A wire with a resistance of 0.5 , Omega , m^{ | vedclass

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Question

(C) The circumference of the circle is $2\pi R$. Given $R = 1 \, m$,the total length is $2\pi \, m$. The resistance per unit length is $0.5 \, \Omega

Vedclass · Accepted Answer

$\frac{2\pi}{\pi + 4} \, \Omega$

Vedclass · Answer

(C) The circumference of the circle is $2\pi R$. Given $R = 1 \, m$,the total length is $2\pi \, m$. The resistance per unit length is $0.5 \, \Omega \, m^{-1}$.The circle is divided into two semicircles by the diameter. The length of each semicircle is $\pi R = \pi \, m$. The resistance of each semicircle is $R_1 = R_2 = (0.5 \, \Omega \, m^{-1}) 	imes (\pi \, m) = 0.5\pi \, \Omega$.The diameter has a length equal to $2R = 2 \, m$. The resistance of the wire along the diameter is $R_3 = (0.5 \, \Omega \, m^{-1}) 	imes (2 \, m) = 1 \, \Omega$.These three resistances $(R_1, R_2, R_3)$ are connected in parallel between the two ends of the diameter ($A$ and $B$).The equivalent resistance $R_{eq}$ is given by:$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{0.5\pi} + \frac{1}{0.5\pi} + \frac{1}{1} = \frac{2}{0.5\pi} + 1 = \frac{4}{\pi} + 1 = \frac{4 + \pi}{\pi}$.Therefore,$R_{eq} = \frac{\pi}{\pi + 4} \, \Omega$. Note: The provided options in the source were slightly inconsistent; based on the calculation,the correct value is $\frac{\pi}{\pi + 4} \, \Omega$.

$A$ wire with a resistance of $0.5 \, \Omega \, m^{-1}$ is bent into a circle of radius $1 \, m$. $A$ similar wire is connected across its diameter. What is the equivalent resistance between the two ends of the diameter?

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