The potentiometer wire `AB` is `600` cm long.
a. At what distance from A should be jockey `J` touch the wire to get zero deflection i the galvanometer.
b. If the jockey touches the wire at a distance `560 cm` from A, what will be the current through the galvanometer.

Answer

`320 cm` b. (3 epsilon)/(22r)`

Explanation

Current in `AB`, `I = (E)/(r + 15 r) = (E)/(16 r)`
Potential gradient, `k = (15 rI)/(600) = (15 r)/(600) xx (E)/(16r) = (E)/(640)`
Let `E//2` be balanced at `l` then
`(E)/(2) = kl = (E)/(640)l` or `l= 320 cm`
b. Let potential at `C` be zero and at `A` be x. Then applying kirchhoff's first law at `C`, we get
`(x - 0)/(14r) + ((x - E)//(2 - 0))/(r) + ((x - E - 0))/(2r) = 0`
or `x = (14 E)/(22)`
`I_(g) = (x - (E)/(2))/(r) = (((14 E)/(22)) - (E)/(2))/(r) = (3E)/(22r)`