A ring is made of a wire having a resistance R0 = 12 Ω . 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 (8/3) Ω

Question

A ring is made of a wire having a resistance R0 = 12 Ω . 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 (8/3) Ω  (A) (l1/l2) = (3/8) (B) (l1/l2) = (1/2) (C) (l1/l2) = (5/8) (D) (l1/l2) = (1/3)

Tardigrade Admin · Accepted Answer

Since R propto l According to problem R1 + R2 = 12 Ω and (R1 × R2/R1 + R2) = (8/3) Ω On solving this we get R1 = 4 Ω and R2 = 8 Ω therefore the ratio (l1/l2) = (1/2)

Q. A ring is made of a wire having a resistance $R_{0} = 12 Ω$ . 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 $\frac{8}{3} Ω$

$\frac{l _{1}}{l _{2}} = \frac{3}{8}$

$\frac{l _{1}}{l _{2}} = \frac{1}{2}$

$\frac{l _{1}}{l _{2}} = \frac{5}{8}$

$\frac{l _{1}}{l _{2}} = \frac{1}{3}$

Since $R \propto l$
According to problem $R_{1} + R_{2} = 12 Ω$
and $\frac{R _{1} \times R _{2}}{R _{1} + R _{2}} = \frac{8}{3} Ω$
On solving this we get $R_{1} = 4Ω$
and $R_{2} = 8 Ω$
therefore the ratio $\frac{l _{1}}{l _{2}} = \frac{1}{2}$

Q. A ring is made of a wire having a resistance R0​=12Ω . 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 38​Ω

l2​l1​​=83​

l2​l1​​=21​

l2​l1​​=85​

l2​l1​​=31​