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: Calculate the current in the primary circuit. The total resistance in the primary circuit is the sum of the potentiometer wire resistance (\(R_w\)) and the series resistance (\(R_s\)). \[ R_{total} = R_w + R_s = 8 \, \Omega + 242 \, \Omega = 250 \, \Omega \] Using Ohm's Law, the current \(I\) flowing through the circuit is: \[ I = \frac{E}{R_{total}} = \frac{2.5 \, \text{V}}{250 \, \Omega} = 0.01 \, \text{A} \]
Step 2: Calculate the potential drop across the potentiometer wire. \[ V_{wire} = I \times R_w = 0.01 \, \text{A} \times 8 \, \Omega = 0.08 \, \text{V} \]
Step 3: Calculate the potential gradient. Potential gradient (\(k\)) is the potential drop per unit length. \[ k = \frac{V_{wire}}{L} = \frac{0.08 \, \text{V}}{2.5 \, \text{m}} = 0.032 \, \text{V/m} \]
Step 4: Calculate the potential difference for length \(l = 20\) cm. Given length \(l = 20 \, \text{cm} = 0.2 \, \text{m}\). \[ \Delta V = k \times l = 0.032 \, \text{V/m} \times 0.2 \, \text{m} \] \[ \Delta V = 0.0064 \, \text{V} = 6.4 \, \text{mV} \]