Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A galvanometer has 30 divisions and a sensitivity $16\ \mu\text{A/div}$ . It can be converted into a voltmeter to read $3\ \text{V}$ by connecting (approximately):

Resistance nearly $6\ \text{k}\Omega$ in series

$6\ \text{k}\Omega$ in parallel

$500\ \Omega$ in series

It cannot be converted

Step-by-Step Solution

Calculate Full Scale Current ( $I_g$ ): The galvanometer has 30 divisions and a current sensitivity (current per division) of $16\ \mu\text{A/div}$ . $I_g = 30 \times 16\ \mu\text{A} = 480\ \mu\text{A} = 480 \times 10^{-6}\ \text{A}$ .
Principle of Conversion: To convert a galvanometer into a voltmeter, a high resistance ( $R$ ) must be connected in series with the galvanometer coil . This ensures the galvanometer draws a very small current while measuring the potential difference.
Calculation: The total resistance ( $R_{eq}$ ) required for the circuit to measure a voltage $V$ with a full-scale current $I_g$ is given by Ohm's Law: $R_{eq} = R + R_g = \frac{V}{I_g}$ where $R_g$ is the internal resistance of the galvanometer. $R + R_g = \frac{3}{480 \times 10^{-6}} = \frac{3000000}{480} = 6250\ \Omega = 6.25\ \text{k}\Omega$ .
Approximation: The added series resistance $R = 6250\ \Omega - R_g$ . Since the galvanometer resistance $R_g$ is typically small (order of $10-100\ \Omega$ ), the required series resistance is approximately $6.25\ \text{k}\Omega$ . The closest option is nearly $6\ \text{k}\Omega$ in series.

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