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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$1 : 9$

Vedclass · Answer

(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 $BC$?

Similar Questions

Write the equation representing the relation between resistivity and temperature.

Conductivity increases in the order of

$A$ metal rod of length $10 \text{ cm}$ and a rectangular cross-section of $1 \text{ cm} \times 0.5 \text{ cm}$ is connected to a battery across opposite faces. The resistance will be . . . . . . .

$A$ resistance $R$ is stretched to four times its length. Its new resistance will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module