Question

Consider a wire 1.00 meters in length with 0.100 mm diameter and resistivity, ? = 1.18              x              10^’(Ωm. Give all calculated answers to 3 significant figures and report voltages in millivolts.

a) (10 pts) Calculate the resistance of the wire.

?_)*+, = ________________________

b) (10 pts) The same wire is placed in an AC voltage divider circuit with a function generator. It follows, in series, a resistor, R=5.00 kΩ. An oscilloscope’s channel one shows a 9.10 peak-to-peak voltage sine wave across the resistor and wire series. If channel two is connected across only the wire, what peak-to-peak voltage is displayed on channel two?

?_)*+, = ________________________

c)     (10 pts) Like the experiment you performed in the lab, this wire has a sliding adjustable contact point which allows you to exclude a segment of the one meter wire, producing a smaller signal to channel two. Using included lengths ?_- = 0.9       ? and ?_. = 0.1   ?, calculate the corresponding output voltages, ?_- and ?_., you expect to measure at those points.

                 At ?_-,   ?_-________________________, and at ?_.,   ?_.________________________

d)    (10 pts) Now suppose that you use a multimeter to check the resistance of the wire. Measuring the resistance of the full 1.00 meter length of wire several times, you find resistance values of 16.5Ω, 16.7Ω, 15.9Ω and 15.7Ω. Perform 1D stats on this set of measurements and report the mean and standard error. Using these values, and your answer to part (a) as the accepted value, compare precision and accuracy of your resistance measurements. Briefly discuss how you interpret the precision vs accuracy test.

Answer 1

1 s=1m, 8- 1.18x10 m , R = 0.05x103 m REFL (lexió? xi) (ax (0.05X163)2) = R-15.02 How resistor & wire are connected in series, then het resistance, R=5KL + 15 r = soisor So, I - V - 1.81 mA Now voltage around wir wire = IXY Vwine = 1.81 mA xist = 27.2 mV To we know that Ral I So, Ri= R XL = 15x0.9 - 13.50 Lo R₂ =R x L₂ = 15001 - 1.500 V =I&R = 1.81MA X 13:52 = 24.4 mv V₂ = IXR₂ = 2.72 mv (d) Rmean = (16.5+ 16.7+ 15.9 + 15.7) – 16. 22 for Standard Error 0=Elli-M2 (n-1) 0= (16.5-16.2)² + (16.7-16.21² + 115.9-16.20² + (15.7-16.212 (4-1) = 0.476 How seat -0.47% - 0.238 JY 54 Now for mean error, ME = [116.5-16.21 + 1 16.7-16.21 + 185.9-16.21 +115.7-16.217 ME=0. you are

Accuracy refers to the closeness of a measured value to a standard or known value.

whereas, Precision refers to the closeness of two or more measurements to each other.

Now here we can see that the values of of different measurements are not too close thus the measurements are not enough precise.

Also, we can see that as we have assumed the calculated value to be accepted value with is 15.0 ohm , there is big difference between the measured values of more than 0.7 ohm thus the measured values are not enough accurate.

Thus we can say that the measured values are not enough precise and accurate as well.