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(C) The circumference of the circle is $2\pi r$. The arc $AB$ (minor arc) has a length of $\frac{1}{4}(2\pi r) = \frac{\pi r}{2}$. The remaining arc $

Vedclass · Accepted Answer

$ \frac{6\pi\lambda r}{3\pi+16} $

Vedclass · Answer

(C) The circumference of the circle is $2\pi r$. The arc $AB$ (minor arc) has a length of $\frac{1}{4}(2\pi r) = \frac{\pi r}{2}$. The remaining arc $AB$ (major arc) has a length of $2\pi r - \frac{\pi r}{2} = \frac{3\pi r}{2}$.The resistance of the minor arc is $R_1 = \lambda \cdot \frac{\pi r}{2}$.The resistance of the major arc is $R_2 = \lambda \cdot \frac{3\pi r}{2}$.The two radial wires $OA$ and $OB$ each have a resistance of $R_3 = \lambda r$.Points $A$ and $B$ are connected by the minor arc $(R_1)$ and the path through the center $(R_3 + R_3 = 2\lambda r)$ in parallel with the major arc $(R_2)$.However, based on the circuit configuration, the minor arc $R_1$ is in parallel with the series combination of the major arc $R_2$ and the two radial wires $2\lambda r$ is incorrect. The correct interpretation is that the minor arc $R_1$ is in parallel with the combination of the major arc $R_2$ and the two radial wires $OA$ and $OB$ in series with each other, but the radial wires are connected to the center $O$. The circuit consists of three parallel branches between $A$ and $B$: the minor arc $(R_1)$, the major arc $(R_2)$, and the path $A-O-B$ $(2\lambda r)$.$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{2\lambda r} = \frac{1}{\lambda \pi r / 2} + \frac{1}{\lambda 3\pi r / 2} + \frac{1}{2\lambda r} = \frac{2}{\lambda \pi r} + \frac{2}{3\lambda \pi r} + \frac{1}{2\lambda r}$.$\frac{1}{R_{eq}} = \frac{1}{\lambda r} [\frac{2}{\pi} + \frac{2}{3\pi} + \frac{1}{2}] = \frac{1}{\lambda r} [\frac{6+2}{3\pi} + \frac{1}{2}] = \frac{1}{\lambda r} [\frac{8}{3\pi} + \frac{1}{2}] = \frac{1}{\lambda r} [\frac{16 + 3\pi}{6\pi}]$.$R_{eq} = \frac{6\pi \lambda r}{3\pi + 16}$.

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