What is the condition for the error in measuring resistance on a Wheatstone bridge to be a minimum? How can it be achieved?

Answer

In the middle of the slid wire.

Explanation

The relative error is `delta=(h_(R))/(R )+h/l+h/(L-i)`, where `h_(R)` is the error in the value of the calibrating resistance, h is the error in the position of the slide wire. We have `delta=(h_(R))/(R )+(h L)/(I (L-l))`
Hence the error is at its minimum when the expression y=l (L-l) is at its maximum. But `y=IL-l^2=(L^2)/4-L^2/4+2l L/2-l^2=(L^2)/4-(l-L/2)^2`
is at its maximum when l=L/2. i.e, when the slide is in hte middle of the scale. This will be the case, if the calibrating resistance is
choosen as close tot the resistance being measured as possible.