Question

The circuit diagram of a balanced meter bridge is shown in the figure. The balanced point is obtained at $40 \, \text{cm}$ from the end A. When a $10 \, \Omega$ resistor is joined in series with $R_1$, the balanced point is obtained at $40 \, \text{cm}$ from the end B. Find the values of $R_1$ and $R_2$.

Hint:

In meter bridge problems, balanced length ratio equals resistance ratio: $\dfrac{R_1}{R_2} = \dfrac{l_1}{l_2}$.

Answer 1

Step 1: Balanced bridge condition.
\[ \frac{R_1}{R_2} = \frac{l_1}{l_2}. \]
Step 2: First case (without extra resistor).
$l_1 = 40 \, \text{cm}, \quad l_2 = 60 \, \text{cm}$. \[ \frac{R_1}{R_2} = \frac{40}{60} = \frac{2}{3}. \quad (1) \]
Step 3: Second case (with 10 $\Omega$ in series with $R_1$).
Balanced point at $40 \, \text{cm}$ from B $\Rightarrow$ $l_1 = 60 \, \text{cm}, l_2 = 40 \, \text{cm}$. \[ \frac{R_1 + 10}{R_2} = \frac{60}{40} = \frac{3}{2}. \quad (2) \]
Step 4: Solve equations (1) and (2).
From (1): \[ R_1 = \frac{2}{3} R_2. \quad (3) \] Substitute in (2): \[ \frac{\frac{2}{3}R_2 + 10}{R_2} = \frac{3}{2}. \] \[ \frac{2R_2}{3R_2} + \frac{10}{R_2} = \frac{3}{2}. \] \[ \frac{2}{3} + \frac{10}{R_2} = \frac{3}{2}. \] \[ \frac{10}{R_2} = \frac{3}{2} - \frac{2}{3} = \frac{9 - 4}{6} = \frac{5}{6}. \] \[ R_2 = \frac{10 \times 6}{5} = 12 \, \Omega. \] Correction check: Let's recalc carefully: \[ \frac{\frac{2}{3}R_2 + 10}{R_2} = \frac{3}{2}. \] \[ \frac{2R_2}{3R_2} + \frac{10}{R_2} = \frac{3}{2}. \] \[ \frac{2}{3} + \frac{10}{R_2} = \frac{3}{2}. \] \[ \frac{10}{R_2} = \frac{3}{2} - \frac{2}{3} = \frac{9 - 4}{6} = \frac{5}{6}. \] \[ R_2 = \frac{10 \times 6}{5} = 12 \, \Omega. \] Then, from (3): \[ R_1 = \frac{2}{3} \times 12 = 8 \, \Omega. \]
Step 5: Final Answer.
\[ R_1 = 8 \, \Omega, \quad R_2 = 12 \, \Omega. \] ---