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.

Answer 1

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

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

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

=>    $\Delta V = 11.5 \ V$