resistance of a tungsten filament?

If it were, we might not have one of the world's largest electronic corporations, Hewlett Packard, now known as HP. The Model 200AB Audio Oscillator was one of their first products. See:

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This RC oscillator uses a lamp in the output amplifier feed-back loop to maintain constant output. I have one of the original HP 200 A B units. The lamp is a GE candelabra based bulb, rated at 3W, 120V. The room temperature resistance is 453 ohms, measured with a high-impedance digital ohmmeter. When measured with a standard moving coil meter it goes rapidly to `900 ohms due to the current flowing through it on the X10 ohms scale. At 120 V the resistance is 1600 ohms to produce 9 W.

Just thought this might be of interest! :-)

Reply to
VWWall
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---------- Since Ohm's Law, as originally expounded and as it is still defined, does not actually involve calculus (except that in terms of calculus, dv/di =constant which is not true for a lamp filament) , and, in particular, does not involve instants in time, what's your beef?

If you disagree with me and the references, cite the points that you disagree on and give a reasoned argument. So far you haven't. Otherwise....

Don Kelly @shawcross.ca remove the X to answer

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Reply to
Don Kelly

A 1.2kW bulb will probably have a resistance of 1/12th that of your 100W tested bulb, about 1.2 ohms which is much closer to zero and so close in fact as to merit my term 'about zero'. And I did say: 'when cold' (bear in mind that ambient temperature is not the same as 'cold' necessarily).

The circuit is purely resistive and ohm's law equations can therefore be applied. My mistake to even mention linearity as it's an irrelevant issue.

Such laws, as Ohm's, can be expressed in two ways, in words or as an equation. With the equation, if you check the relationship between the figures that you supplied, you will find that they obey ohm's law: V = IR At any value of resistance that the filament has at any time, that resistance, the current and voltage will always be exactly in accordance with ohm's law.

Now as for the life of a bulb. The life you refer to is probably not one that takes account of the frequent sudden current surges that take place at switch-on; these are often what kills a bulb when otherwise it's life could go on and on for a much longer time and I think it is that longer time which you referred to.

Paul E. Coughlin

Reply to
PEC

----------- In comparison, the resistance at the normal operating condition is also only

1/12 that of a 100 watt bulb. Shall I make the equally asinine comment that a 100 watt bulb at normal operating voltage will have 12 times the resistance and can be treatred as an open circuit?

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----- In most cases, an ambient temperature, even for an oven light, is essentially cold compared to operating temperature at rated voltage.

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--------------- In the case of a lamp filament- non-linearity is hardly trivial. Justify your reasoning.

--------- V=IR is fine but please look up the definition of "Ohm's Law" and why it became a "Law" V=IR is NOT Ohm's Law. Ohm found that many materials, over the normal operating range and at constant temperature, had R constant- and said so. What advantage does V=RI have over (V=function of I) in the case where "R" depends on I? None- you still have to solve the non-linear relationship.

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--------- At any particular point where a given V and a given I occur, there is an infinite number of V- I relationships which satisfy that relationship. V=RI is not a relationship which, while true at a single point, not particularly useful unless R is independent of the point at which it is measured (i.e. constant as in"Ohm's Law" .

The problem is at other points

That is at V=120V, I =10A gives "R" =12 ohms. but at 130V, I will not be

130/12 =10.83 A and "R" will not be 12 ohms. What use is V=RI if R changes when V changes? Here is an example: 1)A given device has the characteristic V=0.2*I^2 +0.5*I This implies R=0.2I+4 Now suppose that I =10A which leads to R=25 ohms and V=250Volts All that this tells us is that that 250V, the current will be 10A 2)If the resistance was R=25 constant, we get the same result. 3)If R =0.02*I^3 +5 the result will be the same.

Now consider what happens when the voltage is raised to 260V.

1)260 =(0.2I +0.5)I =0.2I^2 +0.5*I (the original relationship) give I =34.8A for an "R" of 7.5 ohms (obviously not a lamp filament) 2) constant R gives 260=25I give I =10.4 A 3) I =6.3A and "R" =41 ohms

Three cases in which: at V=250 volts, I= 10A and "R"=25 ohms. and 3 cases where this information does not determine what happens at another voltage- that is where "R" is numerically useless although conceptionally useful. This example is extreme but the "numerically useless, conceptually useful" concept of resistance exists for many devices including lamps and many electronic devices.

Small signal models for transistors etc, linearise in a range about an operating point and are reasonable within this range. They use R=dv/di which is the slope of the v-i curve at the operating point. This allows the use of such things (within limits) as loop equations, Thevenin, etc which depend on linearity (V/I =constant) but are useless in the case of V/I not constant.

-----------. > Now as for the life of a bulb. The life you refer to is probably not one that takes account of the

---------- I can agree with you on that.

Reply to
Don Kelly

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