Wheatstone Bridge new

Wheatstone Bridge

No1 No2 No3 No4 No5 10 meters 10 meters 20 meters 20 meters 10 meters 22 Copper Wire 28 Copper Wire 22 Copper Wire 28 Copper Wire 22 Copper Nickel Wire Part 1: TABLE ONE: MEASUREMENTS / RESISTANCES WIRE NO. RUN # R 0 L 1 L 2= 100-L1 R X = L1/L2 x R0 Average: < R X > 5 1 20 39 100 – 39= 61 12.7 12.84 2 16 44.5 100 – 44.5= 55.5 12.82 3 12 52 100 – 52= 48 13 4 1 6 46.5 100 – 46.5= 53.5 5.21 4.34 2 5 47.5 100 – 47.5= 52.5 4.52 3 4 43.7 100 – 43.7= 52.7 3.31 3 1 1 51.1 100 – 51.1= 48.9 1.04 1.07 2 2 35.3 100 – 35.3= 64.7 1.09 3 3 27 100 – 27= 73 1.10 2 1 4 35 100 – 35= 65 2.15 2.13 2 3 42 100 – 42= 58 2.17 3 2 51 100 – 51= 49 2.08 1 1 3 27 100 – 27= 73 1.10 0.73 2 2 23 100 – 23= 77 0.59 3 1 34 100 – 34= 66 0.51 TABLE TWO: RESISTIVITIES WIRE NO. METAL LENGTH L DIAMETER D CROSS SECTION AREA A < R x > ρ 5 Copper-Nickel 10 m 0.635 mm 3.17 x 10^-7 m 12.84 4.07 x 10^-7 4 Copper 20 m 0.305 mm 7.31 x 10^-8 m 4.34 1.58 x 10^-8 3 Copper 20 m 0.635 mm 3.17 x 10^-7 m 1.07 1.69 x 10^-8 2 Copper 10 m 0.305 mm 7.31 x 10^-8 m 2.13 1.55 x 10^-8 1 Copper 10 m 0.635 mm 3.17 x 10^-7 m 0.73 2.31 x 10^-8 Average p = 1.78 x 10^-8 Discrepancy of Copper Wires: 3.48%

Case 1: Case 2: Case 3: Question 3 : On the basis of the above table for Question 2, explain the reasons for the suggested values of R 0 in “Procedure”.

A: The reasons for the suggested values of R 0 in procedure is to help decrease the percent error. We need to know the suggested values of R0 from the procedure in order to have a low value for percent error. We can find that the closer we are to 50 cm the lower the percent error is. Conclusion: In the final analysis, we were able to learn how to calculate the resistivity of metals. We were able to comprehend how the Wheatstone Bridge is used to accumulate accurate measurements of resistance. Since the percent discrepancy was 3.48%, we can conclude that there was only a small amount of error when finding the values of the copper samples.