A cell balances against a length of 200 cm on a potentiometer wire, when it is shunted by a resistance of 8Ω. The balancing length reduces by 40 cm, when it is shunted by a resistance of 4 Ω . Calculate the balancing length when the cell is in open circuit. Also calculate the internal resistance of the cell.
Solution
Given: R1 = 8Ω, l2 = 200cm,
R2 = 4Ω, l'2 = 160cm
Balancing length (l1) = ?
Internal resistance of the cell (r) = ?
r = `"R" (("l"_1 - "l"_2)/"l"_2)`
From first condition ,
`"r" = 8 (("l"_1 - 200)/200)` ......(1)
From first condition ,
`"r" = 4 (("l"_1 - 160)/160)` .....(2)
From equation (1) and (2),
we get
`8 (("l"_1 - 200)/200) = 4 (("l"_1 - 160)/160)`
`therefore ("l"_1 - 200)/25 = ("l"_1 - 160)/40`
`therefore 25 "l"_1 - 4000 = 40 "l"_1 - 8000`
`therefore 15 "l"_1 = 4000`
`therefore "l"_1 = 266.67 cm`
From equation (1) we get ,
internal resistance ,
`"r" = 8 (("l"_1 200)/200) = (266.67 - 200)/25`
`therefore "r" = 2.667 Omega`
