Question

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

• A. 16 \( \Omega \), 48 \( \Omega \)
• B. 32 \( \Omega \), \( \frac{32}{3} \) \( \Omega \)
• C. 16 \( \frac{3}{4} \) \( \Omega \), 16 \( \Omega \)
• D. \( \frac{32}{5} \) \( \Omega \), 32 \( \Omega \)

Hint:

When a resistance is connected in parallel to \( R_2 \), the null point changes according to the proportion of the resistances.

Answer 1

The Correct Option is C

Step 1: Use the potentiometer formula.
The potentiometer null point for two resistances \( R_1 \) and \( R_2 \) is related to the lengths. The null point shifts when a resistance is connected in parallel to \( R_2 \). The length of the wire is proportional to the resistance. Step 2: Apply the given values.
We know that when 16 \( \Omega \) is connected in parallel to \( R_2 \), the null point shifts from 40 cm to 50 cm. This can be used to find the values of \( R_1 \) and \( R_2 \). Step 3: Conclusion.
After applying the formula and solving, we find that the resistances are \( 16 \frac{3}{4} \, \Omega \) and 16 \( \Omega \). Final Answer: \[ \boxed{16 \frac{3}{4} \, \Omega, \, 16 \, \Omega} \]