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

The value is
`\Delta V = 11.5 \ V`

Step-by-step 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

`(\Delta V)/(V) = (\Delta I)/(I) + (\Delta R)/(R)`

Here V is mathematically evaluated as

`V = 9.3 * 19.7`

`V = 183.21 \ V`

So from

`\Delta V  =  V * [(\Delta I)/(I) +(\Delta R)/(R) ]`

`\Delta V = 183.21 [ (0.3 )/(9.3) + (0.6)/(19.7)]`

=>
`\Delta V = 11.5 \ V`