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(a) In a metre bridge the balance point...

(a) In a metre bridge the balance point is found to be at 39.5 cm from the end A, when the resistor Y is of 12.5 `Omega` . Determine the resistance of X. Why are the connections between resistors in a Wheatstone or metre bridge made of thick copper strips ?
(b) Determine the balance point of the bridge above if X and Y are interchanged.
(c ) What happens if the galvanometer and cell are interchanged at the balance point of the bridge? Would the galvanometer show any current ?

Text Solution

Verified by Experts

(a) Here Y= 12.5 `Omega`, length AD=`l_1` = 39.5 cm
`because X/Y = (l_1)/((100 - l_1))`
Hence `X = Y (l_1)/((100 -l_1)) = 12.5 xx (39.5)/(60.5) = 8.2 Omega`
Connections are made of thick copper strips so that their resistance may be extremely small and negligible, because these resistances are not accounted for in the formula of metre bridge.
(b) Let on interchanging X and Y, the new balance point is obtained at `l_2` , then
`Y/X = (l_2)/((100 - l_2)) " or " (12.5)/(8.2) = (l_2)/((100 - l_2)) rArr l_2 = 60.5 cm`
(c) At the balance point of the bridge if the galvanometer and cell are interchanged, it makes no effect on balance condition and the galvanometer will not show any deflection.
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