# Percentage uncertatainty

## Homework Statement

In a simple electrical circuit, the current in a resistor is measured as 2.5 ± 0.05 mA. The resistor is marked as having a value of 4.7 Ω ± 2%. If these values were used to calculate the power dissipated in the resistor, what would be the percentage uncertainty in the value obtained?

none

## The Attempt at a Solution

i'm sorry, i post this in the advanced physics too, i'm confused where to post this.. hope you can help me...

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PeterO
Homework Helper

## Homework Statement

In a simple electrical circuit, the current in a resistor is measured as 2.5 ± 0.05 mA. The resistor is marked as having a value of 4.7 Ω ± 2%. If these values were used to calculate the power dissipated in the resistor, what would be the percentage uncertainty in the value obtained?

none

## The Attempt at a Solution

i'm sorry, i post this in the advanced physics too, i'm confused where to post this.. hope you can help me...
I think that when multiplying two quantities, you merely add the percentage errors??

In this case you are multiplying 3 things I x I x R as in P = I2R; so you would add all three percentage errors.
You are given the percentage error in R, and can calculate the percentage error in I from the values given.

Peter

Can you show me the solution? i'm confused... thanks peter...

PeterO
Homework Helper
Can you show me the solution? i'm confused... thanks peter...
If you wish to calculate the area of a rectangle with length 5.0 ±0.1 cm by 10.0 ± 0.1 cm, and give a percentage uncertainty, then

A = l x w so Area = 50 cm2

Now the percentage error.
5.0 ± 0.1 means an uncertainty of 0.1 in 5 or 1 in 50 or 2%
10.0 ± 0.1 mans an uncertainty of 1 in 100 or 1%