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Discussion :: Measurements and Instrumentation

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

    2. A.
    400 ± 0.42 W
    B.
    400 ± 4.6 W
    C.
    400 ± 8.85 W
    D.
    400 ± 10.65 W

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    Answer : Option C

    Explanation :

    % 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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