- #1
daveco-inc
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Hey, this is my first post ever on PF, and I have done my best to follow the guidelines on posts asking for help with homework. Please know that I am not after the answer just to cheat and get the questions right, I am after the method so that I could do a similar question myself in the future.
Three resistors in series are 4.0 Ohms, 9.4 Ohms and again 4 Ohms.
They all have a tolerance or uncertainty of 10%. What is the total resistance including uncertainties?
Total resistance in series = R(1) + R(2) + R(3)
The first part is easy, 4.0+9.4+4.0= 17.4
Now, I believe that when calculating uncertainty in this situation, you are meant to calculate the uncertainty for each resistor, and then add.
10% of 4 = 0.4
10% of 9.4 = .94
10 % of 4 = 0.4
Total uncertainty of all resistors combined = 1.74.
So am I correct in saying that the total resistance in series = 17.4 Ohms + or - 1.74 Ohms, correct? The thing I am worried about is whether it is +- 1.74, or half; 0.84.
Again, thanks for the help, and please let me know if I have broken any rules or how I can improve my post.
Thanks.
Homework Statement
Three resistors in series are 4.0 Ohms, 9.4 Ohms and again 4 Ohms.
They all have a tolerance or uncertainty of 10%. What is the total resistance including uncertainties?
Homework Equations
Total resistance in series = R(1) + R(2) + R(3)
The Attempt at a Solution
The first part is easy, 4.0+9.4+4.0= 17.4
Now, I believe that when calculating uncertainty in this situation, you are meant to calculate the uncertainty for each resistor, and then add.
10% of 4 = 0.4
10% of 9.4 = .94
10 % of 4 = 0.4
Total uncertainty of all resistors combined = 1.74.
So am I correct in saying that the total resistance in series = 17.4 Ohms + or - 1.74 Ohms, correct? The thing I am worried about is whether it is +- 1.74, or half; 0.84.
Again, thanks for the help, and please let me know if I have broken any rules or how I can improve my post.
Thanks.