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When two known resistance R and S are co...

When two known resistance R and S are connected in the left and right gaps of a meter bridge, the balance point is found at a distance `l_(1)` from the zero end of the meter bridge wire. An unknown resistance X is now connected in parallel to the resistance S and the balance point is found at a distance `l_(2)` from the zero end of the meter bridge wire,

Obtain a formula for X in terms of `l_(1), l_(2)` and S.

Text Solution

Verified by Experts

When resistance R and S are connected to the left and right gaps of meter bridge and bridge is balanced at length `l_(1)` from zero end, then
`R/S = (l_(1))/((100-l_(1)))`....(i)
When unknown resistance X is connected in parallel to S, then effective resistance in right gap is
`S' = (SX)/(S+X)`....(ii)
Now, balance point is obtianed at length `l_(2)`,
`:. (R)/(S') = (l_(2))/((100-l_(2)))`
Putting the value of S', we have
`(R ( S + X))/(SX) = (l_(2))/((100-l_(2)))`....(iii)
Dividing (iii) by (i), we get
`(S + X)/(X) = (l_(2))/((100 -l_(2))) xx ((100 - l_(1)))/(l_(1))`
or `S/X + 1= (l_(2)(100-l_(1)))/(l_(2)(100-l_(2)))`
or `S/X = (100 l_(2) - l_(1)l_(2) - 100 l_(1) + l_(1)l_(2))/(l_(2)(100-l_(2))) = (100(l_(2)-l_(1)))/(l_(2)(100-l_(2)))`
or `X = (l_(1)(100-l_(2)))/(100(l_(2)-l_(1))) S`
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