Physics, asked by amritpreetkaurjuneja, 6 months ago

In Ohm's experiment to measure resistance different observation of unknown resistance are 4.12ohm, 4.08ohm,4.22ohmand 4.14ohm .Find average absolute error, relative error and percentage error.​

Answers

Answered by karajini2004
0

Answer:

Absolute error = 0.02, 0.06, 0.08 and 0.00

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

−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}R

mean

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

R_{mean}=4.14R

mean

=4.14

Absolute error,

ΔR₁= R_{mean}-R_{1}R

mean

−R

1

=4.14-4.12=0.02

ΔR₂= R_{mean}-R_{2}R

mean

−R

2

=4.14-4.08=0.06

ΔR₃= R_{mean}-R_{3}R

mean

−R

3

=4.14-4.22=0.08

ΔR₄= R_{mean}-R_{4}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.

ΔR_{mean}R

mean

=(0.02+0.06+0.08+0.00)/4=0.04

Relative error = ΔR_{mean}R

mean

/R_{mean}R

mean

=0.04/4.14=9.66\times10^{-3}9.66×10

−3

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

Percentage error = (ΔR_{mean}R

mean

/R_{mean}R

mean

)*100

\frac{0.04}{4.14} \times100

4.14

0.04

×100 =9.67%

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