{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# Eq 1 and 3 show us that and that therefore and

This preview shows pages 3–5. Sign up to view the full content.

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:

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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.
.521 is very close to .5 or, one half the emf, and results in a 4.2% error. Proven theoretically-
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page3 / 5

Eq 1 and 3 show us that And that Therefore And finally our...

This preview shows document pages 3 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online