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 $\frac{8}{3}\Omega$

Let x is the resistance per unit length then

equivalent resistance $R=\frac{R_1{R}_2}{R_1+R_2}$

$\frac{\left({\mathrm{x}}_1{\mathrm{l}}_1\right)\left({\mathrm{x}}_2{\mathrm{l}}_2\right)}{{\mathrm{xl}}_1+{\mathrm{xl}}_2}$

$\Rightarrow \frac{8}{3}=\mathrm{x}\frac{{\mathrm{l}}_1{\mathrm{l}}_2}{{\mathrm{l}}_1+{\mathrm{l}}_2}$

$\frac{8}{3}=\mathrm{x}\frac{{\mathrm{l}}_1}{\frac{1_1}{1_2}+1}\dots$

$\text{ also }{R}_n=x{l}_1+x{l}_2$

$12=x\left(l_1+l_2\right)$

$12={\mathrm{xl}}_2\left(\frac{{\mathrm{l}}_1}{{\mathrm{l}}_2}+1\right)\dots$

$\frac{\text{ (i) }}{\text{ (ii) }}\Rightarrow \frac{\frac{8}{3}}{\frac{12}{1}}=\frac{\frac{{\mathrm{xl}}_1}{\left(\frac{l_1}{l_2}+1\right)}}{{\mathrm{xl}}_2\left(\frac{{\mathrm{l}}_1}{{\mathrm{l}}_2}+1\right)}=\frac{{\mathrm{l}}_1}{{\mathrm{l}}_2{\left(\frac{{\mathrm{l}}_1}{{\mathrm{l}}_2}+1\right)}^2}$

${\left(\frac{l_1}{l_2}+1\right)}^2\times \frac{8}{36}=\frac{l_1}{l_2}$

$\left(y^2+1+2y\right)\times \frac{8}{36}=y\text{ (wherey }=\frac{1_1}{1_2}\text{ ) }$

$8y^2+8+16y=36y$

$\Rightarrow 2y^2-5y+2=0$

$\Rightarrow 2y^2-4y-y+2=0$

$\Rightarrow 2y(y-2)-1(y-2)=0$

$\Rightarrow (2y-1)(y-2)=0$

$\Rightarrow \mathrm{y}=\frac{l_1}{l_2}=\frac{1}{2}\text{ or }2$