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How is the slope of a voltage vs time graph the current?

  1. Dec 23, 2013 #1

    x86

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    A lot of websites say that if you take the slope of a voltage vs time graph, you get the current. However, the math tells a different story.

    where V = voltage, J = joules, C = coulombs, A = amperes, s = seconds

    A = C / s
    C = A * s
    V = J / C
    J = C / V

    if we take the slope of Voltage vs Time, our unit is:

    V/s = J/(C * s) = J / (A * s^2) = (C * V) / (A * s^2) = (C * V) / (C /s * s^2) = V/s

    No matter what I do, I can never get the unit ampere.

    How is it mathematically possible that the slope of a voltage vs time graph has the unit of the current? I don't get it.
     
  2. jcsd
  3. Dec 23, 2013 #2
    The slope does not have the units of current. The units are V/s (which can be expressed in different ways) as you have shown. Perhaps you misread the information on the websites or you looked at websites giving wrong information.
     
    Last edited: Dec 23, 2013
  4. Dec 26, 2013 #3
    Which websites?

    If V = J / C then J ≠ C / V

    You forgot to consider capacitance.

    First the variables need to be used properly.
    Type______Symbol______Unit
    Voltage...........E...............V
    Current...........I................A
    Charge...........Q................C
    Capacitance....C................F
    time...............t................s

    Q = C * E
    ∴ E = Q / C
    E / t = Q / (C * t)
    E / t = I / C
     
  5. Dec 26, 2013 #4

    mfb

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    Staff: Mentor

    That is true for capacitors if you include capacitance as constant factor.
     
  6. Dec 27, 2013 #5

    psparky

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    I agree with that if you are referring to:

    C*dv/d(t)=i(t)

    Then yes, the slope of the voltage multiplied by the capacitance will give the current at any given time thru the capacitor.
     
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