How Does Error in Radius Affect Resistance Calculations?

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The forum discussion focuses on calculating the percentage error in the radius of a wire when estimating the electrical resistance, R, defined by the formula R = k / r^2, where k is a constant and r is the radius. The user seeks assistance in applying differentials to ensure that the percentage error in R remains within +/- 3%. The correct approach involves using the derivative dR/dr = -2k/r^3, leading to the relationship dR/R = -2(dr/r). By setting dR/R equal to 0.06, the user can derive the necessary percentage error in r.

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Hello,

I'm having trouble with this maths question and was wondering if someone could help me?? The question asks:


The electrical resistance R of a wire is given by

k / r^2 where k = constant, r = radius of wire.

Use differentials to estimate the percentage error in the measured value of r if we want the percentage error in R to be within +/- 3%.


I'm really not sure if this is the right approach... please help me...

Using the method of linear approx. i have...
ΔR = R(r + Δr) - R(r)
~ (dR(r) / dr) * ( Δr )

I know...
R = k / r^2
dR/dr = -2*k*( r^-3 )

and therefore...
ΔR = -2*k*( r^-3 ) * ( Δr )

So I'm thinking to increase R by 3% i need...
ΔR / R(r) = 0.03

but after substituting everything in i get Δr = 0.03??

please i really need help!
 
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caelestis said:
The electrical resistance R of a wire is given by

k / r^2 where k = constant, r = radius of wire.

Use differentials to estimate the percentage error in the measured value of r if we want the percentage error in R to be within +/- 3%.

R = k/r^2 =>
dR = -(2k/r^3)dr =>
dR/R = -2(dr/r).

Now put dR/R = 6/100 to find the corresponding dr/r.
 

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