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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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