Numerical explaning Limiting Error of any Function (09:41)

Lecture Notes :-

Let us solve Numericals explaining limiting error of any function.

Q. A variable w is related to three variables x, y, z as w=\frac{xy}{z} . The variables are measured with meters of accuracy ± 0.5 % reading, ±1% of full scale value and ±1.5% reading. The actual readings of the three meters are 80,20 and 50 with 100 being the full scale value for all three. The maximum limiting error in the measurement of w will be [ Gate-2006]

a) ± 0.5 % reading

b) ± 5.5 % reading

c) ± 6.7 % reading

d) ± 7.0 reading

Ans.

Given that , w=\frac{xy}{z}

Taking log on both sides

log X = log X₁ + log X₂

Differentiate w.r.t x

\frac{1}{X}=\frac{1}{x_{1}}\frac{dx_{1}}{dX}+\frac{1}{x_{2}}\frac{dx_{2}}{dX} \frac{dX}{X}=\pm \left ( \frac{dx_{1}}{x_{1}}+ \frac{dx_{2}}{x_{2}} \right )

It is also given that, full scale reading = 100

Reading of X = 80

Reading of Y = 20

Reading of Z = 50

w=\frac{xy}{z} —(i)

Taking log on both sides,

Log w = log x + log y – log z

\frac{1}{\omega }=\frac{1}{x}\frac{\delta x}{\delta \omega }+\frac{1}{y}\frac{\delta y}{\delta \omega }- \frac{1}{z}\frac{\delta z}{\delta \omega} \frac{\delta\omega }{ \omega }=\frac{\delta x}{x}+\frac{\delta y}{y}-\frac{\delta z}{z }

Here δx,δy and δz are the unknown values.

δx = ± 0.5 % of reading

= \pm \frac{0.5}{100}\times 80=\pm 0.4

δy = ± 1 % of full scale value

= \pm \frac{1}{100}\times 100=\pm 1

δz = ± 1.5 % of reading

= \pm \frac{1.5}{100}\times 50=\pm 0.75 \frac{\delta w}{w}=\frac{\delta x}{x}+\frac{\delta y}{y}-\frac{\delta z}{z} \frac{\delta w}{w}=\pm \left ( \frac{0.4}{80}+\frac{1}{20}+\frac{0.75}{50} \right )\times 100= \pm 7 %

So, option (d) is the correct answer.

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