physics question
posted by ong .
A resistance box has the following components and tolerances:
10 resistors each of 100k§Ù ¡¾ 0.05%
10 resistors each of 10k§Ù ¡¾ 0.05%
10 resistors each of 1k§Ù ¡¾ 0.05%
10 resistors each of 100§Ù ¡¾ 0.1%
10 resistors each of 10§Ù ¡¾ 0.5%
Determine both in ohms and percentage the limit of uncertainty in setting of 453.72k§Ù

first, calculate the additive errors.
4x100k error is 400Kx.0005=200
5x10k error is 50kx.0005=25
3x1k error is 3kx.0005=1.5ohm
7x100 error is 700x.001=.7
2x10 error is 20x.005=.1 ohm
error^2=200^2+25^2+1.5^2+.7^2 +.1^2
do the math, that is the error in ohms. Then figure percent.
limit of uncertainity is harder. Notice your percents are to one significant figure, so 200 in the first one can range from 150 to 250. all the others are withing that error, so you have +250 ohm. Your instructor may have given you other instructions on this. If the percent error was to be considered absolute, then add the errors.