physics lab 7

# Then use this data to determine graphically the

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Then use this data to determine graphically the relationship between voltage and current and the relationship between resistance and power. Data Analysis Displayed here is the data for Resistance, current, voltage, and power obtained during the experiment and on the two pages after the table there are graphs for power (J/s) vs. resistance(ohms) and voltage(V) vs. current (amps). Table data Resistance (ohms) Current (amps) Voltage (V) Power (J/s) 300 0.005 2.2 0.0110 200 0.008 2.4 0.0192 100 0.014 1.9 0.0266 90 0.014 1.9 0.0266 80 0.016 1.8 0.0288 70 0.017 1.7 0.0289 60 0.018 1.65 0.0297 50 0.022 1.65 0.0363 48 0.022 1.65 0.0363 46 0.023 1.61 0.0370 44 0.024 1.6 0.0384 42 0.024 1.58 0.0379 41 0.025 1.52 0.0380 40 0.027 1.5 0.0405 39 0.027 1.58 0.0427 38 0.027 1.52 0.0410 37 0.027 1.55 0.0419 36 0.027 1.5 0.0405 35 0.028 1.48 0.0414 34 0.028 1.48 0.0414 33 0.029 1.5 0.0435

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32 0.030 1.5 0.0450 31 0.031 1.4 0.0427 30 0.031 1.4 0.0427 29 0.032 1.45 0.0464 28 0.033 1.4 0.0462 27 0.034 1.35 0.0459 26 0.035 1.3 0.0455 Figure 1: V as a function of i Figure 2: P as a function of R for values of R up to 100 Ω Problems: 1) Show that Eq. (4) follows from Eq. (1) and (3), and that Eq. (4) with Eq. (5) gives P max =E 2 /4r. 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: 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=29 ohms. Therefore, r=29 ohms. Adapting equation 2: And for comparison, using: Our measured value for P max was .0464 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 1.86 Volts in part 2. At maximum power output, the terminal voltage was measured at 1.45 Volts.

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