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The following measurements are obtained on a single-phase load: V = 220 V ± 1%, I = 5.0 A ± 1% and W = 55 W ± 2%. If the power factor is calculated using these measurements, the worst-case error in the calculated power factor in percent is _________. (Give answer up to once decimal)

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Concept:

Combination of Quantities with Limiting Error:

Two or more quantity each having a limiting error are combined it is advantageous to be able to compute the limiting error of the combination.

The limiting error can easily obtain by considering the relative increment of the function.

Considered y be the final result which is the function of measured quantity u, v, and w.

Product or Quotient of Quantities:

Let,

\(y = uvw\;or\;y = \frac{u}{{vw}}\;or\;y = \frac{1}{{uvw}}\)

The corresponding limiting error can be written as,

\(\frac{{\delta y}}{y} = \pm \left( {\frac{{\delta u}}{u} + \frac{{\delta v}}{v} + \frac{{\delta w}}{w}} \right)\)

Calculation:

Given that,

V = 220 V ± 1%

I = 5 A ± 1%

W = 55 W ± 2%

Power factor = cos ϕ

W = VI cos ϕ

\(\Rightarrow \cos \phi = \frac{W}{{VI}}\)

Hence, the total error in the measurement

= ± (1 + 1 + 2)%

= ± 4%

Sum of Quantities:

Let,

y = u + v

The relative increment of the function is given by,

\(\frac{{dy}}{y} = \frac{{d\left( {u + v} \right)}}{y} = \frac{{du}}{y} + \frac{{dv}}{y}\)

The expression the result in terms of relative increment of the component quantities

\(\frac{{dy}}{y} = \frac{u}{y}\frac{{du}}{u} + \frac{v}{y}\frac{{dv}}{v}\)

The error in the component quantity is represented by ± δu and ± δv.

The corresponding limiting error can be written as,

\(\frac{{\delta y}}{y} = \pm \left( {\frac{u}{y}\frac{{du}}{u} + \frac{v}{y}\frac{{dv}}{v}} \right)\)

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