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Measurements and Instrumentation

Q1:

The current through a resistance R is shown in figure. The computed value of power is

A 400 ± 0.42 W

B 400 ± 4.6 W

C 400 ± 8.85 W

D 400 ± 10.65 W

ANS:C - 400 ± 8.85 W

 error in current is 2% and % error in resistance is 0.2%. Since P = I2R, % error in power = 2 x 2% + 0.2% = 4.2%.



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