Question

In the given figure, there is a circuit of potentiometer of length AB = 10 m. The resistance per unit length is 0.1 $\Omega$ per cm. Across AB, a battery of emf E and internal resistance 'r' is connected. The maximum value of emf measured by this potentiometer is : 

• A. 2.25 V
• B. 2.75 V
• C. 5 V
• D. 6 V

Hint:

A potentiometer cannot measure an EMF greater than the potential drop across its entire wire length.
Check units carefully: resistance was given per cm, but length was in meters.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The maximum EMF that a potentiometer can measure is equal to the total potential difference across its wire \(AB\).
This potential difference is determined by the current flowing through the potentiometer wire from the primary driver battery.
Step 2: Key Formula or Approach:
1. Total resistance of wire \(R_{AB} = \rho \times L\).
2. Current in primary circuit \(I = \frac{V_{\text{driver}}}{R_{AB} + R_{\text{series}}}\).
3. Maximum measurable EMF \(V_{max} = I \times R_{AB}\).
Step 3: Detailed Explanation:
Given:
Length of wire \(L = 10 \text{ m} = 1000 \text{ cm}\).
Resistance per unit length \(\rho = 0.1 \text{ }\Omega\text{/cm}\).
Total resistance of wire \(R_{AB} = 0.1 \times 1000 = 100 \text{ }\Omega\).

Driver battery EMF \(V = 6 \text{ V}\).
Series resistance in primary circuit \(R_s = 20 \text{ }\Omega\).
Current in primary circuit:
\[ I = \frac{6}{100 + 20} = \frac{6}{120} = 0.05 \text{ A} \]
Potential difference across wire \(AB\):
\[ V_{AB} = I \times R_{AB} = 0.05 \times 100 = 5 \text{ V} \]
The maximum EMF that can be measured is the potential drop across the entire length of the wire.
Step 4: Final Answer:
The maximum value of EMF measured is 5 V.