Re: Is zero even or odd?

From: John Fields (jfields_at_austininstruments.com)
Date: 12/27/04 

Date: Mon, 27 Dec 2004 10:30:56 -0600

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>
>> When using Ohm's law:
>>
>> E
>> R = ---
>> I
>>
>> The assumption is made that in order for resistance to be measured, a
>> voltage and a current must exist. Implicit in that assumption is that
>> the voltage must be applied across, and the current forced to flow
>> through, the resistance.
>
>Correct so far.
>
>> 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.
>
>In the above circuit, the current measurement
>is accurate but the voltage is the sum of that
>across R and that across the ammeter. You must
>subtract I times the resistance of the ammeter
>from the voltage measurement to find the voltage
>across the resistor. 

---
The short, according to the OP, is perfect, so the voltage drop across
it is zero.  With no voltage difference across the ammeter, no current
will flow through it _or_ through the resistor.
---
>> The proper circuit:
>>
>>      +---(V)---+
>>      |         |
>> (-)---o---[R]---o---(A)---o---(+)
>>
>> Will yield the proper results if examined using Ohm's law.
>
>In the above circuit, the voltage measurement
>is accurate but the current is the sum of that
>through R and that through the voltmeter. You
>must subtract V divided by the resistance of
>the voltmeter from the current measurement to
>find the current through the resistor.
---
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 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.
---
>Neither circuit gives both readings accurately.
---
Knowing the meter resistance, the second circuit does.
--- 
>
><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.  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.
---   
>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.
---
The very notion of a superconductor renders Ohm's law unfit to
characterized it, so saying that 0/0 can have any value because Ohm's
law doesn't work for superconductors is nonsensical.
-- 
John Fields

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