**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,

$\frac{E_{1}}{E_{2}}=\frac{l_{1}}{l_{2}}$

$\frac{i R}{i X}=\frac{l_{1}}{l_{2}}$

$X=\frac{l_{1}}{l_{2}} \times R$

$=\frac{68.5}{58.3} \times 10=11.749 \Omega$

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.

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