Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A galvanometer of $50~\Omega$ resistance has 25 divisions. A current of $4 \times 10^{-4} \text{ A}$ gives a deflection of one division. To convert this galvanometer into a voltmeter having a range of $25 \text{ V}$ , it should be connected with a resistance of:

$2500~\Omega$ as a shunt

$2450~\Omega$ as a shunt

$2550~\Omega$ in series

$2450~\Omega$ in series

Step-by-Step Solution

Concept: To convert a galvanometer into a voltmeter, a high resistance ( $R$ ) is connected in series with the galvanometer coil. This ensures that the total resistance is high enough to not draw significant current from the circuit being measured .
Calculate Full Scale Deflection Current ( $I_g$ ):

Current per division = $4 \times 10^{-4} \text{ A}$ .
Total divisions = 25.
$I_g = 25 \times 4 \times 10^{-4} = 100 \times 10^{-4} = 10^{-2} \text{ A} = 0.01 \text{ A}$ .

Formula: The potential difference ( $V$ ) across the combination is given by Ohm's law: $V = I_g (G + R)$ Where $G$ is the galvanometer resistance and $R$ is the series resistance.
Calculation:

Given $V = 25 \text{ V}$ and $G = 50~\Omega$ .
Rearranging the formula for $R$ : $R = \frac{V}{I_g} - G$ $R = \frac{25}{0.01} - 50$ $R = 2500 - 50 = 2450~\Omega$

Conclusion: A resistance of $2450~\Omega$ must be connected in series.

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started