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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okpalawalter8 okpalawalter8 · Answer 1 · 2020-09-23T03:31:34+02:00

Answer:

The value is [tex]\Delta V = 11.5 \ V[/tex]

Explanation:

From the question we are told that

The formula for the Electric potential difference is [tex]V = IR[/tex]

The current is [tex]I = 9.3 \ A[/tex]

The uncertainty of the current is [tex]\Delta I = 0.3 \ A[/tex]

The Resistance is [tex]R = 19.7\ Ohms[/tex]

The uncertainty of the Resistance is [tex]\Delta R = 0.6 \ Ohms[/tex]

Taking the log of both sides of the formula

[tex]log V= log(IR)[/tex]

=> [tex]log V= logI+ logR [/tex]

differentiating both sides

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

Here V is mathematically evaluated as

[tex]V = 9.3 * 19.7[/tex]

[tex]V = 183.21 \ V [/tex]

So from

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

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

=> [tex]\Delta V = 11.5 \ V[/tex]