The resistance of a potentiometer wire of length `10 m` is `20 Omega`. A resistance box and a `2` volt accumulator are connected in series with it. What resistance should be introduced in the box to have a potential drop of one microvolt per millimetre of the potentiometer wire ?

Answer

`980 Omega `

Explanation

Here `L = 10 m , r = 20 Omega V = 2 V, R = ?`
`K = 1 mu V//mm = 10^(-6) V//m = 10^(-3)V//m`
Now `K = ((V)/(R + r)) (r )/(L) or 10^(-3) = (2)/((R + 20)) xx (20)/(10)`
`or R + 20 = (4)/(10^(-3)) = 4000`
or `R= 4000 - 20 = 3980 Omega `