In the simple Wheatstone bridge circuit, where the length `AB` of bridge circuit wire is `1m` the resistance `X` and `Y` have value `5 Omega` and `2 Omega` respectively, When `X` is shuted by a length of a wire the balance point to found to be `0.625 m` from A. What is the resistance of the shunt ? If the shunt is `0.75 m` long end 0.25 mm in diameter, what is the respectivity of the material of the wire ?

Answer

`6.54 xx 10^(-7) Omega m `

Explanation

Let `R` be the resistance of shunted wire , the effective resistance of `R and 5 Omega` is parallel
` =- (5 xx R) (% + R)`
As balance pouint
`(5R//(5 + R))/(2) = (0.625)/(1 - 0.625) = (0.625)/(0.375) = (5)/(3)`
On solving we get `R = 10 Omega`
Now `p = (RA)/(1) = (R pir^(2))/(l) `
`= (10 xx (22//7) xx (0.125 xx 10^(-3))^(2))/(0.75)`
`= 6.54 xx 10^(-7) Omega m `