Question

some amount of current is measured by two different ammetres such that I1= 10.04 and I2= 9.88 calculate the relative error and percentage error

Answer 1

Answer:

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Answer 2

relative error=∆a(mean)÷a(mean)

percentage error=∆a(mean)÷a(mean)×100=relative error ×100

so, we will find the a(mean) and ∆a(mean) first and then the relative error and percentage error.

calculating the a(mean):-

$a(mean) = \frac{10.04 \times + 9.88}{2} = 9.96$

we will take the mean as 9.96 because we have to consider the significant figures concept also while calculating the errors.

during addition we consider the least number of decimal digits behind any of the number added for addition in significant figures.

now, we will calculate the ∆a(mean):-

$delta \: a(mean) = \frac{(a1 - a(mean)) +(a2 + a(mean)) }{2}$

$delta \: a(mean) = \frac{(10.04 - 9.96) + (9.88 - 9.96)}{2}$

remember that in the above case one value will be positive and other will be negative but to calculate the maximum possible error we take the modulus of every value.

so, by the above equation ∆a(mean)=0.08

so, the relative error will be=0.08÷9.96=0.008032128514

but now the significant figures concept comes into play.

we will have to take our error such that it has only two significant figures.

in case of division or multiplication, we find the number having least significant figures and the answer must also have the same number of significant figures.

now the relative error becomes =0.01(round off to two significant figures)

the percentage error will be =0.01×100=1%

relative error has no units or symbols, it is just a real number.

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