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Homework Help: Total charge lifted by the charge escalator

  1. Feb 27, 2010 #1
    total charge lifted by the "charge escalator"

    1. The problem statement, all variables and given/known data

    A 1.5 V flashlight battery is connected to a wire with a resistance of 5.00 ohms. The figure shows the battery's potential difference as a function of time. The slope is linear going from (0, 1.5 V) to (2 hrs, 0).

    2. Relevant equations

    I = [tex]\Delta[/tex]V / R
    I = dQ / dt

    3. The attempt at a solution

    I = 1.5 V / 5 ohms = 0.3 A
    0.3 A = dQ / (2 hrs x 60 x 60) --> dQ = 2160 C

    This is not the correct answer, and I don't understand why. Thanks for your help.
  2. jcsd
  3. Feb 27, 2010 #2
    Re: total charge lifted by the "charge escalator"

    When you solve the problem like that, you are assuming the current was .3A for the entire 2 hours, but as the graph shows, the voltage and therefore current steadily increase with time.

    Set the problem up more like this:
    [tex] \frac{V(t)}{R}=\frac{dQ}{dt}[/tex]
    You can solve it with one integration. If you don't know integration, you can solve it the way you did and use the average voltage.
  4. Feb 27, 2010 #3
    Re: total charge lifted by the "charge escalator"

    So I tried this:

    average of the voltage (1.5 V) over 2 hours = 1.5 / (2 x 60 x 60) = 2.1 x 10^-4

    average I = average V / R = 2.1 x 10^-4 / 5 = 4.2 x 10^-5

    average I = dQ / dt --> 4.2 x 10^-5 = dQ / (60 x 2 x 2) --> dQ = 0.01

    but this answer is still incorrect. Any ideas? Thanks.
  5. Feb 27, 2010 #4
    Re: total charge lifted by the "charge escalator"

    The average value of that function is 1.5/2 = .75 volts.

    Edit: You should probably solve it the calculus way since you've been told about the derivative definition of current:
    [tex] \frac{V(t)}{R}=\frac{dQ}{dt}[/tex]
    First we define V(t) as a function of time. It's of the form mx+b since it's linear, and by looking at the graph, you can calculate m and see that b = 0.

    multiply both sides by the differential time:
    [tex] \frac{mt*dt}{R}=dQ[/tex]
    Integrate both sides:
    [tex]\int_{t_i}^{t_F}\frac{mt}{R}\, dt=\int_{0}^{Q_{total}} dQ[/tex]
    [tex]Q_{total}=\int_{t_i}^{t_F}\frac{mt}{R}\, dt[/tex]
    Last edited: Feb 28, 2010
  6. Feb 28, 2010 #5
    Re: total charge lifted by the "charge escalator"

    I don't understand why the average V would be 1.5 / 2. Shouldn't I divide by the time in seconds? 2 hours = 2 x 60 x 60 seconds, that's why I used this number as the denominator.


    What numbers do I put in this? Is m the slope of the graph? Would that be 1.5 / 2 or 1.5 / (2 x 60 x 60)? And what am I using for t, 2 or the time in seconds?

    If I use t = 2, then I have Q = ((1.5/2) x 2^2) / (2 x 5) = 0.3 C
  7. Feb 28, 2010 #6
    Re: total charge lifted by the "charge escalator"

    You wouldn't divide by the time to find average voltage. An easy way to check this is to see the units would be volts per second for average voltage... We want volts. The function's average is the middle value of the linear function, because the higher parts past the halfpoint are exactly balanced by the lower parts before it.

    As for your second question, if you are integrating, you can use time in hours if the slope is in voltage per hour or you can use seconds if the slope is in voltage per second. It's up to how you find your slope. Slope WOULD be V/s unlike average voltage. m = 1.5/(2*3600) for volts per second.

    However, when you're using average voltage, you need to use seconds and cannot use hours or anything due to the definition of current (which is found by V/R). 1.5/2 * 2 * 3600/5 = total charge.
  8. Feb 28, 2010 #7
    Re: total charge lifted by the "charge escalator"

    i did average V / R = (1.5 / 2) / 5 = 0.15
    I = dQ / dt --> 0.15 = dQ / (2 x 60 x 60) --> dQ = 1080 C
    and it's correct! finally! thanks for your help.
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