Question

In Ohm's experiment, the values of an unknown resistance were found to be 4.12 Ω, 4.08Ω, 4.22Ω and 4.14Ω. Calculate absolute error, relative error and percentage error in these measurements.

Answer 1

Answer:

Absolute error = 0.02, 0.06, 0.08 and 0.00

Relative error = $9.66 \times 10^{-3}$

Percentage errror = 9.67%

Explanation:

Absolute error: It is defined as difference between mean value and given individual value.

Given R₁=4.12, R₂=4.08, R₃=4.22,R₄=4.14

Mean value of all resistance=

$R_{mean}$=(R₁+R₂+R₃+R₄)/4=(4.12+4.08+4.22+4.14)/4

$R_{mean}=4.14$        

Absolute error,

ΔR₁= $R_{mean}-R_{1}$=4.14-4.12=0.02

ΔR₂= $R_{mean}-R_{2}$=4.14-4.08=0.06

ΔR₃= $R_{mean}-R_{3}$=4.14-4.22=0.08

ΔR₄= $R_{mean}-R_{4}$=4.14-4.14=0.00

Relative error:It is the ratio of absolute mean error and given individual mean value.

Δ$R_{mean}$=(0.02+0.06+0.08+0.00)/4=0.04

Relative error = Δ$R_{mean}$/$R_{mean}$=0.04/4.14=$9.66\times10^{-3}$

Percentage error = when relative error is expressed in terms of percentage then percentage error is defined.

Percentage error = (Δ$R_{mean}$/$R_{mean}$)*100

                           $\frac{0.04}{4.14} \times100$=9.67%

Answer 2

Answer:

Absolute error = 0.02, 0.06, 0.08 and 0.00

Relative error =

Percentage errror = 9.67%

Explanation:

Absolute error: It is defined as difference between mean value and given individual value.

Given R₁=4.12, R₂=4.08, R₃=4.22,R₄=4.14

Mean value of all resistance=

=(R₁+R₂+R₃+R₄)/4=(4.12+4.08+4.22+4.14)/4

       

Absolute error,

ΔR₁= =4.14-4.12=0.02

ΔR₂= =4.14-4.08=0.06

ΔR₃= =4.14-4.22=0.08

ΔR₄= =4.14-4.14=0.00

Relative error:It is the ratio of absolute mean error and given individual mean value.

Δ=(0.02+0.06+0.08+0.00)/4=0.04

Relative error = Δ/=0.04/4.14=

Percentage error = when relative error is expressed in terms of percentage then percentage error is defined.

Percentage error = (Δ/)*100

                          =9.67%