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Is this a valid Joules formula?

  1. Nov 25, 2013 #1
    Is this a valid Joules formula anyone has heard of? It seems to work for the few examples I have written down. I'm just curious if anyone has seen it before.

    I came up with it trying to equate volts to joules but had to include a function of time (via resistance) to make it work, i.e. this: joules = (V/R * C) * V/C and came up with the shorter one here:

    J=V^2 / R

    J=Joules
    V=Voltage
    R=Resistance
    C=Coulomb


    In otherwords, since we know V = J / C , I was trying to equate Volts and Joules and switch them (J = V / C). But it wouldn't work without a function of time attached to volts. Hence the formula above, thus suggesting we are missing a "times resistance" or "times seconds" somewhere in the basic formulas (as so much confusion arises over the whole 1 second assumed granularity thing).

    Interestingly, when I simplified the above first theoretical formula, Coulomb (mass?) cancelled out and only resistance (time??) was left...

    and another thought..we always see V = I X R but Amps (I) already implies time (per second).
     
    Last edited: Nov 25, 2013
  2. jcsd
  3. Nov 25, 2013 #2
    Basically, I was theorizing that these known equations:

    joules = watt * seconds, watts = joules/second, seconds = joules/ watts

    are basically these same equations:

    volts = amps * resistance, amps = volts / resistance, resistance = volts / amps

    ...showing themselves in a different way... hence... resistance is perhaps really just time


    ... even speed formulas seem to match:
    speed = distance / time, etc. so speed could be amps... etc. Again just a naive theory.

    and so then I tried equating volts to joules...
     
  4. Nov 25, 2013 #3

    davenn

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    No, there's no time factor in Ohms law

    Dave
     
  5. Nov 25, 2013 #4
    I just realized that J=V^2 / R (which is the same as V = sqrt(J*R)) is just the Watts formulas (W=V^2/R and V=sqrt(W*R). So it works because I actually haven't added a component of time as I thought. I guess it is because amps (employed to get the appropriate R as employed here) skews things to 1 second.
     
    Last edited: Nov 25, 2013
  6. Nov 25, 2013 #5
    Right. I just realized that. But it actually does relate it to a "zero" time factor of one second (so to speak)? Yes? Hence why my "formula" works for joules as my examples never calculated out joules over any "real" time (i.e multiple of 1 second).

    I'm trying to relate it to speed = distance / time (if possible). Thus, ohm's law is as if we always said the speed had to be per 1 second. And then we would use speed x secs to get... accumulated speed? Speed used? Joule-speed :)?

    Or perhaps... speed = distance / R and somehow times "time"... where R=0 and is assumed nominal/minimul resistance to speed. You sort of see where I'm trying going with that?

    I guess maybe it would be speed = distance / 1sec? And then the "joule" version speed = distance / time ?? I guess I'm probably just off on a big tangent here.
     
    Last edited: Nov 25, 2013
  7. Nov 26, 2013 #6

    davenn

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    No, you have a lot of misunderstandings there
    you cant intermix formulae like that

    Time and Ohms law don't and don't need to mix
    A fixed value of resistor with a fixed voltage across it will always give a fixed current flow
    Time is irrelevant

    Dave


     
  8. Nov 26, 2013 #7
    So Amps (being the total number of coulombs which are flowing in one second) is not actually time based?

    So would ohm's law be correct in helping to calculate the total coulomb flow for half a second given the Amp value?

    So given this example:
    12vdc with 6ohm resistor circuit.
    I=V/R. So Amps = 12/6 = 2A

    So if I use that 2A in the formula for calculating Coulomb:
    C=A*s

    ... this formula assumes Amps are per 1 second (i.e. coulomb = amps if s=1) which means ohm's law is inherently working with regards to 1 second.

    Ohm's law must be somehow inherently calibrated to 1 second, no? Otherwise how does it work in the coulomb formula? I understand you mean in and of itself it doesn't have a time factor, and that the flow is constant and that certainly helps. But is it truly not time based?
     
    Last edited: Nov 26, 2013
  9. Nov 26, 2013 #8
    How ever they came up with that formula and the corresponding meanings/values for ohms and volts is calibrated such that the Amps get measured as multiples of coulomb per one second. Per one second. I'm guessing that our calibration for ohm values and volt values would likely be much different if it was calibrated to some other time length than 1 second. You see what I'm saying? I could be completely wrong that's why I'm asking. And if I'm not wrong then that was sort of my point in seeing a time relationship potentially in theory between the different formulae.
     
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