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Voltage -> Energy

  1. Sep 1, 2007 #1
    1. The problem statement, all variables and given/known data

    The manufacturer of a 6 V dry-cell flashlight battery says that the battery will deliver 15 mA for 60 continuous hours. During that time, the voltage will drop from 6 V to 4 V. Assume the drop in voltage is linear with time. How much energy does the battery deliver in this 60 h interval?

    2. Relevant equations

    [tex]p=\frac {dw}{dt}=vi[/tex]
    [tex]w=\int vi \; dt[/tex]
    [tex]y-y_1=m(x-x_1)[/tex]


    3. The attempt at a solution

    Okay, so this is a pretty elementary problem...but whatever. I know that at t=0, the battery will be putting out 6 V. I also know that after 60 hours (or 2.6E^5 seconds), the battery will be putting out 4 V. I used this information, along with the equation of a line in point-slope form, to find a function of voltage in terms of time. That function is...

    [tex]V(t)=-9.3E^{-6}t + 6 [/tex]

    Okay, so now I have a function for voltage, and a constant current, enough to use my second equation, [tex]w=\int vi \; dt[/tex], where w is energy.

    [tex]w=\int_0^{2.16E^5}(-9.3E^{-6}t+6)(15E^{-3}\;dt[/tex]
    The units work out right, and I arrive at -3.25E^9 J.

    Does everything look alright? Or have I gone in a totally incorrect direction. The negative value is what is really throwing me off.
     
  2. jcsd
  3. Sep 1, 2007 #2

    Astronuc

    User Avatar

    Staff: Mentor

    Since the current is constant, bring it outside of the integral.

    One can also write the integrand as 6 - 9.26E-6t, which should integrate to _____?

    I get a positive number.


    One can do a quick overcheck with V*I*time, remembering that V*I = power.

    Now since V decreases linearly from 6 to 4 V, the average is simply 5 V (J/coul), then multiply by current 0.015 A (coul/s), and time of 216,000 s.
     
  4. Sep 1, 2007 #3
    Thanks! I'm not exactly sure why I was getting a negative number, but when I went through and re-did it, it came out positive. But hey, atleast I know my method was sound!

    I wasn't positive I could just use the average, although it makes perfect sense. Silly me.


    Once again, thank you very much!
     
  5. Sep 2, 2007 #4

    CEL

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    I also found a negative energy, which is OK. A positive energy is dissipated at the load. The power source delivers a negative energy.
    This goes with the definition of active and passive elements. In a passive element the energy is always nonnegative, while in an active element it can be negative. The power source is an active element.
     
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