A ring is made of a wire having a resistance | vedclass

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(D) Let $r$ be the resistance per unit length of the wire.Let $R_1 = r l_1$ and $R_2 = r l_2$ be the resistances of the two arcs between points $A$ an

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$\frac{l_1}{l_2} = \frac{1}{2}$

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(D) Let $r$ be the resistance per unit length of the wire.Let $R_1 = r l_1$ and $R_2 = r l_2$ be the resistances of the two arcs between points $A$ and $B$.The total resistance of the ring is $R_0 = R_1 + R_2 = 12\,\Omega$.Since the arcs are in parallel,the equivalent resistance $R_{eq}$ is given by:$R_{eq} = \frac{R_1 R_2}{R_1 + R_2} = \frac{8}{3}\,\Omega$.Substituting $R_1 + R_2 = 12$,we get:$\frac{R_1 R_2}{12} = \frac{8}{3} \Rightarrow R_1 R_2 = 32$.We have a system of two equations:$1$) $R_1 + R_2 = 12$$2$) $R_1 R_2 = 32$Substituting $R_2 = 12 - R_1$ into the second equation:$R_1(12 - R_1) = 32 \Rightarrow 12R_1 - R_1^2 = 32 \Rightarrow R_1^2 - 12R_1 + 32 = 0$.Solving the quadratic equation:$(R_1 - 8)(R_1 - 4) = 0$.So,$R_1 = 8\,\Omega$ and $R_2 = 4\,\Omega$ (or vice versa).Since resistance is proportional to length $(R \propto l)$,the ratio of lengths is:$\frac{l_1}{l_2} = \frac{R_1}{R_2} = \frac{8}{4} = 2$ or $\frac{4}{8} = \frac{1}{2}$.

$A$ ring is made of a wire having a resistance $R_0 = 12\,\Omega$. Find the positions of points $A$ and $B$ such that the equivalent resistance of the circuit shown is $8/3\,\Omega$.

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