: Suppose somebody, using the same apparatus which you used, measured I = 45.5 ma, and V = 8.2 volts on some resistor. Using your recorded u

Question

: Suppose somebody, using the same apparatus which you used, measured I = 45.5 ma, and V = 8.2 volts on some resistor. Using your recorded uncertainties for the 50 ma and 10-volt scales, what would be the maximum % uncertainty in R if it were calculated from the Ohm’s Law Equation (1)? Use calculus methods to answer this question if you can.

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Thu Cúc 5 years 2021-08-30T01:30:52+00:00 1 Answers 46 views 0

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  1. Thunguyet
    0
    2021-08-30T01:32:30+00:00
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    Answer:

    R = (18 ± 2) 10¹ Ω

    ΔR = 2 10¹ Ω

    Explanation:

    Ohm’s law relates voltage to current and resistance

               V = i R

                R = \frac{V}{i}V / i

    the absolute error of the resistance is

               ΔR = | | \frac{dR}{DV} | \ \Delta V + | \frac{dR}{di} | \ \Delta i

    the absolute value guarantees the worst case, maximum error

               ΔR = \frac{1}{i} \Delta V+ \frac{V}{i^2} \Delta i

    The error in the voltage let be approximate, if we use a scale of 10 V, in general the scales are divided into 20 divisions, the error is the reading of 1 division, let’s use a rule of direct proportion

              ΔV = 1 division = 10 V / 20 divisions

              ΔV = 0.5 V

    The current error must also be approximate, if we have the same number of divisions

               Δi = 50 mA / 20 divisions

               Δi = 2.5 mA

           

    let’s calculate

              ΔR = \frac{1}{45.5 \ 10^{-3}} \ 0.5 + \frac{8.2}{(45.5 \ 10^{-3})^2 } \ 2.5 \ 10^{-3}

              ΔR = 10.99 + 9.9

              ΔR = 20.9 Ω

    The absolute error must be given with a significant figure

              ΔR = 2 10¹ Ω

    the resistance value is

              R = 8.2 / 45.5 10-3

              R = 180 Ω

    the result should be

              R = (18 ± 2) 10¹ Ω

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