: 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.

R = (18 ± 2) 10¹ Ω

ΔR = 2 10¹ Ω

Explanation:

Ohm's law relates voltage to current and resistance

           V = i R

            R = V / i

the absolute error of the resistance is

           ΔR = |

the absolute value guarantees the worst case, maximum error

           ΔR =

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 =

          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

as we know our atmosphere or air have different levels. every level has other gasses, and every gas has another permeability. when the light comes from lower permeability to higher permeability, it changes its direction. so that means that when the light comes from the moon to the earth, it changes it's direction many more times. it basically means that the shape of the light part of the moon is changed on the earth.

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