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

3980

Explanation

`V = kl (1muV//mm)(10 xx 10^(3) mm) = 10^(-2)V`
`I = V/R =(10^(-2)V)/(20 Omega) = 5 xx 10^(-4) A`
If `R.` is the required resistance to be introduced in the resistance box,
`I = E/(R + R.) rArr 5 xx 10^(-4) =2/(20 Omega + R.)`
`rArr R. = 3980 Omega`