A wire of resistance 5 , Omega is bent in the | vedclass

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(A) The total resistance of the wire is $R = 5 \, \Omega$.When the wire is bent into a circle,the resistance is distributed uniformly along the circum

Vedclass · Accepted Answer

$1.25 \, \Omega$

Vedclass · Answer

(A) The total resistance of the wire is $R = 5 \, \Omega$.When the wire is bent into a circle,the resistance is distributed uniformly along the circumference.When we consider two points at the ends of a diameter,the circle is divided into two equal semicircular parts.Each semicircular part has a resistance of $R' = R / 2 = 5 / 2 = 2.5 \, \Omega$.These two semicircular parts are connected in parallel between the two points.The equivalent resistance $R_{eq}$ is given by the formula for parallel resistors: $1 / R_{eq} = 1 / R' + 1 / R'$.$1 / R_{eq} = 1 / 2.5 + 1 / 2.5 = 2 / 2.5$.$R_{eq} = 2.5 / 2 = 1.25 \, \Omega$.

$A$ wire of resistance $5 \, \Omega$ is bent in the form of a closed circle. What is the resistance between two points at the ends of any diameter of the circle?

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