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. $\frac{{\ell }_1}{{\ell }_2}=\frac{3}{8}$
• B. $\frac{{\ell }_1}{{\ell }_2}=\frac{1}{2}$
• C. $\frac{{\ell }_1}{{\ell }_2}=\frac{5}{8}$
• D. $\frac{{\ell }_1}{{\ell }_2}=\frac{1}{3}$

Answer 1

Answer: (B)

Initially $R_1+R_2=12\Omega$
Across point A & B, both branches will be in parallel connection.
$\frac{R_1{R}_2}{R_1+R_2}=\frac{8}{3}\Omega$
$\Rightarrow R_1{R}_2=32$
Now $R_2-R_1=\sqrt{{\left(R_1+R_2\right)}^2-4R_1{R}_2}$
$\Rightarrow \sqrt{12^2-4\times 32}=4\Omega$
So $R_1=4\Omega$ and $R_2=8\Omega$
For a uniform cross-sectional area, resistance will be proportional to the length of the wire.
Therefore, $\frac{{\ell }_1}{{\ell }_2}=\frac{4}{8}=\frac{1}{2}$