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.

Answer 1

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