The potentiometer wire of length `200 cm` has a resistance of `20 Omega`. It is connected in series with a resistance `10 Omega` and an accumulator of emf `6 V` having negligible internal resistance. A source of `2.4 V` is balanced against length `1` of the potentiometer wire. Find the length `l` of the potentiometer wire. Find the length `l`

Answer

` 120 cm`

Explanation

Here `L = 200 m , r = 20 Omega R = 10 Omega ,V = 6 V`
`epsilon = 2.4 V`
Potential gradient
`K = ((V)/(R + r)) (r )/(L)= ((6)/(20 + 10))= (20)/(200) V//cm `
Balancing length `= (epsilon)/(K) = (2.4)/(0.02) = 120 cm`