Re: Is zero even or odd?
From: John Fields (jfields_at_austininstruments.com)
Date: 12/27/04
- Next message: Michael Mendelsohn: "Re: Is zero even or odd?"
- Previous message: Paul Lalli: "Re: Perl Developer neded."
- In reply to: George Dishman: "Re: Is zero even or odd?"
- Next in thread: George Dishman: "Re: Is zero even or odd?"
- Reply: George Dishman: "Re: Is zero even or odd?"
- Messages sorted by: [ date ] [ thread ] [ subject ] [ author ]
Date: Mon, 27 Dec 2004 13:25:31 -0600
On Mon, 27 Dec 2004 16:51:57 -0000, "George Dishman"
<george@briar.demon.co.uk> wrote:
>
>"John Fields" <jfields@austininstruments.com> wrote in message
>news:h8c0t0dvf5fq2ig0uqgaihf89ovrikmsn2@4ax.com...
>> On Mon, 27 Dec 2004 15:47:32 -0000, "George Dishman"
>> <george@briar.demon.co.uk> wrote:
>>
>>>
>>>"John Fields" <jfields@austininstruments.com> wrote in message
>>>news:2dvvs0ppsmtl1tgcpv4rp4fd8a5cko8bak@4ax.com...
>>>> On Mon, 27 Dec 2004 02:23:11 +0100, Michael Mendelsohn
>>>> <invalid@msgid.michael.mendelsohn.de> wrote:
>>>>
><snip>
>>>> Your circuit:
>>>>
>>>>
>>>> +---------------------(V)----+
>>>> | |
>>>> (-)-----o-------[__R__]---o---(A)----o--------(+)
>>>> |____________________________|
>>>> the short
>>>>
>>>> contrives to hide the resistance while purporting to use Ohms law to
>>>> determine the resistance so, quite clearly, the results obtained will
>>>> be nonsensical.
><snip second circuit>
>
>> In the above circuit, as in the previous, the instrumentation is
>> assumed to be perfect, so there will be no current required to read
>> the voltage.
>
>If the instrumentation is assumed to be perfect, there
>will be no voltage dropped across the ammeter and hence
>the first circuit (above) is equally valid.
--- LOL! In the first circuit, the short is also assumed to be perfect, so there will be no need for instrumentation, perfect or not. --- >> If that's not satisfactory then a wheatstone bridge can >> be used to measure the voltage with no regard given to the impedance >> of the voltmeter. > >Agreed, it is easier to achieve it in reality that >way, but for the purposes of considering 0/0 that >is academic. --- Then so is considering the meter resistance at all. I'm starting to think that you're not so much interested in considering 0/0 as you are in being argumentative and confrontational. If that's the case, then this 'discussion' is over. --- >>>Neither circuit gives both readings accurately. >> >> --- >> Knowing the meter resistance, the second circuit does. > >The point was, knowing the meter resistance, >both circuits do. --- No. In the case of the first circuit, it's not necessary to know anything about the meters since even if they're not there it doesn't matter. In the case of the second circuit it's only necessary to know the resistance of the voltmeter to determine its contribution to the current flowing in the ammeter. --- >>><snip proof that 1=1> >>> >>>If you want to use an Ohms Law example to >>>understand this, realise that when you try to >>>calculate 0/0, you are asking "what resistance >>>will allow zero current to flow when zero voltage >>>is applied. The answer is any resistance. >> >> --- >> Yes, and that's why I wouldn't ordinarily use Ohm's law to try to >> prove that 0/0 = 1. > >Neither would I, but I'm not the OP. --- Which is supposed to mean what, exactly? --- >> In this case, however, to indicate to the OP that >> when the current through the resistor and the voltage across it are >> both numerically equal, the value of the resistance will be one ohm >> and will remain one ohm as the voltage across the resistance goes to >> zero. > >Sure, but as has been pointed out, if you start >by assuming any other value of resistance, the >numerical value of the ratio also remains constant >at that other value. You have simply illustrated a >particular example, not proven any general rule. --- Yes, that's what I was trying to do. (Illustrate a particular example, that is.) By writing: "In this case, however, to indicate to the OP that when the current through the resistor and the voltage across it are both numerically equal, the value of the resistance will be one ohm and will remain one ohm as the voltage across the resistance goes to zero." I thought that I was making it clear that I was talking about a particular instance by writing "In this case"... and "when the current through the resistor and the voltage across it are both numerically equal"... Both rather limiting. Sorry if I confused you. --- >>>Equally, you could ask what current flowing through a >>>superconductor produces zero volts and again the >>>answer is any current. Hence 0/0 can have any >>>value and it is therefore undefined. --- See John Woodgate's reply. --- >> The very notion of a superconductor renders Ohm's law unfit to >> characterized it, > >Not at all, R is defined as V/I and for any current >(below the level that destroys the superconduction >obviously), V=0 hence R=0. --- Yes, I should have thought through that one a little more critically. Thanks. --- >> so saying that 0/0 can have any value because Ohm's >> law doesn't work for superconductors is nonsensical. > >I wouldn't have chosen Ohms Law myself but since that >is the topic, you should at least get the arguments >right. --- Are you _intentionally_ being arrogant, is it something over which you have little control, or am I misreading your tone? -- John Fields
- Next message: Michael Mendelsohn: "Re: Is zero even or odd?"
- Previous message: Paul Lalli: "Re: Perl Developer neded."
- In reply to: George Dishman: "Re: Is zero even or odd?"
- Next in thread: George Dishman: "Re: Is zero even or odd?"
- Reply: George Dishman: "Re: Is zero even or odd?"
- Messages sorted by: [ date ] [ thread ] [ subject ] [ author ]
Relevant Pages
|
