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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
$\frac{l_1}{l_2} = \frac{1}{2}$
B
$\frac{l_1}{l_2} = \frac{5}{8}$
C
$\frac{l_1}{l_2} = \frac{1}{3}$
D
$\frac{l_1}{l_2} = \frac{3}{8}$
Detailed Solution
The two arcs have resistances $R_1$ and $R_2$ (proportional to $l_1$ and $l_2$) and are in parallel between A and B.
$R_1 + R_2 = 12$ ...(i)
$\frac{R_1R_2}{R_1 + R_2} = \frac{8}{3} \Rightarrow R_1R_2 = 32$ ...(ii)
Solving: $R_1 = 4\ \Omega$ and $R_2 = 8\ \Omega$
$\frac{l_1}{l_2} = \frac{R_1}{R_2} = \frac{1}{2}$
$R_1 + R_2 = 12$ ...(i)
$\frac{R_1R_2}{R_1 + R_2} = \frac{8}{3} \Rightarrow R_1R_2 = 32$ ...(ii)
Solving: $R_1 = 4\ \Omega$ and $R_2 = 8\ \Omega$
$\frac{l_1}{l_2} = \frac{R_1}{R_2} = \frac{1}{2}$