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

R = (18 ± 2) 10¹ Ω

ΔR = 2 10¹ Ω

Step-by-step explanation:

Ohm's law relates voltage to current and resistance

V = i R

R =
`(V)/(i)`V / i

the absolute error of the resistance is

ΔR = |
`| (dR)/(DV) | \ \Delta V + | (dR)/(di) | \ \Delta i`

the absolute value guarantees the worst case, maximum error

ΔR =
`(1)/(i) \Delta V+ (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 =
`(1)/(45.5 \ 10^(-3)) \ 0.5 + (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