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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View Full DocumentWhich 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.
.521 is very close to .5 or, one half the emf, and results in a 4.2% error.
Proven theoretically
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 Spring '12
 Dr.Phaneuf
 Physics, Force, Electromotive Force, Volt, PMAX, maximum power output

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