In the given circuit,the lengths of AB and BC | vedclass

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(A) The resistance of a wire is given by $R = \rho \frac{L}{A} = \rho \frac{L}{\pi r^2}$.Since the lengths $L$ and resistivity $\rho$ are the same for

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$1 : 9$

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(A) The resistance of a wire is given by $R = \rho \frac{L}{A} = \rho \frac{L}{\pi r^2}$.Since the lengths $L$ and resistivity $\rho$ are the same for both segments,the resistance $R$ is inversely proportional to the square of the radius: $R \propto \frac{1}{r^2}$.Let $r_{AB} = 3r$ and $r_{BC} = r$. Then the ratio of resistances is $\frac{R_{AB}}{R_{BC}} = \frac{r_{BC}^2}{r_{AB}^2} = \frac{r^2}{(3r)^2} = \frac{1}{9}$.Since the segments $AB$ and $BC$ are in series,the same current $I$ flows through both.According to Ohm's Law,the potential difference $V = IR$.Therefore,the ratio of potential differences is $\frac{V_{AB}}{V_{BC}} = \frac{I R_{AB}}{I R_{BC}} = \frac{R_{AB}}{R_{BC}} = \frac{1}{9}$.

In the given circuit,the lengths of $AB$ and $BC$ are equal. The radius of $AB$ is three times that of $BC$. What is the ratio of the potential difference across $AB$ to that across $BC$?

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