A ring is made of a wire having a resistance | vedclass

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(B) Let the resistances of the two arcs be $R_1$ and $R_2$. Since the total resistance of the ring is $R_0 = 12\,\Omega$,we have $R_1 + R_2 = 12\,\Ome

Vedclass · Accepted Answer

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

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(B) Let the resistances of the two arcs be $R_1$ and $R_2$. Since the total resistance of the ring is $R_0 = 12\,\Omega$,we have $R_1 + R_2 = 12\,\Omega$.When connected in parallel,the equivalent resistance is given by $\frac{R_1 R_2}{R_1 + R_2} = \frac{8}{3}\,\Omega$.Substituting $R_1 + R_2 = 12$,we get $\frac{R_1 R_2}{12} = \frac{8}{3}$,which implies $R_1 R_2 = 32$.We know that $(R_2 - R_1)^2 = (R_1 + R_2)^2 - 4 R_1 R_2 = 12^2 - 4(32) = 144 - 128 = 16$.Thus,$R_2 - R_1 = 4\,\Omega$.Solving the system $R_1 + R_2 = 12$ and $R_2 - R_1 = 4$,we find $R_2 = 8\,\Omega$ and $R_1 = 4\,\Omega$.Since resistance is proportional to length $(R \propto l)$,the ratio of lengths is $\frac{l_1}{l_2} = \frac{R_1}{R_2} = \frac{4}{8} = \frac{1}{2}$.

$A$ ring is made of a wire having a resistance $R_0 = 12\,\Omega$. Find the ratio of the lengths $\frac{l_1}{l_2}$ of the two arcs between points $A$ and $B$ as shown in the figure,such that the equivalent resistance $R$ of the sub-circuit between these points is equal to $\frac{8}{3}\,\Omega$.

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