A wire of resistor R is bent into a circular ring of radius r. Equivalent resistance between two points X and Y on its circumference,

Q: A wire of resistor $R$ is bent into a circular ring of radius $r$. Equivalent resistance between two points $X$ and $Y$ on its circumference, when angle $XOY$ is $\alpha$, can be given by

a (a) Here ${R_{XWY}} = \frac{R}{{2\pi r}} \times (r\alpha ) = \frac{{R\alpha }}{{2\pi }}$ and ${R_{XZY}} = \frac{R}{{2\pi r}} \times r(2\pi - \alpha ) = \frac{R}{{2\pi }}(2\pi - \alpha )$ ${R_{eq}} = \frac{{{R_{XWY}}{R_{XZY}}}}{{{R_{XWY}} + {R_{XZY}}}} = \frac{{\frac{{R\alpha }}{{2\pi }} \times \frac{R}{{2\pi }}(2\pi - \alpha )}}{{\frac{{R\alpha }}{{2\pi }} + \frac{{R(2\pi - \alpha )}}{{2\pi }}}}$$ = \frac{{R\alpha }}{{4{\pi ^2}}}(2\pi - \alpha )$

SECTION - A [PHYSICS MCQ]

GSEB - English Medium

A wire of resistor $R$ is bent into a circular ring of radius $r$. Equivalent resistance between two points $X$ and $Y$ on its circumference, when angle $XOY$ is $\alpha$, can be given by

A$\frac{{R\alpha }}{{4{\pi ^2}}}(2\pi - \alpha )$
B$\frac{R}{{2\pi }}(2\pi - \alpha )$
C${R} (2\pi - \alpha )$
D$\frac{{4\pi }}{{R\alpha }}(2\pi - \alpha )$

Diffcult

STD 11 Science
JEE
STD 12 - 3. current electricity
Physics

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