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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let $r$ be the resistance per unit length of the wire. The total resistance is $R_0 = r(l_1 + l_2) = 12 \,\Omega$.The resistances of the two arcs

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let $r$ be the resistance per unit length of the wire. The total resistance is $R_0 = r(l_1 + l_2) = 12 \,\Omega$.The resistances of the two arcs are $R_1 = r l_1$ and $R_2 = r l_2$.Since these two arcs are in parallel,the equivalent resistance $R$ is given by:$R = \frac{R_1 R_2}{R_1 + R_2} = \frac{(r l_1)(r l_2)}{r(l_1 + l_2)} = \frac{r l_1 l_2}{l_1 + l_2} = \frac{8}{3} \,\Omega$.We know $r(l_1 + l_2) = 12$,so $r = \frac{12}{l_1 + l_2}$.Substituting $r$ into the equation for $R$:$\frac{(\frac{12}{l_1 + l_2}) l_1 l_2}{l_1 + l_2} = \frac{8}{3} \implies \frac{12 l_1 l_2}{(l_1 + l_2)^2} = \frac{8}{3}$.Let $y = \frac{l_1}{l_2}$. Then $l_1 = y l_2$. Substituting this:$\frac{12 (y l_2) l_2}{(y l_2 + l_2)^2} = \frac{8}{3} \implies \frac{12 y l_2^2}{l_2^2 (y + 1)^2} = \frac{8}{3} \implies \frac{12 y}{(y + 1)^2} = \frac{8}{3}$.Cross-multiplying:$36 y = 8(y^2 + 2y + 1) \implies 36 y = 8y^2 + 16y + 8$.$8y^2 - 20y + 8 = 0 \implies 2y^2 - 5y + 2 = 0$.Factoring the quadratic equation:$2y^2 - 4y - y + 2 = 0 \implies 2y(y - 2) - 1(y - 2) = 0$.$(2y - 1)(y - 2) = 0$.Thus,$y = \frac{1}{2}$ or $y = 2$. Therefore,$\frac{l_1}{l_2} = \frac{1}{2}$ or $2$.

$A$ 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 $\frac{8}{3} \,\Omega$.

Similar Questions

Two wires of equal diameters with resistivities ${\rho _1}$ and ${\rho _2}$ and lengths $x_1$ and $x_2$ are joined in series. The equivalent resistivity of the combination is

The current $I$ flowing through the given circuit will be $.....A$.

$A$ wire of resistance $12 \, \Omega$ is bent to form an equilateral triangle. What is the equivalent resistance between any two corners of the triangle?

The effective resistance across $AB$ in the network shown is .......... $\Omega$.

Two identical heater filaments are connected first in parallel and then in series. At the same applied voltage,the ratio of heat produced in the same time for parallel to series will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module