----------------------------
ehsjr said:
The only issue I see is the pedantry causes confusion.
Are you saying, beyond that, that the formula
V = IR does not yield accurate results when you
know two of the values and solve for the third?
----
The cause of confusion is not the pedantry but the assumption that V=IR is
Ohm's Law. This is a common mistake and unfortunately is often repeated and
taught as such. V=IR is a relationship that can be written given as data, 2
of the 3 variables. True you can fix one and vary another and the third can
be found. Ohm would have been laughed into obscurity if his "law" simply
said that V and I were related (and you could give his name to the ratio
V/I).
What he found was that the ratio V/I of the materials he tested was
"constant" within a wide range of voltages (for constant temperature as
temperature effects were known). That "Constant R" is the core of Ohm's Law
and is the basis for it being called (or misnamed) a "Law". That is what
makes Ohm's Law useful. Ohm's Law is not that a change in one quantity
affects another but is based on how it affects the other .
My point is that that formula, V=IR, however useful per se, is NOT Ohms Law.
Ohm's Law implies that if the voltage changes, the current change is
directly proportional and the CONSTANT of proportionality is R . Certainly
you can say V=10V and I=1A so R is 10 ohms - at that particular point.
However except where R is independent of V and I it tells nothing of use at
other points- say when 5V is applied. Ohm's Law is the relationship that
allows us to do that for "linear" circuits.
Real world, change any one of the values and you
may change both of the other values. The voltage
may sag or rise when you change the resistance,
or the current that the supply could provide at
10V might not be available at 20 V. The resistance
changes with delta T and T changes with delta I
and I changes with delta V and delta R, by the
formula V = IR.
-----
Real world -
You are considering the source and the circuit as a whole, not the
individual resistive element. Does Ohm's Law apply to the whole circuit? Not
really but can be extended if all resistances are ohmic.
By the way V+deltaV =(I+deltaI)* (R+deltaR) is quite valid so deltaV
=R*delta I +I*delta R +(delta R)*delta I is also valid. Which do you mean?
R=dv/di =constant is often used for small signal analysis of electronic
devices - essentially assuming linearization over a limited range.
Certainly the resistance changes with temperature and temperature depends on
current among other things. In that case knowing R at temperature T does
not give any indication of what R is at T+deltaT. V=IR is true but it
doesn't necessarily give any useful information when there is a change in
any one of the variables. However, Ohm's Law, strictly speaking, does
consider temperature constant. A light bulb, if the temperature of the
filament could be kept constant would have constant resistance and would be
"ohmic". Temperature is a major factor so the bulb is not "ohmic and knowing
V, I at some given voltage isn't of use for predicting what I (and in
consequence R) will be when the voltage is increased by 10%.
V = IR is useful for fixed or variable resistances.
It's used all the time for both pots and fixed resistances.
--------
I am not denying its usefulness. And these are actually ohmic resistances.
For a fixed resistance R is assumed constant. For a pot, R at any given pot
setting is also considered constant.
Back to the point- V=IR is not Ohm's Law unless R is constant.
Note that you most likely would measure the voltage across the bulb and the
current through it and then look for an ohmic resistor which had the same
ratio of V/I. You would not measure the bulb resistance with an ohmmeter
and expect this to be of any use in selecting a resistor on the assumption
that its cold resistance is the same as the resistance when in use in your
circuit.
Answering the question you ask below: say you have a
load that works exactly how you want it to when it
is in series with a 60 watt light bulb. You measure
V and I, determine R of the bulb, and install a resistor
of that value. Useful and practical information.
----------
Even though you have not actually answered the question with an honest " I
can't" (neither can I say any more than that), you have presented a good
point; but all you are in fact doing is determining the voltage and current
of the bulb in that application and then replacing the bulb with an ohmic
resistor which has the same V/I ratio. (I hope you are not trusting an
ohmmeter ). However, as the bulb is not ohmic (v=function of some power
(<>1)of i) and the resistor is ohmic (v=constant*i), the use of the resistor
will not give the same results as the bulb if anything else in the circuit
changes (unless the current is so low that heating is negligable).
Look at the second question that I gave- with 2 elements, one of which is
linear (call it the load) and the other is non-linear (call it the bulb).
Find the voltage and current for this circuit.
Try it. If you get the answer (R =16.8 ohms) then try it again for 20V and
see if the 16.8 ohm value gives useful information for that situation.
Note that in circuit analysis, series and parallel combinations, loop & node
equations, Thevenin, Norton, and the general ability to "easily" solve
complex circuits, all depend on superposition which depends on linearity.
That is:
v=i*R , v=Ldi/dt, i=Cdv/dt
requires that R, L and C are constant (i.e this includes Ohm's Law) or can
be treated as constant over the range of interest in order to apply any of
these theorems. If this is not so -then it is back to Kirchoff's equations
and really messy solutions.
In many non-linear situations, the model is linearized over a relatively
small region where Ohm's law gives a good approximation. Compare small
signal analysis of an electronic device where one can assume constant R (or
g or beta, etc) rather than large signal load line analysis where this
assumption is not of use.