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3 years ago

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

otential 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.

1 answer:

german3 years ago

6 0

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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