error %tage for resistance in parallel connection
I looked up the solution, it gives a general formula to be remembered for the %tage error without any derivation. I'm not sure how to form the equation for the error percentage. I only know rules for error propagation of addition, subtraction, multiplication and division. Thanks
13 Replies
@Gyro Gearloose
Note for OP
+solved @user to close the thread when your doubt is solved. Mention the user who helped you solve the doubt. This will be added to their stats.Yeah you have to apply the formula just calculate the resistance the perform error correction as useful and multiply then and just take the percentage
The derivation is
Comrade Rock Astley
change the - to + for error analysis
and you get your formula for parallel resistors
Indeed
In most of the cases (like this) the maximum percentage error is asked hence we consider the change in signs..
Also in some questions (very very rare) they would tell you to find the probable answer ,and also hint you to consider the negative.. in those cases we won't be finding maximum error and have to go Accordingly...
Thank you so much
So would it be incorrect to find the relative error by finding it for the numerator and the denominator then finding the total one?
Also, how is ln(R1) = dR1/R1?
The last step is differentiation
We then say that 100 times dR1/R1 is the percentage
dR1 is the amount of error of R1
Doesn't d in differentiation represent a very small change, so that we can find instantaneous rate of change. Wouldn't the amount of error, delta R1, be a different value and much larger than this tiny change?
You’re right
dx must represent a very tiny change in x
But then there is no finite limit as to how small it can be
The only comment we can make is that this method becomes more and more accurate as the error percentage gets smaller and smaller
This is just an approximation for practical purposes, if you wanna think of it that way
Got it, thanks
+solved @Comrade Rock Astley
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