Find the uncertainty in a calculated electrical potential difference from the measurements of current and resistance. Electric potential dif

Question

Find the uncertainty in a calculated electrical potential difference from the measurements of current and resistance. Electric potential difference depends on current and resistance according to this function V(I,R) = IR. Your measured current and resistance have the following values and uncertainties I = 9.3 Amps, 0.3 Amps and R = 19.7 Ohms and 0.6 Ohms. What is the uncertainty in the , ? Units are not needed in your answer.

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RI SƠ 3 years 2021-07-15T15:57:56+00:00 1 Answers 51 views 0

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  1. haidang
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    2021-07-15T15:59:40+00:00
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    Answer:

    The value is   \Delta V = 11.5 \ V

    Explanation:

    From the question we are told that

       The  formula for the Electric potential difference is V = IR

        The  current is  I = 9.3 \ A

         The  uncertainty of the current is   \Delta I = 0.3 \ A

         The  Resistance is   R = 19.7\ Ohms

           The  uncertainty of the  Resistance is  \Delta R = 0.6 \ Ohms

    Taking the log of both sides of the formula  

              log V= log(IR)

    =>       log V= logI+ logR

    differentiating both sides

            \frac{\Delta V}{V} = \frac{\Delta I}{I} + \frac{\Delta R}{R}

    Here V is mathematically evaluated as

              V = 9.3 * 19.7

                V = 183.21 \ V

    So from

           \Delta V = V * [\frac{\Delta I}{I} +\frac{\Delta R}{R} ]

            \Delta V = 183.21 [ \frac{0.3 }{9.3} + \frac{0.6}{19.7}]

    =>    \Delta V = 11.5 \ V

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