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 experiment:
The potential difference across a segment of the wire is directly proportional to its balancing length.
At the null point, no current flows through the galvanometer.
Therefore, the ratio of resistances is equal to the ratio of the corresponding balancing lengths.
Step 1: Initial null condition. Given balancing length: \[ l_1 = 40\ \text{cm} \] Total length of the potentiometer wire: \[ L = 100\ \text{cm} \] Hence, \[ \frac{R_1}{R_2} = \frac{40}{60} = \frac{2}{3} \] \[ R_1 = \frac{2}{3}R_2 \quad \cdots (1) \]
Step 2: After connecting a \(16\,\Omega\) resistor in parallel with \(R_2\). The equivalent resistance of \(R_2\) in parallel with \(16\,\Omega\) is: \[ R_2' = \frac{R_2 \times 16}{R_2 + 16} \] The new null point is obtained at: \[ l_2 = 50\ \text{cm} \] Thus, \[ \frac{R_1}{R_2'} = \frac{50}{50} = 1 \] \[ R_1 = R_2' \quad \cdots (2) \]
Step 3: Substitute from equations (1) and (2). \[ \frac{2}{3}R_2 = \frac{16R_2}{R_2 + 16} \] Canceling \(R_2\) (since \(R_2 \neq 0\)): \[ \frac{2}{3} = \frac{16}{R_2 + 16} \] Cross-multiplying: \[ 2(R_2 + 16) = 48 \] \[ 2R_2 + 32 = 48 \] \[ 2R_2 = 16 \Rightarrow R_2 = 32\,\Omega \]
Step 4: Determine the value of \(R_1\). \[ R_1 = \frac{2}{3} \times 32 = \frac{64}{3} = \frac{32}{3}\,\Omega \]