How Is Percentage Uncertainty Calculated in Power Dissipation?

[moimoi24]

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?

Homework Equations

none

The Attempt at a Solution

please help me, what is the solution for this one? thanks

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

 

[PeterO]

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?

Homework Equations

none

The Attempt at a Solution

please help me, what is the solution for this one? thanks

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

 

[moimoi24]

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

 

[PeterO]

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%

SO the uncertainty in the answer is 3% [add them together]

So Area is 50 cm² ± 3%

That is as close to the solution you seek I will give,

 

[Nickie]

I would approach this problem by first understanding the concept of percentage uncertainty and its impact on calculations. Percentage uncertainty is a measure of the possible error or deviation in a measurement or calculated value. In this case, the current in the resistor is measured with a ± 0.05 mA uncertainty, while the resistor's value is given with a ± 2% uncertainty.

To calculate the power dissipated in the resistor, we use the formula P = I^2R, where I is the current and R is the resistance. Substituting the given values, we get P = (2.5 ± 0.05)^2 x (4.7 Ω ± 2%).

To determine the percentage uncertainty in the calculated power, we can use the following formula:

Percentage uncertainty = (absolute uncertainty / measured value) x 100

In this case, the absolute uncertainty in the current is 0.05 mA and the absolute uncertainty in the resistance is 0.094 Ω. The measured value for the current is 2.5 mA and the measured value for the resistance is 4.7 Ω.

Plugging these values into the formula, we get:

Percentage uncertainty = ((0.05 mA / 2.5 mA) + (0.094 Ω / 4.7 Ω)) x 100 = 4%

Therefore, the percentage uncertainty in the calculated power is 4%. This means that the actual power dissipated in the resistor could be 4% higher or lower than the calculated value.

In conclusion, it is important to consider the percentage uncertainty in all measurements and values used in calculations to better understand the possible error or deviation in the final result. This can help us improve the accuracy and reliability of our experiments and data analysis.