Question

A meter bridge with two resistances \( R_1 \) and \( R_2 \) as shown in figure was balanced (null point) at 40 cm from the point \( P \). The null point changed to 50 cm from the point \( P \), when a \( 16\,\Omega \) resistance is connected in parallel to \( R_2 \). The values of resistances \( R_1 \) and \( R_2 \) are 

• A. \( R_2 = 4\,\Omega,\; R_1 = \frac{4}{3}\,\Omega \)
• B. \( R_2 = 16\,\Omega,\; R_1 = \frac{16}{3}\,\Omega \)
• C. \( R_2 = 8\,\Omega,\; R_1 = \frac{16}{3}\,\Omega \)
• D. \( R_2 = 12\,\Omega,\; R_1 = \frac{12}{3}\,\Omega \)

Hint:

For meter bridge problems, always apply the balance condition \( \dfrac{R_1}{R_2} = \dfrac{l}{100-l} \) before and after any modification in the circuit.

Answer 1

The Correct Option is C

In a meter bridge at balance condition, \[ \frac{R_1}{R_2} = \frac{l}{100-l}. \] Step 1: Initial balance condition.
Null point at 40 cm: \[ \frac{R_1}{R_2} = \frac{40}{60} = \frac{2}{3}. \] \[ R_1 = \frac{2}{3}R_2. \] Step 2: New balance condition after adding \(16\,\Omega\) in parallel with \(R_2\).
Equivalent resistance: \[ R_2' = \frac{16R_2}{16 + R_2}. \] New null point at 50 cm: \[ \frac{R_1}{R_2'} = \frac{50}{50} = 1. \] \[ R_1 = R_2'. \] Step 3: Substitute values.
\[ \frac{2}{3}R_2 = \frac{16R_2}{16 + R_2}. \] \[ \frac{2}{3}(16 + R_2) = 16. \] \[ 32 + 2R_2 = 48. \] \[ R_2 = 8\,\Omega. \] \[ R_1 = \frac{2}{3}\times 8 = \frac{16}{3}\,\Omega. \] Final Answer: \[ \boxed{R_2 = 8\,\Omega,\; R_1 = \frac{16}{3}\,\Omega} \]