Question

(Don't answer of you don't know. I want the explanation not just answer) 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

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Answer 1

$\huge{\underline{\underline{\boxed{\sf{\purple{Answer : }}}}}}$

$\small \underline {\sf{As \: we \: know \: that  \:  Resistance  \: ∝  \: length \: , that's \: why }}$

$\small \underline {\sf{Here  \: RXWY \: = \: 2πrR \: × \: (rα)= \: 2πRα  \: (∵α=rl) }}$

$\small \underline {\sf{ and RXZY \: = \: 2πrR×r(2π−α) \: = \: 2πR(2π−α) }}$

$\small \underline {\sf{Req= \: RXWY+Rxwy \div Rxwy Rxzy= 2πRα \: + \: 2πR \: (2π−α ) \: 2πRα× \: 2πR(2π−α) }}$

$\small \underline {\sf{Ra / 4π(2π-a)}}$

Answer 2

Answer:

Asweknowthat Resistance ∝ length, that

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swhy

\small \underline {\sf{Here \: RXWY \: = \: 2πrR \: × \: (rα)= \: 2πRα \: (∵α=rl) }}

Here RXWY=2πrR×(rα)=2πRα (∵α=rl)

\small \underline {\sf{ and RXZY \: = \: 2πrR×r(2π−α) \: = \: 2πR(2π−α) }}

and RXZY=2πrR×r(2π−α)=2πR(2π−α)

\small \underline {\sf{Req= \: RXWY+Rxwy \div Rxwy Rxzy= 2πRα \: + \: 2πR \: (2π−α ) \: 2πRα× \: 2πR(2π−α) }}

Req=RXWY+Rxwy÷RxwyRxzy=2πRα+2πR(2π−α)2πRα×2πR(2π−α)

\small \underline {\sf{Ra / 4π(2π-a)}}

Ra/4π(2π−a)