Question

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 $\frac{8}{3} \Omega$

• A. $\frac{l_1}{l_2}=\frac{5}{8}$
• B. $\frac{l_1}{l_2}=\frac{1}{3}$
• C. $\frac{l_1}{l_2}=\frac{3}{8}$
• D. $\frac{l_1}{l_2}=\frac{1}{2}$

Answer 1

Answer: (D)

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

Answer 2

Let x be resistance per unit length of the wire. Then, The resistance of the upper portion is $\, \, \, \, R_1 =xl_1$ The resistance of the lower portion is $\, \, \, \, R_2 =xl_2$ Equivalent resistance between A and B is $\, \, \, \, R=\frac{R_1R_2}{R_1+R_2}=\frac{(xl_1)(xl_2)}{xl_1+xl_2}$ $\, \, \, \, \frac{8}{3} =\frac{xl_1l_2}{l_1+l_2}\,$ or $\, \frac{8}{3}=\frac{xl_1l_2}{l_2\bigg(\frac{l_1}{l_2}+1\bigg)}$ or $\frac{8}{3} =\frac{xl_1}{\bigg(\frac{l_1}{l_2}+1\bigg)}$ ...(i) Also $R_0=xl_1+xl_2$ $\, \, \, \, \, \, 12 = x (l_1+l_2)$ $\, \, \, \, \, \, 12 =xl_2\bigg(\frac{l_1}{l_2}+1\bigg)$ ...(ii) Divide (i) by (ii), we get $\, \, \, \, \frac{\frac{8}{3}}{12} =\frac{\frac{xl_1}{\bigg(\frac{l_1}{l_2}+1\bigg)}}{xl_2\bigg(\frac{l_1}{l_2}+1\bigg)}\,$ or $\, \frac{8}{36}=\frac{l_1}{l_2\bigg(\frac{l_1}{l_2}+1\bigg)^2}$ $\bigg(\frac{l_1}{l_2}+1\bigg)^2{}\, \frac{8}{36}= \frac{l_1}{l_2}\,$ or $\, \bigg(\frac{l_1}{l_2}+1\bigg)^2\, \frac{2}{9} =\frac{l_1}{l_2}$ Let $y =\frac{l_1}{l_2}$ $\therefore 2(y+1)^2 = 9y \,$ or $\, 2y^2+2+4y =9y$ or $2y^2 - 5y + 2 =0$ Solving this quadratic equation, we get $\, \, \, \, \, y= \frac{1}{2}\,$ or $2\, \therefore \frac{l_1}{l_2} = \frac{1}{2}$