In the figure shown,for given values of R_1 a | vedclass

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(A) Let the resistance per unit length of the wire be $\sigma$. The total length $AB = 100\, cm$.Case $1$: Balance point is at $40\, cm$ from $A$.Usin

Vedclass · Accepted Answer

$\frac{10}{3}\,\Omega, 5\,\Omega$

Vedclass · Answer

(A) Let the resistance per unit length of the wire be $\sigma$. The total length $AB = 100\, cm$.Case $1$: Balance point is at $40\, cm$ from $A$.Using the Wheatstone bridge principle: $\frac{R_1}{R_2} = \frac{\sigma 	imes 40}{\sigma 	imes (100 - 40)} = \frac{40}{60} = \frac{2}{3}$.So,$R_1 = \frac{2}{3} R_2$  --- $(1)$Case $2$: $R_2$ is shunted by $10\,\Omega$. The new resistance $R_2' = \frac{R_2 	imes 10}{R_2 + 10}$.The balance point shifts to $50\, cm$.Using the principle: $\frac{R_1}{R_2'} = \frac{50}{50} = 1$.So,$R_1 = R_2' = \frac{10 R_2}{R_2 + 10}$ --- $(2)$Equating $(1)$ and $(2)$:$\frac{2}{3} R_2 = \frac{10 R_2}{R_2 + 10}$$2(R_2 + 10) = 30$$2 R_2 + 20 = 30$$2 R_2 = 10 \implies R_2 = 5\,\Omega$.Substituting $R_2 = 5\,\Omega$ in $(1)$:$R_1 = \frac{2}{3} 	imes 5 = \frac{10}{3}\,\Omega$.

In the figure shown,for given values of $R_1$ and $R_2$,the balance point for the jockey is at $40\, cm$ from $A$. When $R_2$ is shunted by a resistance of $10\,\Omega$,the balance point shifts to $50\, cm$. Find $R_1$ and $R_2$ $(AB = 1\, m)$.

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