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. The total resistance is $R_0 = r(l_1 + l_2) = 12 \,\Omega$.The resistances of the two arcs

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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. The total resistance is $R_0 = r(l_1 + l_2) = 12 \,\Omega$.The resistances of the two arcs are $R_1 = r l_1$ and $R_2 = r l_2$.Since these two arcs are in parallel,the equivalent resistance $R$ is given by:$R = \frac{R_1 R_2}{R_1 + R_2} = \frac{(r l_1)(r l_2)}{r(l_1 + l_2)} = \frac{r l_1 l_2}{l_1 + l_2} = \frac{8}{3} \,\Omega$.We know $r(l_1 + l_2) = 12$,so $r = \frac{12}{l_1 + l_2}$.Substituting $r$ into the equation for $R$:$\frac{(\frac{12}{l_1 + l_2}) l_1 l_2}{l_1 + l_2} = \frac{8}{3} \implies \frac{12 l_1 l_2}{(l_1 + l_2)^2} = \frac{8}{3}$.Let $y = \frac{l_1}{l_2}$. Then $l_1 = y l_2$. Substituting this:$\frac{12 (y l_2) l_2}{(y l_2 + l_2)^2} = \frac{8}{3} \implies \frac{12 y l_2^2}{l_2^2 (y + 1)^2} = \frac{8}{3} \implies \frac{12 y}{(y + 1)^2} = \frac{8}{3}$.Cross-multiplying:$36 y = 8(y^2 + 2y + 1) \implies 36 y = 8y^2 + 16y + 8$.$8y^2 - 20y + 8 = 0 \implies 2y^2 - 5y + 2 = 0$.Factoring the quadratic equation:$2y^2 - 4y - y + 2 = 0 \implies 2y(y - 2) - 1(y - 2) = 0$.$(2y - 1)(y - 2) = 0$.Thus,$y = \frac{1}{2}$ or $y = 2$. Therefore,$\frac{l_1}{l_2} = \frac{1}{2}$ or $2$.

$A$ ring is made of a wire having a resistance $R_0 = 12 \,\Omega$. Find the points $A$ and $B$,as shown in the figure,at which a current-carrying conductor should be connected so that the resistance $R$ of the sub-circuit between these points is equal to $\frac{8}{3} \,\Omega$.

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