How Much Energy Does a Linearly Declining Battery Voltage Deliver Over Time?

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SUMMARY

The discussion centers on calculating the energy delivered by a battery with a linear voltage decline from 1.5V to 1.0V over a 40-hour period while providing a continuous current of 9mA. The power function derived is P(t) = 0.0135 - 3.1x10^-8t, leading to an energy calculation of 1622.6 Joules through integration. A discrepancy arises when comparing this result to the area under the power vs. time graph, which yields only 324 Joules, prompting a request for clarification on the graphing method used. The issue is identified as a failure to account for the entire area under the curve, specifically the triangular area representing the power function.

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  • Understanding of basic electrical concepts, specifically Ohm's Law.
  • Familiarity with calculus, particularly integration techniques.
  • Knowledge of power calculations in electrical circuits.
  • Ability to interpret graphical representations of mathematical functions.
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  • Review the principles of Ohm's Law and its application in circuit analysis.
  • Study integration techniques for calculating area under curves in calculus.
  • Learn about power calculations in varying voltage scenarios.
  • Explore graphical methods for accurately determining areas under curves.
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This discussion is beneficial for electrical engineering students, physics learners, and anyone involved in battery technology or energy calculations, particularly those interested in understanding power delivery over time in electrical systems.

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Homework Statement


A battery provides a continuous current of 9mA (.009A) for 40hrs (144,000sec). During that time the voltage drops from 1.5 to 1.0. Assume the drop is linear with time. How much energy does the battery deliver during this 40hr interval



Homework Equations


P=IV


The Attempt at a Solution


First I found an equation for voltage as a function of time (s) V(t)=1.5-3.472x10^-6t.
Then I figured power as a function of time P(t)=IxV(t)=.009(1.5-3.472x10^-6t)
P(t)=.0135-3.1x10^-8t. To find the energy delivered, I integrated P(t) so that E=.0135t-3.1x10^8t^2/2. The limits were from 0 to 144,000. I ended up with 1944-321.408=1622.6J
My problem is that when I just made a graph of power vs time and found the area under it from 0 to 144,000 I got 324J. Could I please get some guidance as to what I'm doing wrong?
 
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Check your graph.

ehild
 
Ok when I tried to find the area using the graph i only took the area of the triangle.
 

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