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 problem describes a meter bridge experiment. The balance condition for a meter bridge is given by $\frac{R_1}{R_2} = \frac{l}{100-l}$, where $R_1$ is the resistance in the left gap, $R_2$ is the resistance in the right gap, and $l$ is the distance of the balance point from the left end (A) of the wire in cm. The total length of the wire is 100 cm (1 m).

In the first case, the balance point is at 40 cm from A. So, $l_1 = 40$ cm. The resistance in the left gap is $R_1$ and in the right gap is $R_2$. Applying the balance condition: $\frac{R_1}{R_2} = \frac{40}{100-40} = \frac{40}{60} = \frac{2}{3}$ This gives us the first equation: $3R_1 = 2R_2$.

In the second case, $R_2$ is shunted by a resistance of 10 $\Omega$. When a resistance $R_s$ is shunted with $R_2$, the equivalent resistance is given by the parallel combination formula: $R_{2,eq} = \frac{R_2 \times R_s}{R_2 + R_s}$. Here, $R_s = 10 \Omega$. So, $R_{2,eq} = \frac{10R_2}{R_2 + 10}$. The balance point shifts to 50 cm from A. So, $l_2 = 50$ cm. The resistance in the left gap is still $R_1$, and in the right gap is $R_{2,eq}$. Applying the balance condition: $\frac{R_1}{R_{2,eq}} = \frac{50}{100-50} = \frac{50}{50} = 1$ This gives us the second equation: $R_1 = R_{2,eq}$.

Substitute the expression for $R_{2,eq}$ into the second equation: $R_1 = \frac{10R_2}{R_2 + 10}$.

Now we have a system of two equations:

1. $3R_1 = 2R_2 \implies R_1 = \frac{2}{3}R_2$
2. $R_1 = \frac{10R_2}{R_2 + 10}$

Substitute the expression for $R_1$ from equation (1) into equation (2): $\frac{2}{3}R_2 = \frac{10R_2}{R_2 + 10}$

Since $R_2$ is a resistance, it cannot be zero. We can divide both sides by $R_2$: $\frac{2}{3} = \frac{10}{R_2 + 10}$

Now, solve for $R_2$: $2(R_2 + 10) = 3 \times 10$ $2R_2 + 20 = 30$ $2R_2 = 30 - 20$ $2R_2 = 10$ $R_2 = 5 \Omega$

Now substitute the value of $R_2$ 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 $R_2 = 5 \Omega$.