Question

In a potentiometer, the null point for two resistances \(R_1\) and \(R_2\) is at \(40\ \text{cm}\) as shown. If a \(16\,\Omega\) resistance is connected in parallel to \(R_2\), the null point shifts to \(50\ \text{cm}\). Find the values of \(R_1\) and \(R_2\) respectively.

• A. \(16\,\Omega,\ 48\,\Omega\)
• B. \(32\,\Omega,\ \dfrac{32}{3}\,\Omega\)
• C. \(\dfrac{16}{3}\,\Omega,\ 8\,\Omega\)
• D. \(\dfrac{32}{3}\,\Omega,\ 32\,\Omega\)

Hint:

In potentiometer problems:
Resistance ratio = balancing length ratio
Always re-calculate equivalent resistance when a resistor is added in parallel
Use successive null conditions to form equations

Answer 1

The Correct Option is D

Concept: In a potentiometer:
Potential difference across a length of wire is directly proportional to its balancing length.
At null point, no current flows through the galvanometer.
Hence, ratio of resistances equals ratio of balancing lengths.
Step 1: Initial null condition. Given null length: \[ l_1 = 40\ \text{cm} \] Total length of potentiometer wire: \[ L = 100\ \text{cm} \] Thus, \[ \frac{R_1}{R_2} = \frac{40}{60} = \frac{2}{3} \] \[ R_1 = \frac{2}{3}R_2 \quad \cdots (1) \]
Step 2: After connecting \(16\,\Omega\) in parallel with \(R_2\). Effective resistance of \(R_2\) in parallel with \(16\,\Omega\): \[ R_2' = \frac{R_2 \times 16}{R_2 + 16} \] New null length: \[ l_2 = 50\ \text{cm} \] Hence, \[ \frac{R_1}{R_2'} = \frac{50}{50} = 1 \] \[ R_1 = R_2' \quad \cdots (2) \]
Step 3: Substitute from equations (1) and (2). \[ \frac{2}{3}R_2 = \frac{16R_2}{R_2 + 16} \] Cancel \(R_2\) (non-zero): \[ \frac{2}{3} = \frac{16}{R_2 + 16} \] Cross-multiplying: \[ 2(R_2 + 16) = 48 \] \[ 2R_2 + 32 = 48 \] \[ 2R_2 = 16 \Rightarrow R_2 = 32\,\Omega \]
Step 4: Find \(R_1\). \[ R_1 = \frac{2}{3} \times 32 = \frac{64}{3} = \frac{32}{3}\,\Omega \]