(adsbygoogle = window.adsbygoogle || []).push({}); 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.

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# Homework Help: Voltage -> Energy

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