Two potentiometer wires `w_(1)` and `w_(2)` of equal length `l` connected to a battery of emf `epsilon_(P)` and internal resistance 'r' as shown through two switches `s_(1)` and `s_(2)`. A battery of emf `epsilon` is balanced on these potentiometer wires. if potentiometer wire `w_(1)` is of resistance 2r and balancing length on `w_(1)` is `l//2` when only `s_(1)` is closed and `s_(2)` is open. On closing `s_(2)` and opening `s_(1)` the balancing length on `w_(2)` is found to be `((2l)/(3))` then find the resistance of potentiometer wire `w_(2)`

Answer

r

Explanation

When `S_(1)` is closed
`i = (epsilon_(rho))/(2r +r) = (epsilon_(rho))/(3r)`
`:. Epsilon - (epsilon_(rho))/(3r). R = (epsilon_(rho))/(3)`….(1)
When `S_(2)` is closed (Let resistance of `w_(2)` be R)
`I = (epsilon_(rho))/(R + r)`
`epsilon = (epsilon_(rho))/(R + r) (R (2)/(3))`....(2)
From (1) & (2) R = r
Alternate solution :
If in second case both `S_(1)` & `S_(2)` are closed
`epsilon=(epsilon_(rho))/(r+2r).(2r)/(l).(l)/(2)=(epsilon_(rho))/(3)`
`epsilon=(epsilon_(rho))/(r+(2rx)/(2r+x)).(2rx)/(2r+x)(2)/(3)`
`rArr (epsilon_(rho))/(3)=(4 epsilon_(rho)x)/(2r+x+2x)`