A ring is made of a wire having a resistance $R_0=12 \Omega$. Find the points $A$ and $B$, as shown in the…
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 $8 / 3 \Omega$
$\frac{l_1}{l_2}=\frac{5}{8}$
$\frac{l_1}{l_2}=\frac{1}{3}$
$\frac{l_1}{l_2}=\frac{3}{8}$
$\frac{l_1}{l_2}=\frac{1}{2}$
Solution
We knows $R \propto l$
Here, $\quad R_1+R_2=12 \Omega$
and $\quad \frac{R_1 \times R_2}{R_1+R_2}=\frac{8}{3} \Omega$
$\Rightarrow \quad R_1 R_2=32 \Omega$
We get, $R_1=8$ and $R_2=4$
Again, $R_1=\frac{12 l_1}{l_1+l_2}$
and $R_2=\frac{12 l_2}{l_1+l_2}$
Hence, $\quad \frac{l_1}{l_2}=\frac{1}{2}$
