#### Question

Figure 3.34 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*?

#### Solution

Resistance of the standard resistor, *R* = 10.0 Ω

Balance point for this resistance, *l*_{1} = 58.3 cm

Current in the potentiometer wire = *i*

Hence, potential drop across *R*, *E*_{1} = *iR*

Resistance of the unknown resistor = *X*

Balance point for this resistor, *l*_{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`

`X=l_1/l_2xxR`

`=68.5/58.3xx10=11.749 Omega`

Therefore, the value of the unknown resistance, *X*, is 11.75 Ω.

If we fail to find a balance point with the given cell of emf, *ε*, 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.