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:
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! :-)
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?
In most cases, an ambient temperature, even for an oven light, is
essentially cold compared to operating temperature at rated voltage.
In the case of a lamp filament- non-linearity is hardly trivial. Justify
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
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.
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 V0V, I A gives "R" ohms. but at 130V, I will not be
130/12 .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 A which leads to R% ohms and V%0Volts
All that this tells us is that that 250V, the current will be 10A
2)If the resistance was R% 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
4.8A for an "R" of 7.5 ohms (obviously not a lamp filament)
2) constant R gives 260%I give I .4 A
3) I =6.3A and "R" A ohms
Three cases in which: at V%0 volts, I= 10A and "R"% 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
Small signal models for transistors etc, linearise in a range about an
operating point and are reasonable <approximations> 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
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