Measuring Devices ( Galvanometer, Voltmeter and Ammeter & Meter Bridge ): Practice Problem & Solution

Solution Explained:

To solve this problem, we apply the core principles of Measuring Devices ( Galvanometer, Voltmeter and Ammeter & Meter Bridge ). Understanding the underlying formula is key to arriving at the correct answer below:

The potential gradient $k = \frac{V}{L} = \frac{6\text{ V}}{3\text{ m}} = 2\text{ V/m}$.nThe potential difference across a $50\text{ cm}$ ($0.5\text{ m}$) segment is $\Delta V = k \times l$.n$\Delta V = 2\text{ V/m} \times 0.5\text{ m} = 1\text{ V}$.