Question

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

• A. $\frac{10}{3}\Omega, 5\Omega$
• B. 20$\Omega$, 30$\Omega$
• C. 10$\Omega$, 15$\Omega$
• D. 5$\Omega$, $\frac{15}{2}\Omega$

Answer 1

Answer: (A)

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

Answer 2

The meter bridge works on the principle of the Wheatstone bridge. At the balance point, the ratio of the resistances in the two gaps is equal to the ratio of the resistances of the two segments of the bridge wire. Let the resistance per unit length of the wire be $\rho$.

In the first case, the balance point is at a distance $l_1 = 40$ cm from A. The resistance of the left segment of the wire is $R_{left\_wire} = \rho l_1 = 40\rho$. The resistance of the right segment of the wire is $R_{right\_wire} = \rho (100 - l_1) = \rho (100 - 40) = 60\rho$. The resistance in the left gap is $R_1$. The resistance in the right gap is $R_2$. At the balance point, we have: $\frac{R_1}{R_2} = \frac{R_{left\_wire}}{R_{right\_wire}} = \frac{40\rho}{60\rho} = \frac{40}{60} = \frac{2}{3}$ This gives the equation: $3R_1 = 2R_2$. (1)

In the second case, $R_2$ is shunted by a resistance of 10 $\Omega$. The equivalent resistance of the right gap is the parallel combination of $R_2$ and 10 $\Omega$. Let $R'_2$ be the equivalent resistance in the right gap. $R'_2 = \frac{R_2 \times 10}{R_2 + 10}$ The balance point shifts to $l_2 = 50$ cm from A. The resistance of the left segment of the wire is $R'_{left\_wire} = \rho l_2 = 50\rho$. The resistance of the right segment of the wire is $R'_{right\_wire} = \rho (100 - l_2) = \rho (100 - 50) = 50\rho$. At the new balance point, we have: $\frac{R_1}{R'_2} = \frac{R'_{left\_wire}}{R'_{right\_wire}} = \frac{50\rho}{50\rho} = \frac{50}{50} = 1$ This gives the equation: $R_1 = R'_2$. Substituting the expression for $R'_2$: $R_1 = \frac{10R_2}{R_2 + 10}$. (2)

Now we solve the system of equations (1) and (2). From equation (1), $R_1 = \frac{2}{3}R_2$. Substitute this into equation (2): $\frac{2}{3}R_2 = \frac{10R_2}{R_2 + 10}$ Since $R_2$ is a resistance, $R_2 \neq 0$. We can divide both sides by $R_2$: $\frac{2}{3} = \frac{10}{R_2 + 10}$ Cross-multiply: $2(R_2 + 10) = 3 \times 10$ $2R_2 + 20 = 30$ $2R_2 = 30 - 20$ $2R_2 = 10$ $R_2 = \frac{10}{2} = 5 \Omega$

Now substitute the value of $R_2$ back into the expression for $R_1$: $R_1 = \frac{2}{3}R_2 = \frac{2}{3} \times 5 = \frac{10}{3} \Omega$

Thus, $R_1 = \frac{10}{3}\Omega$ and $R_2 = 5\Omega$.