The figure shows a potentiometer circuit for comparison of two resistances. The balance point with a standard resistor `R = 10.0Omega` is found to be 58.3cm, while that the unknown resistance 'X' is 68.5 cm. Determine the value of x. What might you do if you failed to find a balance point with the given cell of emf `epsilon`?

Explanation

Given : `l_(1)=53.3cm,l_(2)=68.5 cm,R=10Omega.X=?`
Let .I. pe the current in the potentiometer wire.
Let `V_(1)` and `V_(2)` be potential drops across R and X respectively.
Thus `(V_(1))/(V_(2))=(IR)/(Ix)`
i.e., `(V_(1))/(V_(2))=R/x`
or `x=(RV_(2))/(V_(1))` ....(1)
`therefore(V_(2))/(V_(1))=(l_(2))/(l_(2))` ...(2)
by using (2) in (1) we get,
`x=(10l_(2))/(l_(1))=(10xx68.5)/(58.3)`
`thereforex=11.75Omega`
If we fail to find the balance point with the given cell of emf `epsilon` then the p.d across R or X will be greater than the p.d. across the potentiaometer wire. A suitable resistor of high resistance may be used in series with the cell of emf `.epsilon.`.