when cold. Since none of the bulbs are running at their rated voltage (and wattage), they will be relatively cool - especially the ones rated at 60 watts.<p>If light bulbs were constant resistance, the 60 watt bulbs would be R=V^2/P, or R=240 ohms.
The 10 watt bulb would be 1440 ohms. Therefore, the total resistance would be 1440+240+240=1920 ohms. Since they are highly temperature sensitive, the resistance will be lower.<p>Ron

Dimbulb
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### Re: resistance in series

Possibly the resistance of the filament changes as it warms up and at that temperature the curve is close to your formula.<p>If you use a big nichrome wirewound resistor with just a few milliamps Ohms law fairly linear. Light bulbs have often been used as illustrations but in reality they behave on a curve. <p>A thermistor changes resistance as temperature rises and can be graphed. To a lesser extent an average 1 % metal film resistor can be graphed. On a critical part of a circuit "accuracy is a temperature thing". <p>If you need an ultra precision resistor you will not find an online store as they prefer not to sell to hobbiest. Your order goes into the circular file and a catalog jockey will get back with you someday.<p>[ March 14, 2003: Message edited by: 1206DX ]</p>

Will
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### Re: resistance in series

All tungsten lamps possess negative temerature co-efficients i.e as they get hotter their resistsnce increases - just like a thermistor. So, you apply 110 volts which causes the lamp filament to heat up so that the resistance eventually balances with the voltage at some level which consumes approximately the lamp rated power.
As an example, when I saw your letter I tested the cold resistance of 4 lamps and they were as below<p>Watts Measured Power of 110 Resistance
Rating Resistance volts in this for this
resistance rated power
------ --------- ------------- -----------
60 16.5 733 202
75 13.1 924 161
100 12.5 968 121
100 17.5 691 121<p> The initial resistance means little since materials of manufacture are different from one manufacturer to another. One of the 100 watt bulbs was suppoed to be special 'Soft Lite' or something like that which means little to me
Will
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Ron H
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### Re: resistance in series

Will, you said "All tungsten lamps possess negative temerature co-efficients ".<p>I think you meant to say positive instead of negative.<p>Ron

togabailey
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### Re: resistance in series

Thanks for your responses, it helped alot, but when I put the numbers on paper "theory" I get conflicting answers.
Im trying to build a lab using the the different formulas.
what do you think IM screwing up?

Will
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### Re: resistance in series

Ronh - You are, of course correct, it should have said Positive Temperature Co-Efficient I'm clearly too far gone to be trying to think while I write. As the famed Scottish Bard Robert Burns once was reputed to have said " Is it wise to remain silent and be thought a fool - Or to speak up and dispel all doubt "
Unfortunately the small table of results - which looked perfectly OK on my computer screen - got some how messed in the transmission. It is relatively easy to see which numbers were in which columns but the column headings got so screwed up as to be unreadable.
1) Lamp rated Power (Watts)
2) Lamp Cold Resistance as measured
3) Power which 110 volts AC would dissipate
in this measured resistance
4) Hot resistance of lamp required to
dissipate lamp rated power at 110 volts<p>For the benefit of the original poster - the results were obtained as follows (For the first row) 3rd column - P (Watts = V^2/R = 110*110/16.5 = 733 watts. For the second column the resistance calculated to get the (rated Power) resiatance is V^2/P = 110*110/60 = 202 watts.
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russlk
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### Re: resistance in series

As Ron H said, you have to measure the current, and use Ohm's law.

Dimbulb
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### Re: resistance in series

I guess, and having limited info that the heat measurements method needs a mod on the filament only. Possibly reverse engineering to get in the ballpark with the 3/2 coefficient in the temperature resistance curve. I think this involves a controlled light to heat conversion.

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