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

Answer 1

Answer: (B)

Since $R \, \propto l$
According to problem $R_1 + R_2 = 12\, \Omega$
and $\frac{R_1 \times R_2}{R_1 + R_2} = \frac{8}{3} \Omega$
On solving this we get $R_1 = 4 \Omega$
and $R_2 = 8\, \Omega $
therefore the ratio $ \frac{l_1}{l_2} = \frac{1}{2}$