Ncert Exercises & Solutions

3.23 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?

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{iR}{iX} = \frac{l_{1}}{l_{2}}

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

= \frac{68.5}{58.3} \times 10 = 11.749 Ω

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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