Figure shows a potentiometer circuit for comparison of two resistances. The balance point with a standard resi

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. The value of X is:

Accepted Answer

In a potentiometer arrangement used to compare resistances, the potential drop across a resistance is directly proportional to the balancing length of the wire ( $V \propto l$ ). Assuming the resistors $R$ and $X$ are connected in series such that the same current flows through them, the ratio of resistances is equal to the ratio of their respective balancing lengths: $\frac{X}{R} = \frac{l_X}{l_R}$ .

Substituting the given values: $X = R \times \frac{l_X}{l_R} = 10.0 \, \Omega \times \frac{68.5 \text{ cm}}{58.3 \text{ cm}}$ $X \approx 10.0 \times 1.175 = 11.75 \, \Omega$ .

The closest value among the options is 11.7 \Omega .

Substituting the given values:
$X = R 	imes \frac{l_X}{l_R} = 10.0 \, \Omega 	imes \frac{68.5 	ext{ cm}}{58.3 	ext{ cm}}$
$X \approx 10.0 	imes 1.175 = 11.75 \, \Omega$.

The closest value among the options is 11.7 Ω.

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