Question

In Ohm’s experiments, the values of the unknown resistances were found to be 6.12 Ω, 6.09 Ω, 6.22 Ω, 6.15 Ω. Calculate the (mean) absolute error, relative error, and percentage error in these measurements.

Answer 1

Given: a₁ = 6.12 Ω, a₂ = 6.09 Ω, a₃ = 6.22 Ω, a₄ = 6.15 Ω

To find:

i. Absolute error `(Δ"a"_"mean")` 
ii. Relative error
iii. Percentage error

Formulae: 

1. `"a"_"mean" = ("a"_1 + "a"_2 + "a"_3 + "a"_4)/4`

2. `Delta"a"_"n" = |"a"_"mean" - Delta"a"|`

3. `Delta"a"_"mean" = (Delta"a"_1 + Delta"a"_2 + Delta"a"_3 + Delta"a"_4)/4`

4. Relative error = `(Delta"a"_"mean")/"a"_"mean"`

5. Percentage error = `(Delta"a"_"mean")/"a"_"mean" xx 100%`

Calculation: From formula (i),

`"a"_"mean" = (6.12 + 6.09 + 6.22 + 6.15)/4`

`= 24.58/4 = 6.145` Ω

From formula (ii),

`Delta"a"_1 = |6.145 - 6.12| = 0.025`

`Delta"a"_2 = |6.145 - 6.09| = 0.055`

`Delta"a"_3 = |6.145 - 6.22| = 0.075`

`Delta"a"_4 = |6.145 - 6.15| = 0.005`

From formula (iii),

`Delta"a"_"mean" = (0.025 + 0.055 + 0.075 + 0.005)/4`

`= 0.160/4`

= 0.04 Ω

From formula (iv),

Relative error =`0.04/6.145 = 0.0065` Ω

From formula (v),

Percentage error = 0.0065 × 100 = 0.65%

1. The mean absolute error is 0.04 Ω.
2. The relative error is 0.0065 Ω.
3. The percentage error is 0.65%.