Does anyone know why Ohm's law is writtem V=IR (E=IR)
instead of I=V/R? Of course I know they're equivalent.
I just wondered why - if I is the dependent variable and V
and R are the independent variables - it's basic form
isn't I=V/R.
But I is only the dependant variable in *one* way of looking at it. V=IR is
the appropriate way of writing it if you consider ohm's law to mean " V =
the voltage drop across resistance R with current I passing through it". You
can come up with a similar, equally valid phrase, for the third way of
writing it out (r=v/i).
I was tought to remember ohm's equation in the same way as the
distance=speed.time thing:- picture a triangle like this:-
D
-----
S | T
If you cover up the value you want to find with a finger, the layout of the
other 2 gives you the form of the equation.
In addition to Jon's comments, I would like to add that in the classic
form of the equation, ie E = I x R that the 'E' represents "Electromotive
Force", 'I' represents "Intensity" of the current flowing, and of course
'R' represents "resistance". Just an FYI.
Regards,
Jim
> Does anyone know why Ohm's law is writtem V=IR (E=IR)
> instead of I=V/R? Of course I know they're equivalent.
> I just wondered why - if I is the dependent variable and V
> and R are the independent variables - it's basic form
> isn't I=V/R.
>
> --
> http://www.piclist.com hint: To leave the PICList
> .....piclist-unsubscribe-requestKILLspam@spam@mitvma.mit.edu
> > Does anyone know why Ohm's law is writtem V=IR (E=IR)
> > instead of I=V/R? Of course I know they're equivalent.
> > I just wondered why - if I is the dependent variable and V
> > and R are the independent variables - it's basic form
> > isn't I=V/R.
I was always taught R = V/I.
Ohms law is the *definition* of resistance. Unlike many equations which
are approximations, R always equals V/I.
Now, that doesn't mean that R has to be a constant. Semiconductors
obviously have a non-linear V/I curve. Most resistors are made up
of substances which have a highly linear V/I curve. However, they can
be temperature dependent (usually only slightly for good resistors).
Even then, R is found by measuring V and I and then running it through
the equation.
> Rick,
>
> In addition to Jon's comments, I would like to add that in the classic
> form of the equation, ie E = I x R that the 'E' represents "Electromotive
> Force", 'I' represents "Intensity" of the current flowing, and of course
> 'R' represents "resistance". Just an FYI.
I was wondering why I and E were used for current and voltage. This makes
sense now.
Does anyone know why Ohm's law is writtem V=IR (E=IR) instead of I=V/R?
Of course I know they're equivalent.
I just wondered why - if I is the dependent variable and V and R are the
independent variables - it's basic form isn't I=V/R.
> But I is only the dependant variable in *one* way of looking at it. V=IR
is
> the appropriate way of writing it if you consider ohm's law to mean " V =
> the voltage drop across resistance R with current I passing through it".
You {Quote hidden}
> can come up with a similar, equally valid phrase, for the third way of
> writing it out (r=v/i).
>
> I was tought to remember ohm's equation in the same way as the
> distance=speed.time thing:- picture a triangle like this:-
>
>
> D
> -----
> S | T
>
> If you cover up the value you want to find with a finger, the layout of
> If you cover up the value you want to find with a finger, the layout of
the
> other 2 gives you the form of the equation.
>
>
> V
> -------
> I | R
>
> Ie. Cover up I, and you see V over R.
>
> Jon
Yeah they showed me that in school too. Then there's the 12 equations in a
circle around it. But the one that's always stuck in my mind was the
mnemonic "an 'I'ndian saw an 'E'agle over a 'R'iver".
E
I= ---
R
Mathematically, no form is more right then another, it's just a matter of
what you have and what you need. As long as you can remember one of them
and a little algebra you're good to go.
-Denny
> Does anyone know why Ohm's law is writtem V=IR (E=IR)
> instead of I=V/R? Of course I know they're equivalent.
> I just wondered why - if I is the dependent variable and V
> and R are the independent variables - it's basic form
> isn't I=V/R.
Hehe, good questions, perhaps because people are more "scared" of division
then they are of multiplication? Perhaps it's efficiency: V=IR requires one
less character then I=V/R. :) TTYL
> But I is only the dependant variable in *one* way of
> looking at it. V=IR is the appropriate way of writing it
> if you consider ohm's law to mean " V = the voltage drop
> across resistance R with current I passing through it".
> You can come up with a similar, equally valid phrase,
> for the third way of writing it out (r=v/i).
But aren't R and V the only two of the three variables
that you can manipulate directly? I depends on these,
right? I think writing it I=V/R would be less confusing
to the novice. With the V=IR form, I've seen
descriptions like "if the current doubles the voltage doubles." Mathematically, yes, but it's not like you can
double the current independently of V and R.
----- Original Message -----
From: "Rick Regan" <KILLspamdrrdrspamBeGoneCHARTER.NET>
> But aren't R and V the only two of the three variables
> that you can manipulate directly?
No, not really. There is a current source in the EE lab that maintains a
set current regardless of voltage or resistance (within reason). I think
ordinary transistors do this to some extent also, as they are current
amplifying devices. It just so happens that most people are more acquainted
with measurements in voltage due to the use of batteries, wall warts, etc.
A current source makes I an independent variable. Also there are some
applications where R changes with I, which makes R the dependent variable!
So in terms of which is the "proper" way to write it, I'd say it really
doesn't matter. It's probably just easiest to avoid the fraction, so most
people write E=IR.
I think writing it I=V/R would be less confusing
> to the novice. With the V=IR form, I've seen
> descriptions like "if the current doubles the voltage doubles."
Mathematically, yes, but it's not like you can
> double the current independently of V and R.
>
> --
> http://www.piclist.com hint: To leave the PICList
> EraseMEpiclist-unsubscribe-requestEraseMEmitvma.mit.edu
> > But aren't R and V the only two of the three variables
> > that you can manipulate directly?
>
> No, not really. There is a current source in the EE lab
> that maintains a set current regardless of voltage or
> resistance (within reason). I think ordinary
> transistors do this to some extent also, as they are
> current amplifying devices.
I think I'm getting in over my head here (I'm not an EE).
Could you elaborate on how you can produce current, and
how you can do so keeping R and V constant? So in other
words, are you saying Ohm's law does not apply in certain
cases?
It's not so much Ohm's law does not apply, as you have characteristics
based on the device(s) you are using. Take for example a FET
transistor. Your drain current (when the device is in sat.) is iD = 1/2
K'n (W/L)(Vgs-Vt)^2 assuming no early voltage/effect. Now, as you can
see, if you apply a constant Vgs (gate to source voltage) you can change
the amount of current the device will pass by altering the Width and
Length of the channels in the device. These are all IC characteristics,
but this is how you produce a resistor from a transistor.
So the long and short of it is, Ohm's law is correct, but it really only
applies in specific instances.
If you really want to fold your head inside out, then imagine a wire
becoming a resistor, capacitor and inductor all at the same time. This
is what happens when you run an extremely high frequency signal through
it. So the value of the resistance of a wire changes with frequency.
But if you look in to the physics of it, you'll also see it can change
based on cross sectional area.
So as you can see, it's easiest and typically most useful to write Ohm's
law as V=IR (usually interested in voltage, and it's easier to write).
However, it really is quite simplistic and when it comes to computing
currents, etc. things become a tad convoluted when we move away from
your typical (low f) discrete based design.
I hope this answers some of your questions.
I encourage you to look in to:
BJTs (bipolar junction transistors, typical trans..)
FETs (Field Effect Transistors)
Transmission lines
Anything regarding frequency response in a circuit
All those things should keep you busy for at least a month.
>>>But aren't R and V the only two of the three variables
>>>that you can manipulate directly?
>>
>>No, not really. There is a current source in the EE lab
>>that maintains a set current regardless of voltage or
>>resistance (within reason). I think ordinary
>>transistors do this to some extent also, as they are
>>current amplifying devices.
>
>
> I think I'm getting in over my head here (I'm not an EE).
> Could you elaborate on how you can produce current, and
> how you can do so keeping R and V constant? So in other
> words, are you saying Ohm's law does not apply in certain
> cases?
>
> --
> http://www.piclist.com hint: To leave the PICList
> TakeThisOuTpiclist-unsubscribe-request.....TakeThisOuTmitvma.mit.edu
> > Ohms law is the *definition* of resistance. Unlike many
> > equations which are approximations, R always equals V/I.
>
> So then why isn't R=V/I the common form?
Because if you assume resistance is a constant (which is true for things
generally called "resistors" then the other forms of the statement
make sense.
It's not so much Ohm's law does not apply, as you have characteristics
based on the device(s) you are using. Take for example a FET
transistor. Your drain current (when the device is in sat.) is iD = 1/2
K'n (W/L)(Vgs-Vt)2 assuming no early voltage/effect. Now, as you can
see, if you apply a constant Vgs (gate to source voltage) you can change
the amount of current the device will pass by altering the Width and
Length of the channels in the device. These are all IC characteristics,
but this is how you produce a resistor from a transistor.
So the long and short of it is, Ohm's law is correct, but it really only
applies in specific instances.
If you really want to fold your head inside out, then imagine a wire
becoming a resistor, capacitor and inductor all at the same time. This
is what happens when you run an extremely high frequency signal through
it. So the value of the resistance of a wire changes with frequency.
But if you look in to the physics of it, you'll also see it can change
based on cross sectional area.
So as you can see, it's easiest and typically most useful to write Ohm's
law as V=IR (usually interested in voltage, and it's easier to write).
However, it really is quite simplistic and when it comes to computing
currents, etc. things become a tad convoluted when we move away from
your typical (low f) discrete based design.
I hope this answers some of your questions.
I encourage you to look in to:
BJTs (bipolar junction transistors, typical trans..)
FETs (Field Effect Transistors)
Transmission lines
Anything regarding frequency response in a circuit
All those things should keep you busy for at least a month.
On Thursday, Apr 15, 2004, at 07:38 US/Pacific, D. Jay Newman wrote:
>> 'E' represents "Electromotive Force", 'I' represents "Intensity" of
>> the current flowing, and of course 'R' represents "resistance".
>
> I was wondering why I and E were used for current and voltage. This
> makes sense now.
>
Hmm. Was the original language actually english? Ohm was german...
Am Freitag, 16. April 2004 01:37 schrieben Sie: {Quote hidden}
> On Thursday, Apr 15, 2004, at 07:38 US/Pacific, D. Jay Newman wrote:
> >> 'E' represents "Electromotive Force", 'I' represents "Intensity" of
> >> the current flowing, and of course 'R' represents "resistance".
> >
> > I was wondering why I and E were used for current and voltage. This
> > makes sense now.
>
> Hmm. Was the original language actually english? Ohm was german...
>
> BillW
>
> --
> http://www.piclist.com hint: To leave the PICList
> RemoveMEpiclist-unsubscribe-requestEraseMEspam_OUTmitvma.mit.edu
On Thursday, Apr 15, 2004, at 14:26 US/Pacific, Rick Regan wrote:
> Could you elaborate on how you can produce current, and
> how you can do so keeping R and V constant?
Don't forget that at the time Ohm's law was derived, I don't think that
"real" constant voltage sources were any more common than "real"
constant current sources. Our thinking is a bit skewed by the fact
that voltage regulators are so ubiquitous NOW.
Originally, you'd probably think of I and V as measurable quantities,
and R as the way to relate the measurements. NOT varying V to get a
different current through a resistor...
It seems natural to you to deal with voltage as your base since that's
how electronics is taught. Today's electronics revolve around constant
voltage sources, a/d measurements are always voltage level measurements,
etc.
However, there's a whole world of electronics where current is used as
the 'base' and voltage is not emphasized. There are some applications
where current is easier to use, such as the common current loop
industrial controls, rs-422 (IIRC, a current loop serial port), etc.
It's not difficult to take a simple constant voltage circuit and convert
it to a constant current circuit.
In fact, current is one of the reasons so many people shie away from
analog electronics. Some circuits are /exceptionally/ difficult to
analyze from a voltage point of view, but are practically trivial from a
current point of view. And thanks to this dandy equation it's easy to
find the voltages once you've solved the circuit for current.
Transisters and op-amps both work based on currents.
>>But I is only the dependant variable in *one* way of
>>looking at it. V=IR is the appropriate way of writing it
>>if you consider ohm's law to mean " V = the voltage drop
>>across resistance R with current I passing through it".
>>You can come up with a similar, equally valid phrase,
>>for the third way of writing it out (r=v/i).
>>
>>
>
>But aren't R and V the only two of the three variables
>that you can manipulate directly? I depends on these,
>right? I think writing it I=V/R would be less confusing
>to the novice. With the V=IR form, I've seen
>descriptions like "if the current doubles the voltage doubles." Mathematically, yes, but it's not like you can
>double the current independently of V and R.
>
>--
>http://www.piclist.com hint: To leave the PICList
>@spam@piclist-unsubscribe-requestspam_OUT.....mitvma.mit.edu
>
>
>
>
>
> If you really want to fold your head inside out, then imagine a wire
> becoming a resistor, capacitor and inductor all at the same time.
Not mine. I always insist on Ideal brand Conductors. None of this
messy 'real-world' stuff for me, no sir. When I want a capacitor, I'll
add one myself, thankyouverymuch! :-)
Am Freitag, 16. April 2004 04:09 schrieben Sie:
> Shawn Wilton wrote:
> > If you really want to fold your head inside out, then imagine a wire
> > becoming a resistor, capacitor and inductor all at the same time.
>
> Not mine. I always insist on Ideal brand Conductors. None of this
> messy 'real-world' stuff for me, no sir. When I want a capacitor, I'll
> add one myself, thankyouverymuch! :-)
>
> -Adam
Uh damn, that would be a huge formula to calculate... Quiet ugly...
You'll have a very complex calculation for the capacitance with multiple
E-field calculations.
The inductance will be a little easier, as it can be solved by a single
(complex) integral.
The resistance is the easiest thing, (length/diameter) * conductivity
I could post all the formulas needed, but surely not as pure ASCII...
> It seems natural to you to deal with voltage as your base since that's
> how electronics is taught.
Just to help add some confusion: Even ignoring anything except resistance,
"pure" resistors don't exist - all have SOME dependency on current and
voltage. Fortunately, the affects are small enough to be ignored for all but
the most demanding applications.
Am Freitag, 16. April 2004 04:09 schrieben Sie:
> Shawn Wilton wrote:
> > If you really want to fold your head inside out, then imagine a wire
> > becoming a resistor, capacitor and inductor all at the same time.
>
> Not mine. I always insist on Ideal brand Conductors. None of this
> messy 'real-world' stuff for me, no sir. When I want a capacitor, I'll
> add one myself, thankyouverymuch! :-)
>
> -Adam
Uh damn, that would be a huge formula to calculate... Quiet ugly...
You'll have a very complex calculation for the capacitance with multiple
E-field calculations.
The inductance will be a little easier, as it can be solved by a single
(complex) integral.
The resistance is the easiest thing, (length/diameter) * conductivity
I could post all the formulas needed, but surely not as pure ASCII...
Voltage and pressure are by convention the defining dependent variables,
i.e., V=IR, P=hd. Voltage is by analogy an electrical pressure that depends
upon the product of two independent variables, I and R, Pressure (P) is the
dependent variable that depends upon the product of the two independent
variables height (h) and density (d). Of course we can algebraically
manipulate the variables to solve for anyone of them. However, Occam's
razor suggests the simplicity of direct proportional relationship as opposed
to directly as one and inversely as the other. I hope this helps.
> But I is only the dependant variable in *one* way of looking at it. V=IR
is
> the appropriate way of writing it if you consider ohm's law to mean " V =
> the voltage drop across resistance R with current I passing through it".
You {Quote hidden}
> can come up with a similar, equally valid phrase, for the third way of
> writing it out (r=v/i).
>
> I was tought to remember ohm's equation in the same way as the
> distance=speed.time thing:- picture a triangle like this:-
>
>
> D
> -----
> S | T
>
> If you cover up the value you want to find with a finger, the layout of
> ----- Original Message -----
> From: "Rick Regan" <drrdr@spam@CHARTER.NET>
> > But aren't R and V the only two of the three variables
> > that you can manipulate directly?
>
> No, not really. There is a current source in the EE lab that maintains a
> set current regardless of voltage or resistance (within reason). I think
> ordinary transistors do this to some extent also, as they are current
> amplifying devices. It just so happens that most people are more
acquainted {Quote hidden}
> with measurements in voltage due to the use of batteries, wall warts, etc.
> A current source makes I an independent variable. Also there are some
> applications where R changes with I, which makes R the dependent variable!
>
> So in terms of which is the "proper" way to write it, I'd say it really
> doesn't matter. It's probably just easiest to avoid the fraction, so most
> people write E=IR.
>
>
>
>
> I think writing it I=V/R would be less confusing
> > to the novice. With the V=IR form, I've seen
> > descriptions like "if the current doubles the voltage doubles."
> Mathematically, yes, but it's not like you can
> > double the current independently of V and R.
> >
> > --
> > http://www.piclist.com hint: To leave the PICList
> > EraseMEpiclist-unsubscribe-requestRemoveMESTOPspammitvma.mit.edu
>
> --
> http://www.piclist.com hint: To leave the PICList
> RemoveMEpiclist-unsubscribe-requestKILLspamTakeThisOuTmitvma.mit.edu
>
>-----Original Message-----
>From: Rick Regan [RemoveMEdrrdrspam_OUTCHARTER.NET]
>Sent: 15 April 2004 19:46
>To: PICLISTspamMITVMA.MIT.EDU
>Subject: Re: [EE]: Why isn't Ohm's Law written I=V/R?
>
>
>> But I is only the dependant variable in *one* way of
>> looking at it. V=IR is the appropriate way of writing it
>> if you consider ohm's law to mean " V = the voltage drop across
>> resistance R with current I passing through it". You can
>come up with
>> a similar, equally valid phrase, for the third way of writing it out
>> (r=v/i).
>
>But aren't R and V the only two of the three variables
>that you can manipulate directly? I depends on these,
>right? I think writing it I=V/R would be less confusing
>to the novice. With the V=IR form, I've seen
>descriptions like "if the current doubles the voltage
>doubles." Mathematically, yes, but it's not like you can
>double the current independently of V and R.
>
So how does this apply to any other arrangement? It's not like you can vary
ANY ONE of the parameters without affecting the others. i.e. varying
voltage across a fixed resistance changes the current, changing the current
through a fixed resistance varies the voltage across the resistance. It's
basic maths, given one fixed parameter the Ohms law states the other two are
either proportional or inversely proportional to each other. You can hardly
single out one arrangement of the formulae and say it's not valid.
Regards
Mike
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> ----- Original Message -----
> From: "Rick Regan" <drrdrspamBeGone.....CHARTER.NET>
> > But aren't R and V the only two of the three variables
> > that you can manipulate directly?
>
> No, not really. There is a current source in the EE lab that maintains a
> set current regardless of voltage or resistance (within reason). I think
> ordinary transistors do this to some extent also, as they are current
> amplifying devices.
____________________________________
Actually, the voltage accross a PN junction is equal to the log of the
current flowing through the junction. This was the fundamental concept in
the early analog computation. The voltage across the junction is the
dependent variable.
____________________________________
It just so happens that most people are more acquainted {Quote hidden}
> with measurements in voltage due to the use of batteries, wall warts, etc.
> A current source makes I an independent variable. Also there are some
> applications where R changes with I, which makes R the dependent variable!
>
> So in terms of which is the "proper" way to write it, I'd say it really
> doesn't matter. It's probably just easiest to avoid the fraction, so most
> people write E=IR.
>
>
>
>
> I think writing it I=V/R would be less confusing
> > to the novice. With the V=IR form, I've seen
> > descriptions like "if the current doubles the voltage doubles."
> Mathematically, yes, but it's not like you can
> > double the current independently of V and R.
> >
> > --
> > http://www.piclist.com hint: To leave the PICList
> > KILLspampiclist-unsubscribe-request.....mitvma.mit.edu
>
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> So how does this apply to any other arrangement? It's
> not like you can vary ANY ONE of the parameters without
> affecting the others. i.e. varying voltage across a
> fixed resistance changes the current, changing the
> current through a fixed resistance varies the voltage
> across the resistance. It's basic maths, given one
> fixed parameter the Ohms law states the other two are
> either proportional or inversely proportional to each
> other. You can hardly single out one arrangement of the
> formulae and say it's not valid.
Noone said that any of the arrangements were invalid. I
was looking for the more "natural" way to express it
because I assumed only V and R could be manipulated
directly. However, several have pointed out that I
(current) can also be manipulated directly, which I did
not know. (I still don't understand how but that's ok
for now I guess).
In any case, I still think I=V/R is more intuitive for
the beginner. Different batteries and resistor values can
be plugged in to cause a change in I. At least it helps
my view of the world....
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Ohm's Law states a relationship between three
interdependent variables: voltage (V), current (I), and resistance (R). R depends on I and V, I depends on R and
V, and V depends on I and R. The law does not express an
equation in the sense that you can plug in two fixed values
and get an answer. It is an iterative equation which
converges to a steady state. In that steady state, you
can measure the three quantities or derive the third after
measuring any two.
Is this a reasonable way to look at it?
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>-----Original Message-----
>From: Rick Regan [RemoveMEdrrdrRemoveMEEraseMECHARTER.NET]
>Sent: 16 April 2004 15:19
>To: KILLspamPICLISTspamBeGoneMITVMA.MIT.EDU
>Subject: Re: [EE]: Why isn't Ohm's Law written I=V/R?
>
>The law does not express an equation in the sense that
>you can plug in two fixed values and get an answer.
>In that steady state, you can measure the three quantities
>or derive the third after measuring any two.
>Is this a reasonable way to look at it?
Personaly I think two statements listed contradict each other. If you
cannot plug in two fixed values, how can you possibly ever derive the third
value?
Regards
Mike
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>-----Original Message-----
>From: Rick Regan [KILLspamdrrdrspamBeGoneCHARTER.NET]
>Sent: 16 April 2004 13:42
>To: @spam@PICLISTSTOPspam@spam@MITVMA.MIT.EDU
>Subject: Re: [EE]: Why isn't Ohm's Law written I=V/R?
>
>
>> So how does this apply to any other arrangement? It's
>> not like you can vary ANY ONE of the parameters without
>affecting the
>> others. i.e. varying voltage across a fixed resistance changes the
>> current, changing the current through a fixed resistance varies the
>> voltage across the resistance. It's basic maths, given one
>> fixed parameter the Ohms law states the other two are
>> either proportional or inversely proportional to each
>> other. You can hardly single out one arrangement of the
>> formulae and say it's not valid.
>
>Noone said that any of the arrangements were invalid. I
>was looking for the more "natural" way to express it
>because I assumed only V and R could be manipulated
>directly. However, several have pointed out that I
>(current) can also be manipulated directly, which I did
>not know. (I still don't understand how but that's ok
>for now I guess).
You can only change current by changing voltage or resiatnce, as per Ohms
law! A bench supply in constant current mode simply adjusts voltage to
achieve the desired current through an arbitrary resistance
>
>In any case, I still think I=V/R is more intuitive for
>the beginner. Different batteries and resistor values can
>be plugged in to cause a change in I. At least it helps
>my view of the world....
It's only usefulll in that form when you need to derive current from a known
voltage and resistance. Without wishing to be rude, the fact that you don't
understand how a constant current supply can be realised means that you may
not have fully grasped the funadmentals regarding Ohms law. I would guess
that you find I=V/R intuitive simply because it happens to be the form in
which you most commonly use the law?
Regards
Mike
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>-----Original Message-----
>From: Shawn Wilton [spamBeGoneshawnBLACK9.NET]
>Sent: 16 April 2004 15:45
>To: spam_OUTPICLISTSTOPspamMITVMA.MIT.EDU
>Subject: Re: [EE]: Why isn't Ohm's Law written I=V/R?
>
>
>Because in electronica nothing is ever fixed.
Are you joking? If not could you please expand on this?
Mike
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On the other hand, a bench supply in constant voltage mode simply adjusts
current to keep the voltage level approximately stationary. I think this is
more the point that he is missing.
>----- Original Message -----
>From: "Michael Rigby-Jones" <TakeThisOuTMichael.Rigby-JonesspamRemoveMEBOOKHAM.COM> <snip>
>
>>
>> You can only change current by changing voltage or resiatnce, as per
>> Ohms law! A bench supply in constant current mode simply adjusts
>> voltage to achieve the desired current through an arbitrary
>resistance
Well of course not! My meaning was apparently missed. For a varying R, a
constant voltage source must vary I to maintain some V. (See Ohm's law) :D
I probably should have more explicitly stated the varying R part.
Its just a matter of perspectives, really.
> >----- Original Message -----
> >From: "Michael Rigby-Jones" <KILLspamMichael.Rigby-Jonesspamspam_OUTBOOKHAM.COM> <snip>
> >
> >>
> >> You can only change current by changing voltage or resiatnce, as per
> >> Ohms law! A bench supply in constant current mode simply adjusts
> >> voltage to achieve the desired current through an arbitrary
> >resistance
>
> >{Original Message removed}
On Friday, Apr 16, 2004, at 08:25 US/Pacific, Michael Rigby-Jones wrote:
>> On the other hand, a bench supply in constant voltage mode
>> simply adjusts current to keep the voltage level approximately
>> stationary. I think this is more the point that he is missing.
>>
>
> No it does not. In constant voltage mode, the only thing that can
> adjust the current flowing is the resistance connected. A PSU cannot
> magicaly alter current into a fixed resiatnce without changing it's
> voltage.
>
Well, yes, but that's just Ohms law. The transistors in your power
supply are current-controlling devices. To achieve constant voltage,
the change the amount of current they're willing to provide. This is
effectively done by changing the internal resistance between a
pseudo-ideal voltage source and the load resistance. Of course.
The point is that it's no more valid to think of a constant-current
supply as adjusting the voltage to meet the load than to think of a
constant-voltage supply as adjusting the current to meet the load...
BillW
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> Personaly I think two statements listed contradict each other. If you
> cannot plug in two fixed values, how can you possibly ever derive the third
> value?
I'm saying something more subtle. You can measure two
values, which are "fixed" only in that the circuit has
reached a steady state. I'm taking into account the "real
world" effects that other posters have pointed out.
> It's only usefulll in that form when you need to derive current from a known
> voltage and resistance. Without wishing to be rude, the fact that you don't
> understand how a constant current supply can be realised means that you may
> not have fully grasped the funadmentals regarding Ohms law. I would guess
> that you find I=V/R intuitive simply because it happens to be the form in
> which you most commonly use the law?
I don't consider it rude. In fact, you might be the only
one who agrees with me. I guess I'm just not expressing
myself clearly enough. You keep I constant by keeping V/R
constant, hence I=V/R!
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> You can only change current by changing voltage or resiatnce, as per Ohms
> law! A bench supply in constant current mode simply adjusts voltage to
> achieve the desired current through an arbitrary resistance
Well, we've come full circle, because that's what I said
in my original post. Since then, others appeared to have contradicted that, saying you can manipulate current
directly.
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You CAN manipulate current directly, but of course it effects the voltage
and/or the resistance according to Ohm's law.
Example 1: Set your bench supply to deliver 100 mA through a 100 ohm
resistor. Measure the voltage across the resistor and behold! V = IR = 10
volts.
Example 2: Set your bench supply to deliver 10 volts through a 100 ohm
resistor. Measure the current, and I = V/R = 10/100 = 100mA
Example 3: Imagine you could set your bench supply to output both a
constant voltage and current, say 100mA and 10V. Then connect a wire
between the output and ground. Upon making the connection, you would find
the wire heats up to some temperature at which its total resistance to flow
of current is 100 ohms. (never actually do this.. it would probably destroy
your bench supply)
So you see then that no single variable is always or never the dependent
variable (which is what I think everybody has been saying all along?)
> Well, yes, but that's just Ohms law. The transistors in your power
> supply are current-controlling devices. To achieve constant voltage,
> the change the amount of current they're willing to provide. This is
> effectively done by changing the internal resistance between a
> pseudo-ideal voltage source and the load resistance. Of course.
>
> The point is that it's no more valid to think of a constant-current
> supply as adjusting the voltage to meet the load than to think of a
> constant-voltage supply as adjusting the current to meet the load...
You say "adjusting the current", and according to your
prior paragraph, this is done by adjusting the resistance.
So this seems to support what I said in my original post,
that you can only manipulate R and V directly (not I).
Hence, V and R are the independent variables, and I is the
dependent variable.
So in the "iterative equation" view of a circuit, you
start with an "initial value" for R and V, which results
in a value for I, which influences R and V, etc., until a steady state is reached. No?
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> You CAN manipulate current directly, but of course it effects the voltage
> and/or the resistance according to Ohm's law.
>
> Example 1: Set your bench supply to deliver 100 mA through a 100 ohm
> resistor. Measure the voltage across the resistor and behold! V = IR = 10
> volts.
I understand that you manipulate it in the sense that you
turn the "current knob" to the desired setting, but what
is really happening inside? Isn't it just adjusting the
voltage (and perhaps resistance?) to get to the desired
current? If that is true, then current is not really
being manipulated - it is just a by-product of the V/R
combo.
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> Does anyone know why Ohm's law is writtem V=IR (E=IR)
> instead of I=V/R? Of course I know they're equivalent.
> I just wondered why - if I is the dependent variable and V
> and R are the independent variables - it's basic form
> isn't I=V/R.
The law stated by Simon Ohm was that the voltage drop
across a resistance was linearly related to the current
through the device. At the time this was "discovered",
nonlinear circuit elements were unknown. How the law
is algebraically stated is irrelevant; both of the above
equations are equivalent and therefore they are both
"Ohm's Law". Which one is used depends on one's
personal history.
Now that semiconductor junctions are commonplace,
it is necessary for engineers to admit that Ohm's Law
is not always correct. The cases for which it does hold
true are referred to as "Ohmic", as in "the voltage between
the collector and emitter of a transistor is the sum of the
junction potentials and the ohmic drops in the bonding
wires".
John Power
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On Friday, Apr 16, 2004, at 11:50 US/Pacific, Rick Regan wrote:
> you turn the "current knob" to the desired setting, but what is really
> happening inside? Isn't it just adjusting the voltage (and perhaps
> resistance?) to get to the desired current? If that is true, then
> current is not really being manipulated - it is just a by-product of
> the V/R combo.
No. Like I said before, if your power supply contains bipolar
transistors, then it's all BASED on current, since bipolar transistors
are current-controlling devices.
There's this big transistor on the back, probably on a heat sink. The
power supply allows the base current of that transistor to rise
(increasing the collector current as well, but the beta factor) until
it senses that a reference voltage has reached a correct value. In CV
mode, the reference voltage is just the output voltage In CC mode, the
reference voltage is the voltage across some current-sensing resistor.
Your linear regulators (7805, etc) work the same way. The internal
components are controlling current, rather than voltage directly.
A transistor is a fine example of a device that permits current to be
controlled directly. Bipolar transistors are current-controlled, FETs
are voltage controlled, but in each case it's the current that is being
manipulated. (Come to think of it, tubes work this way too...)
BillW
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> No. Like I said before, if your power supply contains bipolar
> transistors, then it's all BASED on current, since bipolar transistors
> are current-controlling devices.
>
> There's this big transistor on the back, probably on a heat sink. The
> power supply allows the base current of that transistor to rise
> (increasing the collector current as well, but the beta factor) until
> it senses that a reference voltage has reached a correct value. In CV
> mode, the reference voltage is just the output voltage In CC mode, the
> reference voltage is the voltage across some current-sensing resistor.
> Your linear regulators (7805, etc) work the same way. The internal
> components are controlling current, rather than voltage directly.
>
> A transistor is a fine example of a device that permits current to be
> controlled directly. Bipolar transistors are current-controlled, FETs
> are voltage controlled, but in each case it's the current that is being
> manipulated. (Come to think of it, tubes work this way too...)
It's probably my fault for not being clear enough
about what I was asking, so I will make one more attempt
(I promise - this is my last post!).
I am not saying the algebra is wrong. I am not saying
that at any given time it's not convenient to put any one
of the three variables on the left hand side of the
equation. I was just trying to understand how current
could be an entity in its own. This was my understanding
and I guess you're saying I'm wrong:
- You can have V without I
- You can have R without I
- You CAN'T have I without V and R
Thanks for the responses. (I will now unsubscribe from the [EE] tags and stick with [PIC] ;)).
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I mentioned Ockham's razor earlier, also known as "The Law of Parsimony."
Generally, it applies to science[The simplest explanation is most
advantageous], but it is equally valuable in mathematics. And, yes, it can
be applied to Stoke's and Green's theorems. When actually solving real
problems by applied mathematics it usually helps to present the expressions
in the simplest readable form. For example, sometimes you factor
expressions so they are visibly more simple. This is sometimes the case
where you might avoid the table of integrals, or transforms, etc. P=IE,
E=IR, P= hd, or other such linear relationships are in the simplest readable
form by direct proportionality.
That being said, all other valid manipulations will provide equally valid
results. But there is room, here, to accept that some people prefer a
different configuration from the convention. However, before discarding
convention it is worthwhile learning it's value.
> > From: Rick Regan[SMTP:EraseMEdrrdrRemoveMECHARTER.NET]
> > Sent: Thursday, April 15, 2004 10:13 AM
> > To: spamPICLIST.....spamMITVMA.MIT.EDU
> > Subject: [EE]: Why isn't Ohm's Law written I=V/R?
>
> > Does anyone know why Ohm's law is writtem V=IR (E=IR)
> > instead of I=V/R? Of course I know they're equivalent.
> > I just wondered why - if I is the dependent variable and V
> > and R are the independent variables - it's basic form
> > isn't I=V/R.
>
> The law stated by Simon Ohm was that the voltage drop
> across a resistance was linearly related to the current
> through the device. At the time this was "discovered",
> nonlinear circuit elements were unknown. How the law
> is algebraically stated is irrelevant; both of the above
> equations are equivalent and therefore they are both
> "Ohm's Law". Which one is used depends on one's
> personal history.
>
> Now that semiconductor junctions are commonplace,
> it is necessary for engineers to admit that Ohm's Law
> is not always correct. The cases for which it does hold
> true are referred to as "Ohmic", as in "the voltage between
> the collector and emitter of a transistor is the sum of the
> junction potentials and the ohmic drops in the bonding
> wires".
>
> John Power
>
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Hey don't feel bad at all. It's probably best that you get this confusion
cleared up right now or it will haunt your circuits forever. What you are
saying is still basically incorrect though. Voltage can (sort of) exist
without current, but not without resistance. When you're reading a voltage
off two pins, remember that they are effectively separated by an infinite
resistance of air. Without this resistance, there could be no voltage. In
Ohm's law, if R=infinity, then current drops to zero. Resistance can exist
without either current or voltage (as in a resistor sitting there on your
desk), but its arguable whether its actually resistance unless it is
resisting something. Without a current to resist it's no better than a clod
of dirt or something. Keep in mind exactly what current is... It's the
flow of electrons, electricity in it's most fundamental form. Current
exists in all electrical circuits. Without electrons flowing to the battery
terminals, there would be no voltage on them. Without current flowing
through a resistor, there would be no voltage. Current, in its most basic
form, can (sort of) exist without any voltage. All you need is some
organized flow of electrons through a negligible resistance. If you haven't
considered the following analogy to fluid dynamics it might help your
understanding.
Think of voltage as a pressure in a system of pipes, current as the amount
of water flowing, and resistance as a necking down of the pipe. Large
resistance equate to large pipe-area reductions. It should now become clear
that with an infinite resistance the pressure (voltage) assumes some
constant state. For a pressure/voltage drop to occur there must be some
resistance, since if there is no resistance the water/current will just flow
freely. It would be silly to say that there is some pressure in an empty
pipe, and likewise it would be silly to say that there is some voltage
without a current backing it. Voltage is more the willingness of a current
to overcome resistance than a measure of electricity. Every day you incur
thousands of volts upon your body just walking around on the carpet, but
obviously this does not really do anything (besides zap that expensive new
chip you're working on.)
I'm sure there's a better rendition of the analogy out there somewhere since
I just pulled this out of my arse. Search on google if you think it will
help.
Jeeze I hope that helped some, surely there's gotta be something in there
that sparks a lightbulb?
Actually, I would go a tad further and say that none of the three exists
without the other two.
>> - You can have V without I
How would you measure a voltage without having any current flowing -- i.e.
without relocating some charge carriers? If you can't, how can you "have"
it? In order to have a voltage without a current, you would have to
postulate a resistance with the value "infinity" -- something possible in
theory, but not really in practice. (At least nobody's seen one so far...)
And then /knowing/ that it is there definitely would require you to take
out /some/ charge carriers -- and bingo, there's the current you don't want
to have.
>> - You can have R without I
Again, how can you measure a resistance without moving any current (charge
carriers) through the resistance? Keep in mind that the concept of a fixed
resistance (independent of anything else) is a simplification, helping in
quick and dirty engineering, but is not correct. A resistance is only the
proportional constant between the current and the voltage. There is no
resistance without a current (and a voltage), as the resistance is only
defined by the relationship between those two.
A 5% carbon film resistor without a voltage applied to it is not an
electrical resistance. Only after applying a voltage and a current, it
becomes a resistance of a certain value (which is more or less close to the
one that it's supposed to have, depending on the applied voltage and
current, among other variables).
>> - You CAN'T have I without V and R
Now this one is the only one that's correct.
BTW, there's also conductivity: I = G * V :) (Just another form to
express the proportion between current and voltage.)
> Hey don't feel bad at all. It's probably best that you get this confusion
> cleared up right now or it will haunt your circuits forever. What you are
> saying is still basically incorrect though. Voltage can (sort of) exist
> without current, but not without resistance. When you're reading a voltage
> off two pins, remember that they are effectively separated by an infinite
> resistance of air. Without this resistance, there could be no voltage. In
> Ohm's law, if R=infinity, then current drops to zero. Resistance can exist
> without either current or voltage (as in a resistor sitting there on your
> desk), but its arguable whether its actually resistance unless it is
> resisting something. Without a current to resist it's no better than a clod
> of dirt or something. Keep in mind exactly what current is... It's the
> flow of electrons, electricity in it's most fundamental form. Current
> exists in all electrical circuits. Without electrons flowing to the battery
> terminals, there would be no voltage on them. Without current flowing
> through a resistor, there would be no voltage. Current, in its most basic
> form, can (sort of) exist without any voltage. All you need is some
> organized flow of electrons through a negligible resistance. If you haven't
> considered the following analogy to fluid dynamics it might help your
> understanding.
>
> Think of voltage as a pressure in a system of pipes, current as the amount
> of water flowing, and resistance as a necking down of the pipe. Large
> resistance equate to large pipe-area reductions. It should now become clear
> that with an infinite resistance the pressure (voltage) assumes some
> constant state. For a pressure/voltage drop to occur there must be some
> resistance, since if there is no resistance the water/current will just flow
> freely. It would be silly to say that there is some pressure in an empty
> pipe, and likewise it would be silly to say that there is some voltage
> without a current backing it. Voltage is more the willingness of a current
> to overcome resistance than a measure of electricity. Every day you incur
> thousands of volts upon your body just walking around on the carpet, but
> obviously this does not really do anything (besides zap that expensive new
> chip you're working on.)
>
> I'm sure there's a better rendition of the analogy out there somewhere since
> I just pulled this out of my arse. Search on google if you think it will
> help.
>
> Jeeze I hope that helped some, surely there's gotta be something in there
> that sparks a lightbulb?
>
>Hey don't feel bad at all. It's probably best that you get this confusion
>cleared up right now or it will haunt your circuits forever. What you are
>saying is still basically incorrect though. Voltage can (sort of) exist
>without current, but not without resistance.
Where do you get this?
As to "sort of.." you better believe it.
What do you think stores the program in your processor?
Charge (voltage) consisting of electrons quantum-tunneled into the gate of a mosfet.
Voltage, is simply a measurement of the relative charge between two objects.
It is entirely measureable without ANY current flow at all.
Current can exist without voltage at all, in fact it does all the time.
Current is electron flow.
Take a plate of lead, cool with liquid helium, and drop a magnet on it.
The resistance of the lead is literally zero ohms, and the current induced by the magnet sets up an opposing magnetic field, that supports the magnet.
At 10:52 AM 4/17/2004 -0300, Gerhard Fiedler wrote:
>Actually, I would go a tad further and say that none of the three exists
>without the other two.
>
>>> - You can have V without I
>
>How would you measure a voltage without having any current flowing -- i.e.
>without relocating some charge carriers?
Electrostatic force is one method.
>If you can't, how can you "have"
>it? In order to have a voltage without a current, you would have to
>postulate a resistance with the value "infinity" -- something possible in
>theory, but not really in practice.
You better hope it's achievable, the program memory storage in your pic works that way.
>Again, how can you measure a resistance without moving any current (charge
>carriers) through the resistance?
It doesn't stop being a resistance when there's no current flowing.
> Keep in mind that the concept of a fixed
>resistance (independent of anything else) is a simplification, helping in
>quick and dirty engineering, but is not correct. A resistance is only the
>proportional constant between the current and the voltage. There is no
>resistance without a current (and a voltage), as the resistance is only
>defined by the relationship between those two.
No. That's how you MEASURE resistance, but resistance itself is an intrinsic property of the material, the free electron mobility.
>>> - You CAN'T have I without V and R
>
>Now this one is the only one that's correct.
Yes, you can. Messiner effect in superconductors demonstrates this quite nicely.
>>Actually, I would go a tad further and say that none of the three exists
>>without the other two.
(I think when the OP said "you can/can't have x without y and z" I took
that to mean "you can/can't have the value of x without the values of y and
z" and responded that way.)
>>>> - You can have V without I
>>
>>In order to have a voltage without a current, you would have to
>>postulate a resistance with the value "infinity" -- something possible in
>>theory, but not really in practice.
>
> You better hope it's achievable, the program memory storage in your pic works that way.
I suppose you are talking about Flash memory. Are you sure that there is no
current? I mean, not "no current for practical purposes in most cases" but
/no/ current? Why do Flash devices have limited (and for long-term or
high-temperature applications actually relevant) data retention times?
Somehow the charges seem to be slowly moving out of their location (or
charges moving in, depending on the POV). Maybe as slowly as a few
electrons per second, but isn't that more than nothing?
>>Again, how can you measure a resistance without moving any current (charge
>>carriers) through the resistance?
>
> It doesn't stop being a resistance when there's no current flowing.
How do you know? So far, I thought that resistance was being defined by the
ratio between a voltage and a current. What would be the value of 0/0 (if
there really was a 0)?
>> Keep in mind that the concept of a fixed
>>resistance (independent of anything else) is a simplification, helping in
>>quick and dirty engineering, but is not correct. A resistance is only the
>>proportional constant between the current and the voltage. There is no
>>resistance without a current (and a voltage), as the resistance is only
>>defined by the relationship between those two.
>
> No. That's how you MEASURE resistance, but resistance itself is an intrinsic property of the material, the free electron mobility.
Don't you have to measure the resistance in order to know it? I agree that
every material has a resistance. But the value of it under certain
conditions is not known until it has been measured. Assuming it to be
constant is, as we all should know, a simplification that has its limits.
So in order to really know whether that assumption is correct, you'd have
to measure the resistance. And I think that in order to measure resistance,
you would have to measure both voltage and current. You don't measure, you
have an assumption of a resistance value, not a resistance value.
(I'm sure I'm not the only one who ever has forced a current through a
device and measured the voltage across it, to find out whether this device
was still the resistor it maybe once was -- or maybe never was. I wouldn't
trust the "intrinsic properties" when troubleshooting a device -- the last
word is a measurement to find out what is and what is just illusion :)
>>>> - You CAN'T have I without V and R
>>
>>Now this one is the only one that's correct.
>
> Yes, you can. Messiner effect in superconductors demonstrates this quite nicely.
I'm not familiar enough with superconductors to be able to discuss this
with any depth. But in my (engineering and other) experience there is not
really much if anything in nature (which includes technology, in this
context) that's literally 0 or infinity. The two are mathematical concepts,
not physical ones. In physics, they tend to be simplifications that are
valuable and helpful within their limits. Both usually mean "outside the
range of interest (or measurability)".
Actually, if you take it by face value, no scientist would ever say that
there's 0 (or infinity) of anything. In order to be able to say so, you
have to measure. You can't measure neither. The only thing you can say is
that it's less or more than what you can detect with your experiment
configuration. So yes, there are currents less than 1 aA, and generally
they are /considered/ 0 for most practical purposes. But /are/ they 0?
Back to superconductors: The fact that there is no potential difference in
the first place that makes the charge carriers move when applying a
magnetic field to a superconductor doesn't necessarily mean there is no
potential difference once they started moving.
>
>I suppose you are talking about Flash memory. Are you sure that there is no
>current? I mean, not "no current for practical purposes in most cases" but
>/no/ current? Why do Flash devices have limited (and for long-term or
>high-temperature applications actually relevant) data retention times?
>Somehow the charges seem to be slowly moving out of their location (or
>charges moving in, depending on the POV). Maybe as slowly as a few
>electrons per second, but isn't that more than nothing?
The stored charge leaks off by quantum effects, and the wearout mechanism is where the electrons get stuck in the insulator. There is no current flow.
>>>Again, how can you measure a resistance without moving any current (charge
>>>carriers) through the resistance?
>>
>> It doesn't stop being a resistance when there's no current flowing.
>
>How do you know? So far, I thought that resistance was being defined by the
>ratio between a voltage and a current. What would be the value of 0/0 (if
>there really was a 0)?
No, resistance is intrinsic to the material. Don't confuse the thing, with how it's measured. A gallon is not a container.
>Don't you have to measure the resistance in order to know it?
Yes, but it's still whatever value it was, when you aren't measuring it.
> I agree that
>every material has a resistance. But the value of it under certain
>conditions is not known until it has been measured.
Ok, but it still is resistive.. This makes as much sense as putting resistors in a box, and saying that they aren't resistors anymore.
>> Yes, you can. Messiner effect in superconductors demonstrates this quite nicely.
>
>I'm not familiar enough with superconductors to be able to discuss this
>with any depth. But in my (engineering and other) experience there is not
>really much if anything in nature (which includes technology, in this
>context) that's literally 0 or infinity.
Superconductors are like quantum mechanics, they don't follow the rules that you are used to for other things.
>Actually, if you take it by face value, no scientist would ever say that
>there's 0 (or infinity) of anything. In order to be able to say so, you
>have to measure. You can't measure neither.
You can measure the magnetic field produced by a current flowing.
Since the field remains constant, you know the magnitude of the current flow, and that it is not changing.
>Back to superconductors: The fact that there is no potential difference in
>the first place that makes the charge carriers move when applying a
>magnetic field to a superconductor doesn't necessarily mean there is no
>potential difference once they started moving.
Apply ohm's law. When R is zero, E must be zero, even when I is non-zero.
>How would you measure a voltage without having any current flowing --
>i.e. without relocating some charge carriers?
You can't without relocating any charge carriers, so why not relocate some
and put them back immediately ? Like in a vibrating reed voltmeter/charge
meter. You have a moving electrode in an electrical field. The motion is
repetitive and the probe will sense an ac voltage with amplitude equal to
the difference in potential between the two extremes of the motion. It
need not touch anything. The energy comes from the mechanism that does the
moving, by moving the probe charge in the field. A small influence will be
sensed on the measured object but it will be ac. The electrophorus
operates on the same principle.
On Sunday, Apr 18, 2004, at 06:34 US/Pacific, Gerhard Fiedler wrote:
>>> In order to have a voltage without a current, you would have to
>>> postulate a resistance with the value "infinity" -- something
>>> possible in
>>> theory, but not really in practice.
>>
>> You better hope it's achievable, the program memory storage in your
>> pic works that way.
>
> I suppose you are talking about Flash memory. Are you sure that there
> is no current? I mean, not "no current for practical purposes in most
> cases" but /no/ current?
You can have an electric field, measured in volts, without current
flowing. Just like with magnets (there's even an "electret", the
electrical equivalent of a permanent magnet.) Field effect transistors
can sense the electric field (more or less) the way a hall effect
device senses a magnetic field, so you can implement eproms, flash,
etc...
Ohm's law refers to circuits, though. Not static fields or components?
>>>> Again, how can you measure a resistance without moving any current
>>>> (charge carriers) through the resistance?
>>>
>>> It doesn't stop being a resistance when there's no current flowing.
>>
>> How do you know? So far, I thought that resistance was being defined by
>> the ratio between a voltage and a current. What would be the value of
>> 0/0 (if there really was a 0)?
>
> No, resistance is intrinsic to the material. Don't confuse the thing,
> with how it's measured. A gallon is not a container.
I don't think I confused it with how it's measured. I thought resistance
was /defined/ as the quotient of voltage and current. How would you define
resistance (in a way that is independent of current and voltage, so that
the definition works at V=I=0)?
>> I agree that every material has a resistance. But the value of it under
>> certain conditions is not known until it has been measured.
>
> Ok, but it still is resistive.. This makes as much sense as putting
> resistors in a box, and saying that they aren't resistors anymore.
They have a resistance, but of unknown value. The value that's on them is
spec'ed for certain conditions, and literally 0 current is not part of the
specs.
> Superconductors are like quantum mechanics, they don't follow the rules
> that you are used to for other things.
If that's true, then it is at least a possibility that some classic
electric concepts like resistance are not anymore useful in that realm --
in the same way as some classic mechanical concepts are not very useful in
quantum mechanics.
>> Actually, if you take it by face value, no scientist would ever say that
>> there's 0 (or infinity) of anything. In order to be able to say so, you
>> have to measure. You can't measure neither.
>
> You can measure the magnetic field produced by a current flowing. Since
> the field remains constant, you know the magnitude of the current flow,
> and that it is not changing.
Again, "not changing" is something you should be careful with if you really
want to get to the bottom of things. With any experiment, the most you can
say is that the rate of change is smaller than x, with x>0.
>> Back to superconductors: The fact that there is no potential difference
>> in the first place that makes the charge carriers move when applying a
>> magnetic field to a superconductor doesn't necessarily mean there is no
>> potential difference once they started moving.
>
> Apply ohm's law. When R is zero, E must be zero, even when I is
> non-zero.
/When/ R is 0. How do you know that it is? As a thought experiment: how
would you measure 0 Ohm? If it exists, there must be a way to measure it.
(The other question is whether Ohm's law still makes any sense in a realm
that, according to you, doesn't follow normal laws. But your assumption is
that it still makes sense. So let's work with it.)
>>How would you measure a voltage without having any current flowing --
>>i.e. without relocating some charge carriers?
>
> You can't without relocating any charge carriers, so why not relocate some
> and put them back immediately ? Like in a vibrating reed voltmeter/charge
> meter. You have a moving electrode in an electrical field. The motion is
> repetitive and the probe will sense an ac voltage with amplitude equal to
> the difference in potential between the two extremes of the motion. It
> need not touch anything. The energy comes from the mechanism that does the
> moving, by moving the probe charge in the field. A small influence will be
> sensed on the measured object but it will be ac. The electrophorus
> operates on the same principle.
That's a good trick. But you would agree with me that you are measuring
resistance at a current > 0A? One of the questions debated was whether
there is a resistance (or whether it makes sense to speak of one) if there
is no current flowing. In your experiment, current is flowing, forth and
back, if I understand it correctly.
On Sunday, Apr 18, 2004, at 14:58 US/Pacific, Gerhard Fiedler wrote:
>
>> Superconductors are like quantum mechanics, they don't follow the
>> rules that you are used to for other things.
>
> If that's true, then it is at least a possibility that some classic
> electric concepts like resistance are not anymore useful in that realm
Well, yes, of course. Ohm's law doesn't apply particularly well to
anything semiconductor (except piecewise over small ranges, where it is
still useful!) And there are lots of semiconductor devices that
exhibit regions where the meaured resistance will be negative...
>>> Superconductors are like quantum mechanics, they don't follow the
>>> rules that you are used to for other things.
>>
>> If that's true, then it is at least a possibility that some classic
>> electric concepts like resistance are not anymore useful in that realm
>
> Well, yes, of course. Ohm's law doesn't apply particularly well to
> anything semiconductor (except piecewise over small ranges, where it is
> still useful!) And there are lots of semiconductor devices that
> exhibit regions where the meaured resistance will be negative...
The "concept of resistance" is not the same as Ohm's law. As you say, there
are many devices with resistances that are a function of voltage or current
(i.e. that don't follow Ohm's law), yet the concept of resistance still
gets applied to them (as can be seen when you talk about "negative
resistance").
On Monday, Apr 19, 2004, at 03:53 US/Pacific, Gerhard Fiedler wrote:
>
>> Ohm's law refers to circuits, though. Not static fields or
>> components?
>
> My question here is whether "static fields" isn't a simplification for
> saying "/very/ low current situations".
No. Get into things like electron microscopes or particle
accelerators, and electric fields become things very distinct indeed
from currents.
What would your static magnetic field represent a "very low current"
OF, anyway?
Heisenberg's uncertainty principle enters the picture if you actually
want to MEASURE zero, of course. While it's usually thought of in
terms of the uncertainty of a non-zero value, measuring exactly zero
violates it just as ... certainly.
(or... Think of most of electronics as a "bulk" simplification of
electron physics. Things like single electrons clearly have an
electric field, whether they're involved in current flows or not...)
>>> Ohm's law refers to circuits, though. Not static fields or
>>> components?
>>
>> My question here is whether "static fields" isn't a simplification for
>> saying "/very/ low current situations".
>
> No. Get into things like electron microscopes or particle
> accelerators, and electric fields become things very distinct indeed
> from currents.
>
> What would your static magnetic field represent a "very low current"
> OF, anyway?
>
> Heisenberg's uncertainty principle enters the picture if you actually
> want to MEASURE zero, of course. While it's usually thought of in
> terms of the uncertainty of a non-zero value, measuring exactly zero
> violates it just as ... certainly.
>
> (or... Think of most of electronics as a "bulk" simplification of
> electron physics. Things like single electrons clearly have an
> electric field, whether they're involved in current flows or not...)
This seems to say that 0 (voltage, current, resistance) is outside of the
realm of Ohm's law. And that nobody has yet been able to confirm a
resistance of 0 (because that would require to be able to measure a voltage
of 0).
>>Again: has anybody ever measured (literally) 0 of anything anywhere, with
>>the required precision?
>
> I have zero orangutans..
It's probably easier to measure 0 orangutans than it is to measure 0 V (or
0 Ohms). You can think of it as the first stage in the process. Once you
get there, please let me know :)
> It's probably easier to measure 0 orangutans than it is to measure 0 V (or
> 0 Ohms). You can think of it as the first stage in the process. Once you
> get there, please let me know :)
But HOW LONG do you have them for?
Heisenberg says that for very (very very very ...) short periods of time you
will occasionally have non zero quantities of Orangutans. Perhaps that
explains the difficulty in controlling the mess around here.
>
>This seems to say that 0 (voltage, current, resistance) is outside of the
>realm of Ohm's law. And that nobody has yet been able to confirm a
>resistance of 0 (because that would require to be able to measure a voltage
>of 0).
This would seem to state that the entire superconductor industry is wrong?
At 10:29 PM 4/20/2004 +1200, Russell McMahon wrote:
>> > I have zero orangutans..
>
>> It's probably easier to measure 0 orangutans than it is to measure 0 V (or
>> 0 Ohms). You can think of it as the first stage in the process. Once you
>> get there, please let me know :)
>
>But HOW LONG do you have them for?
>
>Heisenberg says that for very (very very very ...) short periods of time you
>will occasionally have non zero quantities of Orangutans. Perhaps that
>explains the difficulty in controlling the mess around here.
Ah, but they won't really have time to get anything done anyway.
>That's a good trick. But you would agree with me that you are measuring
It's not a trick, you can buy these things, they are used in research.
>resistance at a current > 0A? One of the questions debated was whether
>there is a resistance (or whether it makes sense to speak of one) if
>there is no current flowing. In your experiment, current is flowing,
>forth and back, if I understand it correctly.
Consider whence the energy comes. A current means by definition moving
charge. The the moving is done by the probe charge so the current *is* the
charge and *it* moves. The field stays put and minds its own business.
It (its poles) oppose the motion of the charge with a force (electrostatic
attraction/repulsion). As far as it is concerned the probe charge does not
exist (or is too small and too far away to notice). A small perturbation
occurs because the probe impedance is not infinite so it couples a little
ac into the field. Actually it is more complicated because these devices
can measure the charge of a single electron. Millikan's famous experiment
is a version of this method (used as a null instrument).
With superconductors and Josephson loops you have current flowing in a
circuit with zero resistance. There are superconducting laboratory magnets
in common use which can do this for months. You just supply LHe. If the
idea of current in a circuit of zero resistance bothers you imagine the
whole thing is a permanent magnet and it's field comes from the spin
orientation of the atoms in it. No current flowing, right ? Now think of
it again, but this time the atom is the size of the device and the
spin-analog which is current is 'travelling' in it, through the whole
device which is made 'transparent' by the superconducting materials. Still
no voltage. The current is real enough to cause the device to
self-destruct if it comes out of the superconducting state before being
discharged. Search Google for 'quenching superconducting laboratory
magnet'. I do not use such devices but they are real enough.
>>This seems to say that 0 (voltage, current, resistance) is outside of the
>>realm of Ohm's law. And that nobody has yet been able to confirm a
>>resistance of 0 (because that would require to be able to measure a voltage
>>of 0).
>
> This would seem to state that the entire superconductor industry is wrong?
Well, if you take the marketing publications of an /industry/ for real,
you're probably lost anyway. (Think software industry ... or think TV
commercials ... or think /any/ commercial :)
But what is "the superconductor industry" saying that would collide with
the above?
We all know that Ohm's law (that the voltage across and the current through
a piece of material are proportional and that that proportion is constant),
while useful in practice when applied correctly within its limits, has
indeed limitations that we all know of -- no need to bother the
superconductor industry for that one. So what's so revolutionary about what
I wrote?
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>>That's a good trick.
>
> It's not a trick, you can buy these things, they are used in research.
trick (trik) n., adj., v. <tricked, trick-ing>
n.
1. a crafty or underhanded device, maneuver,
or stratagem intended to deceive or
cheat; artifice; ruse; wile.
2. a roguish or mischievous act; practical
joke; prank.
3. a clever or ingenious device or
expedient; adroit technique: the tricks
of the trade.
4. the art or knack of doing something
skillfully: the trick of making others
laugh.
I was using the word along meaning number 3. Anything wrong with "clever or
ingenious device"? I'm sure there are more such "tricks" that one can
buy...
> With superconductors and Josephson loops you have current flowing in a
> circuit with zero resistance. There are superconducting laboratory magnets
> in common use which can do this for months. You just supply LHe. If the
> idea of current in a circuit of zero resistance bothers you [...]
It's not that it's bothering me, it's the fact that none of the proponents
of 0 resistance so far bothered to answer my question about how to in fact
measure 0 Ohm. Sorry for maybe appearing dense here, but I don't understand
a lot about superconductivity. However, I do understand a bit about
measuring and precision of measurements, and every experiment I've seen so
far had a limited precision. (Of course there is David's orang-utan
experiment, but even that is limited, in duration. And I doubt his house is
superconducting for orang-utans :)
For example at http://en.wikipedia.org/wiki/Superconductor they write that
"experiments have in fact demonstrated that currents in superconducting
rings persist for years without any measurable degradation."
Keyword here is "measurable". The /theory/ maybe says that resistance is 0
Ohms (again, I'm not a superconductor expert), but then they try to
/verify/ that theory and say (correctly) "without measurable degradation."
This would indeed indicate a /very/ low resistance, but if you take those
experiments (and not the theory they are trying to verify), you probably
will reach the conclusion that the results are that the resistance is
smaller than a finite value x -- not that the resistance is 0.
The whole discussion was about the difference between "practically 0" --
which means nothing else than "smaller than I'm able to or care to measure"
and is what we mostly deal with --, and the postulated "literally 0" --
which still needs a way to verify its existence that nobody seems to be
able to come up with. Just show me a way (a thought experiment) that
measures 0 Ohms; come up with an experiment, say what you measure, how and
with what precision (all finite, of course), and how this might lead to a
measurement of 0 Ohms. That's all I was/am asking for. Should be easy, if
it's actually as commonplace as stated.
In that superconductor current experiment I cited above, it would require
to measure the current for an infinite amount of time or with an infinitely
low resolution to be able to state 0 Ohms resistance. The moment you
introduce a limited amount of time and a limited resolution of current
measurement, you will get a result of the type "the resistance is smaller
than x" with x being > 0 Ohms.
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I'm not an expert in superconductors by any means, but do they really
violate ohm's law?
If V=IR, and R is 0, then V is zero too, but I doesn't have to be. So, you
could have a current flowing round a ring, and there would be no voltage
(drop). Which fits, cos the thing is superconducting. Furthermore, this
current can do no work, because to do work, it would need an EMF, and ohm's
law states that V=IR, so to do work R must be > 0.
Also, the uncertaintly principle would suggest that we can't actually
measure 0 resistance; our measurement would affect the resistance.
Maybe what I just wrote is complete b******s. I don't know. It's interesting
nonetheless.
>If V=IR, and R is 0, then V is zero too, but I doesn't have
>to be. So, you could have a current flowing round a ring,
>and there would be no voltage (drop). Which fits, cos the
>thing is superconducting.
fair enough
>Furthermore, this current can do no work, because to do work,
>it would need an EMF, and ohm's law states that V=IR, so to
>do work R must be > 0.
Wrong, because the current still creates a magnetic field, and you can take
energy out of the magnetic field. Ask any user of superconducting magnets,
typically devices like MRI scanners, particle accelerator rings etc.
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On Wednesday, Apr 21, 2004, at 03:05 US/Pacific, Gerhard Fiedler wrote:
> It's not that it's bothering me, it's the fact that none of the
> proponents of 0 resistance so far bothered to answer my question about
> how to in fact measure 0 Ohm.
Current flowing without resistance can be sensed via its magnetic
field, in ways similar to sensing voltage (ie in eeprom cells) without
current... Since the field doesn't degrade in a superconducting loop,
one can conclude that the resistance is indeed zero.
Moreover, I think the physicists have both models and experiments along
the lines of "if it behaves this way, the resistance IS zero" in the
same sense that they can say "if the charge is bigger than one
electron, it must be at least two electrons." Electronics is just a
simplification of physics, after all.
BillW
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>> It's not that it's bothering me, it's the fact that none of the
>> proponents of 0 resistance so far bothered to answer my question about
>> how to in fact measure 0 Ohm.
>
> Current flowing without resistance can be sensed via its magnetic
> field, in ways similar to sensing voltage (ie in eeprom cells) without
> current... Since the field doesn't degrade in a superconducting loop,
> one can conclude that the resistance is indeed zero.
Here you say "since the field doesn't degrade". As I see it, you can have a
theory that it doesn't degrade. That's good, and that's one part of what
science is about: creating theories. The other part, however, is verifying
or falsifying those theories. Here you need to think about how to measure
what the theory predicts.
"The field doesn't degrade." This has two components that need to be
measured: the intensity of the field, and then you need to relate that to a
time. I suppose you can't make the time infinite. So you have a finite
time, big but limited (like a few years). Then you need to measure that
field. For all I know, field measurements have limited precision; at least
when I do precision measurements, the result comes always in a form of "the
value is x, and that's within plus y, minus z". I haven't yet seen a
measurement device where either y or z were 0 (except in some cases where
it is impossible to measure something <0, and therefore for a result of 0 z
could be 0). If the two field measurements -- before and after time t --
have in fact the same result, you have a (small, but >0) maximum field
difference 2*z and a (big but finite) minimum time t. This gives a
degradation that may be 0 (or maybe even negative, considering a possible
y), but also may be a small value of 2*z/t, according to your measurements.
Where does this confirm that zero resistance exists?