Figure `6.12` shows a potentiometer circular for comparison of two resistances. The balance point with a standard resistor `R = 10.0 Omeag ` is found to be `58.3 cm`, while that with the unknows resistance `X` is `68.5 cm`. Determine the value of `X`. What would you do if you fail to find a balance point with the given cell `E`?

Explanation

Here, `l_(1) = 58.3 cm`, `l_(2) = 68.5 cm`, `R = 10 Omega`, `X = `? Let `I` be the current in the potentiometre wire and `E_(1)` and `E_(2)` be the potential drops across `R` and `X`, respectively. Then
`(E_(2))/(E_(1)) = (IX)/(IR) = (X)/(R)` or `X = (E_(2))/(E_(1)) R` (i)
But `(E_(2))/(E_(1)) = (l_(2))/(l_(1))`
From Eq. (i),
`X = (l_(2))/(l_(1))R = (68.5)/(58.3) xx 10.0 = 11. 75 Omega`
If there is no balance point with the given cell of emf `E`, it means the potential drop across `R` or `X` is greater than the potential drop across the potentiometer wire `AB`. In order to obtain the balance point, the potential drops across `R and `X` are to be reduced, which is possible by reducing the current. For that, either a suitable resistance should be used. Another possible way is to increase the potential drop acros the potentiometer wire by increasing the voltage if driver cell.