Cells `A` and `B` and a galvanometer `G` are connected to a side wire `OS` by two sliding contacts `C` and `D` as shows in Fig. `6.17`. The slide wire is `100 cm` long and has a resistance of `12 Omega`. With `OD = 75 cm`, the galvanometer gives no deflections when `OC` is `50 cm`. If `D` is moved to touch the end of wire `S`, the value of `OC` for which the galvanometer shows no deflection is `62.5 cm`. The emf of cell `B` is `1.0 V`. Calculate
(i) the potential difference across `O` and `D` when `D` is at `75 cm` mark from `O`
(ii) the potential difference across `OS` when `D` touches `S`
(iii) internal resistance of cell `A`
(iv) the emf of cell `A`

Explanation

Resistance of wire `OD` is
`(12)/(100) xx 75 = 9 Omega`
Let `E` and `r` be the emf and internal resistance of cell `E`, respectively.
(i) Since `1 V` is balanced across `50 cm`, so potential gradient of wire is `1//50 V cm^(-1)`. Therefore, voltage drop across wire `OD` of length `67 cm` is `(1//50) xx 75 = 1.5 V`.
(ii) Now potential gradient of wire is `1//62.5 V cm^(-1)` Therefore, voltage drop across wire `OS` of length `100 cm` is `(1//62.5) xx 100 = 1.6 V`.
(iii) For first case
`((E)/(9 + r)) xx 9= 1.5` (i)
For second case
`((E)/(12 + r)) xx 12 = 1.6` (ii)
(iv) On solving Eqs. (i) and (ii), we get `r = 3 Omega` and `E = 2 V`.