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Eq 1 and 3 show us that and that therefore and

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Eq. (1) and (3) show us that: And that: Therefore: And finally, our result matches Eq. (4): Now, we will prove that Eq. (4) and (5) can be combined to achieve the desired equation: Eq (5) states that R=r when P is at a maximum Therefore:
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Which can be reduced for the desired result: 2) Determine the experimental values for E and r using the data from the first graph and equation (2). From the second graph, read P max and compare with E 2 /4r. Read R for P max and compare with r. Equation 2 states that: Additionally, we know that when P is at a maximum, i 2 r=i 2 R From Table 1 we know that Power is at a maximum when R=50 ohms. Therefore, r=50 ohms. Adapting equation 2: And for comparison, using: Our measured value for P max was .0432 Watts, so to find our percent difference: 3) Show that at maximum power output, the terminal voltage of the battery is half its emf. Proven experimentally- We calculated the emf to be equal to 2.88 Volts in part 2. At maximum power output, the terminal voltage was measured at 1.46 Volts.
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.521 is very close to .5 or, one half the emf, and results in a 4.2% error. Proven theoretically-
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Eq 1 and 3 show us that And that Therefore And finally our...

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