Question

Figure shows a potentiometer circuit for comparison of two resistances. The balance point with a standard resistor R = 10.0 Ω is found to be 58.3 cm, while that with 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 ε ?

Answer 1

Resistance of the standard resistor, `R = 10.0 Omega`
Balance point for this resistance, `l_(1)=58.3` cm
Current in the potentiometer wire `= l`
Hence, potential drio across `R, E_(1)=iR`
Resistance of the unknown resistor = X
Balance point for this resistor, `I_(2)= 68.5` cm
Hence, potential drop across `X, E_(2)= iX`
The relation connecting emf and balance point is,
`(E_(1))/(E_(2))=(l_(1))/(l_(2))`
`(iR)/(iX) = (l_(1))/(l_(2))`
`(iR)/(iX)=(l_(1))/(l_(2))`
`X = (l_(1))/(l_(2))xxR`
`= (68.5)/(58.3) xx 10 = 11.789 Omega`
Therefore, the value of the unknown resistance, X, is `11.75 Omega`
If we fall to find a balance point with the given cell emf `epsilon`, then the potential drop across R and X must be reduced by putting a resistance in series with it. Only if the potential drop across R or X is smaller than the potential drop across the potentiometer wire AB, a balance point is obtained.