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A ring is made of a wire having a resistance R₀ = 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 $\frac{8}{3}\Omega.$

Option 1)
$\frac{\ell_{1}}{\ell_{2}}=\frac{5}{8}$

Option 2)
$\frac{\ell_{1}}{\ell_{2}}=\frac{1}{3}$

Option 3)
$\frac{\ell_{1}}{\ell_{2}}=\frac{3}{8}$

Option 4)
$\frac{\ell_{1}}{\ell_{2}}=\frac{1}{2}$

Answers (1)

best_answer

net n be resistance per unit length of the wire

Upper Portion $R_{1}=xl_{1}$

Lower Portion $R_{2}=xl_{2}$

Equivalent resistance between A and B is

$R=\frac{(xR_{1})(xl_{2})}{xl_{2}+xl_{1}}=\frac{xl_{1}l_{2}}{l_{1}+l_{2}}$

$\frac{8}{3}=\frac{xl_{1}l_{2}}{l_{1}+l_{2}}=>\frac{8}{3}=\frac{xl_{1}}{(l_{1}/l_{2}+1)}$ -------------------------(i)

$R_{o}=xl_{1}+xl_{2}=>12=x(l_{1}+l_{2})$

12= $xl_{2}(1+\frac{l_{1}}{l_{2}})$ -----------------------------------(ii)

$\therefore eqn's (i) and (ii) \frac{8/3}{12}=\frac{xl_{1}/l_{1}/l_{1}+1}{xl_{2}(\frac{l_{1}}{l_{2}+1})}$

or $\frac{8}{36}=\frac{l_{1}}{l_{2}(\frac{l_{1}}{l_{2}}+1)^{2}}$

$(\frac{l_{1}}{l_{2}}+1)^{2}\frac{8}{36}=\frac{l_{1}}{l_{2}}$ or $(\frac{l_{1}}{l_{2}}+1)^{2}$ $\frac{2}{9}$ = $\frac{l_{1}}{l_{2}}$

Let y = $\frac{P_{1}}{l_{2}}$

$2(y+1)^{2}=9y$

=> $2y^{2}+2+4y=9y$

$2y^{2}-5y+2=0$

=> $y=\frac{1}{2}$ or 2

$y=\frac{l_{1}}{l_{2}}=\frac{1}{2}$

Option 1)

$\frac{\ell_{1}}{\ell_{2}}=\frac{5}{8}$

This option is incorrect

Option 2)

$\frac{\ell_{1}}{\ell_{2}}=\frac{1}{3}$

This option is incorrect

Option 3)

$\frac{\ell_{1}}{\ell_{2}}=\frac{3}{8}$

This option is incorrect

Option 4)

$\frac{\ell_{1}}{\ell_{2}}=\frac{1}{2}$

This option is correct

Posted by

Aadil

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