(i) In a meter bridge , the balance point is found to be at `30 cm` from the end A when resistance `R` in left gap of bridge is of `12 Omega`. Find resistance `S` in the right gap of bridge
(ii) If the cell and the galvanometer are interchanged in the balance point, would it effect flow of current through the galvanometer
(iii) Calculate the balance point of the bridge if `R` and `S` are interchanged

Answer

`70 cm`

Explanation

`(R )/(S) = (I)/((100 - 1)) or (12)/(S) = (30)/(100 - 30) = (30)/(70) = (3)/(7)`
or `84 = 3S or S = (84)/(3) = 28 Omega`

When the wheatone bridge is balance the potential of `B` is same as that of `D` ,so `(P)/(Q) =n (R )/(S)`
`or (P)/(R ) = (Q)/(S)` which is the condiction for balanced wheatston bridge if mean potential of A is same as that of C Therefore three will be current flowing through galvanometer
(iii) When `R and S` are interchanged , then
`(S)/(R ) = (1)/((100 - 1)) or (28)/(12) = (1)/((100 - 1))`
or `(7)/(3) = (1)/((100 - 1)) or 1 = 70 cm`