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) = 58.3cm , l_(2) = 68.5cm ,R=10Omega ,X=?`
Let `I` be the current in the potentiometer wire and `Epsilon_(1)` and `Epsilon_(2)` be the potential drops across `R` and `X` respectively. When connected in circuit by closing respective key. Then `epsilon_(2)/epsilon_(1) = (IX)/(IR) =X/R or X = epsilon_(2)/epsilon_(1)R` ...(i)
But `epsilon_(2)/epsilon_(1) = l_(2)/l_(1)`
`:.` From (i), `X=l_(2)/l_(1)R = 68.3/58.3xx10.0 = 11.75Omega`
If there is no balance point with given cell of e.m.f. `epsilon`, it means 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 be reducing the current in `R` and `X`. For that, either a suitable resistance should be put in series with `R` and `X` or a cell of smaller e.m.f. `epsilon` should be used. Another possible way is to increase the potential drop across the potentiometer wire by increasing the voltage of driver cell.