resistance of a tungsten filament?

The light emission from a tungsten filament will depend upon its temperature. Electrical power input will be converted into thermal energy that heats the filament. At the same time, the filament gets cooled by conduction, convection (gas filling), radiation, etc. The filament temperature will be determined by the way these power flows balance.

Bill

-- Ferme le Bush

You are right, mostly. They are right, a little. Resistance will change with temperature. However, it is a small change and only over large temperature ranges. I.e. from room temp. to red hot. This is why incandescent lamps will suddenly brighten when you slowly dim up from zero.

Tell your buddies "Wattage is not a measurement of light!!!!" It is a measure of electrical power. It is routinely used to describe lights because when all else is the same, the output of a lamp varies by the power used.

In a simple circuit like an incandescent lamp> Watts = Volts x Amps Given that the wall voltage doesn't change, the only change is the resistance. Simply > E(voltage) = I(current) x R(resistance)

Note that added resistance will reduce the amperage. see

Therefore, the voltage to the lamp will change when the total resistance changes. That will lower the temperature of the filament and change the light output.

The higher the resistance the LOWER the amperage and therefore wattage as well. I.e. Infinite resistance = zero power (air gap means OFF) Zero resistance = maximum power (short circuit, melted copper, tripped safety devices) Your memory of your old reading is way off. ;-)

If you need to prove it for yourself; Hook two identical lamps in series (wall hot->lamp-> neutral to hot-> lamp-> wall neutral.) The lamps now each have half the voltage they normally have. They will be far cooler (orange-ish) and NOT as bright. (Resistance dimmers work this way. Modern dimmers are VERY different and cut the total load.) You will most likely notice that the total output is down also, as the lamps are not working as efficiently as before. (This is the relationship between voltage and output and it is complex!)

I hope that answers you and your buddies. I tried to cover the bases without getting lost in details or following your interconnected questions. If its not clear just ask.

Richard Reid, LC (that means Lighting Certified by the NCQLP) Luminous Views

PS: I have spent years in theatrical work hooking up 250 - 5000W lamps on 14 - 8 gu. cables, the changes are so small as to hardly be noticeable. The resistance of 500' of 12/3 cable with 10A is minimal especially with stranded cable. The connectors are critical however. For more Google "Ohms law" and "voltage drop"

Thanks for your response and input.

I found a great chart showing the relationship between current, power, resistance, lumens and lumens/watt as percentages on this site:

It's figure 2

at infinite resistance the current is zero therefore power is zero

there are way way too many variables to account for to give a simple answer here.

if the purpose of you discussion is: will my lamp be just as bright on less voltage because the consumed power is the same?" i would say 'probably' no because the difference in resistance between a hot filament and a not so hot filament is far less then between a hot filament and a cold filament.

for the purposes of using AV equipment i think if you will be well served if you stick with the concept: more voltage = brighter, less voltage = dimmer.

on high power lighting circuits for example stage lighting when long cables are employed the wire gage is increased to minimize voltage drop.

how much is a quinch?

resistance is seldom used for high power lighting control as the heat dissipation and cost and size make SCRs and TRIACs (and before that thyratrons and ignitrons) a much better solution.

you question is worded ambiguously: does "they" refer to the lamp the dimmer or both?

its a little difficult to respond as i think the questions you intend to ask are hampered by a lack of experience with electricity. let me give you one real world example: i have a test jig which consists of two 500 watt incandescent lamps wired in parallel. this is driven by a 10 amp VARIAC. 2 ammeters are placed in series with the lamps. one is reference the other is a meter to be checked for calibration. when voltage is increased the current goes up and the lamps get brighter. when voltage is decreased the lamps get dimmer and the current goes down. i could graph out a chart showing v vs i but it would only be accurate for this one setup.

It varies. Whether it is significant or not is subjective. Measure it, and decide if it is "significant".

How to measure: put a 1 ohm 25 watt resistor in series with the load. Measure the voltage across the resistor and the voltage across the bulb, and compute the resistance of the bulb. (The voltage across the 1 ohm resistor equals the current through the bulb.) The nominal resistance, without the

1 ohm resistor, can be computed from P = I^2R. Since the bulb is rated 1200 watts at 120 volts, I is 10 amps and R is 12 ohms.

Measure it, as described above.

Ridiculous. Add a million ohms. How much current flows?

Your long cable adds ~1 ohm.

Yes, a cooler bulb has a little lower resistance, but the effect is not linear.

to

"to compensate" implies it was designed purposely to do that, and also implies an equivalent change. Neither is true.

for the added resistance in the cable thereby maintaining

Wrong.

and the added resistance in the cable adds

Wrong.

Short cable draws more power. But the example chosen and the questions surrounding it will only confuse the issue. A bulb with a cold filament has a lower resistance than a bulb with a hot filament. The relationship between temperature and resistance is not linear. Straining at examples that add 1 ohm resistance to a properly functioning

1200 watt, 120 volt bulb circuit and coming up with the conclusion (regardless of which side came up with it) that the bulb maintains 1200 watts regardless of how much resistance is added proves that the example only serves to confuse.

Measure it.

The circuit with the bright bulb uses more power than the circuit with the dimmed bulb. Modern dimmers found in your home act by completely interrupting the current to the bulb for brief moments. A resistance dimmer won't be found in your home. If you had one, it would act by allowing the current to the bulb 100% of the time, but at a reduced level.

Ed

Resistance dimmers were on their way out when I began my theatre career, but I lit shows on Broadway in the early 70s where they, large "piano boards" were still used. To totally dim a light in series with a resistance dimmer it takes five or six times the resistance of light in series with it. I had stage lighting teachers that would say that lighting circuits with resistance dimmers in series with the light used the same amount of power whether the light was at full brightness or if it was dimmed to no light at all.

I've had considerable experience with electricity and have a graduate (three year) engineering degree that included numerous course in electronics and electricity.

But I was trying to convince others in this discussion who were maintaining that no matter how much cable was added in series with a light that the total wattage consumed by the light would alway remain the same. They are maintaining that the light is self regulating - that the current increases as the voltage decreases and the power it uses stays exactly at its rated wattage. I've found the answers I needed since first posting my original message to show otherwise, but they won't budge. They are film electricians and teachers of stage, film and television lighting among others in the business. I'm shocked by their refusal to open their eyes to anything I provided. They simply refuse to see the light.

Wikipedia did say that in an incandescing tungsten light bulb the power to voltage relationship is such that power varies by V to the 1.6 power. This is pretty close to the chart I found, which indicates that the percentage change of resistance of an incandescing tungsten filament is approxiamtely half of the percentage change of the voltage. If the voltage drops 10%, the resistance drops approximately 5%. When you use the formula P= Vsquared over R, you get a considerable drop in power for a 10% voltage drop.

Anyway, thanks again

Thanks Ed,

I'm not in a position to measure it at the moment. Obvioulsy, it's the best way to show everyone what is actually going on. I don't normally work with large wattage lights and my Amprobe has stopped working. I guess I could do some experiments with smaller wattage lights and long runs of cable. Some of those who disagree with me regularly work as film and television gaffers and are in a position to quickly take some measurements, but they are refusing. One gaffer says that he'll measure the voltage at an outlet then plug a very long run of cable into it and measure the voltage at the female end of the cable run. I tell him that it is meaningless to do that without a load.

Anyway, I've given up on them.

Thanks again.

The 1.2kW bulb is only 1.2kW at the stated V and provided it can draw enough I (current). It's resistance varies from about zero ohms when cold and rises more or less linearly (I think) and therefore Ohm's law applies. Thus, increasing the V from 100 to 130, a 30% rise, will cause a 30% rise in it's resistance (assuming it can draw the current it 'wants'). If you add cable you add resistance which may or may not be significant. Calculate the R of the bulb at diffent I values and do the same for the cable then you can compare the ratio of R to see the effect. (eg if R of cable is 1/1000 of total R then effect is minimal but if it is only 1/10 of total R then you have great effect.).

The difference of opinion between those who say a 1.2kW bulb stays

1.2kW regardless and those who say it varies with current is probably due to the two sides focussing on one kind of power to the exclusion of the other kind. There is the stated or rated power, the 1.2kW; that's fixed even when it's still in it's box and sitting in a shop waiting to be sold. But then there's the power output or actual power which will vary according to the current flowing through it (which in turn depends on the voltage across it).

The greater the resistance the lower the current (Ohms law). The lower the current the lower the power. It's easier to grasp the basics of Ohms law using a resistor rather than a bulb because a resistor doesn't change it's value (theoretically) as the current increases but a bulb's does and so you have 'circular' references which become confusing until you understand the order of events, which in this case are: Switch on, zero R, surge of I (current), sudden heat, bulb blows or lights up, bulb R much higher now, current less, meanwhile V has been playing games too possibly, then there is heat in the cable, it's R changes and that affects total R.

The longer the cable the higher the R and therefore the lower the I. The lower the I the lower the P. There is P also being dissipated, used or 'drawn' if you prefer, by the cable (given off as heat). The actual figures involved are simply down to Ohm's law, E = IR and P = IE and from these two: P = I2R. Three simple but useful equations to make a note of.

BUT... if you have 500 ft of cable all coiled up and its AC not DC then the ball game changes to involve AC theory, induction for example

- don't ask!

With dimmers, there are the old sort, variable resistors , which Ohms law will answer your question about. And you have the modern type which are, I think so anyway, mostly thyristor or similar semiconductor devices and these are different in that they don't share the load but turn on and off at each cycle of the mains, 50 or

60 times per second. Thus only a part of each mains cycle is applied to the load and the dimmer itself is more or less no part of that load (unlike the previous type of dimmer). These semiconductor dimmers use negligible power regardless of the load.

That's my understanding but you should verify it yourselves rather than take my word for it. Tip: Get the 'order of events' clear first; split second by split second, what happens to I, P, R, and even E (or V if you prefer) and why. Knowing what affects what and in what order is 'the name of the game' here.

Paul E. Coughlin

When a tungsten filament is giivng off light, does its resistance vary significantly over a voltage range from 100V to 130V?

On our pro video list we've been having a discussion concerning current flow through a simple circuit containing a large wattage incandescent light and different cable lengths.

This is an example of the circuit:

One 1200 Watt tungsten filament (a light bulb) is plugged into a 120 Volt outlet. The cable used is 12/3 copper. If the cable length increases between the filament and the outlet from 2 feet to 502 feet, does the current in the circuit increase or decrease?

Others are maintaining that the 1200 Watt bulb remains 1200 Watts no matter how much resistance is added in series with it to the circuit. I maintain that the added resistance of the long cable increases the overall resistance in the circuit lowering the current and the power.

They are saying that the lightbulb's filament lowers its resistance to compensate for the added resistance in the cable thereby maintaining the lightbulb at 1200 Watts and the added resistance in the cable adds additional power draw on the circuit.

Obviously, the longer cable run adds resistance to the circuit, which along with the filament acts as a voltage divider. The filament is getting less voltage when in series with the long cable. We're back to my original question.

But which circuit draws more power, the one with the short cable or the one with the long cable.

Then there's a related question about resistance dimmers: Do they use the same amount of power when the lightbulb is dimmed out as when the lightbulb is at full brightness? I maintain that the bright bulb uses more power. I think I remember reading tat it takes something ike 5 to

6 times the resistance of the incandescing bulb in series with it to effectively quinch its glow.

Thanks for your input

Incidently, one source claims that 12 AWG wire isn't "rated" to carry

10 Amps for 500 ft. Anyway, a 1000 ft loop of 12 AWG is 1.7 ohms. Using the equations here: come up with a lamp operating point of about 100 VAC (RMS) and 12.5 Amps.

VL = VS - I * R VL = 120 volts - I * 1.7 ohms

100 volts ~= 120 volts - (12.5 Amps * 1.7 ohms)

Mr. M

Ohm's law doesn't apply. Strictly speaking, Ohms law applies to a constant resistance-not a linear one. A light bulb is definitely classed as non-ohmic because the "resistance" does depend on the current. Is it linear?- not likely- that would imply that the current would be constant. I also hope it is not "near 0 resistance" at 0 volts- I hate having breakers trip every time a light is switched on. (Just measured a 100 watt bulb- 12ohms cold, far from 0, and nominal resistance at 120V =144 ohms. Inrush current max of

10A and nominal current of 0.83A) Over a limited range, resistance of the filament varies linearly with temperature but in a light bulb, the temperature and hence the resistance is not a linear function of voltage. This would imply that the current would not change as voltage increases. Another indication is that the life of a bulb in the nominal operating region varies as roughly the 13th power of the voltage- this implies a considerable change in filament temperature and resistance.

By the way, at any voltage at the bulb, the filament draws the "current it wants". The current is determined by the voltage -as is the power.

Don Kelly @shawcross.ca remove the X to answer If you add cable you add resistance which may or may not

The page indicates that the equations that produce the graph are in figure 5 - but I can't find figure 5. What equations did you find and use from the site?

Per the graph, 12.5 amps is clearly wrong. Look at figure 2 on the link you provided. *Everything* on the chart below 100 percent normal voltage is lower - amps, ohms, watts and lumens per watt. Nominal would be 120 volts across the lamp and

10 amps through it. If resistance is added in the feed such that the voltage goes down to 100 volts across the lamp, current *cannot* increase if the graph is correct.

Your last equation:

100 volts ~= 120 volts - (12.5 Amps * 1.7 ohms)

conclusively proves that to get higher wattage (therfore more light) from a lamp, you should reduce the voltage. (At 100 volts and 12.5 amps, that's 1250 watts in the bulb) That should *scream* at you that something is wrong!

Ed

Not true. Ohm's law ALWAYS applies. The issue is that the operating behavior of the lamp has to be specified before the lamp circuit problem can be solved. Normally when using ohms law, 2 of the parameters are known, and the third can be solved for. It's a limitation of algebra, not ohm's law. To solve an algebraic system of equations, you need at least as many equations as unknowns. For example, the supply cable is a good place to start. If you assume it's temperature is constant, then it's resistance is constant. You also know the voltage, hence there is only one unknown, the current, thus only one equation is needed (ohms law) to solve for the current. To calculate the operating point of the lamp, you MUST use an additional equation which models the lamp.

Read this page, it's excellent:

Try using the 5th equation down which relates voltage and current in a lamp. Think of it as ohms law for a lamp. As you can see, it is non-linear, and this is why the electrical behavior of filament lamps is so non intuitive. The exponent is 0.6 (less than one) so more voltage gives less current. A "resistance" gives more current with more voltage. This is why it can be misleading to model a filament as a pure resistance. Nevertheless, ohms law still holds, just be careful how you use it.

MP

>

If he had used Fig 3 (not 5) correctly- I proportional to (V/Vnominal)^0.6 This gives 120 =0.96*(V^0.6) +V resulting in I =9.2A and power to the lamp =954 watts. Lumen output lower than nominal (61%) and resistance of the lamp

95% of nominal. Life of the bulb is about 5 times the nominal life. Hence a common "long life" bulb is rated at 130V and operated on 120V lasts nearly 3 times as long (with less light output) as a 120V bulb at 120V and a bulb rated at 90V and operated at 120V will give a lot of light at a short life in a photographers flood lamp.

Not quite. Since the exponent is positive, more voltage gives more current. Just not a linear amount. For example, using 110% of rated voltage does not give 110% of rated current, it yields 105.9% (1.10^0.6 = 1.0588).

And so does a lamp. Just not a 'one for one' increase.

daestrom

Trying to solve this sort of problem with the chart on sylvania's web page isn't that simple. We can assume the resistance of the supply cable doesn't change (hopefully, it's temperature varies over a narrow range). But since the current drawn by the lamp changes with voltage, and the voltage applied to the lamp changes with current, you need to iterate through a couple of times to get the answer.

If you start with a 1200 watt bulb (rated at 120 volts), then it should draw

10 amps. But if that is at the end of a 1000 ft loop of 12 AWG with 1.7 ohms, that means that 17 volts are dropped in the line, and so now the bulb is operating at 103 volts. According to fig 2 (or the formula in fig 3), such a bulb would only draw...

amps = 10A * (103/120)^0.6 = 9.12 amps.

So, re-working the voltage drop, we find that the voltage at the lamp (if it draws only 9.12 amps) would be 104.5 volts. And one could then figure out the current of the lamp at *that* voltage, and repeat the process again (and again, and again..) until the change in each subsequent pass is small enough to ignore.

Or, you could do some fancy algebra for this particular case, and solve this....

amps/ 10A = ((120-amps*1.7)/120)^0.6

One solution gives us...

120/(10^1.66667)*amps^1.6667 + 1.7*amps = 120 When solved for amps, yields amps=9.1957

So the lamp would be operating at 120-1.7*9.1957 = 104.37 volts, and have a power of 104.37*9.1957= 959.74 watts.

This agrees pretty much with the third formula of fig 3 that would calculate watts=1200*(104.37/120)^1.6=959.83 watts (probably round-off in my calcs).

Since the lamp is operating at 104.37 volts, the lumen output would be (104.37/120)^3.4 = 0.62. So it is only 62% as bright as normal. It's 'color temperature' would also be lower by a factor of 0.943 (e.g. if it was originally rated as a 2800K, it's new operating 'color temperature' would be

2640K).

But it would last longer, by a factor of (120/104.37)^13 = 6.14 times (I think fig 3's sixth formula is inverted, lower voltage should yield *longer* life, not shorter).

daestrom

This paragraph is NOT TRUE. Ohm's law NEVER applies. It is an APPROXIMATION that can be,and often is, very useful.

Most of the really interesting electrical applications involve nonlinear and time varying elements. Vacuum tubes, transistors, varistors etc., do not obey Ohm's law even in approximation. Often, linearization can be used to obtain some approximation to Ohm's law. This is where the techniques of differential geometry com in.

I reveal my age by talking about vacuum tubes even though the same applies to semiconductors. The plate resistance of a triode or pentode was defined by the derivative of plate voltage vs plate current and certainly is not a ohmic quantity.

Even in ideal metallic resistors, Ohm's law is not obeyed. There are thermal noises that cause fluctuations in current even without fluctuation of applied voltage.

Bill

-- Ferme le Bush

V=I*R is, strictly speaking, not "Ohms Law"

See any text or

other net references.

Ohm's Law implies V/I = constant. This is not true for a lamp filament. It is nearly true enough for practical purposes for many conductors over the normal range of currents- as was found by Ohm (and this is what he stated). However, it is not true for a lamp filament.

One can write I=V/R for any situation but it really is of little use to do so when R is a function of voltage or current- the non-linear fact must be faced so why not deal with it directly? That is, using a quantity R is rather pointless- since the non-linearity has to be dealt with - then using V=R*I and R=some function of I is an indirect way of saying V=function of I (which is directly related to what you actually measure).

In addition, it must be remembered that the usual circuit models , series, parallel, loop and node equations, Thevenin, etc... are garbage if R is not constant as they depend on a linear relation between voltage and current (R constant). Fortunately Kirchoff's equations still hold.

Other than this pedantic complaint, you are on target- just roundabout in trying to use Resistance where the concept, except in a hand waving approach, is of little use.

As for the exponent- it is positive- more voltage--> more current. Daestrom has pointed this out.

That's the biggest piece of hogwash trying to rationalize on a techincality that iv'e seen in some time - congrats on that.

Either you don't know what an instant in time is, or you are a calculus flunky who doesn't understand calculus.

Pop

: > Read this page, it's excellent: : >

>

: > Try using the 5th equation down which relates voltage and current in a : > lamp. Think of it as ohms law for a lamp. As you can see, it is : > non-linear, and this is why the electrical behavior of filament lamps : > is so non intuitive. The exponent is 0.6 (less than one) so more : > voltage gives less current. A "resistance" gives more current with : > more voltage. This is why it can be misleading to model a filament as : > a pure resistance. Nevertheless, ohms law still holds, just be careful : > how you use it. : >

: > MP : >

: :

I wouldn't be that harsh, top poster and all...

Certainly R == E/I, so resistance has meaning even with a light bulb. The question in many cases is does R=E/I or does R=dE/dI. Both are useful to talk about.