# Ohm's Law is not really a "Law"

Discussion in 'Electrical Engineering' started by Doug H., May 23, 2007.

1. ### Doug H.Guest

Relax - you already knew this. Conductor resistance is a function of
conductor temperature and conductor temperature is a function of amperage
which is a function of restance.... Interesting circular influences there.
Fortunately, as temperature increases so does resistance. Otherwise we'd
always be battling a current/thermal runaway scenario. So, Ohm's Law is
"close enough" for constant temperatures within limits. But, if you think
that I = E/R is really an immutable "Law", hook up a 12vdc battery to an
incandescent 12v lamp. With a dual trace o-scope, capture the applied
voltage on one channel and the lamp current on the other during the short
period that the fillament is heating up. Ohm's Law implies a linear curve -
but what you capture is quite different. Of course, to get a current input
you can measure the voltage across a very low valued resistor in series with
the lamp which would also heat up and affect the outcome. But, choosing a
very low resistance value helps minimize the resistor's influence.

Yeah - OK - I'm just killing time and fighting off boredom.

Gatorspark likes this.
2. ### The Great AttractorGuest

AT ANY given instantaneous moment in the observation, Ohm's LAW is 100%
true with your bulb example. You have to think in terms of time, and
relevance.

The COLD measured resistance of the filament, and the HOT measured
resistance are two different values, but the reading at each level STILL
yields the EXACT equation results.

How can you not see that FACT?

If the wires feeding a circuit when cold are 0.1 ohm, and when warm are
0.2 ohms, the values of your simulation of that circuit need to show
those parasitic element differences at those different times of operation
and observation.

The laws still carry thorough when your values are corrected for EVERY
moment in time that you are making your observations. AT NO TIME are they
in error. If you have an error in your measurement, YOU are not
including an element of the operation of the circuit you are observing.

It really is THAT simple.

3. ### The Great AttractorGuest

Not at all. What you capture proves the law for every instance along
the way.

4. ### Don KellyGuest

----------------------------
"The Great Attractor"
--------------------------------
Not really- actually you are wrong (and this is a common error). Doug's
statements are closer to the truth.
Ohms law (as Ohm determined from experimental measurements) is actually only
valid in the linear case where E/I is constant independent of either E or
I. He noted a linear relationship between E and I. In other words, R is
constant so that if E doubles, I doubles.
Hence, Ohm's Law, strictly speaking, is not valid for a lightbulb as any
change in current (or applied voltage) changes the resistance. It is also
not valid for a diode or transistor.
Look at the definition of Ohm's Law. (See below) Note the terms "Linear",
"proportional", constant of proportionality.

Ohm's Law is not V=IR EXCEPT when R is constant.

Yes, in a given situation, you can measure V and I and call the result R. If
the voltage changes, I changes and unless the device is linear, R changes.
In that situation, R is actually quite meaningless as a quantity because it
gives no useful information that you don't already have and cannot be used
to predict anything if V changes(there is an infinite set of V-I curves
going through that point -which one is it?)- all you can say from a
measurement at a point is that I =some unknown function of V for which you
have one point. You can't predict , from data at that point, that a change
in voltage will cause a given change in current.

Ohm's tests indicated that within limits, V=a+b*I (equation of a straight
line) becomes V=R*I for many materials. We are lucky that that is so (or can
be treated as such for small variations) because linearity is the basis of
being able to use series and parallel equivalents, Thevenin, etc.

In fact Ohm's Law and our well beloved circuit theorems (except Kirchoff's
Laws) are so much ratshit in non-linear circuits. We tend to forget that (or
never learned it in the first place).

Here is an excerpt from
http://www.physics.uoguelph.ca/tutorials/ohm/Q.ohm.intro.html

"1. Ohm's Law deals with the relationship between voltage and current in an
ideal conductor. This relationship states that:
The potential difference (voltage) across an ideal conductor is proportional
to the current through it.

The constant of proportionality is called the "resistance", R.

Ohm's Law is given by:

V = I R
where V is the potential difference between two points which include a
resistance R. I is the current flowing through the resistance. For
biological work, it is often preferable to use the conductance, g = 1/R; In
this form Ohm's Law is:
I = g V
2. Material that obeys Ohm's Law is called "ohmic" or "linear" because the
potential difference across it varies linearly with the current. "

5. ### The Great AttractorGuest

That is retarded!

The initial current, and ANY instantaneous current at ANY TIME during
ANY instantaneous observation WILL follow the law to the letter!

That is the whole point. OF COURSE it varies as it heats up. That,
however, does NOT negate the law. It is STILL a "pure resistance".

It matters not that it changes with temperature, only that the observer
KNOWS that it does, and KNOWS that that is the reason for any given
instantaneous reading to differ from any other during the period of
observation. The law remains the same. Your knowledge of what takes
place as a flowing current mutates the value of that resistive element is
what you need to change. As it changes, the current does as well. That
FURTHER proves the law, not disproves it.

SHEESH.

6. ### The Great AttractorGuest

It does NOT refer to a diode or transistor. It refers ONLY to purely
resistive elements, and even states such as being a qualifier for it..

7. ### The Great AttractorGuest

The amount of current flowing in a circuit MADE UP OF PURE RESISTANCES
is directly proportional to the electromotive forces impressed on the
circuit, and inversely proportional to the total RESISTANCE of the
circuit.

The term "linear" does not appear in what Ohm wrote, regardless of
whether it is applicable.

At ANY given instance during the observation of the illumination period
of the light bulb, all equations and readings taken STILL match the law,
and there is nothing you can do about it. It doesn't matter that the
resistance changes. ANY readings taken at any point in time during said
change OR even after all settling has occurred, STILL follows the law.

8. ### J. B. WoodGuest

Hello, and you could try and derive Ohm's law from Maxwell's equations if
you're really bored ;-) We did that in an undergrad EE course.
Sincerely,

John Wood (Code 5550) e-mail:
Naval Research Laboratory
4555 Overlook Avenue, SW
Washington, DC 20375-5337

9. ### ehsjrGuest

In a purely resistive circuit, fix one element E,I or R, vary a
second element, the formula predicts the value of the third element.
Seems pedantic and innaccurate to insist it applies only when R
is fixed.

Great Attractor's viewpoint is a whole bunch more practical,
and it is accurate.

Ed

view.

10. ### Don KellyGuest

"The Great Attractor"
And that is where you have hung yourself.
I could apply a voltage to a diode and measure a current. Then according to
what you have said, I could then say R=E/I and this would agree with what
you have emphatically said. was Ohms Law.
Now I have a PURELY RESISTIVE element where R is not independent of E or
I - that is non-linear as is the case for a light bulb (purely resistive
element) or a diode (again purely resistive ). Non-linear purely resistive
elements exist.
Again- where is Ohm's Law useful? Where Ohm said it was- in the case (and
limited range) where E/I is constant independent of either E or I. If your
lightbulb filament could be kept at a constant temperature, it would obey
Ohm's Law as then the resistance would be constant. Otherwise, as you said
R=E/I holds at any given operating point BUT it is not Ohm's Law nor is it
useful information.

In other words, you have now recognised that there is a qualifier- without
recognising that the qualifier is that the element is operating in the range
where it is linear. That is R is constant- independent of E and I- and we
are back to what I (and the reference that I gave you, said). That is the
qualifier that is generally recognised.

Don Kelly
----------------------------

11. ### Don KellyGuest

----------------------------
--------------------

It is pedantic but happens to be important and is correct.
What is the point of saying R=V/I from observed V and I at a single
measurement? Does it give any useful or practical information as you already
know V and I for that situation?
Suppose that I have measured 1A when I applied 10V so R =10ohms- so if I
apply 10V I know that I=1A (which I already know-ho-hum what else is new?
That is what I measured).

If I apply 20V, I simply don't know that R is still 10 ohms and can't
predict I.

Tell me the practicality of that? I can fix V and vary I and calculate an R
but I have to know both V and I to do that -so??? No useful information is
obtained. If, I did this for several points, I could discern a relationship
between V and I (which are actually measured quantities -R isn't) and come
up with some useful relationship such as V=2I or V=3I^5 or V=2/I etc. Ohm
found that within limits, for common conductors of his day, V=(constant)*I
which is a linear relationship. THAT is Ohm's Law. Be thankful that it does
work very well in most situations.

How does one get a known relationship from one data point? Sure you can do
that- for example you can fix R=10 ohms and then say "for V= 10, 20, 30 V ,
the current is 1, 2, 3 A" but is this meaningful if you don't actually know
that R is constant? To do so means that you have assumed that resistance is
constant as an act of faith. In the lightbulb, you have a purely resistive
element and, if you could keep the filament temperature constant, then you
could "fix" R and do some prediction. However, that situation doesn't occur
so you have a purely resistive element which is non-ohmic -that is , it
does't obey Ohm's Law. In this case, "fixing" the resistance gives a model
which doesn't represent the real device.

You and Attractor seem to be confusing "Purely Resistive" with constant R
(or V linearly proportional to I)- that is "ohmic". Tain't necessarily so.
The usefulness of Ohm's Law is when it can be used in the sense that I have
used it- If R is constant one can then determine what happens for
different voltages across the element.
Here's a little problem for you:

Measurements on an unknown purely resistive element give a current of 1A at
10V
The voltage is raised to 20V, what is the current?

Problem 2

Element 1 I=0.001*V^3 (1A at 10V or 10 ohms at this point) Now you have
useful information on the V-I relationship which is not given by saying the
resistance is 10ohms at 10V)
Element 2 R=5ohm constant
Find I for the series combination.
(it isn't 0.67A)
You will have to do a bit of work here- but consider if the circuit was more
complex.

In fact, "The Great Attractor" is absolutely wrong. I know that V=I*R is
too often incorrectly called Ohm's Law but the only reason that it is of
any importance is that resistance materials often have a linear V-I
relationship (within limits). His "education" seems to have missed that.
The reference I gave indicates that.

Gives several references- all of which emphasise direct proportionality -i.e
linearity.

"The Great Attractor" (could he be a re-incarnation of Fats (Bytesass ??)
simply hasn't done his homework. Even Wikipedia has done much better
(but has a non-sensical statement as Ohm's Law doesn't depend on the
definition of the ohm.)

From what I have seen, you can do better.

12. ### ehsjrGuest

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?

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.

V = IR is useful for fixed or variable resistances.
It's used all the time for both pots and fixed resistances.

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.

Ed

13. ### Don KellyGuest

----------------------------
----
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 -
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%.
--------
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.
----------
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.

14. ### JackShephardGuest

The law specifically states that it ONLY applies to such circuits, and
yes, all elements are to be considered. The wires, the contacts, the
"resistor" load itself... all are PART of the circuit, and is all are
"purely resistive" or as you state "ohmic", then they get included.

15. ### Don KellyGuest

----------------------------
--------------
It appears that we agree. My "not really" implies that if one or more
elements are not "ohmic" then Ohm's law cannot be extended to the whole
circuit. I should have been clearer.

Any circuit theorem (except Kirchoff's laws) or combination of components
(eg. series,parallel) depends on linearity -R constant (or ohmic) in V=IR
for all resistive elements.
It is linearity which makes these tools possible (and is the basis of Ohm's
law).

I avoid "purely resistive" as there are purely resistivive elements for
which Ohms law doesn't apply. For example carbon brushes used in DC machines
often exhibit a nearly constant voltage drop over a range of currents. Other
non-linear resistors exist and do not obey Ohm's Law.

Thank you,

16. ### JackShephardGuest

That's BillShit!

One does not EVER put the rated power through a resistor. They are ALL
meant to be utilzed in designs that only tax them at one half to two
thirds of their rating. Any good designer would not only know this basic
tenet, but would design any resistor location to follow it. The main
reason is that any resistor that one does dissipate the rated power in
will be so hot, any nearby component, or PCB surfaces WILL get damaged by
the heat it exhibits.

If you use a resistor rated for HV, then any voltage you put across it
below that should be below the power rating, and should not stress the
resistor so greatly that the resistance value moves appreciably.

I have used SEVERAL sizes of 10MOhm through 10GOhm HV resistors, and
they don't move much under stress, and any movement they do express can
be compensated for with feed forward capacitance, or other methods, but
that is all on a circuit by circuit basis.

17. ### JackShephardGuest

Damn. You got one right! I might be impressed if I wasn't yawning.

It is called "Voltage induced stress", and it will most certainly shift
the "resistance value" due to all the leakage through the resistance
medium that was use to make the resistor.  