Question

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. $\frac{l_1}{l_2}=\frac{5}{8}$
• B. $\frac{l_1}{l_2}=\frac{1}{3}$
• C. $\frac{l_1}{l_2}=\frac{3}{8}$
• D. $\frac{l_1}{l_2}=\frac{1}{2}$

Answer 1

Answer: (D)

Let $x$ be resistance per unit length of the wire.
Then,
The resistance of the upper portion i$R_1=x{l}_1$
The resistance of the lower portion is $R_2=x{l}_2$
Equivalent resistance between $A$ and $B$ is
$R=\frac{R_1{R}_2}{R_1+R_2}=\frac{\left(x{l}_1\right)\left(x{l}_2\right)}{x{l}_1+x{l}_2}$
$\frac{8}{3}=\frac{x{l}_1{l}_2}{l_1+l_2}$
or $\frac{8}{3}=\frac{x{l}_1{l}_2}{l_2\left(\frac{l_1}{l_2}+1\right)}$
or $\frac{8}{3}=\frac{x{l}_1}{\left(\frac{l_1}{l_2}+1\right)}\dots$ (i)
Also $R_0=x{l}_1+x{l}_2$
$12=x\left(l_1+l_2\right)$
$12=x{l}_2\left(\frac{l_1}{l_2}+1\right)\dots (ii)$
Divide (i) by (ii), we get
$\frac{\frac{8}{3}}{12}=\frac{\frac{x{l}_1}{\left(\frac{l_1}{l_2}+1\right)}}{x{l}_2\left(\frac{l_1}{l_2}+1\right)}$
or $\frac{8}{36}=\frac{l_1}{l_2{\left(\frac{l_1}{l_2}+1\right)}^2}$
${\left(\frac{l_1}{l_2}+1\right)}^2\frac{8}{36}=\frac{l_1}{l_2}$
or ${\left(\frac{l_1}{l_2}+1\right)}^2\frac{2}{9}=\frac{l_1}{l_2}$
Let $y=\frac{l_1}{l_2}$
$\therefore 2(y+1{)}^2=9y$
or $2y^2+2+4y=9y$
or $2y^2-5y+2=0$
Solving this quadratic equation, we get
$y=\frac{1}{2}$ or $2$
$\therefore \frac{l_1}{l_2}=\frac{1}{2}$