Numerical explaning Different methods of Limiting Error of any Function (08:30)

Lecture Notes

Numericals explaining different methods of limiting error of any function.

Q. The following measurement are obtained on a single phase load : V = 220 V ± 1% ; I= 5A ± 1% and W= 555 W ± 2% . If the power factor is calculated using these measurement , the worst case error in the calculated power factor in percent is ___. [ Gate 2017]

Ans. Given that, V= 220 V ± 1%

I = 5A ± 1% , W= 555 W ± 2%

We know that , W = VI cos φ

Power factor = cos φ = \frac{W}{VI}

cos\; \phi =\frac{555}{220\times 5}= 0.504

Power factor = \frac{W}{VI}

Taking log on both sides

log cos φ = log W – (log V + log I)

log cos φ = log W – log V – log I

Now, differentiate w.r.t cos φ

\frac{1}{cos\; \phi }=\frac{1}{W}\frac{\delta W}{\delta cos\; \phi }- \frac{1}{V}\frac{\delta V}{\delta cos\; \phi }-\frac{1}{I}\frac{\delta I}{\delta cos\; \phi } \frac{\delta cos\; \phi }{cos\; \phi }=\frac{\delta W}{W }- \frac{\delta V}{V }-\frac{\delta I}{I }

\frac{\delta cos\; \phi }{cos\; \phi } = (±2 ±1 ±1) x 100

\frac{\delta cos\; \phi }{cos\; \phi } = ±(2 +1 +1) x 100

= ±4 %

P.f = cos φ = 0.504 ± 4 %

P.f = cos φ = \frac{W}{VI}

cos\phi=\frac{(555\pm2\%)}{(220\pm1\%)(5\pm1\%)}

P.f = cos φ = 0.504 ± 4 %

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