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Interpreting an XY Scatter Graph

Interpreting an XY Scatter Graph. Example Ohm’s Law. Data. If we set up a circuit with some resistor and an ammeter (to measure current) in series. And supply voltages 1,2,3,4, and 5 volts respectively, the currents (in milliAmperes) shown were measured.

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Interpreting an XY Scatter Graph

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  1. Interpreting an XY Scatter Graph

  2. Example Ohm’s Law

  3. Data If we set up a circuit with some resistor and an ammeter (to measure current) in series. And supply voltages 1,2,3,4, and 5 volts respectively, the currents (in milliAmperes) shown were measured.

  4. Highlight the data, make an XY Scatter chart and choose Layout #9.

  5. Label axes (including units), eliminate legend, change title.

  6. Current versus Voltage • In our set up, we controlled voltage. Furthermore we used the convention that the variable we controlled is plotted on the x-axis. • Now we want to compare the fit line (Trendline) from Excel with the equation from our theory – Ohm’s Law. Excel: y = 0.2807x + 0.0019 Theory V=IR

  7. Rearranging

  8. Identification (Part 1)

  9. Identification (Part 2)

  10. Right click on the Trendline, choose Format Trendline, check Set intercept

  11. Identification (Part3)

  12. Units of slope • Slope is “rise over run” so its units are mA/V. • The resistance was the reciprocal of slope, so its units are V/mA. • The m means milli means 10-3. • A factor of 10-3 in the denominator corresponds to a factor of 10+3 in the numerator. 10+3 goes with kilo or k. • A volt/ampere is an ohm. Thus we have units for the resistance in kilo-ohms.

  13. Interpolation • Our equation y = 0.2812x or I = 0.2812V allows us to guess what the current would be for voltages that we did not measure. • For instance, we would expect the current corresponding to a voltage of 2.5 to be 0.2812*2.5 = 0.703 mA. • When we use the Trendline to estimate a new value within our range (here 1-5 volts), this is called Interpolation.

  14. Extrapolation • We can also use our results to estimate what the current would be for voltages outside the range we measured. • For instance, we would expect the current corresponding to a voltage of 6 to be 0.2812*6 = 1.6872 mA. • When we use the Trendline to estimate a new value outside our range (here 1-5 volts), this is called Extrapolation.

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