- #1
Luke1294
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Homework Statement
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?
Homework Equations
[tex]p=\frac {dw}{dt}=vi[/tex]
[tex]w=\int vi \; dt[/tex]
[tex]y-y_1=m(x-x_1)[/tex]
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.