Question

The length of a potentiometer wire is 2.5 m and its resistance is 8 Ω. A cell of negligible internal resistance and emf of 2.5 V is connected in series with a resistance of 242 Ω in the primary circuit. The potential difference between two points separated by a distance of 20 cm on the potentiometer wire is

• A. 1.6 mV
• B. 4.8 mV
• C. 6.4 mV
• D. 3.2 mV

Hint:

The potential gradient is the most important quantity in potentiometer problems. Once you find it, you can find the potential difference across any length or find the balancing length for any given EMF.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
A potentiometer works on the principle that the potential drop across any portion of a wire of uniform cross-section is directly proportional to its length, provided a constant current flows through it. The key parameter is the potential gradient (potential drop per unit length).
Step 2: Key Formula or Approach:
1. Calculate the total resistance of the primary circuit: \(R_{total} = R_{wire} + R_{series}\).
2. Calculate the current flowing through the potentiometer wire: \(I = \frac{E}{R_{total}}\).
3. Calculate the total potential drop across the potentiometer wire: \(V_{wire} = I \times R_{wire}\).
4. Calculate the potential gradient (k): \(k = \frac{V_{wire}}{L_{wire}}\).
5. Calculate the potential difference across the given length (l): \(V = k \times l\).
Step 3: Detailed Explanation:
Given values:
- Length of potentiometer wire, \(L_{wire} = 2.5 \text{ m}\).
- Resistance of potentiometer wire, \(R_{wire} = 8 \text{ Ω}\).
- EMF of the driver cell, \(E = 2.5 \text{ V}\).
- Series resistance, \(R_{series} = 242 \text{ Ω}\).
- Length segment, \(l = 20 \text{ cm} = 0.2 \text{ m}\).
Calculation Steps:
1. Total resistance in the primary circuit:
\[ R_{total} = R_{wire} + R_{series} = 8 \text{ Ω} + 242 \text{ Ω} = 250 \text{ Ω} \] 2. Current in the primary circuit:
\[ I = \frac{E}{R_{total}} = \frac{2.5 \text{ V}}{250 \text{ Ω}} = 0.01 \text{ A} \] 3. Potential drop across the entire potentiometer wire:
\[ V_{wire} = I \times R_{wire} = 0.01 \text{ A} \times 8 \text{ Ω} = 0.08 \text{ V} \] 4. Potential gradient along the wire:
\[ k = \frac{V_{wire}}{L_{wire}} = \frac{0.08 \text{ V}}{2.5 \text{ m}} = 0.032 \text{ V/m} \] 5. Potential difference across the 20 cm length:
\[ V = k \times l = 0.032 \text{ V/m} \times 0.2 \text{ m} = 0.0064 \text{ V} \] To match the options, convert the result to millivolts (mV):
\[ V = 0.0064 \text{ V} \times 1000 \text{ mV/V} = 6.4 \text{ mV} \] Step 4: Final Answer:
The potential difference is 6.4 mV. Therefore, option (C) is correct.