A uniform wire of resistance 9 , Omega is ben | vedclass

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(B) The total resistance of the wire is $R = 9 \, \Omega$. When bent into a circle,the resistance is proportional to the length of the arc.The arc $AB

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$2$

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(B) The total resistance of the wire is $R = 9 \, \Omega$. When bent into a circle,the resistance is proportional to the length of the arc.The arc $AB$ subtends an angle of $120^{\circ}$ at the center.The resistance of this arc $AB$ is $R_1 = (120^{\circ} / 360^{\circ}) 	imes 9 \, \Omega = (1/3) 	imes 9 \, \Omega = 3 \, \Omega$.The remaining part of the wire has a resistance of $R_2 = 9 \, \Omega - 3 \, \Omega = 6 \, \Omega$.These two parts,$R_1$ and $R_2$,are connected in parallel between points $A$ and $B$.The equivalent resistance $R_{eq}$ is given by:$1/R_{eq} = 1/R_1 + 1/R_2 = 1/3 + 1/6 = (2+1)/6 = 3/6 = 1/2$.Therefore,$R_{eq} = 2 \, \Omega$.

$A$ uniform wire of resistance $9 \, \Omega$ is bent in the form of a circle. The effective resistance across the points $A$ and $B$ is ............... $\Omega$. The angle subtended by the arc $AB$ at the center is $120^{\circ}$.

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